Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C. https://ncertmcq.com/selina-concise-mathematics-class-10-icse-solutions/

Other Exercises

Question 1.
Find the sum of the first 22 terms of the A.P. : 8, 3, – 2,…..
Solution:
A.P. = 8, 3, – 2,…..
Here, a = 8, d = 3 – 8 = – 5, n = 22
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C Q1.1

Question 2.
How many terms of the A.P. : 24, 21, 18, must be taken so that their sum is 78 ?
Solution:
Let n term of the given A.P. be taken
and A.P. = 24, 21, 18……
Let n be the number of terms.
Here, a = 24, d = 21 – 24 = – 3, Sn = 78
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C Q2.1

Question 3.
Find the sum of 28 terms of an A.P. whose nth term is 8n – 5.
Solution:
nth term (Tn) = 8n – 5
T1 = 8 – 5 = 3
T2 = 8 x 2 – 5 = 16 – 5 = 11
T3 = 8 x 3 – 5 = 24 – 5 = 19
A.P. is 3, 11, 19,
Here, a = 3,d = 11 – 3 = 8 and n = 28
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C Q3.1

Question 4.
Find the sum of :
(i) all odd natural numbers less than 50.
(ii) first 12 natural numbers each of which is a multiple of 7.
Solution:
(i) Sum of all odd natural numbers less then 50
1 + 3 + 5 + 7 +…….+ 49
Here a = 1, d = 3 – 1 = 2
Let n be the number of term, then
49 = a + (n – 1)d
=> 49 = 1 + (n – 1) x 2
=> 49 – 1 = (n – 1) x 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C Q4.2

Question 5.
Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.
Solution:
Sum of first 51 terms of an A.P. in which T2 = 14 and T3 = 18
d = T3 – T2 = 18 – 14 = 4
and T2 = a + d
=> 14 = a + 4
=> a = 14 – 4 = 10
A.P. = 10, 14, 18, 22,….
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C Q5.1

Question 6.
The sum of first 7 terms of an A.P. is 49 and that of first 17 terms of it is 289. Find the sum of first tt terms.
Solution:
S7 = 49,
S17 = 289
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C Q6.2

Question 7.
The first term of an A.P. is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.
Solution:
First term of an AP (a) = 5
Last term = 45
and Sn = 1000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C Q7.1

Question 8.
Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.
Solution:
All natural numbers between 250 and 1000 which are divisible by 9 are
252, 261, 270……999
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C Q8.1

Question 9.
The first and the last terms of an A.P. are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?
Solution:
In an A.P.
T1 = a = 34, l = 700, d = 18
Let n be the number of terms, then
Tn = a + (n – 1 )d
=> 700 = 34 + (n – 1) x 18
=> 700 – 34 = 18(n – 1)
=> 180 – 0 = 666
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C Q9.1

Question 10.
In an A.P. the first term is 25, nth term is – 17 and the sum of n terms is 132. Find n and the common difference.
Solution:
In an A.P.
First term (a) = 25
nth term = – 17
and Sn = 132
Let d be the common difference
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C Q10.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C Q10.2

Question 11.
If the 8th term of an A.P. is 37 and the 15th term is 15 more than the 12th term, find the A.P. Also, find the sum of first 20 terms of this A.P.
Solution:
In an A.P.
8th term = 37
15th term = 12th term + 15
Let a be the first term and d be the common difference, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C Q11.1

Question 12.
Find the sum of all multiples of 7 lying between 300 and 700.
Solution:
Multiples of 7 lying between 300 and 700 are 301, 308, 315, 322,…., 693
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C Q12.1

Question 13.
The sum of n natural numbers is 5n² + 4n. Find its 8th term.
Solution:
Sum of n natural number = 5n² + 4n
Sn = 5n² + 4n
S1(a) = 5 x (1)² + 4 x 1
= 5 + 4 = 9
S2 = 5(2)² + 4 x 2 = 20 + 8 = 28
S2 – S1 = T2 = 28 – 9 = 19
=> a + d = 19 => 9 + d = 19
d = 19 – 9 = 10
a = 9, d = 10
T8 = a + (n – 1 )d = 9 + (8 – 1) x 10
= 9 + 7 x 10 = 9 + 70 = 79

Question 14.
The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.
Solution:
In an A.P.
T= 11
T8 = 2T4 + 5
Now, a + 3d = 11
a + 7d = 2 x 11 + 5 = 22 + 5 = 27
Subtracting, 4d = 16
=> d = \(\\ \frac { 16 }{ 4 } \) = 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C Q14.1

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Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises

Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises.

Sales-Tax (VAT) Chapterwise Revision Exercises

Question 1.
A man purchased a pair of shoes for ₹809.60 which includes 8% rebate on the marked price and then 10% sales tax on the remaining price. Find the marked price of the pair of shoes.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q1.1

Question 2.
The catalogue price of an article is ₹36,000. The shopkeeper gives two successive discounts of 10% each. He further gives an off-season discount of 5 % on the balance. If sales-tax at the rate 10% is charged on the remaining amount, find :
(i) the sales tax charged
(ii) the selling price of the article including sales-tax.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q2.1

Question 3.
A sells an article to B for ₹80,000 and charges sales-tax at 8%. B sells the same article to C for ₹1,12,000 and charges sales- tax at the rate of 12%. Find the VAT paid by B in this transaction.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q3.1

Question 4.
The marked price of an article is ₹1,600. Mohan buys this article at 20% discount and sells it at its marked price. If the sales-tax at each stage is 6%; find :
(i) the price at which the article can be bought,
(ii) the VAT (value added tax) paid by Mohan.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q4.1

Question 5.
A sells an old laptop to B for ₹12,600; B sells it to C for ₹14,000 and C sells the same laptop to D for ₹16,000.
If the rate of VAT at each stage is 10%, find the VAT paid by :
(i) B
(ii) C
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q5.2

Banking Chapterwise Revision Exercises

Question 6.
Ashok deposits ₹3200 per month in a cumulative account for 3 years at the rate of 9% per annum. Find the maturity value of this account
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q6.1

Question 7.
Mrs. Kama has a recurring deposit account in Punjab National Bank for 3 years at 8% p.a. If she gets ₹9,990 as interest at the time of maturity, find:
(i) the monthly instalment.
(ii) the maturity value of the account.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q7.2

Question 8.
A man has a 5 year recurring deposit account in a bank and deposits ₹240 per month. If he receives ₹17,694 at the time of maturity, find the rate of interest.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q8.1

Question 9.
Sheela has a recurring deposit account in a bank of ₹2,000 per month at the rate of 10% per anum. If she gets ₹83,100 at the time of maturity, find the total time (in years) for which the account was held.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q9.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q9.2

Question 10.
A man deposits ₹900 per month in a recurring account for 2 years. If he gets 1,800 as interest at the time of maturity, find the rate of interest .
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q10.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q10.2

Shares And Dividend Chapterwise Revision Exercises

Question 11.
What is the market value of 4 \(\frac { 1 }{ 2 }\) % (₹100) share, when an investment of ₹1,800 produces an income of ₹72 ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q11.1

Question 12.
By investing ₹10,000 in the shares of a company, a man gets an income of ₹800; the dividend being 10%. If the face-value of each share is ₹100, find :
(i) the market value of each share.
(ii) the rate per cent which the person earns on his investment.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q12.1

Question 13.
A man holds 800 shares of ₹100 each of a company paying 7.5% dividend semiannually.
(i) Calculate his annual dividend.
(ii) If he had bought these shares at 40% premium, what percentage return does he get on his investment ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q13.1

Question 14.
A man invests ₹10,560 in a company, paying 9% dividend, at the time when its ₹100 shares can be bought at a premium of ₹32. Find:
(i) the number of shares bought by him;
(ii) his annual income from these shares and
(iii) the rate of return on his investment .
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q14.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q14.2

Question 15.
Find the market value of 12% ₹25 shares of a company which pays a dividend of ₹1,875 on an investment of ₹20,000.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q15.2

Linear Inequations Chapterwise Revision Exercises

Question 16.
The given diagram represents two sets A and B on real number lines.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q16.1
(i) Write down A and B in set builder notation.
(ii) Represent A ∪ B, A ∩ B, A’ ∩ B, A – B and B – A on separate number lines.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q16.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q16.3

Question 17.
Find the value of x, which satisfy the inequation:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q17.1
Graph the solution set on the real number line .
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q17.2

Question 18.
State for each of the following statements whether it is true or false :
(a) If (x – a) (x – b) < 0, then x < a, and x < b.
(b) If a < 0 and b < 0, then (a + b)2 > 0.
(c) If a and b are any two integers such that a > b, then a2 > b2.
(d) If p = q + 2, then p > q.
(e) If a and b are two negative integers such that a < b , then \(\frac { 1 }{ a }\) < \(\frac { 1 }{ b }\)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q18.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q18.2

Question 19.
Given 20 – 5x < 5(x + 8), find the smallest value of x when :
(i) x \(\epsilon\) I
(ii) x \(\epsilon\) W
(iii) x \(\epsilon\) N
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q19.1

Question 20.
If x \(\epsilon\) Z, solve : 2 + 4x < 2x – 5 < 3x. Also, represent its solution on the real number line.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q20.1

Quadratic Equation Chapterwise Revision Exercises

Question 21.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q21.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q21.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q21.3

Question 22.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q22.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q22.2

Question 23.
Find the value of k for which the roots of the following equation are real and equal k2x2 – 2 (2k -1) x + 4 = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q23.1

Question 24.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q24.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q24.2

Question 25.
If -5 is a root of the quadratic equation 2x2 +px -15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, find the value of k.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q25.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q25.2

Problems On Quadratic Equations Chapterwise Revision Exercises

Question 26.
x articles are bought at ₹(x – 8) each and (x – 2) some other articles are bought at ₹(x – 3) each. If the total cost of all these articles is ₹76, how many articles of first kind were bought ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q26.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q26.2

Question 27.
In a two digit number, the unit’s digit exceeds its ten’s digit by 2. The product of the given number and the sum of its digits is equal to 144. Find the number.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q27.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q27.2

Question 28.
The time taken by a person to cover 150 km was 2.5 hours more than the time taken in return journey. If he returned at a speed of 10 km/hour more than the speed of going, what was the speed per hour in each direction ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q28.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q28.2

Question 29.
A takes 9 days more than B to do a certain piece of work. Together they can do the work in 6 days. How many days will A alone take to do the work ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q29.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q29.2

Question 30.
A man bought a certain number of chairs for ₹10,000. He kept one for his own use and sold the rest at the rate ₹50 more than he gave for one chair. Besides getting his own chair for nothing, he made a profit of ₹450. How many chairs did he buy ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q30.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q30.2

Question 31.
In the given figure; the area of unshaded portion is 75% of the area of the shaded portion. Find the value of x.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q31.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q31.2

Ratio And Proportion Chapterwise Revision Exercises

Question 32.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q32.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q32.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q32.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q32.4

Question 33.
If a:b = 2:3,b:c = 4:5 and c: d = 6:7, find :a:b :c :d.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q33.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q33.2

Question 34.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q34.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q34.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q34.3

Question 35.
Find the compound ratio of:
(i) (a-b) : (a+b) and (b2+ab): (a2-ab)
(ii) (x+y): (x-y); (x2+y2): (x+y)2 and (x2-y2)2: (x4-y4)
(iii) (x2– 25): (x2+ 3x – 10); (x2-4): (x2+ 3x+2) and (x + 1): (x2 + 2x)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q35.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q35.2

Question 36.
The ratio of the prices of two fans was 16: 23. Two years later, when the price of the first fan had risen by 10% and that of the second by Rs. 477, the ratio of their prices became 11: 20. Find the original prices of two fans.
Solution:
The ratio of prices of two fans = 16 : 23
Let the price of first fan = 6x
then price of second fan = 23x
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q36.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q36.2

Remainder And Factor Theorems Chapterwise Revision Exercises

Question 37.
Given that x + 2 and x – 3 are the factors of x3 + ax + b, calculate the values of a and b. Also find the remaining factor.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q37.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q37.2

Question 38.
Use the remainder theorem to factorise the expression 2x3 + 9x2 + 7x – 6 = 0 Hense, solve the equation 2x3 + 9x2 + 7 x – 6 = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q38.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q38.2

Question 39.
When 2x3 + 5x2 – 2x + 8 is divided by (x – a) the remainder is 2a3 + 5a2. Find the value of a.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q39.1

Question 40.
What number should be added to x3 – 9x2 – 2x + 3 so that the remainder may be 5 when divided by (x – 2) ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q40.1

Question 41.
Let R1 and R2 are remainders when the polynomials x3 + 2x2 – 5ax – 7 and x3 + ax2 – 12x + 6 are divided by (x +1) and (x – 2) respectively. If 2R1 + R2 = 6; find the value of a.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q41.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q41.2

Matrices Chapterwise Revision Exercises

Question 42.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q42.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q42.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q42.3

Question 43.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q43.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q43.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q43.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q43.4

Question 44.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q44.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q44.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q44.3

Question 45.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q45.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q45.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q45.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q45.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q45.5

Arithmetic Progression (A.P.) Chapterwise Revision Exercises

Question 46.
Find the 15th term of the A.P. with second term 11 and common difference 9.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q46.1

Question 47.
How many three digit numbers are divisible by 7 ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q47.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q47.2

Question 48.
Find the sum of terms of the A.P.: 4,9,14,…….,89.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q48.1

Question 49.
Daya gets pocket money from his father every day. Out of the pocket money, he saves ₹2.75 on first day, ₹3.00 on second day, ₹3.25 on third day and so on. Find:
(i) the amount saved by Daya on 14th day.
(ii) the amount saved by Daya on 30th day.
(iii) the total amount saved by him in 30 days.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q49.1

Question 50.
If the sum of first m terms of an A.P. is n and sum of first n terms of the same A.P. is m. Show that sum of first (m + n) terms of it is (m + n).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q50.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q50.2

Geometric Progression (GP) Chapterwise Revision Exercises

Question 51.
3rd term of a GP. is 27 and its 6th term is 729; find the product of its first and 7th terms.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q51.1

Question 52.
Find 5 geometric means between 1 and 27.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q52.1

Question 53.
Find the sum of the sequence 96 – 48 + 24…. upto 10 terms.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q53.1

Question 54.
Find the sum of first n terms of:
(i) 4 + 44 + 444 + …….
(ii) 0.7 + 0.77 + 0.777 + …..
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q54.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q54.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q54.3

Question 55.
Find the value of 0.4Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q55.1.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q55.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q55.3

Reflection Chapterwise Revision Exercises

Question 56.
Find the values of m and n in each case if:
(i) (4, -3) on reflection in x-axis gives (-m, n)
(ii) (m, 5) on reflection in y-axis gives (-5, n-2)
(iii) (-6, n+2) on reflection in origin gives (m+3, -4)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q56.1

Question 57.
Points A and B have the co-ordinates (-2,4) and (-4,1) respectively. Find :
(i) The co-ordinates of A’, the image of A in the line x = 0.
(ii) The co-ordinates of B’, the image of Bin y-axis.
(iii) The co-ordinates of A”, the image of A in the line BB’,
Hence, write the angle between, the lines A’A” and B B’. Assign a special name to the figure B’ A’ B A”
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q57.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q57.2

Question 58.
Triangle OA1B1 is the reflection of triangle OAB in origin, where A, (4, -5) is the image of A and B, (-7, 0) is the image of B.
(i) Write down the co-ordinates of A and B and draw a diagram to represent this information.
(ii) Give the special name to the quadrilateral ABA1 B1. Give reason.
(iii) Find the co-ordinates of A2, the image of A under reflection in x-axis followed by reflection in y-axis.
(iv) Find the co-ordinates of B2, the image of B under reflection in y-axis followed by reflection in origin.
(v) Does the quadrilateral obtained has any line symmetry ? Give reason.
(vi) Does it have any point symmetry ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q58.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q58.2

Section And Mid-Point Formulae Chapterwise Revision Exercises

Question 59.
In what ratio does the point M (P, -1) divide the line segment joining the points A (1,-3) and B (6,2) ? Hence, find the value of p.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q59.1

Question 60.
A (-4,4), B (x, -1) and C (6,y) are the vertices of ∆ABC. If the centroid of this triangle ABC is at the origin, find the values of x and y.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q60.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q60.2

Question 61.
A (2,5), B (-1,2) and C (5,8) are the vertices of a triangle ABC. Pand Q are points on AB and AC respectively such that AP: PB=AQ: QC = 1:2.
(a) Find the co-ordinates of points P and Q
(b) Show that BC = 3 x PQ.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q61.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q61.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q61.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q61.4

Question 62.
Show that the points (a, b), (a+3, b+4), (a -1, b + 7) and (a – 4, b + 3) are the vertices of a parallelogram.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q62.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q62.2

Equation Of straight Line Chapterwise Revision Exercises

Question 63.
Given points A(l, 5), B (-3,7) and C (15,9).
(i) Find the equation of a line passing through the mid-point of AC and the point B.
(ii) Find the equation of the line through C and parallel to AB.
(iii) The lines obtained in parts (i) and (ii) above, intersect each other at a point P. Find the co-ordinates of the point P.
(iv) Assign, giving reason, a special names of the figure PABC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q63.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q63.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q63.3

Question 64.
The line x- 4y=6 is the perpendicular bisector of the line segment AB. If B = (1,3); find the co-ordinates of point A.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q64.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q64.2

Question 65.
Find the equation of a line passing through the points (7, -3) and (2, -2). If this line meets x- axis at point P and y-axis at point Q; find the co-ordinates of points P and Q.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q65.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q65.2

Question 66.
A (-3,1), B (4,4) and C (1, -2) are the vertices of a triangle ABC. Find:
(i) the equation of median BD,
(ii) the equation of altitude AE.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q66.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q66.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q66.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q66.4

Question 67.
Find the equation of perpendicular bisector of the line segment joining the points (4, -3) and (3,1).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q67.1

Question 68.
(a) If (p +1) x + y = 3 and 3y – (p -1) x = 4 are perpendicular to each other find the value of p.
(b) If y + (2p +1) x + 3 = 0 and 8y – (2p -1) x = 5 are mutually prependicular, find the value of p.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q68.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q68.2

Question 69.
The co-ordinates of the vertex A of a square ABCD are (1, 2) and the equation of the diagonal BD is x + 2y = 10. Find the equation of the other diagonal and the coordinates of the centre of the square.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q69.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q69.2

Similarity Chapterwise Revision Exercises

Question 70.
M is mid-point of a line segment AB; AXB and MYB are equilateral triangles on opposite sides of AB; XY cuts AB at Z. Prove that AZ = 2ZB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q70.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q70.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q70.3

Question 71.
In the given figure, if AC = 3cm and CB = 6 cm, find the length of CR.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q71.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q71.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q71.3

Question 72.
The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersect at point O. If BO : OD = 4:7; find:
(i) ∆AOD : ∆AOB
(ii) ∆AOB : ∆ACB
(iii) ∆DOC : ∆AOB
(iv) ∆ABD : ∆BOC
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q72.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q72.2

Question 73.
A model of a ship is made to a scale of 1 : 160. Find :
(i) the length of the ship, if the length of its model is 1.2 m.
(ii) the area of the deck of the ship, if the area of the deck of its model is 1.2 m2.
(iii) the volume of the ship, if the volume of its model is 1.2m3.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q73.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q73.2

Question 74.
In trapezium ABCD, AB || DC and DC = 2 AB. EF, drawn parallel to AB cuts AD in F and BC in E such that 4 BE = 3 EC. Diagonal DB intersects FE at point G Prove that: 7 EF = 10 AB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q74.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q74.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q74.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q74.4

Loci Chapterwise Revision Exercises

Question 75.
In triangle ABC, D is mid-point of AB and CD is perpendicular to AB. Bisector of ∠ABC meets CD at E and AC at F. Prove that:
(i) E is equidistant from A and B.
(ii) F is equidistant from AB and BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q75.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q75.2

Question 76.
Use graph paper for this questions. Take 2 cm = 1 unit on both axes.
(i) Plot the points A (1,1), B (5,3) and C (2,7)
(ii) Construct the locus of points equidistant from A and B.
(iii) Construct the locus of points equidistant from AB and AC.
(iv) Locate the point P such that PA = PB and P is equidistant from AB and AC.
(v) Measure and record the length PA in cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q76.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q76.2

Circles Chapterwise Revision Exercises

Question 77.
In the given figure, ∠ADC = 130° and BC = BE. Find ∠CBE if AB ⊥ CE.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q77.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q77.2

Question 78.
In the given figure, ∠OAB=30° and ∠OCB= 57°, find ∠BOC and ∠AOC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q78.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q78.2

Question 79.
In the given figure, O is the centre of the circle. If chord AB = chord AC, OP⊥ AB and OQ⊥ AC; show that: PB=QC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q79.1

Question 80.
In the given figure, AB and XY are diameters of a circle with centre O. If ∠APX=30°, find:
(i) ∠AOX
(ii) ∠APY
(iii) ∠BPY
(iv) ∠OAX
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q80.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q80.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q80.3

Question 81.
(a) In the adjoining figure; AB = AD, BD = CD and ∠DBC = 2 ∠ABD.
Prove that: ABCD is a cyclic quadrilateral.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q81.1
(b) AB is a diameter of a circle with centre O, Chord CD is equal to radius OC. AC and BD produced intersect at P. Prove that ∠APB = 60°.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q81.2
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q81.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q81.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q81.5
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q81.6

Tangents And Intersecting Chords Chapterwise Revision Exercises

Question 82.
In the given figure, AC=AB and ∠ABC=72°.
OA and OB are two tangents. Determine:
(i) ∠AOB
(ii) angle subtended by the chord AB at the centre.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q82.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q82.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q82.3

Question 83.
In the given figure, PQ, PR and ST are tangents to the same circle. If ∠P = 40° and ∠QRT = 75°, find a, b and c.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q83.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q83.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q83.3

Question 84.
In the given figure, ∠ABC = 90° and BC is diameter of the given circle. Show that:
(i) AC x AD =AB2
(ii) AC x CD = BC2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q84.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q84.2

Question 85.
In the given figure; AB, BC and CA are tangents to the given circle. If AB = 12 cm, BC = 8 cm and AC=10 cm, find the lengths of AD,BE = CF.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q85.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q85.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q85.3

Question 86.
(a) AB and CD are two chords of a circle intersecting at a point P inside the circle. If:
(i) AB = 24 cm, AP = 4 cm and PD = 8 cm, determine CP.
(ii) AP = 3 cm, PB = 2.5 cm and CD = 6.5 cm determine CP.
(b) AB and CD are two chords of a circle intersecting at a point P outside the circle. If:
(i) PA = 8 cm, PC – 5 cm and PD = 4 cm, determine AB.
(ii) PC = 30 cm, CD = 14 cm and PA = 24 cm, determine AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q86.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q86.2

Construction Chapterwise Revision Exercises

Question 87.
Construct a triangle ABC in which AC = 5 cm, BC = 7 cm and AB = 6 cm.
(i) Mark D, the mid point of AB.
(ii) Construct a circle which touches BC at C and passes through D.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q87.1

Question 88.
Using ruler and compasses only, draw a circle of radius 4 cm. Produce AB, a diameter of this circle up to point X so that BX = 4cm. Construct a circle to touch AB at X and to touch the circle, drawn earlier externally.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q88.1

Mensuration Chapterwise Revision Exercises

Question 89.
A cylindrical bucket 28 cm in diameter and 72 cm high is full of water. The water is emptied into a rectangular tank 66 cm long and 28 cm wide. Find the height of the water level in the tank.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q89.1

Question 90.
A tent is of the shape of right circular cylinder upto height of 3 metres and then becomes a right circular cone with a maximum height of 13.5 metres above the ground. Calculate the cost of painting the inner surface of the tent at Rs. 4 per sq. metre, if the radius of the base is 14 metres.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q90.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q90.2

Question P.Q.
In the given figure, diameter of the biggest semi-circle is 108cm, and diameter of the smallest circle is 36 cm. Calculate the area of the shaded portion.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises QP1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises QP1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises QP1.3

Question 91.
A copper wire of diameter 6 mm is evenly wrapped on the cylinder of length 18 cm and diameter 49 cm to cover the whole surface. Find:
(i) the length
(ii) the volume of the wire
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q91.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q91.2

Question 92.
A pool has a uniform circular cross-section of radius 5 m and uniform depth 1.4m. It is filled by a pipe which delivers water at the rate of 20 litres per sec. Calculate, in minutes, the time taken to All the pool. If the pool is emptied in 42 min. by another cylindrical pipe through which water flows at 2 m per sec, calculate the radius of the pipe in cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q92.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q92.3

Question 93.
A test tube consists of a hemisphere and a cylinder of the same radius. The volume of water required to fill the whole tube is 2849/3cm3 and 2618/3cm3 of water are required to fill the tube to a level which is 2 cm below the top of the tube. Find the radius of the tube and the length of its cylinderical part.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q93.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q93.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q93.3

Question 94.
A sphere is placed in an inverted hollow conical vessel of base radius 5 cm and vertical height 12 cm. If the highest point of the sphere is at the level of the base of the cone, find the radius of the sphere. Show that the volume of the sphere and the conical vessel are as 40 : 81.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q94.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q94.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q94.3

Question 95.
The difference between the outer and the inner curved surface areas of a hollow cylinder, 14cm. long is 88sq. cm. Find the outer and the inner radii of the cylinder given that the volume of metal used is 176 cu. cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q95.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q95.2

Trigonometry Chapterwise Revision Exercises

Question 96.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q96.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q96.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q96.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q96.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q96.5

Question 97.
If tan A = 1 and tan B = √3 ; evaluate :
(i) cos A cos B – sin A sin B
(ii) sin A cos B + cos A sin B
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q97.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q97.2

Question 98.
As observed from the top of a 100 m high light house, the angles, of depression of two ships approaching it are 30° and 45°. If one ship is directly behind the other, find the distance between the two ships.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q98.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q98.2

Question 99.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q99.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q99.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q99.3

Question 100.
From the top of a light house, it is observed that a ship is sailing directly towards it and the angle of depression of the ship changes from 30° to 45° in 10 minutes. Assuming that the ship is sailing with uniform speed; calculate in how much more time (in minutes) will the ship reach to the light house.
Solution:
Let LM be the height of light house = h
Angle of depression changes from 30° to 45° in 10 minutes.
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q100.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q100.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q100.3

Statistics Chapterwise Revision Exercises

Question 101.
Calculate the mean mark in the distribution given below :
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q101.1
Also state (i) median class (ii) the modal class.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q101.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q101.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q101.4

Question 102.
Draw an ogive for the following distribution :
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q102.1
Use the ogive drawn to determine :
(i) the median income,
(ii) the number of employees whose income exceeds Rs. 190.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q102.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q102.3

Question 103.
The result of an examination are tabulated below :
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q103.1
Draw the ogive for above data and from it determine :
(i) the number of candidates who got marks less than 45.
(ii) the number of candidates who got marks more than 75.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q103.2

Probability Chapterwise Revision Exercises

Question 104.
A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. If a ball is drawn from the bag, without looking into it, find the probability that the ball drawn is
(i) yellow
(ii) red
(iii) blue
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q104.1

Question 105.
A bag contains 6 red balls, 8 blue balls and 10 yellow balls, all the balls being of the same size. If a ball is drawn from the bag, without looking into it, find the probability that the ball drawn is
(i) yellow
(ii) red
(iii) blue
(iv) not yellow
(v) not blue
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q105.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q105.2

Question 106.
Two dice are thrown at the same time. Write down all the possible outcomes. Find the probability of getting the sum of two numbers appearing on the top of the dice as :
(i) 13
(ii) less than 13
(iii) 10
(iv) less then 10
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q106.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q106.2

Question 107.
Five cards : the ten, jack, queen, king and ace. of diamonds are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen ?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace ? (b) a queen ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q107.1

Question 108.
(i) A lot of 20 bulbs contains 4 defective bulbs, one bulb is drawn at random, from the lot. What is the probability that this bulb is defective ?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises Q108.1

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C.

Other Exercises

Question 1.
In the given circle with diameter AB, find the value of x. (2003)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q1.2
∠ABD = ∠ACD = 30° (Angle in the same segment)
Now in ∆ ADB,
∠BAD + ∠ADB + ∠DBA = 180° (Angles of a ∆)
But ∠ADB = 90° (Angle in a semi-circle)
∴ x + 90° + 30° = 180°
⇒ x + 120° = 180°
∴ x = 180°- 120° = 60°

Question 2.
In the given figure, O is the centre of the circle with radius 5 cm, OP and OQ are perpendiculars to AB and CD respectively. AB = 8cm and CD = 6cm. Determine the length of PQ.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q2.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q2.2
Radius of the circle whose centre is O = 5 cm
OP ⊥ AB and OQ ⊥ CD, AB = 8 cm and CD = 6 cm.
Join OA and OC, then OA = OC=5cm.
∴ OP ⊥ AB
∴ P is the mid-point of AB.
Similarly, Q is the mid-point of CD.
In right ∆OAP,
OA² = OP² + AP² (Pythagorous theoram)
⇒ (5)² =OP² +(4)² ( ∵ AP = AB = \(\frac { 1 }{ 2 }\) x 8 = 4cm)
⇒ 25 = OP² + 16
⇒ OP3 = 25 – 16 = 9 = (3)²
∵ OP = 3 cm
Similarly, in right ∆ OCQ,
OC2 = OQ2 + CQ2
⇒ (5)2 =OQ2+(3)2 (∵ CQ = CD = \(\frac { 1 }{ 2 }\) x 6 = 3cm)
⇒ 25 = OQ2 + 9
⇒ OQ3 = 25 – 9 = 16 = (4)2
∴ OQ = 4 cm
Hence, PQ = OP + OQ = 3-4 = 7 cm.

Question 3.
The given figure shows two circles with centres A and B; and radii 5cm and 3cm respectively, touching each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q3.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q3.2
Join AP and produce AB to meet the bigger circle at C.
AB = AC – BC = 5 cm – 3 cm = 2 cm.
But, M is the mid-point of AB
∴ AM = \(\frac { 2 }{ 2 }\) = 1cm.
Now in right ∆APM,
AP2 = MP2 + AM2 (Pythagorous theorem)
⇒ (5)2 = MP2 -1- (1 )2
⇒ 25 = MP: + 1
⇒ MP: = 25 – 1 = 24

Question 4.
In the given figure, ABC is a triangle in which ∠BAC = 30°. Show that BC is equal to the radius of the circumcircle of the triangle ABC, whose centre is O.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q4.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q4.2
Given: In the figure ABC is a triangle in winch ∠A = 30°
To Prove: BC is the radius of circumcircle of ∆ABC whose O is the centre.
Const: Join OB and OC.
Proof: ∠BOC is at the centre and ∠BAC is at the remaining part of the circle
∴ ∠BOC = 2 ∠BAC = 2 x 30° = 60°
Now in ∆OBC,
OB = OC (Radii of the same circle)
∴ ∠OBC = ∠OCB
But ∠OBC + ∠OCB + ∠BOC – 180°
∠OBC + ∠OBC + 60°- 180°
⇒ 2 ∠OBC = 180°- 60° = 120°
⇒ ∠OBC = \(\frac { { 120 }^{ circ } }{ 2 }\) = 60°
∴ ∆OBC is an equilateral triangle.
∴ BC = OB = OC
But OB and OC are the radii of the circumcircle
∴ BC is also the radius of the circumcircle.

Question 5.
Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q5.1
Given: In ∆ABC, AB-AC and a circle with AB as diameter is drawn which intersects the side BC and D.
To Prove: D is the mid point of BC.
Const: Join AD.
Proof: ∠ 1 = 90° (Angle in a semi-circle)
But ∠ 1 + ∠ 2 – 180° (Linear pair)
∴ ∠ 2 = 90°
Now, in right ∆ ABD and ∆ ACD,
Hyp. AB – Hyp. AC (Given)
Side AD – AD (Common)
∴ ∆ABD = ∆ACD (RHS criterion of congruency)
∴ BD = DC (c.p.c.t.)
Hence D is he mid point of BC. Q.E.D.

Question 6.
In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q6.1
Solution:
Join OE,
Arc EC subtends ∠EOC at the centre and ∠EBC at the remaining part of the circle.
∴ ∠EOC = 2 ∠EBC = 2 x 65° = 130°
Now, in ∆OEC, OE = OC (radii of the same circle)
∴ ∠OEC = ∠OCE
But ∠ OEC + ∠ OCE + ∠ EOC = 180°
⇒ ∠ OCE + ∠ OCE + ∠ EOC = 180°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q6.2

Question 7.
Chords AB and CD of a circle intersect each other at point P, such that AP = CP. S.how that AB = CD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q7.1
Solution:
Given: Two chords AB and CD intersect each other at P inside the circle with centre O and AP = CP.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q7.2
To Prove: AB = CD.
Proof: ∵ Two chords AB and CD intersect each other inside the circle at P.
∴ AP x PB = CP x PD ⇒ \(\frac { AP}{ CP }\) = \(\frac { AD }{ PB }\)
But AP = CP ….(i) (given)
∴ PD = PB or PB = PD …,(ii)
Adding (i) and (ii),
AP + PB = CP + PD
⇒ AB = CD Q.E.D.

Question 8.
The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.
Solution:
Given: ABCD is a cyclic quadrilateral and PRQS is a quadrilateral formed by the angle bisectors of angle ∠A, ∠B, ∠C and ∠D.
To Prove: PRQS is a cyclic quadrilateral.
Proof: In ∆APD,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q8.1
∠1 +∠2 + ∠P= 180° ,…(i)
Similarly, in ∆BQC.
∠3 + ∠4 + ∠Q= 180° ….(ii)
Adding (i) and (ii), we get:
∠1 + ∠2 + ∠P + ∠3 + ∠4 ∠Q = 180° + 180° = 360°
⇒ ∠1 + ∠2 + ∠3 + ∠4 + ∠P + ∠Q = 360° ..(iii)
But ∠1+∠2+∠3+∠4 = \(\frac { 1 }{ 2 }\) (∠A+∠B+∠C+∠D)
= \(\frac { 1 }{ 2 }\) x 360°= 180°
∴ ∠P + ∠Q = 360° – 180° = 180° [From (iii)]
But these are the sum of opposite angles of quadrilateral PRQS
∴ Quad. PRQS is a cyclic quadrilateral. Q.E.D.

Question 9.
In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:
(I) ∠BDC
(ii) ∠BEC
(iii) ∠BAC (2014)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q9.1
Solution:
∠DBC = 58°
BD is diameter
∴ ∠DCB=90° (Angle in semi circle)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q9.2
(i) In ∆BDC
∠BDC + ∠DCB + ∠CBD = 180°
∠BDC = 180°- 90° – 58° = 32°
(ii) ∠BEC =180°-32°
(opp. angle of cyclic quadrilateral)
= 148°
(iii) ∠BAC = ∠BDC = 32°
(Angles in same segment)

Question 10.
D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q10.1
Given: In ∆ABC, AB = AC and D and E are points on AB and AC such that AD = AE, DE is joined.
To Prove: B,C,E,D. are concyclic.
Proof: In ∆ABC, AB = AC
∴ ∠B = ∠C (Angles opposite to equal sides)
Similarly in ∆ADE, AD = AE (given)
∴ ∠ADE = ∠AED
In ∆ABC,
∵ \(\frac { AP }{ AB }\) = \(\frac { AE }{ AC }\)
∴ DE || BC.
∴ ∠ADE = ∠B (Corresponding angles)
But ∠B = ∠C (Proved)
∴ Ext. ∠ADE = its interior opposite ∠C.
∴ BCED is a cyclic quadrilateral.
Hence B,C, E and D are concyclic.

Question 11.
In the given figure, ABCD is a cyclic quadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ADC = 92°, ∠FAE’= 20°; determine ∠BCD. Give reason in support of your answer.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q11.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q11.2
In cyclic quad. ABCD,
AF || CB and DA is produced to E such that
∠ADC = 92° and ∠FAE – 20°
Now, we have to find the measure of ∠BCD In cyclic quad. ABCD,
∠B – ∠D = 180° ⇒ ∠B + 92° = 180″
⇒∠B = 180°-92° = 88°
∵ AF || CB.
∴∠FAB = ∠B = 88°
But ∠FAE – 20° (Given)
Ext. ∠BAE – ∠BAF + ∠FAE
= 88° + 20° = 108°
But Ext. ∠BAE – ∠BCD
∴ ∠BCD = 108°

Question 12.
If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66° and ∠ABC – 80°, calculate
(i) ∠DBC,
(ii) ∠IBC,
(iii) ∠BIC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q12.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q12.2
Join DB and DC, IB and IC.
∠BAC = 66°, ∠ABC = 80°. I is the incentre of the ∆ABC.
(i) ∵ ∠DBC and ∠DAC are in the same segment
∴ ∠DBC – ∠DAC.
But ∠DAC = \(\frac { 1 }{ 2 }\)∠BAC = \(\frac { 1 }{ 2 }\) x 66° = 33°
∴ ∠DBC = 33°.
(ii) ∵ I is the incentre of ∆ABC.
∴ IB bisect ∠ABC
∴ ∠ IBC = \(\frac { 1 }{ 2 }\) ∠ABC = \(\frac { 1 }{ 2 }\) x 80° = 40°.
(iii) ∴∠BAC = 66° ∠ABC = 80°
∴ In ∆ABC,
∠ACB = 180° – (∠ABC + ∠CAB)
= 180°-(80°+ 66°)= 180°- 156° = 34°
∵ IC bisects the ∠C
∴ ∠ ICB = \(\frac { 1 }{ 2 }\) ∠C = \(\frac { 1 }{ 2 }\) x 34° = 17°.
Now in ∆IBC,
∠ IBC + ∠ ICB + ∠ BIC = 180°
⇒ 40° + 17° + ∠BIC = 180°
⇒ ∠ BIC = 180° – (40° + 17°) = 180° – 57°
= 123°

Question 13.
In the given figure, AB = AD = DC= PB and ∠DBC = x°. Determine in terms of x :
(i) ∠ABD
(ii) ∠APB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q13.1
Hence or otherwise prove that AP is parallel to DB.
Solution:
Given: In figure, AB = AD = DC = PB.
∠DBC = x. Join AC and BD.
To Find : the measure of ∠ABD and ∠APB.
Proof: ∠DAC=∠DBC= x(angles in the same segment)
But ∠DCA = ∠DAC (∵ AD = DC)
= x
But ∠ABD = ∠DAC (Angles in the same segment)
In ∆ABP, ext. ∠ABC = ∠BAP + ∠APB
But ∠ BAP = ∠APB (∵ AB = BP)
2 x x = ∠APB + ∠APB = 2∠APB
∴ 2∠APB = 2x
⇒ ∠APB = x
∵ ∠APB = ∠DBC = x
But these are corresponding angles
∴ AP || DB. Q.E.D.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q13.2

Question 14.
In the given figure, ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠CQE are supplementary.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q14.1
Solution:
Given: In the figure, ABC, AEQ and CEP are straight lines.
To prove: ∠APE + ∠CQE = 180°.
Const: Join EB.
Proof: In cyclic quad. ABEP,
∠APE+ ∠ABE= 180° ….(i)
Similarly in cyclic quad. BCQE
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q14.2
∠CQE +∠CBE = 180° ….(ii)
Adding (i) and (ii),
∠APE +∠ABE + ∠CQE +∠CBE = 180° + 180° = 360°
⇒ ∠APE + ∠CQE + ∠ABE + ∠CBE = 360°
But ∠ABE + ∠CBE = 180° (Linear pair)
∴ ∠APE + ∠CQE + 180° = 360°
⇒ ∠APE + ∠CQE = 360° – 180° = 180°
Hence ∠APE and ∠CQE are supplementary. Q.E.D.

Question 15.
In the given figure, AB is the diameter of the circle with centre O.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q15.1
If ∠ADC = 32°, find angle BOC.
Solution:
Arc AC subtends ∠AOC at the centre and ∠ ADC at the remaining part of the circle
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q15.2
∴ ∠AOC = 2 ∠ADC
= 2 x 32° = 64°
∵ ∠AOC + ∠ BOC = 180° (Linear pair)
⇒ 64° + ∠ BOC = 180°
⇒ ∠ BOC=180° – 64° =116° .

Question 16.
In a cyclic-quadrilateral PQRS, angle PQR = 135°. Sides SP and RQ produced meet at point A : whereas sides PQ and SR produced meet at point B.
If ∠ A : ∠ B = 2 : 1 ; find angles A and B.
Solution:
PQRS is a cyclic-quadrilateral in which ∠ PQR =135°
Sides SP and RQ are produced to meet at A and sides PQ and SR are produced to meet at B.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q16.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q16.2

Question 17.
In the following figure, AB is the diameter of a circle with centre O and CD is the chord with length equal to radius OA. If AC produced and BD produced meet at point P ; show that ∠ APB = 60°.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q17.1
Solution:
Given : In the figure, AB is the diameter of the circle with centre O.
CD is the chord with length equal to the radius OA.
AC and BD are produced to meet at P.
To prove : ∠ APB = 60°
Const : Join OC and OD
Proof : ∵ CD = OC = OD (Given)
∴ ∆OCD is an equilateral triangle
∴ ∠ OCD = ∠ ODC = ∠ COD = 60°
In ∆ AOC, OA = OC (Radii of the same circle)
∴ ∠ A = ∠ 1
Similarly, in ∆ BOD,
OB = OD
∴∠2= ∠B
Now in cyclic quadrilateral ACDB,
∠ A CD + ∠B = 180°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q17.2
∠60°+ ∠ 1 + ∠B = 180°
= ∠ 1 + ∠B = 180° – 60°
⇒∠ 1 + ∠B = 120°
But ∠ 1 = ∠ A
∴ ∠ A + ∠B = 120° …(i)
Now, in ∆ APB,
∠ P + ∠ A + ∠ B = 180° (Sum of angles of a triangle)
⇒ ∠P+120°=180° [From (i)]
⇒ ∠P = 180°- 120°= 60°
Hence ∠ P = 60° or ∠ APB = 60° Hence proved.

Question 18.
In the following figure,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q18.1
ABCD is a cyclic quadrilateral in which AD is parallel to BC.
If the bisector of angle A meets BC at point E and the given circle at point F, prove that :
(i) EF = FC
(ii) BF = DF
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q18.2
Given : ABCD is a cyclic quadrilateral in which AD || BC.
Bisector of ∠ A meets BC at E and the given circle at F. DF and BF are joined.
To prove :
(i) EF = FC
(ii) BF = DF
Proof : ∵ ABCD is a cyclic -quadrilateral and AD || BC
∵ AF is the bisector of ∠ A
∴ ∠ BAF = ∠ DAF
∴ Arc BF = Arc DF (equal arcs subtends equal angles)
⇒ BF = DF(equal arcs have equal chords)
Hence proved

Question 19.
ABCD is a cyclic quadrilateral. Sides AB and DC produced meet at point E ; whereas sides BC and AD produced meet at point F. If ∠ DCF : ∠ F : ∠ E = 3 : 5 : 4, find the angles of the cyclic quadrilateral ABCD.
Solution:
Given : In a circle, ABCD is a cyclic quadrilateral AB and DC are produce to meet at E and BC and AD are produced to meet at F.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q19.2

Question 20.
The following figure shows a circle with PR as its diameter.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q20.1
If PQ = 7 cm, QR = 3 cm, RS = 6 cm. find the perimeter of the cyclic quadrilateral PQRS. (1992)
Solution:
In the figure, PQRS is a cyclic quadrilateral in which PR is a diameter
PQ = 7 cm,
QR = 3 RS = 6cm
∴ 3 RS = 6cm
and RS = \(\frac { 6 }{ 3 }\) = 2cm
Now in ∆ PQR,
∠ Q = 90° (Angle in a semi-circle)
∴ PR2 = PQ2 + QR2 (Pythagoras theorem)
= (7)2 + (6)2 = 49 + 36 = 85
Again, in right ∆ PSQ, PR2 = PS2 + RS2
⇒ 85 = PS2 + (2)2
⇒ 85 = PS2 + 4
⇒ PS2 = 85 – 4 = 81 = (9)2
∴ PS = 9cm
Now, perimeter of quad. PQRS = PQ + QR + RS + SP = (7 + 9 + 2 + 6) cm = 24cm

Question 21.
In the following figure, AB is the diameter of a circle with centre O.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q21.1
If chord AC = chord AD, prove that :
(i)arc BC = arc DB
(ii) AB is bisector of ∠ CAD. Further, if the length of arc AC is twice the length of arc BC, find :
(a) ∠ BAC
(b) ∠ ABC
Solution:
Given : In a circle with centre O, AB is the diameter and AC and AD are two chords such that AC = AD.
To prove : (i) arc BC = arc DB
(ii) AB is the bisector of ∠CAD
(iii) If arc AC = 2 arc BC, then find
(a) ∠BAC (b) ∠ABC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q21.2
Construction: Join BC and BD.
Proof : In right angled A ABC and A ABD
Side AC = AD (Given)
Hyp. AB = AB (Common)
∴ ∆ ABC ≅ ∆ ABD (R.H.S. axiom)
(i) ∴ BC = BD (C.P.C.T)
∴ Arc BC = Arc BD (equal chords have equal arcs)
(ii) ∠ BAC = ∠ BAD (C.P.C.T)
∴ AB is the bisector of ∠CAD.
(iii) If arc AC = 2 arc B
Then ∠ABC = 2 ∠BAC
But ∠ABC – 2 ∠BAC = 90°
∴ 2 ∠BAC + ∠BAC = 90°
⇒ 3 ∠BAC = 90° ⇒ ∠BAC = 30°
and ∠ABC = 2 ∠BAC = 2 * 30° = 60°

Question 22.
In cyclic-quadrilateral ABCD ; AD = BC,
∠ BAC = 30° and ∠ CBD = 70°, find:
(i) ∠ BCD
(ii) ∠ BCA
(iii) ∠ ABC
(iv) ∠ ADC
Solution:
ABCD is a cyclic-quadrilateral and AD = BC
∠ BAC = 30°, ∠ CBD = 70°
∠ DAC = ∠ CBD , (Angles in the same segment)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q22.1
But ∠ CBD = 70°
∴ ∠ DAC = 70°
⇒ ∠ BAD = ∠ BAC + ∠ DAC = 30° + 70° = 100°
But ∠ BAD + ∠ BCD = 180°
(Sum of opposite angles of a cyclic quad.)
⇒100°+ ∠ BCD=180° ⇒ ∠ BCD=180° – 100° = 80°
∴ ∠ BCD = 80°
∵ AD = BC (Given)
∴ ∠ ACD = ∠ BDC
(Equal chords subtends equal angles)
But ∠ ACB = ∠ ADB
(Angles in the same segment)
∴ ∠ ACD + ∠ ACB = ∠ BDC + ∠ ADB
⇒ ∠ BCD = ∠ ADC = 80° (∵ ∠ BCD = 80°)
∴ ∠ ADC = 80°
But in ∆ BCD,
∠ CBD + ∠ BCD + ∠ BDC = 180° (Angles of a triangle)
⇒ 70° + 80° + ∠ BDC = ∠ 180°
⇒ 150°+ ∠ BDC = 180°
∴ ∠ BDC = 180° – 150° = 30°
⇒ ∠ ACD = 30° (∵ ∠ ACD = ∠ BDC)
∴ ∠ BCA = ∠ BCD – ∠ ACD = 80° – 30° = 50°
∠ ADC + ∠ABC = 180°
(Sum of opp. angles of a cyclic quadrilateral)
⇒ 80°+ABC = 180°
⇒ ∠ ABC=180°-80° = 100°

Question 23.
In the given figure, if ∠ ACE = 43° and ∠CAF = 62°. Find the values of a, b and c.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q23.1
Solution:
Now, ∠ ACE = 43° and ∠ CAF = 62° (given)
In ∆ AEC,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q23.2
∠ ACE + ∠ CAE + ∠ AEC = 180°
∴ 43° + 62° + ∠ AEC = 180°
105° + ∠ AEC = 180°
⇒ ∠ AEC = 180°- 105° = 75°
Now, ∠ ABD + ∠AED=180°
(Opposite ∠ s of a cyclic quad, and ∠ AED = ∠ AEC)
⇒ 0 + 75°= 180° a = 180° – 75° = 105°
∠ EDF = ∠ BAE (Angles in the alternate segments)
∴ c = 62°
In ∆BAF, ∠a + 62° + ∠b = 180°
⇒ 105°+ 62°+ ∠b= 180°
⇒ 167° + ∠6 = 180°
⇒ ∠b= 180°-167°= 13°
Hence, a= 105°, 6=13° and c = 62°.

Question 24.
In the given figure, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25° .
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q24.1
Find:
(i) ∠CAD
(ii)∠CBD
(iii) ∠ADC
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q24.2
In the given figure,
ABCD is a cyclic quad, in which AB || DC
∴ ABCD is an isosceles trapezium AD = BC
(i) Join BD
and Ext. ∠BCE = ∠BAD
{ Ext. angle of a cyclic quad, is equal to interior opposite angle}
∴ ∠BAD = 80° (∵ ∠BCE = 80°)
But ∠BAC = 25°
∴ ∠CAD = ∠BAD – ∠BAC = 80° – 25° = 55°
(ii) ∠CBD = ∠CAD (Angles in the same segment)
= 55°
(iii) ∠ADC = ∠BCD (Angles of the isosceles trapezium)
= 180°- ∠BCE =180°- 80° = 100°

Question 25.
ABCD is a cyclic quadrilateral of a circle with centre O such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle. If AJD and BC produced meet at P, show that ∠APB = 60°.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q25.1
Given : In a circle, ABCD is a cyclic quadrilateral in which AB is the diameter and chord CD is equal to the radius of the circle
To prove: ∠APB = 60°
Construction : Join OC and OD
Proof: ∵ chord CD = CO = DO (radii of the circle)
∴ ∆DOC is an equilateral triangle
∠DOC = ∠ODC = ∠OCD – 60°
Let ∠A = x and ∠B = y
∵ OA = OD = OC = OB (radii of the same circle)
∴ ∠ODA = ∠OAD = x and ∠OCB = ∠OBC =y
∴∠AOD = 180° – 2x and ∠BOC = 180° – 2y
But AOB is a straight line
∴ ∠AOD + ∠BOC + ∠COD = 180°
180°- 2x + 180° -2y + 60° = 180°
⇒ 2x + 2y = 240°
⇒ x + y = 120°
But ∠A + ∠B + ∠P = 180° (Angles of a triangle)
⇒ 120° + ∠P = 180°
⇒ ∠P = 180° – 120° = 60°
Hence ∠APB = 60°

Question 26.
In the figure, given alongside, CP bisects angle ACB. Show that DP bisects angle ADB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q26.1
Solution:
Given : In the figure,
CP is the bisector of ∠ACB
To prove : DP is the bisector of ∠ADB
Proof: ∵ CP is the bisector of
∴ ∠ACB ∠ACP = ∠BCP
But ∠ACP = ∠ADP {Angles in the same segment of the circle}
and ∠BCP = ∠BDP
But ∠ACP = ∠BCP
∴ ∠ADP = ∠BDP
∴ DP is the bisector of ∠ADB

Question 27.
In the figure, given below, AD = BC, ∠BAC = 30° and ∠CBD = 70°. Find :
(i) ∠BCD
(ii) ∠BCA
(iii) ∠ABC
(iv) ∠ADB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q27.1
Solution:
In the figure,
ABCD is a cyclic quadrilateral
AC and BD are its diagonals
∠BAC = 30° and ∠CBD = 70°
Now we have to find the measures of ∠BCD, ∠BCA, ∠ABC and ∠ADB
∠CAD = ∠CBD = 70°
(Angles in the same segment)
Similarly ∠BAC = ∠BDC = 30°
∴ ∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°
(i) Now ∠BCD + ∠BAD = 180° (opposite angles of cyclic quad.)
⇒ ∠BCD + 100°= 180°
⇒ ∠BCD = 180°-100° = 80°
(ii) ∵ AD = BC (given)
∴ ∠ABCD is an isosceles trapezium
and AB || DC
∴ ∠BAC = ∠DCA (alternate angles)
⇒ ∠DCA = 30°
∠ABD = ∠D AC = 30° (Angles in the same segment)
∴ ∠BCA = ∠BCD – ∠DAC = 80° – 30° = 50°
(iii) ∠ABC = ∠ABD + ∠CBD = 30° + 70° = 100°
(iv) ∠ADB = ∠BCA = 50° (Angles in the same segment)

P.Q.
In the given below figure AB and CD are parallel chords and O is the centre.
If the radius of the circle is 15 cm, find the distance MN between the two chords of length 24 cm and 18 cm respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Qp1.1
Solution:
Given : AB = 24 cm, CD = 18 cm
⇒AM = 12 cm, CN = 9 cm
Also, OA = OC = 15 cm
Let MO = y cm, and ON = x cm
In right angled ∆AMO
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Qp1.2
(OA)2 = (AM)2 + (OM)2
(15)2 = (12)2 + (y2
⇒ (15)2-(12)2
⇒ y2 = 225-144
⇒ y2 = 81 = 9 cm
In right angled ∆CON
(OC)2 = (ON)2 + (CN)2
⇒(15 )2= x2 + (9)2
⇒ x2 = 225-81
⇒x2= 144
⇒ x = 12 cm
Now, MN = MO + ON =y + x = 9 cm + 12 cm = 21 cm

Question 28.
In the figure given below, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°. Find :
(i) ∠OBD
(ii) ∠AOB
(iii) ∠BED (2016)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q28.1
(i) AD is parallel to BC, that is, OD is parallel to BC and BD is transversal.
∴ ∠ODB = ∠CBD = 32° (Alternate angles)
In ∆OBD,
OD = OB (Radii of the same circle)
⇒ ∠ODB = ∠OBD = 32°
(ii) AD is parallel to BC, that is, AO is parallel to BC and OB is transversal.
∴∠AOB = ∠OBC (Alternate angles)
∠OBC = ∠OBD + ∠DBC
⇒ ∠OBC = 32° + 32°
⇒ ∠OBC = 64°
∴ ∠AOB = 64°
(iii) In AOAB,
OA = OB(Radii of the same circle)
∴ ∠OAB = ∠OBA = x (say)
∠OAB + ∠OBA + ∠AOB = 180°
⇒ x + x + 64° = 180°
⇒ 2x = 180° – 64°
⇒ 2x= 116°
⇒ x = 58°
∴ ∠OAB = 58°
That is ∠DAB = 58°
∴ ∠DAB = ∠BED = 58°
(Angles inscribed in the same arc are equal)

Question 29.
In the given figure PQRS is a cyclic quadrilateral PQ and SR produced meet at T.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q29.1
(i) Prove ∆TPS ~ ∆TRQ
(ii) Find SP if TP = 18 cm, RQ = 4 cm and TR = 6 cm.
(iii) Find area of quadrilateral PQRS if area of ∆PTS = 27 cm2.(2016)
Solution:
(i) Since PQRS is a cyclic quadrilateral ∠RSP + ∠RQP = 180°
(Since sum of the opposite angles of a cyclic quadrilateral is 180°)
⇒ ∠RQP = 180° – ∠RSP …(i)
∠RQT + ∠RQP = 180°
(Since angles from a linear pair)
⇒ ∠RQP = 180° – ∠RQT …(ii)
From (i) and (ii),
180° – ∠RSP = 180° – ∠RQT
⇒ ∠RSP = ∠RQT …(iii)
In ∆TPS and ∆TRQ,
∠PTS = ∠RTQ (common angle)
∠RSP = ∠RQT [From (iii)]
∴ ATPS ~ ATRQ (AA similarity criterion)
(ii) Since ∆TPS ~ ∆TRQ implies that corresponding sides are proportional that
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q29.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C Q29.3

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C.

Other Exercises

Question 1.
Prove that, of any two chords of a circle, the greater chord is nearer to the centre.
Solution:
Given: In circle with centre O and radius r.
OM ⊥ AB and ON ⊥ CD and AB > CD
To Prove: OM < ON
Construction: Join OA, OC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q1.1

Question 2.
OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O.
(i) If the radius of the circle is 10 cm, find the area of the rhombus.
(ii) If the area of the rhombus is 32√3 cm² find the radius of the circle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q2.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q2.2

Question 3.
Two circles with centres A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q3.2
Two circles with centres A and B touch each other at C internally.
PQ is the perpendicular bisector of AB meeting the bigger circles at P and Q. Join AP.
Radius AC = 5 cm.
and radius BC = 3 cm.
AB = AC – BC = 5-3 = 2 cm.

Question 4.
Two chords AB and AC of a circle are equal. Prove that the centre of the circle, lies on the bisector of angle BAC.
Solution:
Given: A circle in which two chords AC and AB are equal in length. AL is the bisector of ∠ BAC.
To Prove: O lies on the bisector of ∠ BAC
Proof: In ∆ ADC and ∆ ADB,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q4.1
AD = AD (Common)
AB = AC (Given)
∠ BAD = ∠ CAD (Given)
∴ ∆ ADC = ∆ ADB (SAS postulate)
∴ BD = DC (C.P.C.T.)
and ∠ ADB = ∠ ADC (C.P.C.T.)
But ∠ ADB + ∠ ADC = 180° (Linear pair)
∴ ∠ ADB = ∠ ADC = 90°
∴ AD is the perpendicular bisector of chord BC.
∵ The perpendicular bisector of a chord passes through the centre of the circle.
∴ AD is the bisector of ∠ BAC passes through the centre O of the circle. Q.E.D.

Question 5.
The diameter and a chord of circle have a common end-point. If the length of the diameter is 20 cm and the length of the chord is 12 cm, how far is the chord from the centre of the circle ?
Solution:
AB is the diameter and AC is the chord
∴ AB = 20 cm and AC = 12 cm
Draw OL ⊥ AC
∵ OL ⊥ AC and hence it bisects AC, O is the center of the circle.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q5.1

Question 6.
ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = 110° and angle BAC = 50°. Find angle DAC and angle DCA.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q6.1
ABCD is a cyclic quad, in which AD || BC
∠ ADC = 110°, ∠ BAC = 50°
∠B + ∠D= 180° (Sum of opposite angles of a cyclic quad.)
⇒∠B + 110°= 180°
∴ ∠ B or ∠ ABC = 180° – 110° – 70″
Now. in ∆ ABC,
∠ BAC + ∠ ABC – ∠ ACB = 180°
⇒ 50° + 70° + ∠ ACB – 180°
⇒ 120° – ∠ ACB = 180°
∴ ∠ ACB = 180° – 120° = 60″
OL ⊥ AC and hence it bisects AC, O is the center of the circle.
OL ⊥ AC and hence it bisects AC, O is the center of the circle.
∵ AD || BC (Given)
∴ ∠ DAC = ∠ ACB (Alternate angles)
= 60°
Now, in ∆ ADC,
∠ DAC + ∠ ADC + ∠ DCA -= 180°
60°+ 110° + ∠ DCA = 180°
170° + ∠ DCA – 180°
∴ ∠ DCA = 180″ – 170° = 10°

Question 7.
In the given figure, C and D arc points on the semi-circle described on AB as diameter. Given angle BAD = 70° and angle DBC = 30°. calculate angle BDC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q7.1
Solution:
∴ ABCD is a cyclic quad.
∠ BAD ∠ BCD 180 (Sum of opposite angles)
⇒ 70° + ∠ BCD = 180°
⇒ ∠ BCD = 180°- 70° = 110°
Now in ∆ BCD,
∠ BCD + ∠ DBC + ∠ BDC = 180°
⇒ 30°+ 110° + ∠ BDC = 180°
⇒ 140°+ ∠ BDC = 180°
∴ ∠ BDC = 180°- 140° = 40°

Question 8.
In cyclic quadrilateral ABCD, ∠ A = 3 ∠ C and ∠ D = 5 ∠ B. Find the measure of each angle of the quadrilateral.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q8.2

Question 9.
Show that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q9.1
Given: ∆ABC in which AB=AC and AB as diameter, a circle is drawn which intersects BC at D.
To Prove: BD = DC
Construction: Join AD
Proof: AB is the diameter
∴ ∠ ADB = 90° (Angle in a semi-circle)
But ∠ ADB + ∠ ADC =180° (A linear pair)
∴ ∠ ADC = 90°
Now in right angled ∆s ABD and ACD,
Hypotenuse AB = AC (given)
Side AD = AD (Common)
∴ ∆ ABD ≅ ∆ ACD (RHS postulate)
BD = DC (C.P.C.T.)
Hence the circle bisects base BC at D. Q.ED.

Question 10.
Bisectors of vertex angles A, B and C of a triangle ABC intersect its circumcircle at the points D, E and F respectively. Prove that.
angle EDF = 90° – \(\frac { 1 }{ 2 }\)∠A.
Solution:
Given: A ∆ ABC whose bisectors of angles A, B and C intersect the circumcircle at D, E and F respectively. ED, EF and DF are joined.
∠ EDF = 90° – \(\frac { 1 }{ 2 }\)∠A
Construction: Join BF, FA, AE and EC.
Proof: ∠ EBF = ∠ ECF = ∠ EDF ……..(i)
(Angles in the same segment)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q10.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q10.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q10.3

Question 11.
In the flgure; AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠ C = 20°, find angle AOD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q11.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q11.2
Join OB
In ∆ OBC, BC = OD = OB (radii of the circle)
∴ ∠ BOC = ∠ BCO = 20°
and Ext. ∠ ABO = ∠ BCO + ∠ BOC
= 20°+ 20° = 40° ….(i)
In ∆ OAB, OA = OB (radii of the circle)
∴ ∠ OAB = ∠ OBA = 40° [from (i)]
∠ AOB = 180°- ∠ OAB – ∠ OBA
= 180°-40°-40°= 100°
∵ DOC is a line
∴ ∠ AOD + ∠ AOB + ∠ BOC = 180°
⇒ ∠ AOD + 100° + 20° = 180°
⇒ ∠ AOD + 120° = 180°
∴ ∠ AOD = 180° – 120° = 60°

Question 12.
Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q12.1
Given: In ∆ ABC, ∠ B = 90° which is inscribed in a circle and O is the incentre of the incircle of ∆ABC.
D and d are the diameters of circumcircle and incircle of ∆ ABC.
To Prove: AB +BC + CA = 2D + d.
Construction: Join OL, OM and ON.
Proof: In circumcircle of ∆ ABC,
∠ B = 90° (given)
∴ AB is the diameter of circumcircle i.e. AB = D.
Let radius of incircle = r
∴ OL = OM = ON = r
Now from B, BL, BM are the tangents to the incircle
∴ BL = OM = r
Similarly we can prove that:
AM = AN and CL = CN = R (radius)
(Tagents from the point outside the circle)
Now AB + BC + CA = AM + BM + BL + CL + CA
= AN + r + r + CN + CA
= AN + CN + 2r + CA
= AC + AC + 2r = 2 AC + 2r = 2D + d Q.E.D.

Question 13.
P is the mid point of an arc APB of a circle. Prove that the tangent drawn at P will be parallel to the chord AB.
Solution:
Given: A circle with centre O, AB is an arc whose mid point is P and AB is chord. TPS is the tangent at P.
To Prove: TPS ||AB.
Construction: Join AP and BP.
Prove: TPS is tangent and PA is chord of the circle
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q13.1
∠ APT = ∠ PBA (angles in the alternate segment)
But ∠ PBA = ∠ PAB (∵ PA = BP)
∴ ∠ APT = ∠ PAB
But these are alternate angles
∴ TPS || AB Q.E.D

Question 14.
In the given figure, MN is the common chord of two intersecting circles and AB is their common tangent Prove that the line NM produced bisects AB at P.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q14.1
Prove that the line NM produced bisects AB at P.
Solution:
Given: Two circles intersect each other at M and N. AB is their common tangent, chord MN intersect the tangent at P.
To Prove: P is mid point of AB.
Proof: From P, AP is the tangent and PMN is the secant of first circle.
∴ AP2 = PM x PN ….(i)
Again from P, PB is the tangent and PMN is the secant of the second circle.
PB2 = PM x PN ….(ii)
from (i) and (ii)
AP2 = PB2 ⇒ AP = PB
∴ P is the mid point of AB. Q.E.D.

Question 15.
In the given figure, ABCD is a cyclic- quadrilateral, PQ is tangent to the circle at point C and BD is its diameter. If ∠ DCQ = 40° and ∠ ABD = 60°, find :
(i) ∠ DBC
(ii) ∠ BCP
(iii) ∠ ADB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q15.1
Solution:
(i) PQ is the tangent and CD is the chord
∴ ∠ DCQ = ∠ DBC (angles in the alternate segment)
∴ ∠ DBC = 40° (∵ ∠ DCQ = 40°)
(ii) ∠ DCQ + ∠ DCB + ∠ BCP = 180°
⇒ 40° + 900 + ∠ BCP = 180°(∵ ∠ DCB = 90°)
⇒ 130°+ ∠ BCP = 180°
∵ ∠ BCP =180° -130° = 50°
(iii) In Δ ABD, ∠ BAD = 90° (Angle in a semi circle) and ∠ ABD = 60°
∴ ∠ ADB = 180°- (60° + 90°)
⇒ 1800- 150° = 30°

Question 16.
The given figure shows a circle with centre O and BCD is tangent to it at C. Show that:
∠ ACD + ∠ BAC = 90°.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q16.1
Solution:
Given: A circle with centie O and BCD is a tangent at C.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q16.2
To Prove: ∠ ACD + ∠ BAC = 90°
Construction: Join OC.
Proof: BCD is the tangent and OC is the radius
∴ OC ⊥ BD
⇒ ∠ OCD = 90°
⇒ ∠ OCA + ∠ ACD = 90° ….(i)
But in ∆ OCA
OA = OC (radii of the same circle)
∴ ∠ OCA = ∠ OAC [from (i)]
∠OAC + ∠ACD = 90°
⇒ ∠ BAC + ∠ ACD = 90° Q.E.D.

Question 17.
ABC is a right triangle with angle B = 90°. A circle with BC as diameter meets hypotenuse AC at point D. Prove that 
(i) AC x AD = AB2
(ii) BD2 = AD x DC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q17.1
Solution:
Given: A circle with BC as diameter meets the hypotenuse of right ∆ ABC with ∠ B = 90° meets at D. BD is joined.
To Prove:
(i) AC x AD = AB2
(ii) BD2 = AD x DC
Proof:
(i) In ∆ABC, ∠ B = 90° and BC is the diameter of the circle.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q17.2

Question 18.
In the given figure. AC = AE.
Show that :
(i) CP = EP
(ii) BP = DP.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q18.1
Solution:
Given: In the figure, AC = AE
To Prove: (i) CP = EP (ii) BP = DP
Proof: In ∆ ADC and ∆ ABE,
AC = AE (given)
∠ ACD = ∠ AEB (Angles in the same segment)
∠ A = ∠ A (Common)
∆ ADC ≅ ∆ ABE (ASA postulate)
AB = AD (C.P.C.T.)
But AC = AE (given)
∴ AC – AB = AE – AD
⇒ BC = DE
Now in ∆ BPC and ∆ DPE,
BC = DE (proved)
∠ C = ∠ E (Angles in the same segment)
∠ CBP = ∠ CDE (Angles in the same segment)
∴ ∆ BPC ≅ ∆ DPE (S.A.S. postulate)
∴ BP=DP (C.P.C.T.)
CP = PE (C.P.C.T.) Q.E.D.

Question 19.
ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC = 120°.
Calculate :
(i) ∠ BEC
(ii) ∠ BED
Solution:
(i) In cyclic pentagon, O is the centre of circle.
Join OB, OC
AB = BC = CD (given)
and ∠ ABC = 120°.
∴ ∠ BCD = ∠ ABC = 120°
OB and OC are the bisectors of ∠ ABC and ∠ BCD respectively.
∴ ∠ OBC = ∠ BCO = 60°
In ∆ BOC,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q19.2

Question 20.
In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If ∠ACO=30°, find
(i) ∠BCO
(ii) ∠AOB
(iii) ∠APB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q20.1
Solution:
Given: In the fig. O is the centre of the circle CA and CB are the tangents to the circle from C. ∠ACO = 30°
P is any point on the circle. PA and PB are joined.
To find:
(i) ∠BCO
(ii) ∠AOB
(iii) ∠APB
Proof:
(i) In ∆ OAC and ∆OBC,
OC=OC (common)
OA = OB (radius of the circle)
CA = CB (tangents to the circle)
∴ ∆OAC ≅ ∆OBC (SSS axion)
∴ ∠ACO = ∠BCO = 30°
(ii) ∴ ∠ACB = 30° + 30° = 60°
∴ ∠AOB + ∠ACB = 180°
⇒ ∠AOB+ 60° =180°
∴ ∠AOB = 180° – 60° = 120°
(iii) Arc AB, subtends ∠AOB at the centre and ∠APB is in the remaining part of the circle
∴ ∠APB = \(\frac { 1 }{ 2 }\) ∠AOB = \(\frac { 1 }{ 2 }\) x 120° = 60°

Question 21.
ABC is triangle with AB = 10 cm, BC = 8 cm and AC = 6 cm (not drawn to scale). Three circles are drawn touching each other with the vertices as their centres. Find the radii of the three circles. (2011)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q21.1
Solution:
Given: ABC is a triangle with AB = 10 cm, BC = 8 cm, AC = 6 cm. Three circle are drawn with centre A, B and C touch each other at P, Q and R respectively
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q21.2

Question 22.
In a square ABCD, its diagonals AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and the bisector of angle ABD meets AC at N and AM at L.
Show that :
(i) ∠ ONL + ∠ OML = 180°
(ii) ∠ BAM = ∠ BMA
(iii) ALOB is a cyclic quadrilateral
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q22.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q22.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q22.3

Question 23.
The given figure shows a semi-circle with centre O and diameter PQ. If PA = AB and ∠ BCQ = 140°; find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q23.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q23.2
Join PB
In cyclic quad. PBCQ,
∠ BPQ + ∠ BCQ = 180°
⇒ ∠ BPQ + 140° = 180°
∴ ∠ BPQ = 180° – 140° = 40°
Now, in ∆ PBQ,
∠ PBQ + ∠ BPQ + ∠ BQP = 180°
⇒ 90° + 40° + ∠ BQP = 180°
(∠ PBQ = 90° angle in a semicircle)
⇒ 130° + ∠ BQP = 180°
∴ ∠ BQP = 180° – 130°= 50°
Nowin cyclic quad. PQBA,
∠ PQB + ∠ PAB = 180°
⇒ 50° + ∠ PAB = 180°
∴ ∠ PAB = 180° – 50° = 130°
(ii) Now, in ∆ PAB,
∠ PAB + ∠ APB + ∠ ABP = 180°
⇒ 130° + ∠ APB + ∠ ABP = 180°
⇒ ∠ APB + ∠ ABP = 180° – 130° = 50°
But ∠ APB = ∠ ABP = 25°
∴ (PA = AB)
∠ BAQ = ∠ BPQ = 40°
(Angles in the same segment)
Now, in ∆ ABQ,
∠ AQB + ∠ QAB + ∠ ABQ = 180°
⇒ ∠ AQB + 40° + 115° = 180°
⇒ ∠ AQB + 155° = 180°
⇒ ∠ AQB = 180° – 155° = 25°
(iii) Arc AQ subtends ∠ AOQ at the centre and ∠APQ at the remaining part of the circle,
∠ AOQ = 2 ∠ APQ = 2 x 65° = 130°
Now, in ∆ AOQ,
∠ OAQ = ∠ OQA
(∵ OA = OQ radii of the same circle)
But ∠ OAQ + ∠ OQA + ∠ AOQ = 180°
⇒ ∠ OAQ + ∠ OAQ + 130° = 180°
2 ∠ OAQ = 180° – 130° = 50°
∴ ∠ OAQ = 25°
∵ ∠ OAQ = ∠ AQB (each = 25°)
But these are alternate angles.
∴ AO || BQ.
Q.E.D.

Question 24.
The given figure shows a circle with centre O such that chord RS is parallel to chord QT, angle PRT = 20° and angle POQ = 100°. Calculate
(i) angle QTR
(ii) angle QRP
(iii) angle QRS
(iv) angle STR
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q24.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q24.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q24.3

Question 25.
In the given figure, PAT is tangent to the circle with centre O, at point A on its circumference and is parallel to chord BC. If CDQ is a line segemcnt, show that :
(i) ∠ BAP = ∠ ADQ
(ii) ∠ AOB = 2 ∠ ADQ
(iii) ∠ ADQ = ∠ ADB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q25.1
Solution:
Given: PAT is the tangent to the circle with centre O, at A. Chord BC || PAT is drawn.
CDQ is a line segment which intersects the circle at C and D and meets the tangent PAT at Q.
To Prove:
(i) ∠ BAP = ∠ ADQ (ii)∠ AOB = 2 ∠ ADQ
(iii) ∠ ADQ = ∠ ADB
Proof:
(i) ∵ PAT || BC
∴ ∠ PAB = ∠ ABC (Alternate angles) ….(i)
In cyclic quad. ABCD,
Ext. ∠ADQ = ∠ABC ….(ii)
∴ ∠ PAB = ∠ ADQ (from (i) and (ii))
(ii) Arc AB subtends ∠ AOB at the centre and ∠ADB at the remaining part of the circle.
∴ ∠ AOB = 2 ∠ ADB
= 2 ∠ PAB (In the alt. segment)
= 2 ∠ ADQ [proved in (i)]
(iii) ∵ ∠ BAP = ∠ ADB
(Angles in the alt. segment)
Bui ∠ BAP = ∠ ADQ (Proved in (i))
∴ ∠ ADQ = ∠ ADB Q.E.D.

Question 26.
AB is a line segment and M is its mid-point. Three semi-circles are drawn with AM, MB and AB as diameters on the same side of the line AB. A circle with radius r unit is drawn so that it touches all the three semi-circles. Show that: AB = 6 x r
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q26.1
Given: A line segment AB whose mid-point is M. Three circles are drawn on AB, AM and MB as diameler A circle with radius r is drawn which touches (he three circles externally at L, R and N respectively. M, P, Q are the centres of the three circles.
To Prove: AB= 6r
Construction: Join OP and OQ.
Proof: OM = ON = r
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q26.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q26.3

Question 27.
TA and TB are tangents to a circle with centre O from an external point T. OT intersects the circle at point P. Prove that AP bisects the angle TAB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q27.1
Given: A circle with centre C. From a point T outside th circle, TA and TB are two tangent to the circle OT intersects the circle at P, AP and AB are joined.
To Prove: AP is the bisector of ∠ TAB
Construction: Join PB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q27.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q27.3

Question 28.
Two circles intersect in points P and Q. A secant passing through P intersects the circles in A and B respectively. Tangents to the circles at A and B intersect at T. Prove that A, Q, B and T lie on a circle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q28.1
Given: Two circles intersect each other at P and Q. From P, a secant intersects the circles at A and B respectively. From A and B tangents are drawn which intersect each other at T.
To Prove: A. Q, B and T on a circle.
Construction: Join PQ.
Proof: AT is the tangent and AP is chord
∴ ∠ TAP = ∠ AQP (Angles in all. segment) …(i)
Similarly ∠ TBP = ∠ BQP ….(ii)
Adding (i) and (ii),
∠ TAP + ∠ TBP = ∠ AQP + ∠ BQP = ∠ AQB …..(iii)
Now, in ∆ TAB,
∠ ATB + ∠ TAP + ∠ TBP = 180°
⇒ ∠ ATB + ∠ AQB = 180° (from (iii)
∴ AQBT is a cyclic quadrilateral.
Hence A, Q, B and T lie on the same circle. Q.E.D.

Question 29.
Prove that any four vertices of a regular pentagon are concyclic (lie on the same circle).
Solution:
ABCDE is a regular pentagon.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q29.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q29.2
To Prove: Any four vertices lie on the same circle.
Construction: Join AC.
Proof: Each angle of a regular pentagon

Question 30.
Chords AB and CD of a circle when extended meet at point X. Given AB = 4 cm, BX = 6 cm and XD = 5 cm, calculate the length of CD. [2000]
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q30.1
Let CD = x
∴ chords AB and CD intersect each other at outside the circle.
∴ AX.XB = CX.XD
⇒(4+6)x6 = (x + 5)x5
⇒ 10 x 6 = 5x + 25
⇒ 60 = 5x + 25
⇒ 5x = 60 – 25 = 35
∴ x = \(\frac { 35 }{ 5 }\) = 7
CD = 7 cm

Question 31.
in the given figure. find TP if AT = 16 cm AB = 12 cm.
Solution:
In the figure.
PT is the tangent and TBA is the secant of the circle.
∴ TP2 = TA x TB = 16 x (16 – 12) = 16 x 4 = 64 = (8)2
Hence, TP = 8 cm.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q31.1

Question 32.
In the following figure, a circle is inscribed in the quadrilateral ABCD. If BC = 38 cm. QB = 27 cm, DC = 25 cm and that AD is perpendicular to DC, find the radius of the circle. (1990)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q32.1
Solution:
A circle with centre is inscribed (see the fig.)
in a quadrilateral ABCD. BC = 38 cm, QB = 27cm,
DC = 25 cm and AD ⊥ BC.
Join OP and QS.
∵ OP and OS are the radii of the circle
∴ OP ⊥ AD and OS ⊥ CD
∴ OPDS is a square
∴ OP = OS – DP = DS.
Let length of radius of the circle = r
then DP = DS = r
∴ CS = 25 – r
∵ EQ = BR = 27 cm (tangents to the circle from B)
∴ CR = BC – BR = 38 – 27 = 11 cm
Similarly CR = CS
∴ 25 – r = 11 ⇒ r = 25 – 11 = 14
∴ Radius of the circle = 14 cm

Question 33.
In the given figure, XY is the diameter of the circle and PQ is a tangent to the circle at Y.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q33.1
If ∠AXB = 50° and ∠ABX = 70°, find ∠BAY and ∠APY.
Solution:
In the above figure,
XY is a diameter of the circle PQ is tangent to the circle at Y.
∠AXB = 50° and ∠ABX = 70°
(i) In ∆AXB,
∠XAB + ∠ABX + ∠AXB = 180° (Angles of a triangle)
⇒ ∠XAB + 70° + 50° = 180°
⇒ ∠XAB + 120° =180°
⇒ ∠XAB = 180° – 120° = 60°
But, ∠XAY = 90° (Angle in a semicircle)
∴ ∠BAY = ∠XAY-∠XAB=90°- 60° = 30°
(ii) Similarly ∠XBY=90°(Angle in a semicircle) and ∠CXB = 70°
∴ ∠PBY = ∠XBY-∠XBA =90° – 70° = 20°
∵ ∠BYA = 180° – ∠AXB ( ∵ ∠BYA + ∠AYB = 180°) = 180°- 50° = 130°
∠PYA =∠ABY (Angles in the alternate segment) = ∠PBY = 20°
and ∠PYB = ∠PYA + ∠AYB
= 20° + 130° = 150°
∴ ∠APY = 180°-(∠PYA + ∠ABY)
= 180° -(150° +20°) =180° – 170° = 10°

Question 34.
In the given figi QAP is the tangent point A and PBD is a straight line. If ∠ACB = 36° and ∠APB = 42°, find:
(i) ∠ BAP
(ii) ∠ABD
(iii) ∠ QAD
(iv) ∠ BCD
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q34.1
Solution:
In the given figure, QAP is the tangent to the circle at A and PBD is a B straight line.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q34.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q34.3

Question 35.
In the given figure, AB is a diameter. The tangent at C meets AB produced at Q. If ∠ CAB = 34°,
find :
(i) ∠CBA
(ii) ∠CQB
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q35.1
In ∆ ABC,
we have ∠ ACB = 90°
[Angle in a semicircle is 90°]
(i) Also ∠ CBA + ∠ CAB + ∠ ACB = 180° [Angle sum property of a ∆ ]
⇒ ∠ CBA =180°- ∠ CAB – ∠ ACB = 180°-34°-90° = 180°-124° = 56°
(ii) CQ is a tangent at C and CB is a chord of the circle.
⇒ ∴ ∠ QCB = ∠ BAC = 34° [Angles in the alternate segments]
∠ CBQ =180°- ∠ ABC [Linear pair]
⇒ ∠ CBQ = 180°- 56° = 124° [From (i)]
In ∆ BCQ, we have
⇒ ∠ CQB = 180° – (∠ QCB + ∠ CBQ) [Angle sum property of a ∆ ]
= 180° -(34° + 124°) = 180°- 158° = 22°
Hence, ∠CQB = 22°

Question 36.
In the given figure, A O is the centre of the circle. The tangents at B and D intersect each other at point P.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q36.1
If AB is parallel to CD and ∠ ABC = 55
find :
(i) ∠ BOD
(ii) ∠ BPD
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q36.2
In the given figure,
O is the centre of the circle AB || CD,
∠ ABC = 55° tangents at B and D are drawn which meet at P.
∵ AB || CD
∴ ∠ ABC = ∠ BCD (Alternate angles)
∴ ∠ ABC = 55° (Given)
(i) Now arc BD subtands ∠ BOD at the centre and ∠ BCD at the remaining part of the circle.
∴ ∠BOD = 2∠BCD = 2 x 55° = 110°
(ii) In quad. OBPD,
∠OBP = ∠ODP = 90° (∵ BP and DP are tangents)
∴ ∠BOD + ∠BPD = 180°
⇒ 110° +∠BPD =180°
⇒ ∠BPD =180°-110°= 70°
Hence, ∠BOD = 110° and ∠BPD = 70°

Question 37.
In the figure given below PQ = QR, ∠RQP = 68°, PC and CQ are tangents to the circle with centre O. Calculate the values of:
(i) ∠QOP
(ii) ∠QCP
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q37.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q37.2

Question 38.
In two concentric circles, prove that all chords of the outer circle, which touch the inner circle, are of equal length.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q38.1
Solution:
Given: Two concentric circles with centre O AB and CD are two cords of outer circle which touch the inner circle at P and Q respectively
To prove: AB = CD
Construction : Join OA, OC, OP and OQ
Proof: ∵ OP and OQ are the radii of the inner circle and AB and CD are tangents
∴ OP ⊥ AB and OQ ⊥ CD
and P and Q are the midpoints of AB and CD Now in right AOAP and OCQ,
Side OP = OQ (radii of the inner circle)
Hyp. OA = OC (radii of the outer circle)
∴ ∆OAP = ∆OCQ (R.H.S. axiom)
∴ AP = CQ (c.p.c.t.)
But AP = \(\frac { 1 }{ 2 }\) AB and CQ = \(\frac { 1 }{ 2 }\) CD
∴ AB = CD Hence proved.

Question 39.
In the figure, given below, AC is a transverse common tangent to two circles with centres P and Q and of radii 6 cm and 3 cm respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q39.1
Given that AB = 5 cm , Calculate PQ.
Solution:
In the figure, two circles with centres P and Q and radii 6 cm and 3 cm respectively
ABC is the common transverse tangent to the two circles. AB = 8 cm
Join AP and CQ
∵ AC is the tangents to the two circles and PA and QC are the radii
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q39.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q39.3

Question 40.
In the figure, given below, O is the centre of the circumcircle of triangle XYZ. Tangents at X and Y intersect at point T. Given ∠XTY = 80° and ∠XOZ = 140°, calculate the value of ∠ZXY.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q40.1
Solution:
In the figure, a circle with centre O, is the circumcircle of ∆XYZ.
∠XOZ =140° (given)
Tangents from X and Y to the circle meet at T such that ∠XTY = 80°
∵ ∠XTY = 80°
∴ ∠XOY= 180°-80°= 100°
But ∠XOY + ∠YOZ + ∠XOZ = 360° (Angles at a point)
⇒ 100°+∠YOZ+ 140o = 360o
⇒ 240o+∠YOZ =360°
⇒ ∠YOZ =360°- 240°
⇒∠YOZ =120°
Now arc YZ subtends ∠YOZ at the centre and ∠YXZ at the remaining part of the circle
∴ ∠YOZ = 2 ∠YXZ
⇒ ∠YXZ= \(\frac { 1 }{ 2 }\) ∠YOZ ⇒ ∠YXZ = \(\frac { 1 }{ 2 }\) x 120° = 60°

Question 41.
In the given figure, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, findAE.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q41.1
Solution:
In the given circle,
Chords AE and BC interesect each other at D at right angle i.e., ∠CDE = 90°, AB is joined AB = 5cm, BD = 4 cm, CD = 9 cm
Now we have to find AE.
Let DE=xm
Now in right ∆ABD,
AB= AD2 + BD2 (Pythagoras Theorem)
⇒ (5)2=AD2 + (4)2
⇒ 25 = AD2 + 16
⇒ AD2 = 25-16 = 9 = (3)2
∴AD = 3cm
∵ Chords AE and BC intersect eachothcr at D inside the circle
∴ AD x DE = BD x DC
⇒ 3 x x = 4 x 9
⇒ x= \(\frac { 4×9 }{ 3 }\) = 12cm;
∴ AE=AD + DE = 3 + 12 = 15 cm

Question 42.
In the given circle with centre O, ∠ABC = 100°, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q42.1
Solution:
ABCD is a cyclic quadrilateral
∴ ∠ABC + ∠ADC = 180°
100° +∠ADC = 180°
∠ADC = 180°- 100° = 80°
In ∆ADC
∠ACD + ∠CDA + ∠D AC =180°
40° + 80° +∠D AC = 180°
∠D AC = 180° – 80° – 40° = 60°
Now ∠DAC = 60°
⇒ ∠DCT = 60° [angle in alt. segment]

Question 43.
In the figure given below, O is the centre of the circle and SP is a tangent. If ∠SRT = 65°, find the values of x,y and z. (2015)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C Q43.1
Solution:
In the given figure,
O is the centre of the circle.
SP is tangent ∠SRT =65°.
To find the values of x, y and z
(i) In ∆STR,
∠S = 90° (∵ OS is radius and ST is tangent)
∴ ∠T + ∠R = 90°
⇒ x + 65° = 90°
⇒ x = 90° – 65° = 25°
(ii) Arc CQ subtends ∠SOQ at the centre and
∠STQ at the remaining part of the circle.
∠y = ∠QOS = 2∠T = 2∠x = 2 x 25° = 50°
(iii) In ∆OSP,
∠S = 90°
∴ ∠SOQ or ∠SOP + ∠P = 90°
⇒ y+z=90o
⇒ 50° + z = 90°
⇒ z = 90°-50° = 40°
Hence x = 25°, y = 50° and z = 40°

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C  are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A 

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A 

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A.

Other Exercises

Question 1.
The radius of a circle is 8cm. Calculate the length of a tangent drawn to this circle from a point at a distance of 10cm from its centre.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q1.1
OP = 10 cm,
raduis OT = 8 cm
∵ OT ⊥ PT
∴ In right ΔOTP,
OP= OT2+PT2
⇒  (10)2 =(8)2+PT2
⇒  100 = 64+PT2
⇒ PT2 = 100-64 = 36
∴ PT = \( \sqrt{36} \) = 6 cm

Question 2.
In the given figure O is the centre of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q2.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q2.2
∠OBA = 90° (Radius through the point of contact is perpendicular to the tangent)
⇒  OB2 = OA2 – AB2 ⇒ r2 = (r + 7.5)2 – 152
⇒ r2 = r2 + 56.25 + 15r – 225 168.75
⇒   15r= 168.75
⇒  r =\(\frac { 168.75 }{ 75 }\) ⇒ r=11.25
Hence, radius of the circle = 11.25 cm

Question 3.
Two circles touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q3.1
Solution:
Given: Two circles with centre O and O’ touches at P externally. Q is a point on the common tangent through P.
QA and QB are tangents from Q to the circles respectively.
To Prove: QA=QB.
Proof: From Q, QA and QP are the tangents to the circle with centre O
∴  QA=QP ….(i)
Similarly, QP and QB are the tangents to the circle with centre O’
∴ QP=QB   ….(ii)
From (i) and (ii)
QA=QB                           Q.ED.

Question 4.
Two circles touch each other internally. Show that the tangents drawn to the two circles from any point on the common tangent, are equal in length.
Solution:
Given: Two circles with centre O and O’ touch each other internally at P. Q is a point on the common tangent through P. QP an QB are tangents from Q to the circles respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q4.1

Question 5.
Two circles of radii 5 cm and 3 cm are concentric. Calculate the length of a chord of the outer circle which touches the inner.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q5.1
Solution:
Given: Two concentric circles with radius 5 cm and 3 cm with centre O. PQ is the chord of the outer circle which touches the inner circle at L. Join OL and OP.
OL=3 cm, OP = 5 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q5.2

Question 6.
Three circles touch each other externally. A triangle is formed when the centres of these circles are j oined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.
Solution:
Three circles touches each other externally
Δ ABC is formed by joining the centres A, B and C of the circles.
AB = 6 cm, AC = 8 cm and BC = 9 cm
Let radii of the circles having centres A, B and C be r1, r2, r3 respectively
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q6.2

Question 7.
If the sides of a quadrilateral ABCD touch a circle, prove that: AB + CD = BC + AD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q7.1
Solution:
Given: A circle touches the sides AB, BC, CD and DA of quad. ABCD at P,Q,R and S respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q7.2
To Prove: AB + CD = BC + AD
Proof: Since AP and AS are the tangents to the circle from external point A.
∴ AP = AS     ….(i)
Similarly, we can prove that,
BP=BQ               ….(ii)
CR=CQ             …(iii)
DR=DS          ….(iv)
Adding, we get:
AP + BP + CR + DR = AS + BQ + CQ + DS
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence AB + CD = BC + AD.     Q.E.D.

Question 8.
If the sides of a parallelogram touch a circle (refer figure of Q/7) prove that the parallelogram is a rhombus.
Solution:
Given : The sides AB, BC, CD and DA of ||gm ABCD touches the circle at P, Q, R and S respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q8.1
To Prove : ABCD is a rhombus.
Proof : From A, AP and AS are the tangents to the circle.
∴ AP = AS    ….(i)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q8.2

Question 9.
From the given figure, prove that:
AP + BQ + CR = BP + CQ + AR.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q9.1
Solution:
Given: In the figure, sides of Δ ABC touch a circle at P, Q, R.
To Prove:
(i) AP + BQ + CR = BP + CQ + AR
(ii) AP + BQ + CR = \(\frac { 1 }{ 2 }\) Perimeter of Δ ABC.
Proof :
∵  From B, BQ and BP are the tangents to the circle.
∴ BQ = BP   ….(i)
Similarly we can prove that
AP= AR    ……… (ii)
and CR = CQ  …..(iii)
Adding we get
AP + BQ + CR = BP + CQ + AR ….(iv)
Adding AP + BQ + CR both sides,
2(AP + BQ + CR) = AP + PQ + CQ + QB + AR+CR.
⇒  2 (AP + BQ + CR) = AB + BC + CA
∴ AP + BQ + CR = \(\frac { 1 }{ 2 }\) (AB + BC + CA)
= \(\frac { 1 }{ 2 }\) Perimeter of Δ ABC.            Q.E.D.

Question 10.
In the figure of Q.9 if AB = AC then prove that BQ = CQ.
Solution:
Given:  A circle touches the sides AB, BC, CA of Δ ABC at P, Q and R respectively, and AB = AC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q10.1

Question 11.
Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centres if 
(i) they touch each other externally,
(ii) they touch each other internally.
Solution:
Radius of bigger circle = 6.3 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q11.1
and raduis of smaller circle = 3.6 cm.
(i) Two circles touch each other at P externally. 0 and O’ are the centres of the circles.
Join OP and OP’
OP = 3.6 cm, O’P = 6.3 cm.
Adding we get
OO’ = OP + O’P = 3.6 + 6.3 = 9.9 cm
(ii) If the circles touch each other internally at P.
OP = 3.6 cm and O’P = 6.3 cm.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q11.2
∴ OO’ = O’P – OP
= 6.3 – 3.6 = 2.7 cm

Question 12.
From a point P outside a circle, with centre O, tangents PA and PB are drawn. Prove that:
(i) ∠AOP = ∠BOP,
(ii) OP is the ⊥ bisector of chord AB.
Solution:
Given:  A circle with centre 0. A point P out side the circle. From P, PA and PB are the tangents to the circle, OP and AB are joined.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q12.1
To prove:
(i) ∠AOP = ∠BOP
(ii) OP is the perpendicular bisector of chord AB.
Proof : In ∆ AOP and ∆ BOP,
AP = BP (Tangents from P to the circle.)
OP = OP (Common)
OA = OB (Radii of the same circle)
∴ ∆ AOP s ∆ BOP (SSS postulate)
∴∠AOP = ∠BOP (C.P.C.T.)
Now in ∆ OAM and ∆ OBM,
OA = OB (Radii of the same circle)
OM = OM (Common)
∠AOM = ∠BOM (Proved ∠AOP = ∠BOP)
∴ ∆ OAM = ∆ OBM (S.A.S. Postulate)
∴ AM = MB (C.P.C.T.)
and ∠OMA = ∠OMB (C.P.C.T.)
But ∠OMA + ∠OMB = 180° (Linear pair)
∴ ∠OMA = ∠OMB = 90°
Hence OM or OP is the perpendicular bisector of AB. Q.E.D.

Question 13.
In the given figure, two circles touch each other externally at point P, AB is the direct common tangent of these circles. Prove that:
(i) tangent at point P bisects AB.
(ii) Angle APB = 90°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q13.1
Solution:
Given : Two circles with centre O and O’ touch each other at P externally. AB is the direct common tangent touching the circles at A and B respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q13.2
AP, BP are joined. TPT’ is the common tangent to the circles.
To Prove : (i) TPT’ bisects AB (ii) ∠APB = 90°
Proof :
∵ TA and TP arc the tangents to the circle
∴ TA = TP …(i)
Similarly TP. = TB ….(ii)
From (i) and (ii)
TA = TB
∴ TPT’ is the bisector of AB.
Now in ∆ ATP
TA = TP
∴ ∠TAP = ∠TPA
Similarly in A BTP.
∠TBP = ∠TPB
Adding we get.
∠TAP + ∠TBP = ∠APB
But ∠TAP + ∠TBP + ∠APB = 180°
∴ ∠APB = ∠TAP + ∠TBP = 90°. Q.E.D.

Question 14.
Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that:
∠PAQ = 2∠OPQ
Solution:
Given: A circle with centre O. two tangents PA and QA are drawn from a point A out side the circle OP, OQ. OA and PQ are joined.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q14.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q14.2

Question 15.
Two parallel tangents of a circle meet a third tangent at points P and Q. Prove that PQ subtends a right angle at the centre.
Solution:
Given: A circle with centre O, AP and BQ are two parallel tangents. A third tangent PQ intersect them at P and Q. PO and QO are joined
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q15.2

Question 16.
ABC is a right angled triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q16.1
Calculate the value of x, the radius of the inscribed circle.
Solution:
In ∆ ABC, ∠B = 90°
OL ⊥ AB, OM ⊥ BC and
ON ⊥ AC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q16.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q16.3

Question 17.
In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate :
(i) ∠ QOR
(ii) ∠ QPR given that ∠ A = 60°.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q17.1

Question 18.
In the following figure, PQ and PR are tangents to the circle, with centre O. If ∠ QPR = 60°, calculate:
(i) ∠ QOR
(ii) ∠ OQR
(iii) ∠ QSR.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q18.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q18.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q18.3

Question 19.
In the given figure, AB is the diameter of the circle, with centre O, and AT is the tangent. Calculate the numerical value of x.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q19.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q19.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q19.3

Question 20.
In quadrilateral ABCD; angle D = 90°, BC = 38 cm and DC = 25 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 27 cm. Find the radius of the circle. |1990]
Solution:
BQ and BR arc the tangenls from B to the circle.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q20.1
∴ BR = BQ = 27 cm.
∴ RC = 38-27 = 11 cm.
Since CR and CS are the tangents from C to the circle
∴ CS = CR= 11 cm.
∴ DS = 25 – 11 = 14 cm.
DS and DP are the tangents to the circle
∴ DS = DP
∴ ∠ PDS = 90° (given)
and OP ⊥ AD, OS ⊥ DC
∴ Radius = DS = 14 cm

Question 21.
In the given, PT touches the circle with centre O at point R. Diameter SQ is produced to meet the tangent TR at P.
Given ∠ SPR = x° and ∠ QRP = y°;
Prove that
(i) ∠ ORS = y°
(ii) Write an expression connecting x and y. [1992]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q21.1
Solution:
∠ QRP = ∠ OSR = y (Angles in the alternate segment)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q21.2
But OS = OR (radii of the same circle)
∴ ∠ ORS = ∠ OSR = y°
∴ OQ = OR (radii of the same circle)
∴ ∠ OQR = ∠ ORQ = 90° – y° ….(i) (OR ⊥ PT)
But in ∆ PQR,
Ext. ∠ OQR = x° + y° ….(ii)
from (i) and (ii)
x° + y° = 90° – y°
⇒ x° + 2y° = 90°

Question 22.
PT is a tangent to the circle at T. If ∠ ABC = 70° and ∠ ACB = 50°; calculate :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q22.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q22.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q22.3

Question 23.
In the given figure. O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle AOB = 140° and angle APC 80°; find the angle BAC. [1996]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q23.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A Q23.2
Join OC
∴ PA and PC are the tangents
∴ OA ⊥ PA and OC ⊥ PC
In quad APCO,
∴ ∠ APC + ∠ AOC = 180°
⇒ 80° + ∠ AOC = 180°
∴ ∠ AOC = 180° – 80° = 100°
∠ BOC = 360° – (∠ AOB + ∠ AOC)
= 360° – (140° + 100°)
= 360° – 240° = 120°
Now arc BC subtends ∠ BOC at the centre and ∠BAC at the remaining part of the circle.
∴ ∠ BAC = \(\frac { 1 }{ 2 }\)∠BOC
= \(\frac { 1 }{ 2 }\) x 120° = 60°

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A  are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B 

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B 

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B.

Other Exercises

Question 1.
(i) In the given figure 3 x CP = PD = 9 cm and AP = 4.5 cm. Find BP.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q1.1
(ii) In the given figure, 5 x PA = 3 x AB = 30 cm and PC = 4 cm. Find CD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q1.2
(iii) In the given figure, tangent PT = 12.5 cm and PA = 10 cm; find AB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q1.3
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q1.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q1.5

Question 2.
In the figure given below, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find:
(i) AB.
(ii) the length of tangent PT. (2014)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q2.1
Solution:
PT is tangent and PDC is secant out to the circle
∴ PT² = PC x PD
PT² = (5 + 7.8) x 5 = 12.8 x 5
PT² = 64 ⇒ PT = 8 cm
In ΔOTP
PT² + OT² = OP²
8²+x² = (x + 4)²
⇒ 64 +x² = x² + 16 + 8x
64- 16 = 8x
⇒ 48 = 8x
x = \(\frac { 48 }{ 8 }\) = 6 cm
∴ Radius = 6 cm
AB = 2 x 6 = 12 cm

Question 3.
In the following figure, PQ is the tangent to the circle at A, DB is the diameter and O is the centre of the circle. If ∠ ADB = 30° and ∠ CBD = 60°, calculate
(i) ∠ QAB
(ii) ∠ PAD
(iii) ∠ CDB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q3.1
Solution:
(i) PAQ is a tangent and AB is the chord, ∠ QAB = ∠ ADB (Angles in the alternate segment)
= 30°
(ii) OA = OD (Radii of the same circle)
∴ ∠ OAD = ∠ ODA = 30°
But OA ⊥ PQ
∴ ∠ PAD = ∠ OAP – ∠OAD = 90° – 30° = 60°
(iii) BD is diameter
∴ ∠ BCD = 90° (Angle in semi circle)
Now in ∆ BCD,
∠ CDB + ∠ CBD + ∠ BCD = 180°
⇒ ∠ CDB + 60° + 90° = 180°
⇒ ∠ CDB = 180°- (60° + 90°) = 180° – 150° = 30°

Question 4.
If PQ is a tangent to the circle at R; calculate:
(i) ∠ PRS
(ii) ∠ ROT.

Given O is the centre of the circle and angle TRQ = 30°.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q4.1
PQ is tangent and OR is the radius
∴ OR ⊥ PQ
∴ ∠ORT = 90° =
∠TRQ = 90° – 30° = 60°
But in ∆ OTR, OT = OR (Radii of the circle)
∴ ∠ OTR = 60° or ∠STR = 60°
But ∠PRS = ∠STR (Angles in the alternate segment) = 60°
In ∆ ORT, ∠ OTR = 60°, ∠ TOR = 60°
∴ ∠ROT= 180°-(60°+ 60°)= 180°-120° = 60°

Question 5.
AB is the diameter and AC is a chord of a circle with centre O such that angle BAC = 30°. The tangent to the circle at C intersect AB produced in D. Show that BC = BD.
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q5.1
Given: In a circle, O is the centre,
AB is the diameter, a chord AC such that ∠ BAC = 30°
and a tangent from C, meets AB in D on producing. BC is joined.
To Prove: BC = BD
Construction: Join OC
Proof : ∠ BCD = ∠ BAC =30°(Angle in alternate segment)
Arc BC subtends ∠ DOC at the centre of the circle and ∠ BAC at the remaining part of the circle.
∴ ∠ BOC = 2 ∠ BAC = 2 x 30° = 60°
Now in ∆ OCD,
∠ BOC or ∠ DOC = 60° (Proved)
∠ OCD = 90° (∵ OC ⊥ CD)
∴ ∠ DOC + ∠ ODC = 90°
⇒ 60° + ∠ ODC = 90°
∴ ∠ ODC = 90°- 60° = 30°
Now in ∆BCD,
∵∠ ODC or ∠ BDC = ∠ BCD (Each = 30°)
∴ BC = BD Q.E.D.

Question 6.
Tangent at P to the circumcircle of triangle PQR is drawn. If this tangent is parallel to side QR, show that ∆PQR is isosceles.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q6.1
In a circumcircle of ∆PQR, a tangent TPS is drawn through P which is parallel to QR
To prove : ∆PQR is an isosceles triangle.
Proof:
∵ TS \(\parallel\) QR
∠TPQ = ∠PQR (Alternate angles) ….(i)
∵ TS is tangent and PQ is the chord of the circle
∴ ∠TPQ = ∠RP (Angles in the alternate segment) ….(ii)
From (i) and (ii),
∠PQR = ∠QRP
∴ PQ = PR (Opposite sides of equal angles)
∴ ∆PQR is an isosceles triangle
Hence proved

Question 7.
Two circles with centres O and O’ are drawn to intersect each other at points A and B. Centre O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with centre O’ at A. Prove that OA bisects angle BAC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q7.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q7.2
Given: Two circles with centre O and O’ intersect each other at A and B, O lies on the circumference, of the other circle. CD is a tangent at A to the second circle. AB, OA are joined.
To Prove: OA bisects ∠ BAC.
Construction: Join OB, O’A, O’B and OO’
Proof: CD is the tangent and AO is the chord
∠ OAC = ∠ OBA …(i)
(Angles in alt. segment)
In ∆ OAB, OA = OB (Radii of the same circle)
∴ ∠ OAB = ∠ OBA ….(ii)
From (i) and (ii),
∠ OAC = ∠ OAB
∴ OA is the bisector of ∠ BAC Q.E.D.

Question 8.
Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that: ∠ CPA = ∠DPB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q8.1
Solution:
Given:Two circles touch each other internally at P. A chord AB of the bigger circle intersects the smaller circle at C and D. AP, BP, CP and DP are joined.
To Prove: ∠ CPA = ∠ DPB
Construction: Draw a tangent TS at P to the circles given.
Proof:
∵ TPS is the tangent, PD is the chord.
∴ ∠ PAB = ∠ BPS …(i) ( Angles in alt. segment)
Similarly we can prove that
∠ PCD = ∠ DPS …(ii)
Subtracting (i) from (ii), we gel
∠ PCD – ∠ PAB = ∠ DPS – ∠ BPS
But in ∆ PAC,
Ext. ∠ PCD = ∠ PAB + ∠ CPA
∴ ∠ PAB + ∠ CPA – ∠ PAB = ∠ DPS – ∠ BPS
∠ CPA = ∠ DPB           Q.E.D.

Question 9.
In a cyclic quadrilateral ABCD, the diagonal AC bisects the angle BCD. Prove that the diagonal BD is parallel to the tangent to the circle at point A.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q9.1
Given: ABCD is a cyclic quadrilateral and diagonal AC bisects ∠ BCD. AT A. a tangent TAS is drawn. BD is joined.
To Prove: TS || BD.
Proof: ∠ ADB = ∠ ACB …….(i) (Angles in the same segment)
Similarly ∠ ABD = ∠ ACD ……..(ii)
But ∠ ACB = ∠ ACD (AC is the bisector of ∠ BCD)
∴ ∠ ADB = ∠ ABD |From (i) and (ii)]
TAS is a tangent and AB is chord
∴ ∠ BAS = ∠ ADB (Angles in all segment)
But ∠ ADB = ∠ ABD (Proved)
∴ ∠ BAS = ∠ ABD
But these are alternate angles.
∴ TS || BD.      Q.E.D.

Question 10.
In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If ∠ BCG = 108° and O is the centre of the circle,
Find:
(i) angle BCT
(ii) angle DOC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q10.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q10.2
Join OC, OD and AC.
(i) ∠ BCG + ∠ BCD = 180° (Linear pair)
⇒ 108° + ∠ BCD = 180°
(∵∠ BCG = 108° given)
∴∠ BCD = 180° – 108° = 72°
BC = CD (given)
∴ ∠ DCP = ∠ BCT
But ∠ BCT + ∠ BCD + ∠ DCP = 180°
∴∠ BCT + ∠ BCT + 72° = 180°
(∵∠ DCP = ∠ BCT)
2 ∠ BCT = 180° – 72° = 108°
∴∠ BCT = \(\frac { { 108 }^{ \circ } }{ 2 }\) = 54°
(ii) PCT is the tangent and CA is chord
∴ ∠ CAD = ∠ BCT = 54°
But arc DC subtends ∠ DOC at the centre and
∠ CAD at the remaining part of the circle
∴ ∠ DOC = 2 ∠ CAD = 2 x 54° = 108°.

Question 11.
Two circles intersect each other at points A and B. A straight line PAQ cuts the circles at P and Q. If the tangents at P and Q intersect at point T; show that the points P, B, Q and T arc concyclic.
Solution:
Given: Two circles intersect each other at point A and B. PAQ is a line which intersects circles at P, A and Q. At P and Q, tangents are drawn to the circles which meet at T.
To Prove: P, B, Q, T are concyclic.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q11.1
Construction: Join AB, BP and BQ.
Proof: TP is the tangent and PA. a chord
∴ ∠ TPA = ∠ ABP
(angles in alt. segment)
Similarly we can prove that
∠ TQA = ∠ ABQ …,(ii)
Adding (i) and (ii), we get
∠ TPA + ∠ TQA = ∠ ABP + ∠ ABQ
But in ∆ PTQ,
∠ TPA + ∠ TQA + ∠ PTQ = 180°
⇒ ∠ TPA + ∠ TQA = 180° – ∠ PTQ
⇒ ∠ PBQ = 180°- ∠ PTQ
⇒ ∠ PBQ + ∠PTQ = 180°
But there are the opposite angles of the quadrilateral
∴ Quad. PBQT is a cyclic
Hence P, B. Q and T are concyclic     Q.E.D.

Question 12.
In the figure; PA is a tangent to the circle. PBC is secant and AD bisects angle BAC.
Show that triangle PAD is an isosceles triangle. Also shaw that:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q12.1
Solution:
Given: In a circle PA is the tangent, PBC is the secant and AD is the bisector of ∠BAC which meets the secant at D.
To Prove:
(i) ∆ PAD is an isosceles triangle.
(ii) ∠CAD = \(\frac { 1 }{ 2 }\) [(∠PBA – ∠PAB)]
Proof:
(i) PA is the tangent and AB is chord.
∠PAB = ∠C ….(i)
(Angles in the alt. segment)
AD is the bisector is ∠BAC
∴ ∠1 = ∠2 ….(ii)
In ∆ ADC,
Ext. ∠ADP = ∠C + ∠1
= ∠PAB + ∠2 = ∠PAD
∴ ∆ PAD is an isosceles triangle.
(ii) In A ABC,
Ext. ∠PBA = ∠C + ∠BAC
∴∠BAC = ∠PBA – ∠C
⇒ ∠1 + ∠2 = ∠PBA – ∠PAB [from (i)]
⇒ 2 ∠1 = ∠PBA – ∠PAB
⇒ ∠1 = – [∠PBA – ∠PAB]
⇒ ∠CAD = – [∠PBA – ∠PAB] Q.E.D.

Question 13.
Two circles intersect each other at points A and B. Their common tangent touches the circles v at points P and Q as shown in the figure . Show that the angles PAQ and PBQ arc supplementary. [2000]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q13.1
Solution:
Given: Two circles intersect each other at A and
B. A common tangent touches the circles at P and
Q. PA. PB, QA and QB are joined.
To Prove: ∠ PAQ + ∠ PBQ = 180°
or ∠ PAQ and ∠ PBQ are supplementary.
Construction: Join AB.
Proof: PQ is the tangent and AB is the chord
∴ ∠ QPA = ∠ PBA (alternate segment) ,…(i)
Similarly we can prove that
∠ PQA = ∠ QBA ,…(ii)
Adding (i) and (ii), we get
∠ QPA + ∠ PQA = ∠ PBA + ∠ QBA
But ∠ QPA + ∠ PQA = 180° – ∠ PAQ ,…(iii) (In ∆ PAQ)
and ∠ PBA + ∠ QBA = ∠ PBQ ,…(iv)
from (iii) and (iv)
∠ PBQ = 180° – ∠ PAQ
⇒ ∠ PBQ + ∠ PAQ = 180°
= ∠ PAQ + ∠ PBQ = 180°
Hence ∠ PAQ and ∠ PBQ arc supplementary Q.E.D.

Question 14.
In the figure, chords AE and BC intersect each other at point D.
(i) If ∠ CDE = 90°.
AB = 5 cm, BD = 4 cm and CD 9 cm;
Find DE.
(ii) If AD = BD, show that AE = BC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q14.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q14.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q14.3

Question 15.
Circles with centres P and Q intersect at points A and B as shown in the figure. CBD is a line segment and EBM is tangent to the circle, with centre Q, at point B. If the circles arc congruent; show that CE = BD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q15.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q15.2
Given: Two circles with centre P and Q intersect each other at A and B. CBD is a line segment and EBM is tangent to the circle with centre Q, at B. Radii of the cirlces are equal.
To Prove: CE = BD
Construction: Join AB and AD.
Proof: EBM is the tangent and BD is the chord
∴ ∠ DBM = ∠ BAD (Anglesi in alt. segment)
But ∠ DBM = ∠ CBE (Vertically opposite angles)
∴ ∠ BAD = ∠ CBE
∵ In the same circle or congruent circles, if angles are equal, then chords opposite to them are also equal.
∴ CE = BD Q.E.D.

Question 16.
In the following figure O is the centre of ti.e circle and AB is a tangent to it at point B. ∠BDC = 65°, Find ∠BAO
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B Q16.1
Solution:
Given: ∠BDC = 65° and AB is tangent to circle with centre O.
∵ OB is radius
⇒ OB ⊥ AB
In ∆BDC
∠DBC + ∠BDC + ∠B CD = 180°
90° + 65° + ∠BCD =180°
⇒ ∠BCD = 25°
∵ OE = OC = radius
⇒ ∠OEC = ∠OCE
⇒ ∠OEC = 25°
Also, ∠BOE = ∠OEC + ∠OCE
[Exterior angle = sum of opposite interior angles in a ∆]
⇒ ∠BOE = 25°+ 25°
⇒∠BOE = 50°
⇒ ∠BOA = 50°
In ∆AOB
∠AOB + ∠BAO + ∠OBA = 180°
50° + ∠BAO + 90° = 180°
⇒ ∠BAO = 40°

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B  are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19

Question 1.
Draw a circle of radius 3 cm. Mark a point P at a distance of 5cm from the centre of the circle drawn. Draw two tangents PA and PB to the given circle and measure the length of each tangent.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q1.1
Steps of Construction:
(i) Draw a circle with centre O and radius 3 cm.
(ii) From O, take a point P such that OP = 5 cm.
(iii) Draw the bisector of OP which intersects OP at M.
(iv) With centre M, and radius OM. draw’ a circle which intersects the given circle at A and B.
(v) Join AP and BP.
AP and BP are the required tangents.
On measuring them, AP = BP = 4 cm.

Question 2.
Draw a circle of diameter 9 cm. Mark a point at a distance of 7.5 cm from the centre of the circle. Draw tangents to the given circle from this exterior point. Measure the length of each tangent.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q2.1
Steps of Construction:
(i) Draw a line segment AB = 9 cm.
(ii) Draw a circle with centre O and AB as diameter.
(iii) Take a point P from the centre at a distance of 7.5 cm.
(iv) Draw an other circle OP as diameter which intersects the given circle at T and S.
(v) Join TP and SP.
TP and SP are are required tangents.
On measuring their lengths, TP = SP = 6 cm.

Question 3.
Draw a circle of radius 5 cm. Draw two tangents to this circle so that the angle between the tangents is 45°.
Solution:
Steps of Construction:
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw two arcs making an angle of 180° – 45° = 135°
so that ∠AOB = 135°.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q3.1
(iii) At A and B, draw two rays making an angle of 90° at each point which meet each other at P, out side the circle.
Then AP and BP are the required tangents which make an angle of 45° at P.

Question 4.
Draw a circle of radius 4.5 cm. Draw two tangents to this circle so that the angle between the tangents is 60°.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q4.1
Steps of Construction:
(i) Draw a circle with centre O and radius 4.5 cm.
(ii) Draw two arcs making an angle of 180° – 60° = 120° i.e. ∠AOB = 120°.
(iii) At A and B draw rays making an angle of 90° at each point which meet each other at P outside the circle.
AP and BP are the required tangents which makes an angle of 60° at P.

Question 5.
Using ruler and compasses only, draw an equilateral triangle of side 4.5 cm and draw its circumscribed circle. Measure the radius of the circle.
Solution:
Steps of Construction:
(i) Draw a line segment BC = 4.5 cm.
(ii) With centres B and C, draw two arcs of radius 4.5 cm. which intersect each other at A.
(iii) Join AB and AC,
(iv) Draw the perpendicular bisectors of AB and BC intersecting each other at O.
(v) With centre O, and radius OA or OB or OC draw a circle which will passes through A, B and C.
This is the required circumcircle of ∆ ABC.
Measuring OA = 2.6 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q5.1

Question 6.
Construct triangle ABC, having given = 7 cm, AB – AC = 1 cm and ∠ABC = 45°.
(ii) Inscribe a circle in the ∆ ABC constructed in (i) above,
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q6.1
Steps of Construction:
(i) Draw a line segment BC = 7 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = AB – AC = 1 cm.
(iii) Join EC and draw the perpendicular bisector of EC intersecting BX at A.
(iv) Join AC
∆ ABC is the required triangle.
(v) Draw angle bisectors of ∠ABC and ∠ACB intersecting each other at O.
(vi) From O, draw perpendicular OL to BC.
(vii) O as centre and OL as radius draw circle which touches the sides of the A ABC. This is the required in-circle of ∆ ABC.
On measuring radius OL = 1.8 cm (approx.).

Question 7.
Using ruler and compasses only, draw an equilateral triangle of side 5 cm. Draw its inscribed circle. Measure the radius of the circle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q7.1
Steps of Construction:
(i) Draw a line segment BC = 5 cm.
(ii) With centre B and C, draw two arcs of 5 cm radius each which intersect each other at A.
(iii) Join AB and AC.
(iv) Draw angle bisectors of ∠B and ∠C intersecting each other at O.
(v) From O, draw OL ⊥ BC.
(vi) Now with centre O and radius OL, draw a circle which will touch the sides of the ∆ ABC. Measuring OL =1.4 cm. (approx.).

Question 8.
Using ruler and compasses only,
(i) Construct a triangle ABC with the following data:
Base AB = 6 cm, BC = 6.2 cm and ∠CAB = 60°.
(ii) In the same diagram, draw a circle which passes through the points A, B and C and mark its centre O.
(iii) Draw a perpendicular from O to AB which meets AB in D.
(iv) Prove that AD = BD.
Solution:
Steps of Construction:
(i) Draw a line segment AB = 6 cm.
(ii) At A, draw a ray making an angle of 60° with BC.
(iii) B as centre and 6.2 cm as radius draw an arc which intersect the AX rays at C.
(iv) Join CB.
∆ ABC is the required triangle.
(v) Draw the perpendicular bisectors of AB and AC intersecting each other at O.
(vi) With centre O, and radius as OA or OB or OC, draw a circle which will pass through A, B and C.
(vii) From O, draw OD ⊥ AB.
Proof: In right ∆ OAD and ∆ ODB
Hyp, OA = OB (radii of the saine circle)
Side OD = OD (Common)
∴ OAD ≅ OBD (R.H.S.)
∴ AD = BD (C.P.C.T.)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q8.1

Question 9.
Using ruler and compasses only construct a triangle ABC in which BC = 4 cm, ∠ACB = 45° and perpendicular from A on BC is 2.5 cm. Draw a circle circumscribing the triangle ABC and measure its radius.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q9.1
Steps of Construction:
(i) Draw a line segment BC = 4 cm.
(ii) At C, draw a perpendicular line CX and from it, cut off CE = 2.5 cm.
(iii) From E, draw another perpendicular line EY.
(iv) From C, draw a ray making an angle of 45° with CB, which intersects EY at A.
(v) JoinAB.
∆ ABC is the required triangle.
(vi) Draw perpendicular bisectors of sides AB and BC intersecting each other at O.
(vii) With centre O, and radius OB, draw a circle which passes through A, B and C.
Measuring the radius OB = OC = OA = 2 cm

Question 10.
Perpendicular bisectors of the sides AB and AC of a triangle ABC meet at O.
(i) What do you call the point O ?
(ii) What is the relation between the distances OA, OB and OC?
(iii) Does the perpendicular bisector of BC pass through O ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q10.1
(i) Perpendicular bisectors of sides AB and AC intersect each other at O.
(ii) O is called the circum centre of circumcircle of ∆ ABC.
(iii) OA, OB and OC are the radii of the circumcircle.
(iv) Yes, the perpendicular bisector of BC will also pass through O.

Question 11.
The bisectors of angles A and B of a scalene triangle ABC meet at O.
(i) What is the point O called ?
(ii) OR ancLOQ are drawn perpendiculars to AB and CA respectively. What is the relation between OR and OQ ?
(iii) What is the relation between angle ACO and angle BCO ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q11.1
(i) ∆ ABC is a scalene triangle.
(ii) Angle bisectors of ∠A and ∠B intersect each other at O. O is called the incentre of the incircle of ∆ ABC.
(iii) Through O, draw perpendiculars to AB and AC which meet AB and AC at R and Q respectively.
(iv) OR and OQ are the radii of the in circle and OR =OQ.
(v) OC is the bisector of ∠C
∴∠ACO = ∠BCO

Question 12.
(i) Using ruler and compasses only, construct a triangle ABC in which AB = 8 cm, BC = 6 cm and CA = 5 cm.
(ii) Find its incentre and mark it I.
(iii) With I as centre, draw a circle which will cut off 2 cm chords from each side of the triangle. What is the length of the radius of this circle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q12.1
Steps of Construction:
(i) Draw a line segment BC = 6 cm.
(ii) With centre B and radius 8 cm draw ah arc.
(iii) With centre C and radius 5 cm, draw another arc which intersects the first arc at A.
(iv) Join AB and AC.
∆ ABC is the given triangle.
(v) Draw the angle bisectors of ∠B and ∠A intersecting each other at I.
Then I is the incentre of incircle of ∆ ABC.
(vi) Through I, draw ID ⊥ AB.
(vii) Now from D, cut off DP = DQ = \(\frac { 2 }{ 2 }\) = 1 cm.
(viii) With centre I, and radius IP or IQ, draw a circle which will intersect each side of ∆ ABC cuting chords of 2 cm each.

Question 13.
Construct an equilateral triangle ABC with side 6cm. Draw a circle circumscribing the triangle ABC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q13.1
Steps of construction:
(i) Draw a line segment BC = 6cm.
(ii) With centre B and C, draw arcs with radius 6 cm each which intersect each other at A.
(iii) Join AB and AC,
then ∆ABC is the equilateral triangle.
(iv) Draw the perpendicular bisectors of BC and AB which intersect each other at O.
(v) Join OB and OC and OA.
(vi) With centre O, and radius OA or OB or OC, draw a circle which will pass through A, B and C.
This is the required circle.

Question 14.
Construct a circle, inscribing an equilateral triangle with side 5.6 cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q14.1
Steps of construction:
(i) Draw a line segment BC = 5.6 cm
(ii) With centre B and C,
draw two arcs of radius 5.6cm each which intersect each other at A.
(iii) Join AB and AC, then ∆ABC is an equilateral triangle.
(iv) Draw the angle bisectors of ∠B and ∠C which intersect each other at I.
(v) From I, draw ID ⊥ BC.
(vi) With centre I and radius ID, draw circle which touches the sides of the ∆ABC. This is the required circle.

Question 15.
Draw a circle circumscribing a regular hexagon of side 5cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q15.1
Steps of construction:
(i) Draw a regular hexagon ABCDEF whose each side is 5cm.
(ii) Join its diagonals AD, BE and CF intersecting each other at O.
(iii) With centre O and radius OA, draw a circle which will pass through the vertices of the hexagon A, B, C, D, E and F. This is the required circle.

Question 16.
Draw an inscribing circle of a regular hexagon of side 5.8 cm.
Solution:
Steps of construction:
(i) Draw a line segment AB = 5.8cm.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q16.1
(ii) At A and B, draw rays making an angle of 120° each and cut off AF = BC = 5.8 cm.
(iii) Again at F and C, draw rays making an angle of 120° each and cut off FE = CD = 5.8 cm.
(iv) JoinDE. Then ABCDEF is the regular hexagon.
(v) Draw the bisectors of ∠A and ∠B intersecting each other at O.
(vi) From O, draw OL J. AB.
(vii) With centre O and radius OL, draw a circle which touches the sides of the hexagon. This is the required incircle of the hexagon.

Question 17.
Construct a regular hexagon of side 4 cm. Construct a circle circumscribing the hexagon.
Solution:
Steps of construction:
(i) Draw a circle of radius 4 cm with centre O.
(ii) Since regular hexagon \(\frac { { 360 }^{ \circ } }{ 6 }\) = 60°, draw radii
OA and OB, such that ∠AOB = 60°.
(iii) Cut off arcs BC, CD, DE, EF and each equal to arc AB on given circle.
(iv) Join AB, BC, CD, DE, EF, FA to get required regular hexagon ABCDEF in a given circle.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q17.1

Question 18.
Draw a circle of radius 3.5 cm. Mark a point P outside the circle at a distance of 6 cm from the centre. Construct two tangents from P to the given circle. Measure and write down the length of one tangent (2011).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q18.1
Steps of construction:
(i) Draw a line segment OP = 6 cm
(ii) With centre O and radius 3.5 cm, draw a circle
(iii) Draw the mid point of OP.
(iv) With centre M and diameter OP, draw a circle which intersect the circle at T and S.
(v) Join PT and PS.
PT and PS are the required tangent on measuring the length of PT = PS = 4.8 cm

Question 19.
Construct a triangle ABC in which base BC=5.5 cm,AB = 6cmand ∠ABC = 120°.
(i) Construct a circle circumscribing the triangle ABC.
(ii) Draw- a cyclic quadrilateral ABCD so that D is equidistant from B and C.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q19.1
Steps of construction:
(i) Draw BC = 6 cm. x
(ii) At B, draw ∠XBC= 120°.
(iii) From BX, cut off AB = 6 cm.
(iv) Join AC to get ∆ ABC.
(v) Draw the perpendicular bisector of BC and AB. These bisectors meet at O. With O as centre and radius equal tb OA, draw a circle, which passes through A, B and C. This is the required circumcircle of ∆ABC.
(vi) Produce the perpendicular bisector of BC so that it meets the circle at D. Join CD and AD to _ get the required cyclic quadrilateral ABCD.

Question 20.
Using a ruler and compasses only :
(i) Construct a triangle ABC with the following data : AB = 3.5 cm, BC = 6 cm and ∠ABC = 120°
(ii) In the same diagram, draw a circle with BC as diameter. Find a point P on the circumference of the circle which is equidistant from AB and BC.
(iii) Measure ∠BCP
Solution:
Steps of construction:
(i) Draw AB = 3.5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q20.1
(ii) At B, draw ∠ABX = 120°.
(iii) With B as center draw an arc of radii 6 cm at C.
(iv) Join A and C.
(v) Draw the perpendicular bisector of line BC and draw a circle with BC as diameter.
(vi) Draw angle bisector of ∠B.
Meets the circle at P
∴ P is the required point ∠BCP = 30°

Question 21.
Construct a ∆ABC with BC = 6.5 cm, AB = 5.5 cm, AC = 5 cm. Construct the incircle of the triangle. Measure and record the radius of the incircle. (2014)
Solution:
Steps of construction:
(i) Draw a line segment BC = 6.5 cm.
(ii) From B, draw an arc of radius of 5.5 cm and from C, another arc of 5 cm radius which intersect each other at A.
(iii) Join AB and AC.
∆ABC is required triangle.
(iv) Draw the angle bisectors of ∠B and ∠C which intersect each other at O.
(v) Through O, draw OL ⊥ BC.
(vi) With centre O and radius OL, draw a circle which touches the sides of ∆ABC.
(vii) On measuring, OL = r = 1.5 cm.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q21.1

Question 22.
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°. Hence:
(i) Construct the locus of points equidistant from BA and BC.
(ii) Construct the locus of points equidistant from B and C.
(iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC. (2015)
Solution:
Steps of construction:
(i) Draw a line segment AB = 5.5 cm.
(ii) At A, draw a ray AX making an angle of 105°.
(iii) Cut off AC from AX =6 cm.
(iv) JoinCB.
∆ABC is required triangle.
(v) Draw angle bisector CX of ∠C.
CX is the locus of points equidistant from BA and BC.
(vi) Draw the perpendicular bisector of BC which is the locus of points equidistant from the points B and C.
These two loci intersect each other at P.
Join PC and on measuring it, it is 4.8 cm (approx).
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q22.1

Question 23.
Construct a regular hexagon of side 5 cm. Hence construct all its lines of symmetry and name them. (2016)
Solution:
Steps of construction :
(i) Draw AF measuring 5 cm using a ruler.
(ii) With A as the centre and radius equal to AF, draw an arc above AF.
(iii) With F as the centre, and same radius cut the previous arc at O.
(iv) With O as the centre, and same radius draw a circle passing through A and F.
(v) With A as the centre and same radius, draw an arc to cut the circle above AF at B.
(vi) With B as the centre and same radius, draw an arc to cut the circle at C.
(vii) Repeat this process to get remaining vertices of the hexagon at D and E
(viii) Join consecutive arcs on the circle to form the hexagon.
(ix) Draw the perpendicular bisectors of AF, EF and DE.
(x) Extend the bisectors of AF, EF and DE to meet CD, BC and AB at X, L and O respectively.
(xi) Join AD, CF and EB.
(xii) These are the 6 lines of symmetry of the regular hexagon.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q23.1

Question 24.
Draw a line AB = 5 cm. Mark a point C on AB such that AC = 3 cm. Using a ruler and a compass only, construct:
(i) A circle of radius 2.5 cm, passing through A and C.
(ii) Construct two tangents to the circle from the external point B. Measure and record the length of the tangents. (2016)
Solution:
Steps of construction :
(i) Draw AB = 5 cm using a ruler.
(ii) With A as the centre cut an arc of 3 cm on AB to obtain C.
(iii) With A as the centre and radius 2.5 cm, draw an arc above AB.
(iv) With same radius, and C as the centre draw an arc to cut the previous arc and mark the intersection as O.
(v) With O as the centre and radius 2.5 cm, draw a circle so that points A and C lie on the circle formed
(vi) Join OB.
(vii) Draw the perpendicular bisector of OB to obtain the mid-point of OB, M.
(viii) With M as the centre and radius equal to OM, draw a circle to cut the previous circle at points P and Q.
(ix) Join PB and QB. PB and QB are the required tangents to the given circle from exterior point B.
QB = PB = 3 cm
That is, length of the tangents i.e. 3.2 cm.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles) Ex 19 Q24.1

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere (Surface Area and Volume) Ex 20G

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 20G.

Other Exercises

Question 1.
What is the least number of solid metallic spheres, each of 6 cm diameter, that should be melted and recast to form a solid metal cone whose height is 45 cm and diameter 12 cm ?
Solution:
Diameter of cone = 12cm
∴ Radius (r1) = \(\frac { 12 }{ 2 }\)= 6cm
Height (A) = 45 cm
∴  Volume = \(\frac { 1 }{ 3 }\)πR2h
= \(\frac { 1 }{ 3 }\) x π x 6 x 6 x 45 cm3
= 540 π cm2
∴ Volume of solid spheres = 540 π cm3
Diameter of one sphere = 6cm
∴  Radius = \(\frac { 6 }{ 2 }\) = 3 cm
∴ Volume of one spherical ball
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q1.1

Question 2.
A largest sphere is to be carved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere. (Answer correct to the nearest integer).
Solution:
Radius of cylinder = 7 cm
and height 14 cm
By carving a largest sphere from it,
the radius of the sphere will be = 7 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q2.1

Question 3.
A right circular cylinder having diameter 12 cm and height 15 cm is full of ice-cream. The ice-cream is to be filled in identical cones of height 12 cm and diameter 6 cm having a hemi­spherical shape on the top. Find the number of cones  required.
Solution:
Diameter of cylinder = 12 cm
∴ Radius (r) = \(\frac { 12 }{ 2 }\) = 6 cm
Height (h) = 15 cm
Volume of ice-cream in cylinder = πr²h = π x 6 x 6 x 15 = 540π cm3
∴  Volume of ice-cream = 540π cm3
Now diameter of cone = 6 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q3.1

Question 4.
A solid is in the form of a cone standing on a hemi-sphere with both their radii being equal to 8 cm and the height of cone is equal to its radius. Find, in terms of π , the volume of the solid.
Solution:
Radius of each cone and hemi-sphere (r) = 8 cm
Height of cone (h) = r = 8 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q4.2

Question 5.
The diameter of sphere is 6 It is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire.
Solution:
Diameter of a sphere = 6 cm.
∴ 
Radius = \(\frac { 6 }{ 2 }\) = 3 cm.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q5.2

Question 6.
Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube.
Solution:
Let edge of the cube = a
∴ Volume of cube = a x a x a = a3
The sphere, which exactly fits in the cube, has radius = \(\frac { a }{ 2 }\)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q6.1

Question 7.
An iron pole consisting of a cylindrical por­tion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that 1cm3 of iron has 8 gm of mass (approx).
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q7.1
Solution:
Radius of the base of poles (r) = \(\frac { 12 }{ 2 }\) = 6cm.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q7.3

P.Q.
When a metal cube is completely submerged in water contained in a cylindrical vessel with diameter 30 cm, the level of water rises by 1 \(\frac { 41 }{ 99 }\) cm. Find:
(i) the length of the edge of the cube.
(ii) the total surface area of the cube.
Solution:
Diameter of cylinderical vessel = 30 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Qp1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Qp1.2

Question 8.
In the following diagram a rectangular plat­form with a semi-circular end on one side is 22 metres long from one end to the other end. If the length of the half circumference is 11 metres, find the cost of constructing the platform, 1.5 metres high at the rate of Rs. 4 per cubic metres.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q8.1
Solution:
Length of platform = 22m
Circumference of semicircle = 11m
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q8.3

Question 9.
The cross-section of a tunnel is a square of side 7 m surmounted by a semi-circle as shown in the following figure. The tunnel is 80 m long. Calculate :
(i) Its volume
(ii) The surface area of the tunnel (excluding the floor) and
(iii) its floor area.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q9.1
Solution:
Side of square = 7m
Radius of semicircle = \(\frac { 7 }{ 2 }\) m
Length of tunnel = 80 m.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q9.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q9.3

Question 10.
A cylindrical water tank of diameter 2.8 m and height 4.2 m is being fed by a pipe of diam­eter 7 cm through which water flows at the rate of 4 ms-1. Calculate, in minutes, the time it takes to fill the tank.
Solution:
Diameter of cylindrical tank = 2.8 m 2.8
∴ Radius (r) = \(\frac { 2.8 }{ 2 }\) = 1.4 m
and height (h) = 4.2 m
Volume of water filled in it = πr2h
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q10.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q10.2

Question 11.
Water flows, at 9 km per hour, through a cylindrical pipe of cross-sectional area 25 cm2. If this water is collected into a rectangular cistern of dimensions 7.5 m by 5 m by 4 m; calculate the rise in level in the cistern in 1 hour 15 minutes.
Solution:
Rate of flow of water = 9km / hr
∴ Water flow in 1 hr 15 minutes i.e. in \(\frac { 5 }{ 4 }\)  hr
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q11.1

Question 12.
The given figure shows the cross-section of a cone, a cylinder and a hemisphere all with the same diameter 10 cm, and the other dimensions are as shown. Calculate :
(a) the total surface area,
(b) the total volume of the solid and
(c) the density of the material if its total weight is 1.7 kg.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q12.1
Solution:
Diameter = 10 cm
∴  Radius (r) =  \(\frac { 10 }{ 2 }\)   = 5 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q12.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q12.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q12.4

Question 13.
A solid, consisting of a right circular cone, standing on a hemisphere, is placed upright, in a right circular cylinder, full of water, and touches the bottom. Find the volume of water left in the cylinder, having given that the radius of the cylinder is 3 cm and its height is 6 cm; the radius of the hemisphere is 2 cm and the height of cone is 4 cm. Give your answer to the nearest cubic centimetre. [1998]
Solution:
Radius of cylinder = 3 cm
and height = 6 cm
Radius of hemisphere = 2 cm
and height of cone = 4 cm

Volume of water in the cylinder when it is full
= πr2h = π x (3)2 x 6 cm3 = 54π cm3
Volume of water displaced = Volume of cone + volume of hemisphere

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q13.1

Question 14.
A metal container in the form of a cylinder is surmounted by a hemi-sphere of the same radius. The internal height of the cylinder is 7 m and the internal radius is 3.5 m. Calculate
(i) he total area of the internal surface, exclud­ing the base;
(ii) the internal volume of the container in m3 [1999]
Solution:
Radius of cylinder = 3.5 m
and height = 7 m.
(i) Total surface area of container excluding the base = curved surface area of cylinder + area of hemisphere = 2πrh + 2πr2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q14.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q14.2

Question 15.
An exhibition tent is in the form of a cylin­der surmounted by a cone. The height of the tent above the ground is 85 m and height of the cylindrical part is 50 m. If the diameter of the base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for fold and for stitching. Give your answer to the nearest m2.     [2001]
Solution:
Total height of the tent = 85 m.
Daimeter of the base = 168 m.
∴ Radius (r) = \(\frac { 168 }{ 2 }\) = 84 m
Height of cylindrical part = 50 m
Then height of conical part (h) = 85 – 50 = 35 m
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q15.2

P.Q.
The total surface area of a hollow cylinder, which is open from both sides is 3575 cm2; area of the base ring is 357.5 cm2 and height is 14 cm. Find the thickness of the cylinder.
Solution:
Total surface area of a hollow cylinder = 3575 cm2
Area of base ring = 357.5 cm2
Height = 14 cm.
Let external radius = R
and internal radius = r
and let thickness of cylinder = (R – r) = d
∴  Total surface area = 2πRh + 2πrh + 2π (R2– r2)
= 2πh (R + r) + 2π (R + r) (R – r)
= 2π (R + r) [h + R – r]
= 2π (R + r) (h + d)
= 2π (R + r) (14 + d)
But 2π (R + r) (14 + d) = 3575                  ….(i)
and area of the base = π (R2 – r2) = 357.5
⇒  π (R + r) ( R – r) = 357.5
⇒ π (R + r)d = 357.5                                 ….(ii)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Qp2.1

Question 16.
A test tube consists of a hemisphere and a cylinder of the same radius. The volume of the water required to fill the whole tube  is \(\frac { 5159 }{ 6 }\)cm3, and \(\frac { 4235 }{ 6 }\) cm3 of water is required to fill the tube to a level which is 4 cm below the top of the tube. Find the radius of the tube and the length of its cylindrical part.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q16.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q16.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q16.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q16.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q16.5

Question 17.
A solid is in the form of a right circular cone mounted on a hemisphere. The diameter of the base of the cone, which exactly coincides with hemisphere, is 7cm and its height is 8cm. The solid is placed in a cylindrical vessel of internal radius 7 cm and height 10cm. How much water, in cm3, will be required to fill the vessel completely.
Solution:
Diameter of hemisphere = 7cm
Diameter of the base of a cone = 7cm
∴ Radius (r) = \(\frac { 7 }{ 2 }\) cm = 3.5 cm
Height (h) = 8cm.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q17.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q17.2

Question 18.
Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed.   (2015)
Solution:
Radius of first sphere = 2 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q18.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q18.2

Question 19.
A certain number of metallic cones, each of radius 2 cm and height 3 cm are melted and recast into a solid sphere of radius 6 cm. Find the number of cones used.             (2016)
Solution:
Let the number of cones be n,
Let radius of the sphere be rs, radius of a cone be rc and h be the height of the cone.
Volume of sphere = n(Volume of a metallic cone)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G Q19.1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20G are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere (Surface Area and Volume) Ex 20F

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities Ex 21F

Other Exercises

Question 1.
From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base is removed. Find the volume of the ramaining solid.
Solution:
Height of the cylinder (h) = 10 cm
and radius of base (r) = 6 cm.
∴ Volume of cylinder = πr2h
Height of cone = 10 cm
and radius of base of cone = 6 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q1.1

Question 2.
From a solid cylinder whose height is 16 cm and radius is 12 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out Find the volume and total surface area of the remaining solid.
Solution:
Radius of solid cylinder (R) = 12 cm
and height (H) = 16 cm.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q2.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q2.2

Question 3.
A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 80 m, calculate the total area of canvas required. Also, find the total cost of canvas used at Rs. 15 per metre if the width is 1.5 m.
Solution:
Radius of the cylindrical part of tent (r) = \(\frac { 105 }{ 2 }\)m
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q3.2

Question 4.
A circus tent is cylindrical to a height of 8 m surmounted by a conical part. If total height of the tent is 13 m and the diameter of its base is 24 m; calculate:
(i) total surface area .of the tent,
(ii) area of canvas, required to make this tent allowing 10% of the canvas used for folds and stitching.
Solution:
Total height = 13 m
Diameter of base of the tent = 24 m
∴ Radius (r) = \(\frac { 24 }{ 2 }\) = 12 m
Height of cylindrical part h1 = 8 m
and height of conical part (h2) = 13 – 8 = 5 m
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q4.2

Question 5.
A cylindrical boiler, 2 m high, is 3.5 m in diameter. It has a hemi-spherical lid. Find the volume of its interior, including the part covered by the lid.
Solution:
Diameter of cylinderical boiler = 3.5 m
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q5.2

Question 6.
A vessel is a hollow cylinder fitted with a hemispherical bottom of the same base. The depth of the cylindrical part is 4 \(\frac { 2 }{ 3 }\) m and the diameter of hemisphere is 3.5 m. Calculate the capacity and the internal surface area of the vessel.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q6.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q6.3

Question 7.
A wooden toy is in the shape of a cone mounted on a cylinder as shown alongside.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q7.1
If the height of the cone is 24 cm, the total height of the toy is 60 cm and the radius of the base of the cone = twice the radius of the base of the cylinder = 10 cm ; find the total surface area of the toy.  [Take π = 3.14]
Solution:
Height of the conical part (h1)= 24 cm
total height of the toy = 60 cm
∴ Height of cylinderical part (h) = 60-24 = 36 cm
Radius of the cone (r) = twice the radius of the cylinder = 10 cm
∴ Radius of cylinder (r1) = 5 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q7.2

Question 8.
A cylindrical container with diameter of base 42 cm contains sufficient water to sub­merge a rectangular solid of iron with dimensions 22 cm x 14 cm x 10.5 cm. Find the rise in level of the water when the solid is submerged.
Solution:
Diameter of cylinderical container = 42cm
∴  Radius (r) = \(\frac { 42 }{ 2 }\) = 21 cm.
Dimension of a rectangular solid = 22 cm x 14cm x 10.5 cm
∴ Volume of solid
= 22 x 14 x 10.5 cm3        ….(i)
Let the height of water = h
∴ Volume of water in the container
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q8.1

Question 9.
Spherical marbles of diameter 1.4 cm are dropped into beaker containing some water and are fully submerged. The diameter of the beaker is 7 cm. Find how many marbles have been dropped in it if the water rises by 5.6 cm.
Solution:
Diameter of spherical marble = 1.4 cm.
∴ Radius = \(\frac { 1.4 }{ 2 }\) = 0.7 cm.
Volume of one ball = \(\frac { 4 }{ 3 }\) πr3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q9.1

Question 10.
The cross-section of a railway tunnel is a rectangle 6 m broad and 8 m high surmounted by a semi-circle as shown in the figure. The tun­nel is 35 m long. Find the cost of plastering the internal surface of the tunnel (excluding the floor) at the rate of Rs. 2.25 per m2.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q10.1
Solution:
Breadth of’tunnel = 6 m
and height = 8m
Length of tunnel = 35 m
Radius of semicircle = \(\frac { 6 }{ 2 }\) = 3 m.
Circumference of semicircle = πr = \(\frac { 22 }{ 7 }\) x 3 = \(\frac { 66 }{ 7 }\) m
∴ Internal surface area of the tunnel
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q10.2

Question 11.
The horizontal cross-section of a water tank is in the shape of a rectangle with semi-circle at one end, as shown in the following figure. The water is 2.4 metres deep in the tank. Calculate the volume of water in the tank in gallons.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q11.1
(Given : 1 gallon = 4.5 litres)
Solution:
Length = 21m
breadth = 7 m
Depth of water = 2.4 m
∴ Radius of semicircle = \(\frac { 7 }{ 2 }\) m.
Area of the cross section = Area of rectangle + area of semicircle
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q11.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q11.3

Question 12.
The given figure shows the cross-section of a water channel consisting of a rectangle and a semi-circle. Assuming that the channel is always full, find the volume of water discharged through it in one minute if water is flowing at the rate of 20 cm per second. Give your answer in cubic metres correct to one place of decimal.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q12.1
Solution:
Rate of water flow = 20 cm/sec.
Period = 1 min. = 60 sec.
Radius of semi-circular part (r) = \(\frac { 21 }{ 2 }\)  cm
Height of channel (h) = 7 cm
Length of channel = 20 x 60 = 1200 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q12.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q12.3

Question 13.
An open cylindrical vessel of internal diameter 7 cm and height 8 cm stands on a horizontal table. Inside this is placed a solid metallic right circular cone, the diameter of whose base is 3 \(\frac { 1 }{ 2 }\) cm and height 8 cm. Find the volume of water required to fill the vessel.
If this cone is replaced by another cone, whose height is 1 \(\frac { 3 }{ 4 }\) cm and the radius of whose base is 2 cm, find the drop in the water level.  [1993]
Solution:
Diameter of cylinder = 7 cm
∴ 
Radius (R) = \(\frac { 7 }{ 2 }\) cm
Height (h) = 8 cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q13.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q13.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q13.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q13.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q13.5

Question 14.
A cylindrical can, whose base is horizontal and of radius 3.5 cm, contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate:
(i) the total surface area of the can in contact with water when the sphere is in it;
(ii) the depth of water in the can before the sphere was put into the can. [1997]
Solution:
Radius of the cylindrical can = 3.5 cm
∴  Radius of the sphere which fits in it
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q14.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q14.2

Question 15.
A hollow cylinder has solid hemisphere inward at one end and on the other end it is closed with a flat circular plate. The height of water is 10 cm when flat circular surface is downward. Find the level’of water, when it is inverted upside down, common diameter is 7 cm and height of the cylinder is 20  cm.
Solution:
(i) Diameter of the cylinder = 7 cm
∴ Radius (r) = \(\frac { 7 }{ 2 }\) cm
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q15.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q15.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F Q15.4

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere Ex 20F are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions

Other Exercises

Question 1.
Find the sum of n terms of the series :
(i) 4 + 44 + 444 + ….
(ii) 0.8 + 0.88 + 0.888 + ….
Solution:
(i) 4 + 44 + 444 + ….
= \(=\frac { 4 }{ 9 } \left[ 9+99+999+…. \right] \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q1.2
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 2 Estimation 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q1.4

Question 2.
Find the sum of infinite terms of each of the following geometric progression:
(i)\(1+\frac { 1 }{ 3 } +\frac { 1 }{ 9 } +\frac { 1 }{ 27 } +…\)
(ii)\(1-\frac { 1 }{ 2 } +\frac { 1 }{ 4 } -\frac { 1 }{ 8 } +…\)
(iii)\(\frac { 1 }{ 3 } +\frac { 1 }{ { 3 }^{ 2 } } -\frac { 1 }{ { 3 }^{ 3 } } +…\)
(iv)\(\sqrt { 2 } -\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2\sqrt { 2 } } -\frac { 1 }{ 4\sqrt { 2 } } +…\)
(v)\(\sqrt { 3 } +\frac { 1 }{ \sqrt { 3 } } +\frac { 1 }{ 3\sqrt { 3 } } +\frac { 1 }{ 9\sqrt { 3 } } +… \)
Solution:
(i)\(1+\frac { 1 }{ 3 } +\frac { 1 }{ 9 } +\frac { 1 }{ 27 } +…\) upto infinity
Sn = \(\\ \frac { a }{ 1-r } \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q2.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q2.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q2.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q2.4

Question 3.
The second term of a G.P. is 9 and sum of its infinite terms is 48. Find its first three terms.
Solution:
In a G.P.
T2 = 9, sum of infinite terms = 48
Let a be the first term and r be the common ratio, therefore,
S = \(\\ \frac { a }{ 1-r } \)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q3.1

Question 4.
Find three geometric means between \(\\ \frac { 1 }{ 3 } \) and 432.
Solution:
Let G1, G2 and G3 be three means between
\(\\ \frac { 1 }{ 3 } \) and 432, then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q4.2

Question 5.
Find :
(i) two geometric means between 2 and 16
(ii) four geometric means between 3 and 96.
(iii) five geometric means between \(3 \frac { 5 }{ 9 } \) and \(40 \frac { 1 }{ 2 } \).
Solution:
(i) Two G.M. between 2 and 16
Let G1 , and G1 be the G.M.,
then 2, G1, G2, 16
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q5.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q5.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q5.4

Question 6.
The sum of three numbers in G.P. is \(\\ \frac { 39 }{ 10 } \) and their product is 1. Find numbers
Solution:
Sum of three numbers in G.P = \(\\ \frac { 39 }{ 10 } \)
and their product = 1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q6.2

Question 7.
Find the numbers in G.P. whose sum is 52 and the sum of whose product in pairs is 624.
Solution:
Sum of 3 numbers in G.P. = 52
and their product in pairs = 624
Let numbers be a, ar, ar²
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q7.3

Question 8.
The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers.
Solution:
Sum of three numbers in G.P. = 21
Sum of their squares = 189
Let three numbers be a, ar, ar², then
a + ar + ar² = 21
=> a( 1 + r + r²) = 21….(i)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions Q8.3

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression Additional Questions are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A.

Other Exercises

Question 1.
In the given figure, O is the centre of the circle. ∠OAB and ∠OCB are 30° and 40° respectively. Find ∠AOC. Show your steps of working.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q1.2
In circle with centre O, ∠BAO = 30°, ∠BCO = 40°.
Join BO.
OA = OB = OC (Radii of the circle)
∠OBA = ∠OAB = 30° and ∠OBC = ∠OCB = 40°
∠ABC = 30° + 40° = 70°
Now, AOC is at the centre and ∠ABC is on the remaining part of the circle.
∠AOC = 2 ∠ABC = 2 x 70° = 140°.

Question 2.
In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
(i) Prove that AC is a diameter of the circle.
(ii) Find ∠ACB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q2.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q2.2
(i) In ΔABD
65° + 70° + ∠ADB = 180°
∠ADB = 180° – 65° – 70° = 45°
∠ADC = 45° + 45° = 90°
AC is diameter [Angle in semi circle is 90°]
(ii) ∠ACB = ∠ADB = 45° [angle in same segment]

Question 3.
Given O is the centre of the circle and ∠AOB = 70°. Calculate the value of:
(i) ∠OCA,
(ii) ∠OAC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q3.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q3.2
O is the centre of the circle,
∠AOB = 70°
arc AB subtends ∠AOB at the centre
and ∠ OCA is at the remaining part of circle
∠AOB = 2 ∠OCA
or ∠OCA = \(\frac { 1 }{ 2 }\) ∠AOB = \(\frac { 1 }{ 2 }\) x 70° = 35°
In ΔOAC,
OC = OA (Radii of the same circle)
∠OAC = ∠OCA = 35°

Question 4.
In each of the following figures, O is the centre of the circle. Find the values of a, b, and c.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q4.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q4.2
(i) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part
∠AOB = 2 ∠ACB
or ∠ACB = \(\frac { 1 }{ 2 }\) ∠AOB = \(\frac { 1 }{ 2 }\) x 130° = 65°
or b = 65°
But a + b = 180° (Opposite angles of a cyclic quad.)
a = 180° – b = 180° – 65° = 115°
a = 115°, b = 65°
(ii) Arc AB subtends ∠ AOB at the centre and ∠ACB at the remaining part.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q4.3
Reflex ∠AOB = 2 ∠ACB
or ∠ACB = \(\frac { 1 }{ 2 }\) (reflex ∠AOB) = \(\frac { 1 }{ 2 }\) [360°- 112°]
= \(\frac { 1 }{ 2 }\) x 248° = 124°
Hence, c = 124°.

Question 5.
In each of the following figures, O is the centre of the circle. Find the value of a, b, c and d.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q5.1
Solution:
(i) BOD is a diameter
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q5.2
∠BAD = 90° (Angle in a semi-circle)
∠ADB = 180° – (90° + 35°) = 180° – 125° = 55°
But ∠ACB = ∠ADB = 55° (Angles in the same segment)
a = 55°.
(ii) In ΔEBC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q5.3
Ext. 120° = 25° + ∠BCE
∠BCE = 120° – 25° = 95°
But ∠ADB = ∠ACB = 95° (Angles in the same segment)
b = 95°.
(iii) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q5.4
∠ AOB = 2 ∠ ACB = 2 x 50° = 100°
In ΔAOB,
OA = OB (Radii of the same circle)
∠OAB = ∠OBA
But ∠OAB + ∠OBA = 180° – 100° = 80°
c = ∠OAB = ~ x 80° = 40°.
(iv) In the given figure, O is the centre of the circle.
AOB is its diameter and ∠ABP = 45°
Q is any point and BQ, PQ are joined
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q5.5
In ΔABP,
∠APB = 90° (Angle in a semicircle)
∠PAB + ∠PBA = 90°
⇒ ∠PAB + 45° = 90°
⇒ ∠PAB = 90° – 45°
⇒ ∠PAB = 45°
Now ∠PAB = ∠PQB (Angle in the same segment)
BPQB = 45°
⇒ d = 45°

Question 6.
In the figure, AB is common chord of the two circles. If AC and AD are diameters; prove that D, B and C are in a straight line. O1 and 02 are the centres of two circles.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q6.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q6.2
Given- Two circles with centre O1 and O2 intersect each other at A and B.
AC and AD are the diameters of the circles.
To Prove- D, B, C are in the same straight line.
Construction- Join AB.
Proof- AO1C is diameter.
∠ABC = 90°. (Angle in a semi-circle)
Similarly ∠ABD = 90°,
Adding, we get:
∠ABC + ∠ABD = 90° + 90° = 180°
DBC is a straight line.
or D, B, C are in the same line.

Question 7.
In the figure given beow, find :
(i) ∠BCD,
(ii) ∠ADC,
(iii) ∠ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q7.1
Show steps of your workng.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q7.2
ABCD is a cyclic quadrilateral
∠A + ∠C = 180°.
∠C = 180° – ∠A = 180° – 105° = 75°
or ∠BCD = 75°
DC || AB
∠ADC + ∠DAB = 180° (Angles on the same side of the transversal of || lines)
∠ADC = 180° – ∠DAB = 180° – 105° = 75°
But ∠ADC + ∠ABC = 180° (opposite angles of a cyclic quad.)
∠ABC = 180° – ∠ADC = 180° – 75° = 105°

Question 8.
In the given figure, O is centre of the circle. If ∠ AOB = 140° and ∠ OAC = 50°; find :
(i) ∠ ACB,
(ii) ∠OBC,
(iii) ∠OAB,
(iv) ∠CBA
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q8.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q8.2
O is the centre of circle ∠AOB = 140° and ∠OAC = 50°.
AB is joined
Reflex ∠AOB = 360° – 140° = 220°
But ∠ACB = \(\frac { 1 }{ 2 }\) Reflex ∠AOB = \(\frac { 1 }{ 2 }\) x 220° = 110°
In quad. OACB,
∠AOB + ∠OAC + ∠ACB + ∠OBC = 360°
⇒ 140° + 50° + 110° + ∠OBC = 360°
⇒ 300° + ∠OBC = 360°
⇒ ∠OBC = 360° – 300° = 60°
In ∆OAB,
∠AOB + ∠OAB + ∠OBA = 180°
But ∠OBA = ∠OAB (Angles opposite to equal sides)
140° + ∠OAB + ∠OAB = 180°
2 ∠OAB = 180° – 140° = 40°
∠OAB = \(\frac { 40 }{ 2 }\) = 20°
∠OAB = ∠OBA = 20°.
⇒ ∠OBC = ∠CBA + ∠ABO
⇒ 60° = ∠CBA + 20°
⇒ ∠CBA = 40°

Question 9.
Calculate :
(i) ∠ CDB,
(ii) ∠ ABC,
(iii) ∠ ACB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q9.1
Solution:
∠CDB = ∠BAC (Angles is the same segment) = 49°
∠ABC = ∠ADC (Angles in the same segment) = 43°
∠ADB = ∠ADC + ∠BDC = 43° + 49° = 92°
and ∠ADB + ∠ACB = 180° (opposite angles of a cyclic quad.)
∠ACB = 180° – ∠ADB = 180° – 92° = 88°.

Question 10.
In the figure given below, ABCD is a eyclic quadrilateral in which ∠BAD = 75°; ∠ABD = 58° and ∠ADC = 77°. Find:
(i) ∠BDC,
(ii) ∠BCD,
(iii) ∠BCA.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q10.1
Solution:
In cyclic quad. ABCD,
∠BAD = 75°, ∠ABD = 58° and ∠ADC = 77°.
∠A + ∠C = 180° (opposite angles of a cyclic quad.)
∠C = 180° – ∠A = 180° – 75° = 105° or ∠BCD = 105°
In ΔABD,
∠BAD + ∠ABD + ∠ADB = 180°
⇒ 75° + 58° + ∠ ADB = 180°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q10.2
⇒ 133° + ∠ADB = 180°
⇒ ∠ADB = 180° – 133° = 47°
∠BDC = 77° – ∠ADB = 77° – 47° = 30°
But ∠BCA = ∠BDA (Angles in the same) = 47°

Question 11.
In the following figure, O is centre of the circle and ΔABC is equilateral. Find :
(i) ∠ADB
(ii) ∠AEB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q11.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q11.2
∠ACB and ∠ADB are in the same segment.
∠ADB = ∠ACB. = 60°. (Angle of an equilateral triangle)
AEBD is a cyclic quadrilateral
∠AEB + ∠ADB = 180°
⇒ ∠AEB + 60° = 180°
⇒ ∠AEB = 180° – 60° = 120°.

Question 12.
Given- ∠CAB = 75° and ∠CBA = 50°. Find the value of ∠DAB + ∠ABD
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q12.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q12.2
In ΔABC, ∠CBA = 50°, ∠CAB = 75°,
∠ACB = 180° – (∠CBA + ∠CAB) = 180° – (50° = 75°) = 180° – 125° = 55°
Bui ∠ADB = ∠ACB = 55° (Angles in the same segment)
Now in ΔABD,
∠DAB + ∠ABD + ∠ADB = 180°.
⇒ ∠DAB + ∠ABD + 55° = 180°
⇒ ∠DAB + ∠ABD = 180° – 55° = 125°.

Question 13.
ABCD is a cyclic quadrilateral in a circle with centre O. If ∠ADC = 130°; find ∠BAC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q13.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q13.2
ABCD is a cyclic quadrilateral and ∠ADC = 130°
O is centre of the circle, AOB is diameter.
∠ABC = 180° – 130° = 50°.
In ΔABC,
∠ACB = 90° (angle in semicircle)
∠BAC + ∠CBA = 90°.
∠BAC + 50° = 90°
∠BAC = 90° – 50° = 40°.

Question 14.
In the figure given below, AOB is a diameter of the circle and ∠AOC = 110°. Find ∠BDC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q14.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q14.2
∠AOC + ∠COB = 180° (Linear pair)
∠COB = 180° – ∠AOC = 180° – 110° = 70°
Arc BC subtends ∠COB at the centre and x at the remaining part of circle
∠COB = 2x
⇒ x = \(\frac { 1 }{ 2 }\) ∠COB = \(\frac { 1 }{ 2 }\) x 70° = 35°

Question 15.
In the following figure, O is centre of the circle, ∠AOB = 60° and ∠BDC = 100°. Find ∠OBC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q15.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q15.2
Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of circle,
∠AOB = 2 ∠ACB
or ∠ACB = \(\frac { 1 }{ 2 }\) ∠AOB = \(\frac { 1 }{ 2 }\) x 60° = 30°
Now in ΔDBC,
∠DBC + ∠ACB + ∠BDC = 180°
⇒ ∠DBC + 30° + 100° = 180°
⇒ ∠DBC = 180° – 130° = 50°
or ∠OBC = 50°.

Question 16.
ABCD is a cyclic quadrilateral in which ∠DAC = 27°; ∠DBA = 50° and ∠ADB = 33°. Calculate :
(i) ∠DBC,
(ii) ∠DCB,
(iii) ∠CAB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q16.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q16.2
(i) ∠ CBD = ∠ DAC = 27° (Angles in the same segment)
(ii) In ΔADB,
∠ABD + ∠BAD + ∠BDA = 180°.
⇒ 50° + ∠BAD + 33° = 180°
⇒ ∠BAD + 83° = 180°
⇒ ∠BAD = 180° – 83° = 97°
In cyclic quad. ABCD,
∠BAD + ∠DCB = 180°
⇒ 97° + ∠DCB = 180°
⇒ ∠DCB = 180°- 97° = 83°
(iii) ∠BAD = 97°
⇒ ∠BAC + ∠CAD = 97°
⇒ ∠BAC + 27° = 97°
⇒ ∠BAC = 97° – 27° = 70°
∠CAB = 70°.

Question 17.
In the figure given below, AB is diameter of the circle whose centre is O. Given that:
∠ECD = ∠EDC = 32°. Show that ∠COF = ∠CEF.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q17.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q17.2
Given- AB is the diameter of a circle with centre O
and ∠ECD = ∠EDC = 32°
To Prove- ∠COF = ∠CEF
Proof- Arc CF subtends ∠COF at the centre and ∠CDF at the remaining part of the circle.
∠COF = 2 ∠CDF = 2 x ∠EDC = 2 x 32° = 64° ….. (i)
In ΔCED,
Ext. ∠CEF = ∠CDF + ∠DCE = ∠EDC + ∠ECD = 32° + 32° = 64° ….(ii)
from (i) and (ii)
∠CDF = ∠CEF

Question 18.
In the figure given below, AB and CD arc straight lines through the centre O of a circle. If ∠AOC = 80° and ∠CDE = 40°, find the number of degrees in:
(i) ∠DCE,
(ii) ∠ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q18.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q18.2
In circle, COD is the diameter.
∠CED = 90° (Angle in a semi circle)
In right A CDE,
∠ DCE + ∠ EDC = 90°
⇒ ∠ DCE + 40° = 90°
∠ DCE = 90° – 40° = 50°
In ΔOBC.
Ext. ∠COA = ∠OBC + ∠OCB
⇒ 80° = ∠OBC + 50°
⇒ ∠OBC = 80° – 50° = 30°
or ∠ABC = 30°

Question 19.
In the given figure, AC is a diameter of a circle, whose centre is O. A circle is described on AO as diameter. AE, a chord of the larger circle, intersects the smaller circle at B. Prove that AB = BE.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q19.1
Solution:
Given- AC is a diameter of a circle with centre O.
AE is a chord which intersects the smaller circle with AO as diameter at B.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q19.2
To Prove- AB = BE.
Construction- Join OB.
Proof- ∠ABO = 90° (Angle in a semi circle)
OB ⊥ AE
OB bisects chord AE
Hence, AB = BE.

Question 20.
In the following figure,
(i) if ∠ BAD = 96°, find ∠BCD and ∠BFE,
(ii) Prove that AD is parallel to FE.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q20.1
Solution:
Given- In the figure, ∠BAD = 96°
To Prove-
(i) Find ∠BCD and ∠BFE
(ii) AD || EF
Proof- ABCD is a cyclic quadrilateral.
∠BAD + ∠BCD = 180°
⇒ 96° + ∠BCD = 180°
⇒ ∠BCD = 180° – 96° = 84°
Again BCEF is a cyclic quadrilateral,
Ext. ∠BCD = Int. opposite ∠BFE
∠BFE = 84°.
∠BAD + ∠BFE = 96° + 84° = 180°
But these are on same side of the transversal.
AD || FE.

Question 21.
Prove that
(i) the parallelogram, inscribed in a circle, is a rectangle.
(ii) the rhombus, inscribed in a circle, is a square.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q21.1
(i) ABCD is a parallelogram in a circle with centre O.
To Prove- ABCD is a rectangle.
Proof- ABCD is a cyclic parallelogram.
∠A + ∠C = 180°.
But ∠A = ∠C (opposite angles of a ||gm)
∠A = ∠C = 90°
Similarly we can prove that
∠B = ∠D = 90°
Each angle of a ||gm is right angle
Hence ABCD is a rectangle.
(ii) Given- ABCD is a cyclic rhombus.
To Prove- ABCD is a square.
Proof- ABCD is cyclic rhombus
∠A + ∠C = 180°
But ∠A = ∠C (opposite angles of rhombus)
∠A = ∠C = 90°
Similarly we can prove that ∠B = ∠D = 90°
Each angle of a rhombus is a right angle
ABCD is a square.

Question 22.
In the following figure, AB = AC. Prove that DECB is an isosceles trapezium.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q22.1
Solution:
Given- In the figure, AB = AC.
To Prove- DECB is an isosceles trape∠ium.
Proof- In ∆ABC,
AB = AC
∠B = ∠C
DECB is a cyclic quadrilateral.
∠B + ∠DEC = 180°
∠C + ∠DEC = 180°
But this is the sum of interior angles on one side of a transversal.
DE || BC ….(i)
But ∠ADE = ∠B
and ∠AED = ∠C (Corresponding angles)
∠ADE = ∠AED (∠B = ∠C)
AD = AE (Opposite to equafangles)
But AB = AC (Given)
AB – AD = AC – AE
⇒ DB = EC ….(ii)
From (i) and (ii)
DECB is an isosceles trape∠ium.

Question 23.
Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are draw n. Show that the points A, Q and B are collinear.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q23.1
Given- Two circles with centres O and O’ intersect each other at P and Q.
From P, PA and PB are two diameters are drawn.
To Prove- A, Q and B are collinear.
Construction-
Join PQ, AQ and BQ.
Proof- In first circle.
∠PAQ = 90° (Angle in a semi circle) ….(i)
Similarly, in second circle, ∠PBQ = 90° ….(ii)
Adding (i) and (ii).
∠PAQ + ∠PBQ = 90° + 90° = 180°
But, there are adjacent angles
AQB is a straight line.
Hence A, Q and B are collinear.
Hence proved.

Question 24.
ABCD is a quadrilateral inscribed in a circle, having ∠A = 60°; O is the centre of the circle. Show that:
∠OBD + ∠ODB = ∠CBD + ∠CDB.
Solution:
Given- ABCD is a cyclic quadrilateral in which ∠A = 60°
and O is the centre of the circle.
BD, OB and OD are joined.
To Prove- ∠OBD + ∠ODB = ∠CBD + ∠CDB
Proof- Arc BCD subtends ∠BOD at the centre and ∠BAD at remaining part of the circle.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q24.1
∠BOD = 2 ∠BAD = 2 x 60° = 120°
In ∆BOD,
∠BOD = 120°
∠OBD + ∠ ODB = 180° – 120° = 60° …. (i)
ABCD is a cyclic quadrilateral
∠A + ∠C = 180°
⇒ 60° + ∠C = 180°
⇒ ∠C = 180° – 60° = 120°
In ∆BCD,
∠CBD + ∠CDB + ∠ C = 180°.
∠CBD + ∠CDB + 120° = 180°
∠CBD + ∠CDB = 180° – 120° = 60° ….(ii)
From (i) and (ii),
∠OBD + ∠ODB = ∠CBD + ∠CDB

Question 25.
The figure given below, shows a circle with centre O.
Given- ∠AOC = a and ∠ABC = b.
(i) Find the relationship between a and b. :
(ii) Find the measure of angle OAB, if OABC is a parallelogram.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q25.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q25.2
(i) ∠AOC = a, ∠ABC = b
Reflex ∠AOC = 360° – a
Now arc AC subtends reflex ∠AOC at the centre and ∠ABC at the remaining part of the circle
∠ABC = \(\frac { 1 }{ 2 }\) ref. ∠AOC
b = \(\frac { 1 }{ 2 }\) (360° – a)
⇒ 2b = 360° – a
⇒ a + 2b = 360° ….(i)
(ii) If OABC is a || gm,
then ∠AOC = ∠ABC
⇒ a = b
Substituting the value of a, in ….(i)
b + 2b = 360°
⇒ 3b = 360°
⇒ b = 120°
But ∠OAB + ∠ABC = 180° (Angles in a || gm)
⇒ ∠OAB + b = 180°
⇒ ∠OAB + 120° = 180°
⇒ ∠OAB = 180° – 120° = 60°.

Question 26.
Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD at the centre O is equal to twice the angle APC.
Solution:
Given- Two chords AB and CD intersect each other at P inside the circle, OA, OB, OC and OD are joined.
To Prove- ∠AOC + ∠BOD = 2 ∠APC.
Construction- Join AD
Proof- Arc AC subtends ∠AOC at the centre and ∠ADC at the remaining pari of the circle
∠AOC = 2 ∠ADC ….(i)
Similarly, ∠BOD = 2 ∠BAD ….(ii)
Adding (i) and (ii),
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q26.1
∠AOC + ∠BOD = 2 ∠ADC
⇒ 2 ∠ BAD = 2 (∠ADC + ∠BAD) ….(iii)
But in ∆PAD,
Ext. ∠APC = ∠PAD + ∠ADC = ∠ADC + ∠BAD …(iv)
from (iii) and (iv)
∠AOC + ∠BOD = 2 ∠APC

Question 27.
In the given figure, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°.
Calculate:
(i) ∠RNM,
(ii) ∠NRM.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q27.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q27.2
Join RN and MS. .
(i) RS is the diameter
∠RMS = 90° (Angle in semi circle)
∠RSM + ∠MRS = 90°
∠RSM = 90° – 29° = 61°
But ∠RSM + ∠RNM = 180° (Angles in a cyclic quad.)
61° + ∠RNM = 180°
⇒ ∠RNM = 180° – 61 = 119°
NM || RS
∠NMR = ∠MRS = 29° (Alt. angles)
In ∆RNM,
∠NRM + ∠RNM + ∠NMR = 180°
⇒ ∠NRM + 119° + 29° = 180°
⇒ ∠NRM + 148° = 180°
⇒ ∠NRM = 180° – 148° = 32°.

Question 28.
In the figure given alongside, AB // CD and O is the centre of the circle. If ∠ADC = 25°; find the angle AEB. Give reasons in support of your answer.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q28.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q28.2
AB || CD.
∠BAD = ∠ADC (Alternate angles) = 25° (∠ADC = 25° given)
Join AC and BD.
∠CAD = 90° (Angle in semi circle)
∠CAB = ∠CAD + ∠DAB = 90° + 25° = 115°
Now in cyclic quad. CABD.
∠CAB + ∠BDC = 180°
⇒ ∠CAB + ∠BDA + ∠ADC = 180°
⇒ 115° + ∠BDA + 25° = 180°
⇒ ∠BDA + 140° = 180°
⇒ ∠BDA = 180° – 140° = 40°
∠AEB and ∠BDA are in tire same segment of a circle
∠AEB = ∠BDA = 40° (proved)
Hence ∠AEB = 40°.

Question 29.
Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight fine is drawn to meet the circles at C and D. Prove that AC is || to BD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q29.1
Solution:
Given- Two circles intersect each other at P and Q. Through P, a line APB is drawn to meet the circles in A and B. Through Q, another straight line CQD is drawn meeting the circles in C and D.
AC, BD are joined.
To Prove- AC || BD.
Construction- Join PQ
Proof- APQC is a cyclic quadrilateral.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q29.2
∠A + ∠PQC = 180° …… (i)
In cyclic quad. PBDQ,
Ext. ∠PQC = ∠ B …… (ii)
from (i),
∠A + ∠B = 180°.
But these are interior angles on the same side of a transversal.
AC || BD.

Question 30.
ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q30.1
Given- The sides AB and DC of a cyclic quad. ABCD are produced to meet at P and PA = PD.
To Prove- AD || BC
Proof- In ∆PAD,
PA = PD (given)
∠A = ∠D (angles opposite to equal sides)
ADCB is a cyclic quad.
Ext. ∠PCB = ∠A = ∠D
But these are corresponding angles. ,
BC || AD or AD || BC.

Question 31.
AB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines. Find:
(i) ∠PRB,
(ii) ∠ PBR,
(iii) ∠ BPR.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q31.1
Solution:
∠PRB = ∠BAP (Angles in the same segment)
∠PRB = 35°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q31.2
In ∆ABP,
∠APB = 90° (Angle in semi circle)
∠BPQ = 90°.
In ∆PQR,
∠R + ∠Q + ∠RPQ = 180°
⇒ 35° + 25° + ∠RPQ = 180°
⇒ ∠RPQ = 180° – 60° = 120°
⇒ ∠BPR = ∠RPQ – ∠BPQ = 120° – 90° = 30°
In ∆PBR,
∠PBR = 180° – (∠R + ∠BPR) = 180° – (35° + 30°) = 180° – 65° = 115°

Question 32.
In the given figure SP is bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that SQ = SR.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q32.1
Solution:
Given- SP is the bisector of ∠RPT and PQRS is a cyclic quadrilateral.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q32.2
To Prove SQ = SR.
Proof- In cyclic quad. PQRS,
Ext. ∠SPT = ∠QRS
But ∠RPS = ∠SPT (PS is the bisector of ∠RPT)
∠QRS = ∠RPS ….(i)
But ∠RPS = ∠RQS (Angles in the same segment)
∠QRS = ∠RQS
Now in ∆QRS,
∠QRS = ∠RQS (proved)
SQ = SR (Sides opposite to equal angles)

Question 33.
In the figure, O is the centre of the circle, ∠AOE = 150°, ∠DAO = 51°. Calculate the si∠es of the angles CEB and OCE.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q33.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q33.2
In the figure, ∠AOE = 150°, ∠DAO = 51°
Now in cyclic quad. ADEB,
Ext. ∠CEB = Int. Opp ∠DAO = 51°.
In ∆OEB,
Ext. ∠AOE = ∠OBE + ∠OEB
= ∠OBE + ∠OBE (OB = OE) = 2 ∠OBE
2 ∠OBE = 150°
⇒ ∠ OBE = 75°
∠EBC = 180° – 75° = 105°
Now in ∆EBC,
∠CEB + ∠OCE + ∠EBC = 180°
⇒ 51° + ∠OCE + 105° = 180°
⇒ ∠OCE + 156° = 180°
⇒ ∠OCE = 180° – 156° = 24°.

Question 34.
In the figure, given below, P and Q arc the centres of two circles intersecting at B and C. ACD is a straight line. Calculate the numerical value of x.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q34.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q34.2
In circle with centre P,
Arc AB subtends ∠APB at the centre and ∠ACB at the remaining part of the circle.
∠APB = 2 ∠ACB
⇒ ∠ACB = \(\frac { 1 }{ 2 }\) ∠APB = \(\frac { 1 }{ 2 }\) x 150° = 75°
But ∠ACB + ∠DCB = 180° (Linear pair)
∠DCB = 180° – ∠ACB = 180° – 75° = 105°
In circle with centre O,
Arc BD subtends ∠ BQD at the centre and ∠ DCB at the remaining part of the circle
∠BQD = 2 ∠DCB = 2 x 105° = 210°
But x + ∠BQD = 360° (Angles at a point)
⇒ x + 210° = 360°
⇒ x = 360° – 210° = 150°.

Question 35.
The figure shows two circles which intersect at A and B. The centre of the smaller circle is O and lies on the circumference of the larger circle. Given ∠APB = a°. Calculate, in terms of a°, the value of :
(i) obtuse ∠AOB,
(ii) ∠ACB,
(iii) ∠ADB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q35.1
Give reasons for your answers clearly.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q35.2
Arc AB in small circle subtends ∠AOB at the centre and ∠APB at the remaining part of the circle.
(i) ∠AOB = 2 ∠APB = 2a° (∠APB = a°)
∠APB = (2a)°
(ii) In larger circle, AOBC is a cyclic quad.
∠AOB + ∠ACB = 180°.
⇒ 2a° + ∠ACB = 180°
∠ACB = 180° – 2a° = (180° – 2a°)
(iii) But ∠ ACB and ∠ ADB are in the same segment
∠ADB = ∠ ACB = (180° – 2a°)

Question 36.
In the given figure, O is the centre of the circle and ∠ABC = 55°. Calculate the values of x and y.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q36.1
Solution:
In ∆OBC,
OB = OC (radii of the same circle)
∠OBC = ∠BCO or ∠ABC = ∠BCO
∠BCO = ∠ABC = 55°
Now in ∆OBC,
Ext. AOC = ∠OBC + ∠BCO = 55° + 55° = 110°
x = 110°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q36.2
Now in cyclic quad, ABCD,
∠ADC + ∠ABC = 180°
⇒ y + 55° = 180°
⇒ y = 180° – 55° = 125°

Question 37.
In the given figure, A is the centre of the circle, ABCD is a parallelogram and CDE is a straight line. Prove that: ∠BCD = 2 ∠ABE.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q37.1
Solution:
Given- A is the centre of the circle and ABCD is a parallelogram.
CDE is a straight line.
To Prove- ∠BCD = 2 ∠ABE.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q37.2
Proof- AB || DC (opposite sides of a || gm)
∠ABE = ∠BED (Alternate angles) ….(i)
ABCD is a || gm (given)
∠BAD = ∠BCD (opposite angles of a ||gm) ….(ii)
Now arc BD subtends ∠BAD at the centre and ∠BED at the remaining part of the circle
∠BAD = 2 ∠BED
from (i) and (ii)
∠BCD = 2 ∠ABE

Question 38.
ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given ∠BED = 65°; calculate :
(i) ∠DAB,
(ii) ∠BDC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q38.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q38.2
∠DAB and ∠BED are in the same segment of the circle.
∠DAB = ∠BED = 65° (∠BED = 65° given)
DC || AB (Given)
∠BDC = ∠DBA (Alternate angles)
In ∆ADB,
AOB is the diameter
∠ADB = 90° (Angle in semi circle)
∠DAB + ∠DBA = 90°
⇒ 65° + ∠DBA = 90°
⇒ ∠DBA = 90° – 65° = 25°
But ∠DBA = ∠BDC (proved)
∠BDC = 25°

Question 39.
In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and ∠EAB = 63°. Calculate:
(i) ∠EBA,
(ii) ∠BCD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q39.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q39.2
AOB is the diameter,
∠AEB = 90°
and ∠EAB + ∠EBA = 90°
⇒ 63° + ∠EBA = 90°
⇒ ∠EBA = 90° – 63° = 27°
ED || AB (given)
∠DEB = ∠EBA (Alternate angles) = 27°
In cyclic quad. EBCD,
∠DEB + ∠BCD = 180° (opposite angles of a cyclic quad.)
⇒ 27° + ∠BCD = 180°
⇒ ∠BCD = 180° – 27° = 153°.

Question 40.
The sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E; the sides DA and CB arc produced to meet at F. If ∠BEG = 42° and ∠BAD = 98°; calculate :
(i) ∠AFB,
(ii) ∠ADC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q40.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q40.2
In ∆AED,
∠ADE + ∠AED + ∠EAD = 180° (Angles of a triangle)
⇒ ∠ADE + 42° + 98° = 180°
⇒ ∠ADE + 140° = 180°
⇒ ∠ ADE = 180° – 140° = 40° or ∠ADC = 40°
In cyclic quad. ABCD.
∠BAD + ∠BCD = 180°
⇒ 98° + ∠BCD = 180°
⇒ ∠BCD = 180° – 98° = 82°
Now in ∆FCD,
∠DFC + ∠FDC + ∠FCD = 180°
⇒ ∠AFB + ∠ADC + ∠BCD = 180°
⇒ ∠AFB + 40° + 82° = 180°
⇒ ∠AFB + 122° = 180°
⇒ ∠AFB = 180° – 122° = 58°

Question 41.
In the following figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°. Calculate :
(i) ∠DAB,
(ii) ∠DBA,
(iii) ∠DBC,
(iv) ∠ADC. Also, show that the ∆AOD is an equilateral triangle.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q41.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q41.2
(i) ABCD is a cyclic quadrilateral,
∠DCB + ∠DAB = 180°
⇒ 120° + ∠DAB = 180°
∠DAB =180° – 120° = 60°
(ii) AOB is a diameter.
∠ADB = 90° (Angle in a semi circle)
∠ DAB + ∠DBA = 90°
60° + ∠ DBA = 90°
∠DBA = 90° – 60° = 30°
(iii) In ∆OBD,
OD = OB (radii of the same circle)
∠ ODB = ∠ OBD
or ∠ABD = 30° (from ii)
But DO || CB (given)
∠ODB = ∠DBC (Alternate angles)
⇒ 30° = ∠DBC or ∠DBC = 30°
(iv) ∠ABD + ∠DBC = 30° + 30° = 60°
⇒ ∠ABC = 60°
Again in cyclic quad. ABCD,
∠ADC + ∠ABC = 180°
∠ADC + 60° = 180°
∠ ADC = 180° – 60° = 120°
In ∆AOD,
OA = OD (radii of the same circle)
∠ AOD = ∠ DAO or ∠ DAB = 60° (proved in (i))
∠ADO = 60° (Third angle)
∠ADO = ∠AOD = ∠DAO = 60°
∆AOD is an equilateral triangle.

Question 42.
In the given figure, I is the incentre of ∆ABC. BI when produced meets the circum circle of ∆ABC at D.
Given ∠BAC = 55° and ∠ACB = 65°; calculate:
(i) ∠DCA,
(ii) ∠DAC,
(iii) ∠DCI,
(iv) ∠AIC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q42.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q42.2
Join AD, DC, AI and Cl,
In ∆ABC,
∠BAC = 55°, ∠ACB = 65°
∠ABC = 180° – (∠BAC + ∠ACB) = 180°- (55° + 65°) = 180° – 120° = 60°
In cyclic quad. ABCD,
∠ABC + ∠ADC = 180°
⇒ 60° + ∠ADC = 180°
∠ADC = 180° – 60°= 120°
In ∆ADC,
∠ DAC + ∠ DCA + ∠ ADC = 180°
⇒ ∠ DAC+ ∠ DCA + 120° = 180°
⇒ ∠ DAC+ ∠ DCA = 180° – 120° = 60°
But ∠ DAC = ∠ DCA (I lies on the bisector of ∠ ABC)
∠ DAC = ∠ DCA = 30°
DI is perpendicular bisector of AC
∠ AIC = ∠ ADC= 120°
IC is the bisector of ∠ ACB
∠ ICA = \(\frac { 65 }{ 2 }\) = 32.5°
∠DCI = ∠DCA + ∠ACI = 30° + 32.5° = 62.5° = (62.5)° = 60° 30′.

Question 43.
A triangle ABC is inscribed in a circle. The bisectors of angles BAC, ABC and ACB meet the circum circle of the triangle at points P, Q and R respectively. Prove that:
(i) ∠ ABC = 2 ∠ APQ,
(ii) ∠ ACB = 2 ∠ APR,
(iii) ∠ QPR = 90° – \(\frac { 1 }{ 2 }\) ∠BAC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q43.1
Solution:
Given- ∆ABC is inscribed in a circle. Bisectors of ∠BAC, ∠ABCand ∠ACB meet the circumcircle of the∆ABC at P, Q and R respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q43.2
To Prove-
(i) ∠ ABC = 2 ∠APQ.
(ii) ∠ ACB = 2 ∠ APR.
(iii) ∠ QPR = 90° – \(\frac { 1 }{ 2 }\) ∠BAC.
Construction-
Join PQ and PR.
Proof- ∠ABQ and ∠APQ are in the same segment of the circle.
∠ ABQ = ∠ APQ
But ∠ ABQ = – ∠ ABC (BQ is the angle bisector of ∠ ABC)
\(\frac { 1 }{ 2 }\) ∠ ABC = ∠ APQ
Or ∠ ABC = 2 ∠APQ …,(i)
Similarly, ∠ APR and ∠ ACR are in the same segment of the circle.
∠ APR = ∠ ACR
But ∠ ACR = \(\frac { 1 }{ 2 }\) ∠ ACB (CR is the angles bisector of ∠ ACB)
\(\frac { 1 }{ 2 }\) ∠ ACB = ∠ APR
Or ∠ ACB = 2 ∠ APR ….(ii)
Adding (i) and (ii)
∠ ABC + ∠ ACB = 2 ∠ APQ + 2∠ APR = 2 (∠ APQ + ∠ APR) = 2 ∠ PQR
Or 2 ∠ PQR = ∠ ABC + ∠ ACB
∠ PQR = \(\frac { 1 }{ 2 }\) (∠ ABC + ∠ ACB) ….(iii)
But ∠ ABC + ∠ ACB + ∠ BAC = 180° (Angles of a triangle)
∠ ABC + ∠ ACB = 180° – ∠ BAC ….(iv)
from (iii) and (iv) we get,
∠ PQR = \(\frac { 1 }{ 2 }\) (180° – ∠ BAC) = 90° – \(\frac { 1 }{ 2 }\) ∠ BAC

Question 44.
Calculate the angles x, y and z if :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q44.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q44.2
Ext. ∠ ADC = x + z ….(i)
and in ΔBPC,
Ext. ∠ ABC = y + x ….(ii)
(∠ BCP = ∠ DCQ = x vertically opposite angles)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q44.3
Adding (i) and (ii),
x + z + y + x = ∠ ADC + ∠ ABC.
But ∠ ADC + ∠ ABC = 180° (opposite angles of a cyclic quad)
2x + y + z = 180°
⇒ 2 x 3k + 4k + 5k = 180°
⇒ 6k + 4k + 5k = 180°
⇒ 15k = 180°
⇒ k = 12°
x = 3k = 3 x 12° = 36°= x = 36°
y = 4k = 4 x 12° = 48°= y = 48°
z = 5k = 5 x 12° = 60° = z = 60°

Question 45.
In the given figure, AB = AC = CD and ∠ADC = 38°. Calculate :
(i) Angle ABC
(ii) Angle BEC [1995]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q45.1
Solution:
In the figure, AB = AC = CD, ∠ ADC = 38°
BE is joined.
In ΔACD, AC = CD
∠CAD = ∠CDA = 38°
Ext. ∠ ACB = ∠ CAD + ∠ CDA = 38° + 38° = 76°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q45.2
But in ΔABC,
AB = AC (given)
∠ ABC = ∠ ACB = 76°
and ∠ BAC =180° – (76° + 76°) = 180° – 152° = 28°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q45.3
But ∠ BEC = ∠ BAC (Angles in the same segment)
∠BEC = 28°.

Question 46.
In the given figure. AC is the diameter of circle, centre O. Chord BD is perpendicular to AC. Write down the angles p, q and r in terms of x. [1996]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q46.1
Solution:
Arc subtends ∠ AOB at the centre and ∠ACB at the remaining part of the circle.
∠ AOB = 2 ∠ ACB
⇒ x = 2q ⇒ q = \(\frac { x }{ 2 }\)
But ∠ ADB and ∠ ACB are in the same segment
∠ ADB = ∠ ACB – q
Now in ΔAED,
p + q + 90° = 180° (sum of angles of a Δ)
⇒ p + q = 90°
⇒ p = 90° – q
⇒ p = 90° – \(\frac { x }{ 2 }\)
Arc BC subtends ∠ BOC at the centre and ∠ ADC at the remaining part of the circle
∠BOC = 2 ∠BDC = 2r.
r = \(\frac { 1 }{ 2 }\) ∠ BOC = \(\frac { 1 }{ 2 }\) (180° – x)
(∠ AOB + ∠ BOC = 180°)
r = 90° – \(\frac { 1 }{ 2 }\) x. = 90° – \(\frac { x }{ 2 }\)

Question 47.
In the given figure, AC is the diameter of the circle with centre O. CD and BE are parallel. Angle ∠ AOB = 80° and ∠ ACE = 10°. Calculate:
(i) Angle BEC,
(ii) Angle BCD,
(iii) Angle CED. [1998]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q47.1
Solution:
Arc AB subtends ∠ AOB at the centre and ∠ACB at the remaining part of the circle.
∠ ACB = \(\frac { 1 }{ 2 }\) ∠AOB = \(\frac { 1 }{ 2 }\) x 80° = 40°
But ∠ BOC + ∠ AOB = 180° (A linear pair)
∠ BOC + 80° = 180°
⇒ ∠ BOC = 180° – 80° = 100°
(i) Arc BC subtends ∠ BOC at the centre and ∠ BEC at the remaining part of the circle
∠ BEC = \(\frac { 1 }{ 2 }\) ∠ BOC = \(\frac { 1 }{ 2 }\) x 100° = 50°
(ii) EB || DC
∠ DCE = ∠ BEC (Alternate angles) = 50°
∠ BCD = ∠ BCA + ∠ ACE + ∠ ECD = 40° + 10° + 50° = 100°
(iii) In cyclic quad. CDE,
∠ BED + ∠ BCD = 180°
⇒ ∠ BEC + ∠ CED + ∠ BCD = 180°
⇒ 50° + ∠ CED + 100° = 180° (Proved in (i) and (ii))
⇒ ∠ CEb + 150° = 180°
∠ CED = 180° – 150° = 30°.

Question 48.
In the given figure, AE is the diameter of the circle. Write down the numerical value of ∠ ABC + ∠ CDE. Give reasons for your answer. [1998]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q48.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q48.2
Join OA, OB, OC, OD.
In ΔOAB,
OA = OB (Radii of the same circle)
∠ 1 = ∠ 2
Similarly we can prove that
∠3 = ∠4,
∠5 = ∠6,
∠7 = ∠8
In A OAB,
∠1 + ∠2 + ∠a = 180° (Angles of a triangle)
Similarly ∠3 + ∠4 + ∠b = 180°
∠5 + ∠6 + ∠c = 180°
∠7 + ∠8 + ∠d = 180°
Adding we get
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 + ∠a + ∠b + ∠c + ∠d = 4 x 180° = 720°
⇒∠2 + ∠2 + ∠3 + ∠3 + ∠6 + ∠6+ ∠ 7 + ∠7 + ∠a + ∠b + ∠c + ∠d = 720°
⇒ 2 ∠2 + 2 ∠3 + 2 ∠6 + 2 ∠7 + ∠a + ∠ b + ∠ c + ∠ d = 720°
⇒ 2 [∠2 + ∠3] + 2 [∠6 + ∠7| + 180° = 720° ( ∠a + ∠b + ∠c + ∠d = 180°)
⇒ 2 ∠ ABC + 2 ∠ CDE = 720° – 180° = 540°
⇒ 2 (∠ ABC + ∠ CDE) = 540°
⇒ ∠ ABC + ∠ CDE = 270°

Question 49.
In the given figure, AOC is a diameter and AC is parallel to ED. If ∠ CBE = 64°, calculate ∠ DEC. [1991]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q49.1
Solution:
Join AB.
AOC is the diameter
∠ ABC = 90° (Angle in a semi circle)
⇒ ∠ ABE + ∠ CBE = 90°
⇒ ∠ ABE + 64° = 90°
∠ ABE = 90° – 64° = 26° …(i)
AC || ED
∠ DEC = ∠ ACE (alternate angles)
But ∠ ACE = ∠ ABE (Angles in the same segment)
∠ DEC = ∠ ABE = 26° [from (i)]

Question 50.
Use the given figure to find :
(i) ∠ BAD,
(ii) ∠ DQB. [1987]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q50.1
Solution:
In ΔAPD,
∠ ADP + ∠ DPA + ∠ PAD = 180°
85° + 40° + ∠ PAD = 180°
∠ PAD = 180° – (85° + 40°) = 180° – 125° = 55° or ∠ BAD = 55°
In cyclic quad. ABCD,
∠ ADC + ∠ ABC = 180°
85° + ∠ ABC = 180°
∠ ABC = 180° – 85° = 95°
Now, in ΔAQB,
∠ QAB + ∠ ABC + ∠ BQA = 180°
⇒ 55° + 95° + ∠ BQA = 180°
⇒ 150° + ∠ BQA = 180°
⇒ ∠DQB = ∠ BQA = 180° – 150° = 30°

Question 51.
In the given figure, AOB is a diameter and DC is parallel to AB. If ∠ CAB = x°; find (in terms of x) the values of:
(i) ∠ COB,
(ii) ∠DOC,
(iii) ∠DAC,
(iv) ∠ADC. [1991]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q51.1
Solution:
Join CB.
In ΔAOC,
OA = OC (radii of the same circle)
∠ OCA = 4 OAC = x
Ext. ∠ COB = ∠ OAC + ∠ OCA = x + x = 2x
In ΔACB, ∠ ACB = 90° (Angle in semi circle)
∠ OBC = 90° – ∠ OAC = 90° – x
In cyclic quad. ABCD,
∠ ABC + ∠ ADC = 180°
⇒ ∠OBC + ∠ADC =180°
⇒ (90 – x) + 4 ADC = 180°
∠ADC = 180° – 90° + x = 90° + x
DC || AB
∠ DCO = ∠ COB = 2x (alternate angle)
And ∠ DCA = ∠ CAB = x (alternate angles)
In ΔADC,
∠ DAC + ∠ DCA + ∠ ADC = 180°
∠ DAC + x + 90 + x = 180°
2x + 90° + ∠ DAC = 180°
∠ DAC = 180° – 90° – 2x = 90° – 2x
In ΔOCD,
∠ DOC + ∠ OCD + ∠ CDO = 180°
∠ DOC + 2x + 2x = 180°
∠ DOC = 180° – 4x
Hence ∠ COB = 2x,
∠ DOC = 180° – 4x
∠ DAC = 90° – 2x
and ∠ ADC = 90° + x

Question 52.
In the given figure, AB is the diameter of a circle with centre O. ∠BCD = 130°. Find :
(i) ∠DAB
(ii) ∠DBA
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q52.1
Solution:
Join DB
(i) ∠DAB + ∠DCB = 180° [Opposite angles of a cyclic quadrilateral are supplementary]
∠DAB = 180° – 130° = 50°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q52.2
(ii) In ΔADB,
∠ADB = 90° [Angle in a semi-circle is 90°]
So, ∠DBA = 180° – (∠DAB + ∠ADB) = 180° – (50° + 90°) = 40°

Question 53.
In the given figure, PQ is a diameter of the circle whose centre is O. Given ∠ROS = 42°, calculate ∠RTS. [1992]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q53.1
Solution:
In ΔOPR, OR = OP (radii of the same circle)
∠ OPR = ∠ ORP = x (Say)
∠POR = 180° – 2x
Similarly in ΔOQS,
OS = OQ
∠ OSQ = ∠ SQO = y (say)
∠ SOQ = 180° – 2y
POQ is a straight line,
∠ POR + ∠ ROS + ∠ SOQ = 180°
⇒ 180° – 2x + 42° + 180° – 2y = 180°
⇒ 222° – 2x – 2y = 0
⇒ 2 (x + y) = 222°
x + y = 111° ….(i)
In. ΔPQT,
⇒ ∠P + ∠Q + ∠T = 180°
⇒ ∠ OPR + ∠ SQO + ∠ RTS = 180°
⇒ x + y + ∠RTS = 180°
⇒ ∠ RTS = 180° – (x + y) = 180° – 1110 [From(i)] = 69°

Question 54.
In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠ PQR = 58°, Calculate:
(i) ∠RPQ,
(ii) ∠STP. [1989]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q54.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q54.2
Join PR,
In ΔPQR,
∠ PRQ = 90° (Angle in a semi-circle)
∠ RPQ + ∠ RQP = 90°
⇒ ∠RPQ + 58° = 90°
⇒ ∠RPQ = 90° – 58° = 32°
SR || PQ (given)
∠ SRP = ∠ RPQ (Alternate angles) = 32° [from(i)]
Now, in cyclic quad. PRST,
∠ STP + ∠ SRP = 180°
⇒ ∠STP + 32° = 180°
⇒ ∠STP = 180° – 32° = 148°

Question 55.
AB is the diameter of the circle with centre O. OD is parallel to BC and ∠ AOD = 60°. Calculate the numerical values of: [1987]
(i) ∠ ABD
(ii) ∠ DBC
(iii) ∠ ADC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q55.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q55.2
Join BD,
Arc AD, subtends ∠ AOD at the centre and ∠ ABD at the remaining part of the circle
∠ ABD = \(\frac { 1 }{ 2 }\) ∠ AOD = \(\frac { 1 }{ 2 }\) x 60° = 30°
In ΔOBD,
OB = OD (Radii of the same circle)
∠ ODB = ∠ OBD = ∠ ABD = 30°
OD||BC (given)
∠ ODB = ∠ DBC (Alternate angles)
∠ DBC = ∠ ODB = 30° .
Again OD || BC
∠ AOD = ∠ OBC (Corresponding angles)
⇒ ∠ OBC = ∠ AOD = 60°
Now, in cyclic quad. ABCD,
∠ ADC+ ∠ ABC = 180°
⇒ ∠ ADC + 60° = 180°
⇒ ∠ ADC = 180° – 60° = 120°

Question 56.
In the given figure, the centre O of the small circle lies on the circumference of the bigger circle. If ∠ APB = 75° and ∠ BCD = 40°. find:
(i) ∠AOB,
(ii) ∠ACB,
(iii) ∠ABD,
(iv) ∠ADB. [1984]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q56.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q56.2
Join AB, AD.
(i) Arc AB of the smaller circle subtends ∠ AOB at the centre and ∠ APB at the remaining part of the circle.
∠ AOB = 2 ∠ APB = 2 x 75° = 150°
(ii) OACB is a cyclic quad.
∠AOB + ∠ACB = 180°
⇒ 150° + ∠ACB = 180°
⇒ ∠ACB = 180° – 150° = 30°
(in) Again, ABDC is a cyclic quad,
∠ ABD + ∠ ACD = 180°
⇒ ∠ABD + (30° + 40°) = 180° (∠ ACD = ∠ ACB + ∠ BCD)
⇒ ∠ ABD + 70° = 180°
⇒ ∠ ABD = 180° – 70° = 110°
(iv) ∠ ACB and ∠ADB are in the same segment
∠ ADB = ∠ ACB = 30°

Question 57.
In the given figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°. Find :
(i) ∠BCD
(ii) ∠ACB
Hence, show that AC is a diameter.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q57.1
Solution:
In circle ABCD is a cyclic quadrilateral AC andBD are joined.
∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°
ABCD is a cyclic quadrilateral
65° + ∠BCD = 180°
⇒ ∠BCD = 180° – 65° = 115°
Arc AB subtends ∠ABD and ∠ACD in the same segment
∠ACD = ∠ABD = 70° (∠ABD = 70°)
∠ACB = ∠BCD – ∠ACD = 115° – 70° = 45°
But arc AB subtends ∠ADB and ∠ACD in the same segment
∠ADB = ∠ACB = 45°
∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°
Now in segment ADC,
∠ADC = 90°
Segment ADC is a semi-circle
Hence AC is the diameter of the circle.
Hence proved.

Question 58.
In a cyclic quadrilateral ABCD, ∠A : ∠C = 3 : 1 and ∠B : ∠D = 1 : 5; find each angle of the quadrilateral.
Solution:
In a cyclic quadrilateral ABCD,
∠A : ∠C = 3 : 1 and ∠B : ∠D = 1 : 5
Let ∠A = 3x, ∠C = x,
But ∠A + ∠C = 180° (Opposite angles of a cyclic quadrilateral)
⇒ 3x + x = 180°
⇒ 4x = 180°
⇒ x = 45°
∠A = 3x = 3 x 45° = 135°
∠C = x = 45°
∠B = ∠D = 1 : 5
Similarly, Let ∠B = y and ∠D = 5y
But ∠B + ∠D = 180°
y + 5y = 180°
⇒ 6y = 180°
⇒ y = 30°
∠B = y = 30°
and ∠D = 5y = 5 x 30° = 150°
Hence ∠A = 135°, ∠B = 30°, ∠C = 45° and ∠D = 150°
Hence Proved.

Question 59.
The given figure shows a circle with centre O and ∠ABP = 42°. Calculate the measure of:
(i) ∠PQB
(ii) ∠QPB + ∠PBQ
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q59.1
Solution:
In the figure
∠ABP = 42°.
Join PO, QO
Arc PA subtends ∠POA at the centre and ∠PBA at the remaining part.
∠POA = 2 ∠PBA = 2 x 42° = 84°
But ∠AOP + ∠BOP = 180° (Linear pair)
⇒ ∠POA+ ∠POB = 180°
⇒ 84° + ∠POB = 180°
⇒ POB = 180° – 84° = 96°
Similarly, arc BP subtends ∠BOP on the centre and ∠PQB at the remaining part of the circle
∠PQB = \(\frac { 1 }{ 2 }\) ∠POB = \(\frac { 1 }{ 2 }\) x 96° = 48°
But in ΔPBQ,
∠QPB + ∠PBQ + ∠PQB = 180° (Angles of a triangle)
∠QPB + ∠PBQ + 48° =180°
⇒ ∠QPB + ∠PBQ = 180°
⇒ ∠QPB + ∠PBQ = 180° – 48° = 132
Hence (i) PQB = 48° and
(ii) ∠QPB + ∠PBQ = 132°

Question 60.
In the given figure, M is the centre of the circle. Chords AB and CD are perpendicular to each other.
If ∠MAD = x and ∠BAC = y:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A Q60.1
(i) express ∠AMD in terms of x.
(ii) express ∠ABD in terms of y.
(iii) prove that: x = y.
Solution:
In the figure, M is the centre of the circle chords AB and CD are perpendicular to each other at L.
∠MAD = x and ∠BAC = y
(i) In ΔAMD,
AM = DM (Radii of the same circle)
∠MDA = ∠MAD (Angles opposite to equal sides) = x
But, in ΔAMD,
∠MAD + ∠MDA + ∠AMD = 180° (Sum of angles of a triangle)
⇒ x + x + ∠AMD = 180°
⇒ 2x + ∠AMD = 180°
⇒ ∠AMD = 180°
∠AMD = 180° – 2x
(ii) Arc AD subtends ∠AMD at the circle and ∠ABD at the remaining part of the circle
∠AMD = 2∠ABD
⇒ ∠ABD = \(\frac { 1 }{ 2 }\) ∠ABD = \(\frac { 1 }{ 2 }\) [180°- 2x] = 90° – x
AB ⊥s CD
∠ALC = 90°
In ΔALC,
∠LAC + ∠LCA = 90°
⇒ ∠BAC + ∠DAC = 90°
⇒ y = ∠DAC = 90°
⇒ ∠DAC = 90° – y
But ∠DAC ∠ABD (Angles in the same segment)
∠ABD = 90° – y
But ∠ABD = 90° – x, Proved
90°- x = 90°- y
⇒ ∠x = yHence proved.

Question 61.
In a circle, with centre O, a cyclic quadrilateral ABCD is drawn with AB as a diameter of the circle and CD equal to radius of the circle. If AD and BC produced meet at point P; show that ∠APB = 60°.
Solution:
Given : In circle with centre O,
ABCD is cyclic quadrilateral in which CD is equal radius of the circle and AB is diameter.
CD = \(\frac { 1 }{ 2 }\) AB
AD and BC are produced to meet at P.
To prove: ∠APB = 60°
Construction : Join DO, CO and PB
In ΔDOC,
DO = CO = DC (Radii of the circle)
ΔDOC is an equilateral triangle
∠DOC = 60° ….(i)
Now, arc DC subtends ∠DOC at the centre arc ∠DBC at the remaining part of the circle
∠DBC = \(\frac { 1 }{ 2 }\) ∠DOC = \(\frac { 1 }{ 2 }\) x 60° = 30° …(ii)
But ∠ADB = 90° (Angle in a semi-circle)
∠PDB = 90° (∠PDB x ∠ADB = 180°, Linear pair)
Now in ΔPDB,
∠PDB = 90° (Proved)
⇒ ∠DPB = ∠DBC = 90°
⇒ ∠DPB + 30° = 90°
⇒ ∠DPB = 90° – 30° = 60°
⇒ APB = 60°
Hence proved.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A are helpful to complete your math homework.

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