Raju – Page 6 – NCERT MCQ

RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2

March 31, 2023

RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.2

Other Exercises

Solve each of the following equations and also check your result in each case :
Question 1.
\(\frac { 2x + 5 }{ 3 }\) = 3x – 10
Solution:
\(\frac { 2x + 5 }{ 3 }\) = \(\frac { 3x – 10 }{ 1 }\)
By cross multiplication
⇒ 2x + 5 = 3 (3x – 10)
⇒ 2x + 5 = 9x – 30
⇒ 5 + 30 = 9x – 2x (By transposition)
⇒ 35 = 7x
⇒ x = 5
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 1

Question 2.
\(\frac { a – 8 }{ 3 }\) = \(\frac { a – 3 }{ 2 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 2

Question 3.
\(\frac { 7y + 2 }{ 5 }\) = \(\frac { 6y – 5 }{ 11 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 3

Question 4.
x – 2x + 2 – \(\frac { 16 }{ 3 }\) x + 5 = 3 – \(\frac { 7 }{ 2 }\) x.
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 5

Question 5.
\(\frac { 1 }{ 2 }\) x + 7x – 6 = 7x + \(\frac { 1 }{ 4 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 7

Question 6.
\(\frac { 3 }{ 4 }\) x + 4x = \(\frac { 7 }{ 8 }\) + 6x – 6
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 8

Question 7.
\(\frac { 7 }{ 2 }\) x – \(\frac { 5 }{ 2 }\) x = \(\frac { 20 }{ 3 }\) x + 10
Solution:
\(\frac { 7 }{ 2 }\) x – \(\frac { 5 }{ 2 }\) x = \(\frac { 20 }{ 3 }\) x + 10
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 10

Question 8.
\(\frac { 6x + 1 }{ 2 }\) + 1 = \(\frac { 7x – 3 }{ 3 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 11

Question 9.
\(\frac { 3a – 2 }{ 3 }\) + \(\frac { 2a + 3 }{ 2 }\) = a + \(\frac { 7 }{ 6 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 13

Question 10.
x – \(\frac { x – 1 }{ 2 }\) = 1 – \(\frac { x – 2 }{ 3 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 15

Question 11.
\(\frac { 3x }{ 4 }\) – \(\frac { x – 1 }{ 2 }\) = \(\frac { x – 2 }{ 3 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 17

Question 12.
\(\frac { 5x }{ 3 }\) – \(\frac { x – 1 }{ 4 }\) = \(\frac { x – 3 }{ 5 }\)
Solution:
\(\frac { 5x }{ 3 }\) – \(\frac { x – 1 }{ 4 }\) = \(\frac { x – 3 }{ 5 }\)
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 18

Question 13.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 20
Solution:

Question 14.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 23
Solution:

Question 15.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 26
Solution:

Question 16.
0.18 (5x – 4) = 0.5x + 0.8
Solution:
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 29

Question 17.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 30
Solution:

Question 18.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 33
Solution:

Question 19.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 36
Solution:

Question 20.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 39
Solution:

Question 21.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 42
Solution:

Question 22.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 46
Solution:

Question 23.
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 50
Solution:

Question 24.
(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
Solution:
(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
⇒ (9x² + 6x – 24x – 16) – (8x² + 4x – 22x – 11) = x² + 7x – 3x – 21
⇒ 9x² + 6x – 24x – 16 – 8x² – 4x + 22x + 11 = x² + 4x – 21
⇒ 9x² – 8x² – x² + 6x – 24x + 22x – 4x – 4x = -21 + 16 – 11
⇒ 28x – 32x = -32 + 16
⇒ -4x = -16
⇒ x = 4
Verification:
L.H.S. = (3x – 8) (3x + 2) – (4x – 11) (2x + 1)
= (3 x 4 – 8) (3 x 4 + 2) – (4 x 4 – 11) (2 x 4 + 1)
= (12 – 8) (12 + 2) – (16 – 11) (8 + 1)
= 4 x 14 – 5 x 9 = 56 – 45 = 11
R.H.S. = (x – 3) (x + 7) = (4 – 3) (4 + 7) = 1 x 11 = 11
L.H.S. = R.H.S.

Question 25.
[(2x + 3) + (x + 5)]² + [(2x + 3) – (x + 5)]² = 10x² + 92
Solution:
[(2x + 3) + (x + 5)]² + [(2x + 3) – (x + 5)]² = 10x² + 92
⇒ (2x + 3 + x + 5)² + (2x + 3 – x – 5)² = 10x² + 92
⇒ (3x + 8)² + (x – 2)² = 10x² + 92
⇒ 9x² + 2 x 3x x 8 + 64 + x² – 2 x x x 2 + 4 = 10x² + 92
⇒ 9x² + 48x + 64 + x² – 4x + 4 = 10x² + 92
⇒ 9x² + x² – 10x² + 48x – 4x = 92 – 64 – 4
⇒ 44x = 24
RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2 53

Hope given RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.2 are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B

March 31, 2023

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B.

Other Exercises

Question 1.
A man buys 75, Rs. 100 shares paying 9 per cent dividend. He buys shares at such a price that he gets 12 per cent of his money. At what price did he buy the shares?
Solution:
No. of shares = 75
Value of each share = Rs. 100
Rate of dividend = 9%
Let the market value of each share = x
Thus. 12% of x = 9% of 100
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 1.1
Market value of each share = Rs. 75

Question 2.
By purchasing Rs. 25 gas shares for Rs. 40 each, a man gets 4 per cent profit on his investment. What rate per cent is the company paying ? What is his dividend if he buys 60 shares?
Solution:
Face value of each share = Rs. 25
Market value of each share = Rs. 40
Profit = 4% of his investment
Let the rate of dividend = x %
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 2.1

Question 3.
Hundred rupee shares of a company are available in the market at a premium of Rs. 20. Find the rate of dividend given by the company, when a man’s return on his investment is 15 percent.
Solution:
Face value of each share = Rs. 100
Market value of each share = Rs. 120
Total dividend = 15% on his investment
Let the rate of dividend = x %
Then x % of 100 = 15 % of 120
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 3.1
x = 18%
Rate of dividend = 18%

Question 4.
Rs. 50 shares of a company are quoted at a discount of 10%. Find the rate of dividend given by the company, the return on the investment on these shares being 20 percent.
Solution:
Face value of each share = Rs. 50
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 4.1
Rate of dividend = 18%

Question 5.
A company declares 8 per cent dividend to the share holders. If a man receives Rs. 2,840 as his dividend, find the nominal value of his shares.
Solution:
Rate of dividend = 8%
Total dividend = Rs. 2.840
Let Nominal value of shares = x
Then 8% of x = Rs. 2840
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 5.1
Nominal value of his shares = Rs. 35,500

Question 6.
How much should a man invest in Rs. 100 shares selling at Rs. 110 to obtain an annual income of Rs. 1,680, if the dividend declared is 12% ?
Solution:
Face value of each share = Rs. 100
Market value of each share = Rs. 110
Total annual income = Rs. 1,680.
Rate of dividend = 12%
Let total amount of shares = x
Then x x 12% = 1,680
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 6.1

Question 7.
A company declares a dividend of 11.2% to all its share-holders. If its Rs. 60 share is available in the market at a premium of 25%, how much should Rakesh invest in buying the shares of this company in order to have an annual income of Rs. 1,680 ?
Solution:
Face value of each share = Rs. 60
Market value = Rs. 60 x \(\frac { 125 }{ 100 }\) = Rs. 75
Rate of dividend = 11.2%
Annual income = Rs. 1,680
Let the face value of shares = x
Dividend = x x 11.2%
x x 11.2% = 1680
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 7.1

Question 8.
A man buys 400, twenty rupee shares at a premium of Rs. 4 each and receives a dividend of 12%. Find :
(i) the amount invested by him.
(ii) his total income from the shares.
(iii) percentage return on his money.
Solution:
No. of shares = 400
Face value of one share = Rs. 20
Market value of one share = Rs. 20 + 4 = Rs. 24
Rate of dividend = 12%
(i) Amount invested by the man = Rs. 24 x 400 = Rs. 9600
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 8.1

Question 9.
A man buys 400, twenty-rupee shares at a discount of 20% and receives a return of 12% on his money. Calculate
(i) the amount invested by him.
(ii) the rate of dividend paid by the company.
Solution:
No. of shares = 400
Face value of one share = Rs. 20
Market value of one share = Rs. 20 x \(\frac { 80 }{ 100 }\) = Rs. 16
Amount of shares = Rs. 20 x 400 = Rs. 8,000
(i) Amount invested = Rs. 16 x 400 = Rs. 6,400
(ii) Total dividend = Rs. 6,400 x \(\frac { 12 }{ 100 }\) = Rs. 768
Rate of dividend = \(\frac { 768 x 100 }{ 8000 }\)= 9.6 %

Question 10.
A company with 10,000 shares of Rs. 100 each, declares an annual dividend of 5%.
(i) What is the total amount of dividend paid by the company ?
(ii) What should be the annual income of a man who has 72 shares, in the company ?
(iii) If he received only 4% of his investment, find the price he paid for each share.
Solution:
No. of shares = 10000
Face value of each share = 100
Rate of dividend = 5%
Amount of shares = Rs. 100 x 10,000 = Rs. 10,00,000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 10.1

Question 11.
A lady holds 1800, Rs. 100 shares of a company that pays 15 % dividend annually. Calculate her annual dividend. If she had bought these shares at 40% premium, what is the return she gets as percent on her investment. Give your answer to the nearest integer.
Solution:
No. of shares = 1800
Face value of each share = Rs. 100
Rate of dividend = 15 %
Market value of each share Rs. 140
Total value of shares = Rs. 1800 x 100 = Rs. 1,80,000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 11.1

Question 12.
A man invests Rs. 11,200 in a company paying 6 percent per annum when its Rs. 100 shares can be bought for Rs. 140 find:
(i) his annual dividend.
(ii) his percentage return on his investment.
Solution:
Investment = Rs. 11,200
Rate of dividend = 6%
Market value of a share = Rs. 140
No. of shares = Rs. 11,200 ÷ Rs. 140 = 80
Face value of 80 shares = 80 x Rs. 100 = Rs. 8,000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 12.1

Question 13.
Mr. Sharma has 60 shares of N.V. ₹ 100 and sells them when they are at a premium of 60%. He invests the proceeds in shares of nominal value ₹ 50, quoted at 4% discount, and paying 18% dividend annually. Calculate:
(i) the sale proceeds ;
(ii) the number of shares he buys and
(iii) his annual dividend from the shares.
Solution:
(i) No. of shares = 60
Face value of each share = Rs. 100
Total amount = Rs. 100 x 60 = Rs. 6,000
Market value = Rs. 160
His sale proceed = Rs. 160 x 60 = Rs. 9,600
(ii) In second case :
Nominal value of each share = Rs.50
and Market value = Rs. 50 x \(\frac { 96 }{ 100 }\) = Rs. 48
Rate of dividend = 18%
No. of shares he purchased = \(\frac { 9600 }{ 48 }\) = 200
(iii) Face value of 200 shares = 200 x Rs. 50 = Rs. 10,000
Dividend = Rs. 10000 x \(\frac { 18 }{ 100 }\) = Rs. 1,800

Question 14.
A company with 10,000 shares of nominal value ₹ 100 declares an annual dividend of 8% to shareholders.
(i) Calculate the total amount of dividend paid by the company.
(ii) Ramesh had bought 90 shares of the company at ₹ 150 per share. Calculate the dividend he receives and the percentage of return on his investment.
Solution:
(i) No. of shares = 10,000
Nominal value of each share = Rs. 100
Dividend = 8%
Total face value of 10,000 shares = Rs. 100 x 10,000 = Rs. 10,00,000
and amount of dividend = Rs. \(\frac { 1000000 x 8 }{ 100 }\) = Rs. 8000
(ii) In second case :
Ramesh bought = 90 shares
Market value of each share = Rs. 150
His investment = Rs. 150 x 90 = Rs. 13,500
Face value of 90 shares = Rs. 100 x 90 = Rs. 9,000
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 14.1

Question 15.
Which is the better investment:
16% Rs. 100 shares at 80 or 20% Rs. 100 shares at 120 ?
Solution:
In first case :
Income on Rs. 80 = Rs. 16
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 15.1
From above, it is clear that first investment is better.

Question 16.
A man has a choice to invest in hundred rupee shares of two firms at Rs. 120 or at Rs. 132. The first firm pays r. dividend of 5% per annum and the second firm pays a dividend of 6% per annum. Find :
(i) Which company is giving a better return.
(ii) If a man invests Rs. 26,400 with each firm, how much will be the difference between the annual returns from the two firms.
Solution:
In first case :
Market value of share = Rs. 120
and dividend = 5%
Income on Rs. 120 = Rs. 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 16.1
= Rs. 1,200
Difference = Rs. 1,200 – Rs. 1,100 = Rs. 100

Question 17.
A man bought 360, ten rupee shares of a company paying 12 percent per annum. He sold the shares when their price rose to Rs. 21 per share and invested the proceeds in five rupee shares paying 4.5 per cent per annum at Rs 3.50 per share. Find the annual change in his income.
Solution:
No. of shares bought = 360
Face value of each share = Rs. 10
Dividend = 12%
Cost price of 360 shares = Rs. 360 x 10 = Rs. 3,600
Market value = Rs. 21
Selling price = Rs. 21 x 360 = Rs. 7,560
In second case :
Face value of each share = Rs. 5
Market value of each share = Rs. 3.5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 17.1
Difference in his income = Rs. 486 Rs. 432 = Rs. 54

Question 18.
A man sold 400 (Rs. 20) shares of a company paying 5% at Rs. 18 and invested the proceeds in (Rs. 10) shares of another company paying 7% at Rs. 12. How many (Rs. 10) shares did he buy and what was the change in his income?
Solution:
In first case :
No. of shares sold = 400
Face value of each share = Rs. 20
Market value = Rs. 18
Income = 5%
Amount of his investment = Rs. 18 x 400 = Rs. 7,200
and amount of shares = Rs. 20 x 400 = Rs. 8000
In second case :
Market value of each share = Rs. 12
Face value of each share = Rs. 10
Rate of dividend = 7%
No. of shares purchased = \(\frac { 7200 }{ 12 }\) = 600
Face value of 600 shares = Rs. 10 x 600 = Rs. 6,000
Now, income in first case = Rs. 8000 x \(\frac { 5 }{ 100 }\) = Rs. 400
and income in second case = Rs. 6000 x \(\frac { 7 }{ 100 }\) = Rs. 420
Increase in income = Rs. 420 – 400 = Rs. 20

Question 19.
Two brothers A and B invest Rs. 16,000 each in buying shares of two companies. A buys 3% hundred-rupee shares at 80 and B buys ten rupee shares at par. If they both receive equal dividend at the end of the year, find the rate percent of the dividend received by B.
Solution:
A’s investment = Rs. 16,000
Face value of each share = Rs. 100
Market value of each share = Rs. 80
and rate of dividend = 3%
No. of shares purchased = \(\frac { 16000 }{ 80 }\) = 200
Amount of shares = 200 x Rs. 100 = Rs. 20,000
and amount of dividend = 20,000 x \(\frac { 3 }{ 100 }\) = Rs. 600
B’s investment = Rs. 16,000
Face value of each share = Rs. 10
and amount of dividend = Rs. 600
Rate of dividend = \(\frac { 600 }{ 16000 }\) x 100 = 3.75 %

Question 20.
A man invests Rs. 20,020 in buying shares of nominal value Rs. 26 at 10% premium. The dividend on the shares is 15% per annum Calculate :
(i) The number of shares he buys.
(ii) The dividend he receives annually.
(iii) The rate of interest he gets on his money [2003]
Solution:
Total investment = Rs. 20,020
Nominal value of each share = Rs. 26.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B 20.1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 3 Shares and Dividend Ex 3B are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3

March 30, 2023

RD Sharma Class 8 Solutions Chapter 20 Mensuration I (Area of a Trapezium and a Polygon) Ex 20.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3

Other Exercises

Question 1.
Find the area of the pentagon shown in the figure if AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm and BF = 5 cm, CG = 7 cm and EH = 3
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 1
Solution:
In the figure, here are three triangles and one trapezium.

Question 2.
Find the area enclosed by each of the following figures as the sum of the areas of a rectangle and a trapezium:
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 5
Solution:
(i) In the figure ABCDEF,
Join CF, then, the figure consists one square and one trapezium ABCF is a square whose side = 18 cm

Area of the square = 18 x 18 cm² = 324 cm²
Area of trapezium FCDE = \(\frac { 1 }{ 2 }\) (CF + ED) x 8 cm²
= \(\frac { 1 }{ 2 }\) (18 + 7) x 8
= \(\frac { 1 }{ 2 }\) x 25 x 8 cm²
= 100 cm²
Total area of fig. ABCDEF = 324 + 100 = 424 cm²
(ii) In the figure ABCDEF,
Join BE.
The figure consists of one rectangle BCDE and one trapezium ABEF
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 7
Area of rectangle BCDE = BC x CD = 20 x 15 = 300 cm²
Area of trapezium ABEF,
= \(\frac { 1 }{ 2 }\) (BE + AF) x height
= \(\frac { 1 }{ 2 }\) (15 + 6) x 8 cm²
= \(\frac { 1 }{ 2 }\) x 21 x 8 cm²
= 84 cm²
Area of the figure ABCDEF = 300 + 84 = 384 cm²
(iii) In the figure ABCDEFGH,
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 8
HC= AB = EF = 6 cm
AH = BC = 4 cm
DE = GF = 5 cm
Join HC.
In right ∆CDE,
ED² = CD² + CE²
⇒ (5)² = (4)² + (CE)²
⇒ 25 = 16 + (CE)²
⇒ (CE)² = 25 – 16 = 9 = (3)²
CE = 3 cm
The figure consist a rectangle and a trapezium
Area of rectangle ABCH = AB x BC = 6 x 4 = 24 cm²
Area of trapezium GDEF,
= \(\frac { 1 }{ 2 }\) (GD + EF) x CE 1
= \(\frac { 1 }{ 2 }\) (GH + HC + CD + EF) x CE
= \(\frac { 1 }{ 2 }\) (4 + 6 + 4 + 6) x 3 cm²
= \(\frac { 1 }{ 2 }\) x 20 x 3 cm²
= 30 cm²
Total area of the figure ABCDEFGH = 24 + 30 = 54 cm²

Question 3.
There is a pentagonal shaped park as shown in the figure. Jyoti and Kavita divided it in two different ways.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 9

Find the area of this park using both ways. Can you suggest some another way of finding its area ?
Solution:
In first case, the figure ABCDE is divided into 2 trapezium of equal area.
Now area of trapezium DFBC

Total area of the pentagon ABCDE = 2 x 168.75 = 337.5 m²
In second case, the figure ABCDE is divided into two parts, namely one square and other triangle.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 12
Total area of pentagon ABCDE = 225 + 112.5 = 337.5 m²

Question 4.
Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 cm, AO = 60 cm and AD = 90 cm.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 13
Solution:
In the figure ABCDEF,
AD = 90 cm
BL = 30 cm
AO = 60 cm
CN = 40 cm
AN = 50 cm
EO = 60 cm
AM = 20 cm
FM = 20 cm
AL = 10 cm

Area of ABCDEF = (150 + 800 + 900 + 200 + 1400 + 1600) cm² = 5050 cm²

Question 5.
Find the area of the following regular hexagon:
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 17
Solution:
In the regular hexagon MNOPQR There are two triangles and one rectangle.
Join MQ, MO and RP
NQ = 23 cm,
NA = BQ = \(\frac { 10 }{ 2 }\) = 5 cm
MR = OP = 13 cm

In right ∆BDQ,
PQ² = BQ² + BP²
⇒ (13)² = (5)² + BP²
⇒ 169 = 25 + BP²
⇒ BP² = 169 – 25 = 144 = (12)²
BP = 12 cm
PR = MO = 2 x 12 = 24 cm
Now area of rectangle RPOM = RP x PO = 24 x 13 = 312 cm²
Area of ∆PRQ = \(\frac { 1 }{ 2 }\) x PR x BQ
= \(\frac { 1 }{ 2 }\) x 24 x 5 = 60 cm²
Similarly area ∆MON = 60 cm²
Area of the hexagon MNOPQR = 312 + 60 + 60 = 432 cm²

Hope given RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 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.

RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2

March 30, 2023

RD Sharma Class 8 Solutions Chapter 20 Mensuration I (Area of a Trapezium and a Polygon) Ex 20.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2

Other Exercises

Question 1.
Find the area, in square metres, of the trapezium whose bases and altitude are as under:
(i) bases = 12 dm and 20 dm, altitude =10 dm
(ii) bases = 28 cm and 3 dm, altitude = 25 cm
(iii) bases = 8 m and 60 dm, altitude = 40 dm
(iv) bases = 150 cm and 30 dm, altitude = 9 dm
Solution:
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 1

Question 2.
Find the area of trapezium with base 15 cm and height 8 cm. If the side parallel to the given base is 9 cm long.
Solution:
In the trapezium ABCD,
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 4

Question 3.
Find the area of a trapezium whose parallel sides are of length 16 dm and 22 dm and whose height is 12 dm.
Solution:
Length of parallel sides of a trapezium are 16 dm and 22 dm i.e.
b₁ = 16 dm, b₂ = 22 dm
and height (h) = 12 dm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 6

Question 4.
Find the height of a trapezium, the sum of lengths of whose bases (parallel sides) is 60 cm and whose area is 600 cm²
Solution:
Sum of parallel sides (b₁ + b₂) = 60 cm
Area of trapezium = 600 cm²
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 7

Question 5.
Find the altitude of a trapezium whose area is 65 cm² and whose bases are 13 cm and 26 cm.
Solution:
Area of a trapezium = 65 cm²
Bases are 13 cm and 26 cm
i.e. b₁ = 13 cm, b₂ = 26 cm.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 8

Question 6.
Find the sum of the lengths of the bases of trapezium whose area is 4.2 m² and whose height is 280 cm.
Solution:
Area of trapezium = 4.2 m²
Height (h) = 280 cm = 2.8 m.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 9

Question 7.
Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm from each other. Calculate the area as
(i) the sum of the areas of two triangles and one rectangle.
(ii) the difference of the area of a rectangle id the sum of the areas of two triangles.
Solution:
In trapezium ABCD, parallel sides or bases are 10 cm and 15 cm and height = 6 cm
Area of trapezium
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 10

Area of trapezium = 90 – 15 = 75 cm²
= area of rectangle – areas of two triangles.

Question 8.
The area of a trapezium is 960 cm². If the parallel sides are 34 cm and 46 cm, find the distance between them:
Solution:
Area of trapezium = 960 cm²
Parallel sides are 34 cm and 46 cm
b₁ + b₂ = 34 + 46 = 80 cm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 12
Distance between parallel sides = 24 cm

Question 9.
Find the area of the figure as the sum of the areas of two trapezium and a rectangle.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 13
Solution:
In the figure,
One rectangle is ABCD whose sides are 50 cm and 10 cm.
Two trapezium of equal size in which parallel sides are 30 cm and 10 cm and height

Question 10.
Top surface of a table is trapezium in shape. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 15
Solution:
Top of a table is of trapezium in shape whose parallel sides are 1 m and 1.2 m and distance between them (h) = 0.8 m
Area of trapezium = \(\frac { 1 }{ 2 }\) (Sum of parallel sides) x height

Question 11.
The cross-section of a canal is a trapezium in shape. If the canal is 10 m wide at the top 6 m wide at the bottom, and the area of the cross-section is 72 m², determine its depth.
Solution:
Area of cross-section = 72 m²
Parallel sides of the trapezium = 10 m and 6 m
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 17

Question 12.
The area of a trapezium is 91 cm² and its height is 7 cm. If one of the parallel sides is longer than the other by 8 cm, find the two parallel sides.
Solution:
Area of trapezium = 91 cm²
Height (h) = 7 cm.
Sum of parallel sides
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 18
One parallel side = 9 cm
and second side = 9 + 8 = 17 cm
Hence parallel sides are 17 cm, 9 cm

Question 13.
The area of a trapezium is 384 cm². Its parallel sides are in the ratio 3 : 5 and the perpendicular distance between them is 12 cm. Find the length of each one of the parallel sides.
Solution:
Area of trapezium = 384 cm²
Perpendicular distance (h) = 12 cm
Sum of parallel sides
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 19
First parallel side = 8 x 3 = 24 cm
Second side = 8 x 5 = 40 cm

Question 14.
Mohan wants to buy a trapezium shaped field. Its side along the river is parallel and twice the side along the road. If the area of this field is 10500 m² and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 20
Solution:
Area of the trapezium shaped field = 10500 m²
and perpendicular distance between them (h) = 100 m.

Question 15.
The area of trapezium is 1586 cm² and the distance between the parallel sides is 26 cm. If one of the parallel sides is 38 cm, find the other.
Solution:
Area of a trapezium = 1586 cm²
and distance between the parallel sides (h) = 26
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 22

Question 16.
The parallel sides of a trapezium are 25 cm and 13 cm ; Its nonparallel sides are equal each being 10 cm, find the area of the trapezium.
Solution:
Parallel sides of a trapezium ABCD are 25 cm and 13 cm
i.e. AB = 25 cm, CD = 13 cm
and each non-parallel side = 10 cm
i.e., AD = BC = 10 cm
From C, draw CE || DA and draw CL ⊥ AB
CE = DA = CB = 10 cm
and EB = AB – AE = AB – DC = 25 – 13 = 12 cm
Perpendicular CL bisects base EB of an isosceles ∆CED
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 23

Question 17.
Find the area of a trapezium whose parallel sides are 25 cm, 13 cm and other sides are 15 cm each.
Solution:
In trapezium ABCD, parallel sides are AB and DC.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 24
AB = 25 cm, CD = 13 cm
and other sides are 15 cm each i.e. AD = CB = 15 cm
From C, draw CE || DA and CL ⊥ AB
AE = DC = 13 cm
and EB = AB – AE = 25 – 13 = 12 cm
Perpendicular CL bisects the base EB of the isosceles triangle CEB
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 25

Question 18.
If the area of a trapezium is 28 cm² and one of its parallel sides is 6 cm, find the other parallel side if its altitude is 4 cm.
Solution:
Area of trapezium = 28 cm²
Altitude (h) = 4 cm.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 26
One of the parallel side = 6 cm
Second parallel side = 14 – 6 = 8 cm

Question 19.
In the figure, a parallelogram is drawn in a trapezium the area of the parallelogram is 80 cm², find the area of the trapezium.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 27
Solution:
Area of parallelogram (AECD) = 80 cm²
Side AE (b) = 10 cm

Question 20.
Find the area of the field shown in the figure by dividing it into a square rectangle and a trapezium.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 29
Solution:
Produce EF to H to meet AB at H and draw DK || EH
HF = 4 cm, KD = HE = 4 + 4 = 8 cm
HK = ED = 4 cm,
KB = 12 – (8) = 4 cm
Now, area of square AGFH = 4 x 4 = 16 cm²
area of rectangle KDEH = l x b = 8 x 4 = 32 cm²
and area of trapezium BCDK.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2 30
Total area of the figure = 16 + 32 + 22 = 70 cm²

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RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1

March 30, 2023

RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II (Quadrilaterals) Ex 16.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1

Question 1.
Define the following terms:
(i) Quadrilateral
(ii) Convex Quadrilateral.
Solution:
(i) Quadrilateral: A closed figure made of four line segments is called a quadrilateral such that:
RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1 1
(a) no three points of them are collinear
(b) the line segments do not intersect except at their ends points.
(ii) Convex quadrilateral: A quadrilateral is called a convex quadrilateral of the line containing any side of the quadrilateral has the remaining vertices on the same side of it. In the figure, quadrilateral ABCD is a convex quadrilateral.
RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1 2

Question 2.
In a quadrilateral, define each of the following:
(i) Sides
(ii) Vertices
(iii) Angles
(iv) Diagonals
(v) Adjacent angles
(vi) Adjacent sides
(vii) Opposite sides
(viii) Opposite angles
(ix) Interior
(x) Exterior
Solution:
(i) Sides: In a quadrilateral ABCD, form line segments AB, BC, CD and DA are called sides of the quadrilateral.
(ii) Vertices : The ends points are called the vertices of the quadrilateral. Here in the figure, A, B, C and D are its vertices.
(iii) Angles: A quadrilateral has four angles which are at their vertices. In the figure, ∠A, ∠B, ∠C and ∠D are its angles.
(iv) Diagonals: The line segment joining the opposite vertices is called diagonal. A quadrilateral has two diagonals.
(v) Adjacent Angles : The angles having a common arm (side) are called adjacent angles.
(vi) Adjacent sides : If two sides of a quadrilateral have a common end-point, these are called adjacent sides.
(vii) Opposite sides: If two sides do not have a common end-point of a quadrilateral, they are called opposite sides.
(viii) Opposite angles : The angles which are not adjacent are called opposite angles.
(ix) Interior: The region which is surrounded by the sides of the quadrilateral is called its interior.
(x) Exterior : The part of the plane made up by all points as the not enclosed by the quadrilateral, is called its exterior.
RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1 3

Question 3.
Complete each of the following, so as to make a true statement:
(i) A quadrilateral has ………… sides.
(ii) A quadrilateral has ………… angles.
(iii) A quadrilateral has ……….. vertices, no three of which are …………
(iv) A quadrilateral has …………. diagonals.
(v) The number of pairs of adjacent angles of a quadrilateral is ………….
(vi) The number of pairs of opposite angles of a quadrilateral is ……………
(vii) The sum of the angles of a quadrilateral is …………
(viii) A diagonal of a quadrilateral is a line segment that joins two ………. vertices of the quadrilateral.
(ix) The sum of the angles of a quadrilateral is …………. right angles.
(x) The measure of each angle of a convex quadrilateral is …………. 180°.
(xi) In a quadrilateral the point of intersection of the diagonals lies in ………….. of the quadrilateral.
(xii) A point is in the interior of a convex quadrilateral, if it is in the ……….. of its two opposite angles.
(xiii) A quadrilateral is convex if for each side, the remaining …………. lie on the same side of the line containing the side.
Solution:
(i) A quadrilateral has four sides.
(a) A quadrilateral has four angles.
(iii) A quadrilateral has four vertices, no three of which are collinear .
(iv) A quadrilateral has two diagonals.
(v) The number of pairs of adjacent angles of a quadrilateral is four .
(vi) The number of pairs of opposite angles ot a quadrilateral is two.
(vii) The sum of the angles of a quadrilateral is 360°.
(viii) A diagonal of a quadrilateral is a line segment that join two opposite vertices of the quadrilateral.
(ix) The sum of the angles of a quadrilateral is 4 right angles.
(x) The measure of each angle of a convex quadrilateral is less than 180°.
(xi) In a quadrilateral the point of intersection of the diagonals lies in interior of the quadrilateral.
(xii) A point is in the interior of a convex quadrilateral, if it is in the interior of its two opposite angles.
(xiii) A quadrilateral is convex if for each side, the remaining vertices lie on the same side of the line containing the side.

Question 4.
In the figure, ABCD is a quadrilateral.
(i) Name a pair of adjacent sides.
(ii) Name a pair of opposite sides.
(iii) How many pairs of adjacent sides are there?
(iv) How many pairs of Opposite sides are there ?
RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1 4
(v) Name a pair of adjacent angles.
(vi) Name a pair of opposite angles.
(vii) How many pairs of adjacent angles are there ?
(viii) How many pairs of opposite angles are there ?
Solution:
In the figure, ABCD is a quadrilateral
(i) Pairs of adjacent sides are AB, BC, BC, CD, CD, DA, DA, AB.
(ii) Pairs of opposite sides are AB and CD; BC and AD.
(iii) There are four pairs of adjacent sides.
(iv) There are two pairs of opposite sides.
(v) Pairs of adjacent angles are ∠A, ∠B; ∠B, ∠C; ∠C, ∠D; ∠D, ∠A.
(vi) Pairs of opposite angles are ∠A and ∠C; ∠B and ∠D.
(vii) There are four pairs of adjacent angles.
(viii) There are two pairs of opposite angles.

Question 5.
The angles of a quadrilateral are 110°, 72°, 55° and x°. Find the value of x.
Solution:
Sum of four angles of quadrilateral is 360°
110° + 12° + 55° + x° = 360°
⇒ 237° + x° = 360°
⇒ x° = 360° – 237° = 123°
x = 123°

Question 6.
The three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle.
Solution:
The sum of four angles of a quadrilateral = 360°
Three angles are 110°, 50° and 40°
Let fourth angle = x
Then 110° + 50° + 40° + x° = 360°
⇒ 200° + x° = 360°
⇒ x = 360° – 200° = 160°
x = 160°

Question 7.
A quadrilateral has three acute angles each measures 80°. What is the measure of fourth angle ?
Solution:
Sum of four angles of a quadrilateral = 360°
Sum of three angles having each angle equal to 80° = 80° x 3 = 240°
Let fourth angle = x
Then 240° + x = 360°
⇒ x° = 360° – 240°
⇒ x° = 120°
Fourth angle = 120°

Question 8.
A quadrilateral has all its four angles of the same measure. What is the measure of each ?
Solution:
Let each equal angle of a quadrilateral = x
4x° = 360°
⇒ x° = \(\frac { 360 }{ 4 }\) = 90°
Each angle will be = 90°

Question 9.
Two angles of a quadrilateral are of measure 65° and the other two angles are equal. What is the measure of each of these two angles ?
Solution:
Measures of two angles each = 65°
Sum of these two angles = 2 x 65°= 130°
But sum of four angles of a quadrilateral = 360°
Sum of the remaining two angles = 360° – 130° = 230°
But these are equal to each other
Measure of each angle = \(\frac { 230 }{ 2 }\) = 115°

Question 10.
Three angles of a quadrilateral are equal. Fourth angle is of measure 150°. What is the measure of equal angles ?
Solution:
Sum of four angles of a quadrilateral = 360°
One angle = 150°
Sum of remaining three angles = 360° – 150° = 210°
But these three angles are equal
Measure of each angle = \(\frac { 210 }{ 3 }\) = 70°

Question 11.
The four angles of a quadrilateral are as 3 : 5 : 7 : 9. Find the angles.
Solution:
Sum of four angles of a quadrilateral = 360°
and ratio in angles = 3 : 5 : 7 : 9
Let first angles = 2x
Then second angle = 5x
third angle = 7x
and fourth angle = 9x
3x + 5x + 7x + 9x = 360°
⇒ 24x = 369°
⇒ x = \(\frac { 360 }{ 24 }\) = 15°
First angle = 3x = 3 x 15° = 45°
second angle = 5x = 5 x 15° = 75°
third angle = 7x = 7 x 15° = 105°
and fourth angle = 9x = 9 x 15° = 135°

Question 12.
If the sum of the two angles of a quadrilateral is 180°, what is the sum of the remaining two angles ?
Solution:
Sum of four angles of a quadrilateral = 360°
and sum of two angle out of these = 180°
Sum of other two angles will be = 360° – 180° = 180°

Question 13.
In the figure, find the measure of ∠MPN.
RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1 5
Solution:
In the figure, OMPN is a quadrilateral in which
∠O = 45°, ∠M = ∠N = 90° (PM ⊥ OA and PN ⊥ OB)
Let ∠MPN = x°
∠O + ∠M + ∠N + ∠MPN = 360° (Sum of angles of a quadrilateral)
⇒ 45° + 90° + 90° + x° = 360°
⇒ 225° + x° = 360°
⇒ x° = 360° – 225°
⇒x = 135°
∠MPN = 135°

Question 14.
The sides of a quadrilateral are produced in order. What is the sum of the four exterior angles ?
Solution:
The sides of a quadrilateral ABCD are produced in order, forming exterior angles ∠1, ∠2, ∠3 and ∠4.
RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1 6
Now ∠DAB + ∠1 = 180° (Linear pair) ……(i)
Similarly,
∠ABC + ∠2 = 180°
∠BCD + ∠3 = 180°
and ∠CDA + ∠4 = 180°
Adding, we get
∠DAB + ∠1 + ∠ABC + ∠2 + ∠BCD + ∠3 + ∠CDA + ∠4 = 180° + 180° + 180° + 180° = 720°
⇒ ∠DAB + ∠ABC + ∠CDA + ∠ADC + ∠1 + ∠2 + ∠3 + ∠4 = 720°
But ∠DAB + ∠ABC + ∠CDA + ∠ADB = 360° (Sum of angles of a quadrilateral)
360° + ∠1 + ∠2 + ∠3 + ∠4 = 720°
⇒ ∠l + ∠2 + ∠3 + ∠4 = 720° – 360° = 360°
Sum of exterior angles = 360°

Question 15.
In the figure, the bisectors of ∠A and ∠B meet at a point P. If ∠C = 100° and ∠D = 50°, find the measure of ∠APB.
Solution:
In quadrilateral ABCD,
RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1 7
∠D = 50°, ∠C = 100°
PA and PB are the bisectors of ∠A and ∠B.
In quadrilateral ABCD,
∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)
⇒ ∠A + ∠B + 100° + 50° = 360°
⇒ ∠A + ∠B + 150° = 360°’
⇒ ∠A + ∠B = 360° – 150° = 210°
and \(\frac { 1 }{ 2 }\) ∠A + \(\frac { 1 }{ 2 }\) ∠B = \(\frac { 210 }{ 2 }\) = 105°
(PA and PB are bisector of ∠A and ∠B respectively)
∠PAB + ∠PBA = 105°
⇒ ∠PAB + ∠PBA + ∠APB = 180° (Sum of angles of a triangle)
⇒ 105° + ∠APB = 180°
⇒ ∠APB = 180° – 105° = 75°
∠APB = 75°

Question 16.
In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5. Find the measure of each angle of the quadrilateral.
Solution:
Sum of angles A, B, C and D of a quadrilateral = 360°
i.e. ∠A + ∠B + ∠C + ∠D = 360°
But ∠A = ∠B = ∠C = ∠D = 1 : 2 : 4 : 5
RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1 8
Let ∠A = x,
Then ∠B = 2x
∠C = 4x
∠D = 5x
x + 2x + 4x + 5x = 360°
⇒ 12x = 360°
⇒ x = \(\frac { 360 }{ 12 }\) = 30°
∠A = x = 30°
∠B = 2x = 2 x 30° = 60°
∠C = 4x = 4 x 30° = 120°
∠D = 5A = 5 x 30° = 150°

Question 17.
In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively. Prove that ∠COD = \(\frac { 1 }{ 2 }\) (∠A + ∠B).
Solution:
In quadrilateral ABCD,
RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1 9

Question 18.
Find the number of sides of a regular polygon when each of its angles has a measures of
(i) 160°
(ii) 135°
(iii) 175°
(iv) 162°
(v) 150°.
Solution:
In a n-sided regular polygon, each angle
RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1 11

Question 19.
Find the number of degrees in each exterior angle of a regular pentagon.
Solution:
In a pentagon or a polygon, sum of exterior angles formed by producing the sides in order, is four right angles or 360°
Each exterior angle = \(\frac { 360 }{ 5 }\) = 72°

Question 20.
The measure of angles of a hexagon are x°, (x – 5)° (x – 5)°, (2x – 5)°, (2x – 5)°, (2x + 20)°. Find the value of x.
Solution:
We know that the sum of interior angels of a hexagon = 720° (180° x 4)
⇒ x + x – 5 + x – 5 + 2x – 5 + 2x – 5 + 2x + 20 = 720°
⇒ 9x – 20 + 20 = 720
⇒ 9x = 720
⇒ x = \(\frac { 720 }{ 9 }\) = 80°
x = 80°

Question 21.
In a convex hexagon, prove that the sum of all interior angles is equal to twice the sum of its exterior angles formed by producing the sides in the same order.
Solution:
In a convex hexagon ABCDEF, its sides AB, BG, CD, DE, EF and FA are produced in order forming exterior angles a, b, c, d, e, f
RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1 15
∠a + ∠b + ∠c + ∠d + ∠e + ∠f = 4 right angles (By definition)
By joining AC, AD, and AE, 4 triangles ABC, ACD, ADE and AEF are formed
In ∆ABC,
∠1 + ∠2 + ∠3 = 180° = 2 right angle (Sum of angles of a triangle) …… (i)
Similarly,
In ∆ACD,
∠4 +∠5 + ∠6 = 180° = 2 right angles
In ∆ADE,
∠1 + ∠8 + ∠9 = 2 right angles …(iii)
In ∆AEF,
∠10 + ∠11 + ∠12 = 2 right angles …(iv)
Joining (i), (ii), (iii) and (iv)
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 + ∠9 + ∠10 + ∠11 + ∠12 = 8 right angles
⇒ ∠2 + ∠3 + ∠5 + ∠6 + ∠8 + ∠9 + ∠11 + ∠12 + ∠1 + ∠4 + ∠7 + ∠10 = 8 right angles
⇒ ∠B + ∠C + ∠D + ∠E +∠F + ∠A = 8 right angles
⇒ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 2 (∠a + ∠b + ∠c + ∠d + ∠e + ∠f)
Sum of all interior angles = 2(the sum of exterior angles)
Hence proved.

Question 22.
The sum of the interior angles of a polygon is three times the sum of its exterior angles. Determine the number of sides of the polygon.
Solution:
Let number of sides of a regular polygon = n
Each interior angle = \(\frac { 2n – 4 }{ n }\) right angles
Sum of all interior angles = \(\frac { 2n – 4 }{ n }\) x n
right angles = (2n – 4) right angles
But sum of exterior angles = 4 right angles
According to the condition,
(2n – 4) = 3 x 4 (in right angles)
⇒ 2n – 4 = 12
⇒ 2n = 12 + 4 = 16
⇒ n = 8
Number of sides of the polygon = 8

Question 23.
Determine the number of sides of a polygon whose exterior and interior angles are in the ratio 1 : 5.
Solution:
Ratio in exterior angle and interior angles of a regular polygon = 1 : 5
But sum of interior and exterior angles = 180° (Linear pair)
RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1 16
By cross multiplication:
6n – 12 = 5n
⇒ 6n – 5n = 12
⇒ n = 12
Number of sides of polygon is 12

Question 24.
PQRSTU is a regular hexagon. Determine each angle of ∆PQT.
Solution:
In regular hexagon, PQRSTU, diagonals PT and QT are joined.
RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1 17
In ∆PUT, PU = UT
∠UPT = ∠UTP
But ∠UPT + ∠UTP = 180° – ∠U = 180° – 120° = 60°
∠UPT = ∠UTP = 30°
∠TPQ = 120° – 30° = 90° (QT is diagonal which bisect ∠Q and ∠T)
∠PQT = \(\frac { 120 }{ 2 }\) = 60°
Now in ∆PQT,
∠TPQ + ∠PQT + ∠PTQ = 180° (Sum of angles of a triangle)
⇒ 90° + 60° + ∠PTQ = 180°
⇒ 150° + ∠PTQ = 180°
⇒ ∠PTQ = 180° – 150° = 30°
Hence in ∆PQT,
∠P = 90°, ∠Q = 60° and ∠T = 30°

Hope given RD Sharma Class 8 Solutions Chapter 16 Understanding Shapes II Ex 16.1 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.

RD Sharma Class 8 Solutions Chapter 15 Understanding Shapes I Ex 15.1

March 30, 2023

RD Sharma Class 8 Solutions Chapter 15 Understanding Shapes I (Polygons) Ex 15.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 15 Understanding Shapes I Ex 15.1

Question 1.
Draw rough diagrams to illustrate the following:
(i) Open curve
(ii) Closed curve
Solution:
RD Sharma Class 8 Solutions Chapter 15 Understanding Shapes I Ex 15.1 1

Question 2.
Classify the following curves as open or closed.
RD Sharma Class 8 Solutions Chapter 15 Understanding Shapes I Ex 15.1 2
Solution:
Open curves : (i), (iv) and (v) are open curves.
(ii) , (iii), and (vi) are closed curves.

Question 3.
Draw a polygon and shade its interior. Also draw its diagonals, if any.
Solution:
In the given polygon, the shaded portion is its interior region AC and BD are the diagonals of polygon ABCD.
RD Sharma Class 8 Solutions Chapter 15 Understanding Shapes I Ex 15.1 3

Question 4.
Illustrate, if possible, each one of the following with a rough diagram:
(i) A closed curve that is not a polygon.
(ii) An open curve made up entirely of line segments.
(iii) A polygon with two sides.
Solution:
(i) Close curve but not a polygon.
RD Sharma Class 8 Solutions Chapter 15 Understanding Shapes I Ex 15.1 4
(ii) An open curve made up entirely of line segments.

(iii) A polygon with two sides. It is not possible. At least three sides are necessary

Question 5.
Following are some figures : Classify each of these figures on the basis of the following:
RD Sharma Class 8 Solutions Chapter 15 Understanding Shapes I Ex 15.1 6

(i) Simple curve
(ii) Simple closed curve
(iii) Polygon
(iv) Convex polygon
(v) Concave polygon
(vi) Not a curve
Solution:
(i) It is a simple closed curve and a concave polygon.
(ii) It is a simple closed curve and convex polygon.
(iii) It is neither a curve nor polygon.
(iv) it is neither a curve not a polygon.
(v) It is a simple closed curve but not a polygon.
(vi) It is a simple closed curve but not a polygon.
(vii) It is a simple closed curve but not a polygon.
(viii) It is a simple closed curve but not a polygon.

Question 6.
How many diagonals does each of the following have ?
(i) A convex quadrilateral
(ii) A regular hexagon
(iii) A triangle.
Solution:
(i) A convex quadrilateral
Here n = 4

RD Sharma Class 8 Solutions Chapter 15 Understanding Shapes I Ex 15.1 9

Question 7.
What is a regular polygon ? State the name of a regular polygon of:
(i) 3 sides
(ii) 4 sides
(iii) 6 sides.
Solution:
A regular polygon is a polygon which has all its sides equal and so all angles are equal,
(i) 3 sides : It is an equilateral triangle.
(ii) 4 sides : It is a square.
(iii) 6 sides : It is a hexagon.

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RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1

March 30, 2023

RD Sharma Class 8 Solutions Chapter 20 Mensuration I (Area of a Trapezium and a Polygon) Ex 20.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1

Other Exercises

Question 1.
A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m² ?
Solution:
Area of floor = 1080 m²
Base of parallelogram shaped tile (b) = 24 cm
and corresponding height (h) = 10 cm
Area of one tile = b x h = 24 x 10 = 240 cm²
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 1

Question 2.
A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in Fig. If AB = 60 m and BC = 28 m, find the area of the plot.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 2
Solution:
Length of rectangular portion (l) = 60 m
and breadth (b) = 28 m
Area of the rectangular plot = l x b = 60 x 28 m² = 1680 m²
Radius of semicircular portion (r) = \(\frac { b }{ 2 }\) = \(\frac { 28 }{ 2 }\) = 14 m
Area = \(\frac { 1 }{ 2 }\) πr²
= \(\frac { 1 }{ 2 }\) x \(\frac { 22 }{ 7 }\) x 14 x 14 m²
= 308 m²
Total area of the plot = 1680 + 308 = 1988 m²

Question 3.
A playground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take π = \(\frac { 22 }{ 7 }\)).
Solution:
Length of rectangular portion (l) = 36 m
and breadth (b) = 24.5 m
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 3
= \(\frac { 22 }{ 7 }\) x 150.0625 m²
= 471.625 m²
Total area of the playground = 471.625 + 882 = 1353.625 m²

Question 4.
A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 have been cut. Find the area of the remaining part.
Solution:
Length of rectangular piece (l) = 20 m
breadth (b) = 15 m
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 4
Area of rectangular piece = l x b = 20 x 15 = 300 m²
Radius of each quadrant (r) = 3.5 m
Total area of 4 quadrants

Area of the remaining portion = 300 – 38.5 m² = 261.5 m²

Question 5.
The inside perimeter of a running track (shown in Fig.) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If track is everywhere 14 m wide, find the area of the track. Also, find the length of the outer running track.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 6
Solution:
Inner perimeter = 400 m.
Length (l) = 90 m.
Perimeter of two semicircles = 400 – 2 x 90 = 400 – 180 = 220 m

Question 6.
Find the area of the Figure in square cm, correct to one place of decimal. (Take π = \(\frac { 22 }{ 7 }\))
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 8
Solution:
Length of square (a) = 10 cm.
Area = a² = (10)² = 100 cm²
Base of the right triangle AED = 8 cm
and height = 6 cm

Question 7.
The diameter of a wheel of a bus is 90 cm which makes 315 revolutions per minute. Determine its speed in kilometres per hour. (Take π = \(\frac { 22 }{ 7 }\))
Solution:
Diameter of the wheel (d) = 90 cm.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 10

Question 8.
The area of a rhombus is 240 cm² and one of the diagonal is 16 cm. Find another diagonal.
Solution:
Area of rhombus = 240 cm²
Length of one diagonal (d₁) = 16 cm
Second diagonal (d₂)
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 11

Question 9.
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
In rhombus, diagonal (d₁) = 7.5 cm
and diagonal (d₂) = 12 cm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 12

Question 10.
The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
Solution:
In quadrilateral shaped field ABCD,
diagonal AC = 24 m
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 13
and perpendicular BL = 13 m
and perpendicular DM on AC = 8 m
Area of the field ABED = \(\frac { 1 }{ 2 }\) x AC x (BL + DM)
= \(\frac { 1 }{ 2 }\) x 24 x (13 + 8) m²
= 12 x 21 = 252 m²

Question 11.
Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Solution:
Side of rhombus (b) = 6 cm
Altitude (h) = 4 cm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 14

Question 12.
The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m² is Rs 4.
Solution:
Number of rhombus shaped tiles = 300
Diagonals of each tile = 45 cm and 130 cm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 15
Rate of polishing the tiles = Rs 4 per m²
Total cost = 202.5 x 4 = Rs 810

Question 13.
A rectangular grassy plot is 112 m long and 78 broad. It has a gravel path 2.5 m wide all around it on the side. Find the area of the path and the cost of constructing it at Rs 4.50 per square metre.
Solution:
Length of rectangular plot (l) = 112 m
and breadth (b) = 78 m
Width of path = 2.5 m
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 16
Inner length = 112 – 2 x 2.5 = 112 – 5 = 107 m
and inner breadth = 78 – 2 x 2.5 = 78 – 5 = 73 m
Area of path = outer area – inner area
= (112 x 78 – 107 x 73) m² = 8736 – 7811 = 925 m²
Rate of constructing = Rs 4.50 per m²
Total cost = 925 x Rs 4.50 = Rs 4162.50

Question 14.
Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.
Solution:
Side of rhombus = 20 cm.
One diagonal (d₁) = 24 cm
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 17
Diagonals of a rhombus bisect each other at right angle
AB = 20 cm
and OA = \(\frac { 1 }{ 2 }\) AC = \(\frac { 1 }{ 2 }\) x 24 cm = 12 cm
In right-angled ∆AOB,
AB² = AO² + BO² (Pythagoras theorem)
⇒ (20)² = (12)² + BO²
⇒ 400 = 144 + BO²
⇒ BO² = 400 – 144 = 256 = (16)²
⇒ BO = 16 cm
and diagonal BD = 2 x BO = 2 x 16 = 32 cm
Now area of rhombus ABCD
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 18

Question 15.
The length of a side of a square field is 4 m. What will be the altitude of the rhombus if the area of the rhombus is equal to the square field and one of its diagonal is L m ?
Solution:
Side of square = 4 m
Area of square = (a)² = 4 x 4 =16 m²
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 19
Diagonals of a rhombus bisect each other at right angles.
In right ∆AOB
AB² = QA² + BO² (Pythagoras theorem)
= (8)² + (1)² = 64 + 1 = 65
AB = √65 m.
Now, length of perpendicular AL (h)

Question 16.
Find the area of the field in the form of a rhombus, if the length of each side be 14 cm and the altitude be 16 cm.
Solution:
Length of each side of rhombus = 14 cm.
Length of altitude = 16 cm
Area = Base x altitude = 14 x 16 cm² = 224 cm²

Question 17.
The cost of fencing a square field at 60 paise per metre is Rs 1,200. Find the cost of reaping the field at the rate of 50 paise per 100 sq. metres.
Solution:
Cost of fencing the square field = Rs 1,200
Rate = 60 paise per m.
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 21

Question 18.
In exchange of a square plot one of whose sides is 84 m, a man wants to buy a rectangular plot 144 m long and of the same area as of the square plot. Find the width of the rectangular plot.
Solution:
Side of a square plot = 84 m
Area = (a)² = (84)² = 84 x 84 m² = 7056 m²
Area of rectangular field = 7056 m²
Length (l) = 144 m
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 22

Question 19.
The area of a rhombus is 84 m². If its perimeter is 40 m, then find its altitude.
Solution:
Area of rhombus = 84 m²
Perimeter = 40 m
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 23

Question 20.
A garden is in the form of a rhombus whose side is 30 metres and the corresponding altitude is 16 m. Find the cost of levelling the garden at the rate of Rs 2 per m².
Solution:
Side of rhombus garden (b) = 30 m.
Altitude (h) = 16 m
Area = Base x altitude = 30 x 16 = 480 m²
Rate of levelling the garden = Rs 2 per m²
Total cost = Rs 480 x 2 = Rs 960

Question 21.
A field in the form of a rhombus has each side of length 64 m and altitude 16 m. What is the side of a square field which has the same area as that of a rhombus ?
Solution:
Length of side of rhombus (b) = 64 m
and altitude (h) = 16 m
Area = b x h = 64 x 16 m² = 1024 m²
Now area of square = 1024 m²
Side of the square = √Area = √1024 m = 32 m

Question 22.
The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonals of the rhombus is 22 cm, find the length of the other diagonal.
Solution:
Base of triangle (b) = 24.8 cm
and altitude (h) = 16.5 cm
Area of triangle = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x bh= \(\frac { 1 }{ 2 }\) x 24.8 x 16.5 cm² = 204.6 cm²
Area of rhombus = 204.6 cm²
Length of one diagonal (d₁ = 22 cm
Second diagonal (d₂)
RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 24

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RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5

March 30, 2023

RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5

Other Exercises

Question 1.
Mr. Cherian purchased a boat for Rs 16,000. If the total cost of the boat is depreciating at the rate of 5% per annum, calculate its value after 2 years.
Solution:
Cost of boat = Rs 16,000
Rate of depreciating = 5% p.a.
Period = 2 years
Value of boat after 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 1

Question 2.
The value of a machine depreciates at the rate of 10% per annum. What will be its value 2 years hence, if the present value is Rs 1,0,000 ? Also, find the total depreciation during this period.
Solution:
Present value of machine = Rs 1,00,000
Rate of depreciation = 10% p.a.
Period (n) = 2 years
Value of machine after 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 2

Question 3.
Pritam bought a plot of land for Rs 6,40,000. Its value is increasing by 5% of its previous value after every six months. What will be the value of the plot after 2 years ?
Solution:
Present value of plot = Rs 6,40,000
Increase = 5% per half year
Period (n) = 2 years or 4 half years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 3

Question 4.
Mohan purchased a house for Rs 30,000 and its value is depreciating at the rate of 25% per year. Find the value of the house after 3 years.
Solution:
Present value of the house (P) = Rs 30,000
Rate of depreciation = 25% p.a.
Period (n) = 3 years
Value of house after 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 5

Question 5.
The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 43,740, find its purchased price.
Solution:
Let the purchase price of machine = Rs P
Rate of depreciation = 10% p.a.
Period (n) = 3 years.
and present value = Rs 43,740
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 6

Question 6.
The value of a refrigerator which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs 9,680, for how much was it purchased ?
Solution:
Let the refrigerator was purchased for = Rs P
Rate of depreciation (R) = 12% p.a.
Period (n) = 2 years
and present value (A) = Rs 9,680
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 8

Question 7.
The cost of a TV set was quoted Rs 17,000 at the beginning of 1999. In the beginning of2000, the price was hiked by 5%. Because of decrease in demand the cost was reduced by 4% in the beginning of 2001. What is the cost of the TV set in 2001 ?
Solution:
List price of TV set in 1999 = Rs 17,000
Rate of hike in 2000 = 5%
Rate of decrease in 2001 = 4%
Price of TV set in 2001
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 9

Question 8.
Ashish started the business with an initial investment of Rs 5,00,000. In the first year, he incurred a loss of 4%. However, during the second year he earned a profit of 5% which in third year, rose to 10%. Calculate the net profit for the entire period of 3 years.
Solution:
Initial investment = Rs 5,00,000
In the first year, rate of loss = 4%
In the second year, rate of gain = 5%
and in the third year, rate of gain = 10%
Investment after 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 10

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RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.2

March 30, 2023

RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.2

Other Exercises

Question 1.
Which among the following are nets for a cube ?
RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.2 1
Solution:
Nets for a cube are (ii), (iv) and (vi)

Question 2.
Name the polyhedron that can be made by folding each net:
RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.2 2
Solution:
(i) This net is for a square

(ii) This net is for triangular prism.

(iii) This net is for triangular prism.

(iv) This net is for hexagonal prism.

(v) This net is for hexagon pyramid.

(vi) This net is for cuboid.
RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.2 8

Question 3.
Dice are cubes where the numbers on the opposite faces must total 7. Which of the following are dice ?
RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.2 9

Solution:
Figure (i) shows the net of cube or dice.

Question 4.
Draw nets for each of the following polyhedrons:
RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.2 11

Solution:
(i) Net for cube is given below :

(ii) Net of a triangular prism is as under :

(iii) Net of hexagonal prism is as under :

(iv) The net for pentagonal pyramid is as under:

Question 5.
Match the following figures:
RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.2 17

Solution:
(a) (iv)
(b) (i)
(c) (ii)
(d) (iii)

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RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.1

March 30, 2023

RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.1

Other Exercises

Question 1.
What is the least number of planes that can enclose a solid ? What is the name of the solid ?
Solution:
The least number of planes that can enclose a solid is called a Tetrahedron.

Question 2.
Can a polyhedron have for its faces :
(i) three triangles ?
(ii) four triangles ?
(iii) a square and four triangles ?
Solution:
(i) No, polyhedron has three faces.
(ii) Yes, tetrahedron has four triangles as its faces.
(iii) Yes, a square pyramid has a square as its base and four triangles as its faces.

Question 3.
Is it possible to have a polyhedron with any given number of faces ?
Solution:
Yes, it is possible if the number of faces is 4 or more.

Question 4.
Is a square prism same as a cube ?
Solution:
Yes, a square prism is a cube.

Question 5.
Can a polyhedron have 10 faces, 20 edges and 15 vertices ?
Solution:
No, it is not possible as By Euler’s formula
F + V = E + 2
⇒ 10 + 15 = 20 + 2
⇒ 25 = 22
Which is not possible

Question 6.
Verify Euler’s formula for each of the following polyhedrons :
RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.1 1
Solution:
(i) In this polyhedron,
Number of faces (F) = 7
Number of edges (E) = 15
Number of vertices (V) = 10
According to Euler’s formula,
F + V = E + 2
⇒ 7 + 10 = 15 + 2
⇒ 17 = 17
Which is true.
(ii) In this polyhedron,
Number of faces (F) = 9
Number of edges (E) = 16
Number of vertices (V) = 9
According to Euler’s formula,
F + V = E + 2
⇒ 9 + 9 = 16 + 2
⇒ 18 = 18
Which is true.
(iii) In this polyhedron,
Number of faces (F) = 9
Number of edges (E) =18
Number of vertices (V) = 11
According to Euler’s formula,
F + V = E + 2
⇒ 9 + 11 = 18 + 2
⇒ 20 = 20
Which is true.
(iv) In this polyhedron,
Number of faces (F) = 5
Number of edges (E) = 8
Number of vertices (V) = 5
According to Euler’s formula,
F + V = E + 2
⇒ 5 + 5 = 8 + 2
⇒ 10 = 10
Which is true.
(v) In the given polyhedron,
Number of faces (F) = 9
Number of edges (E) = 16
Number of vertices (V) = 9
According to Euler’s formula,
F + V = E + 2
⇒ 9 + 9 = 16 + 2
⇒ 18 = 18
Which is true.

Question 7.
Using Euler’s formula, find the unknown:
RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.1 2
Solution:
We know that Euler’s formula is
F + V = E + 2
(i) F + 6 = 12 + 2
⇒ F + 6 = 14
⇒ F = 14 – 6 = 8
Faces = 8
(ii) F + V = E + 2
⇒ 5 + V = 9 + 2
⇒ 5 + V = 11
⇒ V = 11 – 5 = 6
Vertices = 6
(iii) F + V = E + 2
⇒ 20 + 12 = E + 2
⇒ 32 = E + 2
⇒ E = 32 – 2 = 30
Edges = 30

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RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5

March 30, 2023

RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5

Other Exercises

Question 1.
Construct a quadrilateral ABCD given that AB = 4 cm, BC = 3 cm, ∠A = 75°, ∠B = 80° and ∠C = 120°.
Solution:
Steps of construction :
(i) Draw a line segment AB = 4 cm.
RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5 1
(ii) At A draw a ray AX making an angle of 75°.
(iii) At B draw another ray BY making an angle of 80° and cut off BC = 3 cm.
(iv) At C, draw another ray CZ making an angle of 120° which intersects AX at D.
Then ABCD is the required quadrilateral.

Question 2.
Construct a quadrilateral ABCD where AB = 5.5 cm, BC = 3.7 cm, ∠A = 60°, ∠B = 105° and ∠D = 90°.
Solution:
∠A = 60°, ∠B = 105° and ∠D = 90°
But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)
⇒ 60° + 105° + ∠C + 90° = 360°
⇒ 255° + ∠C = 360°
⇒ ∠C = 360° – 255° = 105°
Steps of construction :
(i) Draw a line segment AB = 5.5 cm.
(ii) At A, draw a ray AX making an angle of
RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5 2
(iii) At B, draw another ray BY making an angle of 105° and cut off BC = 3.7 cm.
(iv) At C, draw a ray CZ making an angle of 105° which intersects AX at D.
Then ABCD is the required quadrilateral.

Question 3.
Construct a quadrilateral PQRS where PQ = 3.5 cm, QR = 6.5 cm, ∠P = ∠R = 105° and ∠S = 75°.
Solution:
∠P = 105°, ∠R = 105° and ∠S = 75°
But ∠P + ∠Q + ∠R + ∠S = 360° (Sum of angles of a quadrilateral)
⇒ 105° + ∠Q + 105° + 75° = 360°
⇒ 285° + ∠Q = 360°
⇒ ∠Q = 360° – 285° = 75°
Steps of construction :
(i) Draw a line segment PQ = 3.5 cm.
RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5 3
(ii) At P, draw a ray PX making an angle of 105°.
(iii) At Q, draw another ray QY, making an angle of 75° and cut off QR = 6.5 cm.
(iv) At R, draw a ray RZ making an angle of 105° which intersects PX at S.
Then PQRS is the required quadrilateral.

Question 4.
Construct a quadrilateral ABCD when BC = 5.5 cm, CD = 4.1 cm, ∠A = 70°, ∠B = 110° and ∠D = 85°.
Solution:
∠A = 70°, ∠B = 110°, ∠D = 85°
But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)
⇒ 70° + 110° + ∠C + 85° = 360°
⇒ 265° + ∠C = 360°
⇒ ∠C = 360° – 265° = 95°
Steps of construction:
(i) Draw a line segment BC = 5.5 cm.
(ii) At B, draw a ray BX making an angle of 110°.
(iii) At C, draw another ray CY making an angle of 95° and cut off CD = 4.1 cm.
(iv) At D, draw a ray DZ making an angle of 85° which intersects BX at A.
RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5 4
Then ABCD is the required quadrilateral.

Question 5.
Construct a quadrilateral ABCD, where ∠A = 65°, ∠B = 105°, ∠C = 75°, BC = 5.7 cm and CD = 6.8 cm.
Solution:
∠A = 65°, ∠B = 105°, ∠C = 75°
But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)
⇒ 65° + 105° + 75° + ∠D = 360°
⇒ 245° + ∠D = 360°
⇒ ∠D = 360° – 245° = 115°
Steps of construction:
(i) Draw a line segment BC = 5.7 cm.
(ii) At B, draw a ray BX making an angle of
RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5 5
(iii) At C draw a another ray CY making an angle of 75° and cut off CD = 6.8 cm.
(iv) At D, draw a ray DZ making an angle of 115° which intersects BX at A.
Then ABCD is the required quadrilateral.

Question 6.
Construct a quadrilateral PQRS in which PQ = 4 cm, QR = 5 cm, ∠P = 50°, ∠Q = 110° and ∠R = 70°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 4 cm.
(ii) At P, draw a ray PX making an angle of 50°.
(iii) At Q, draw another ray QY making an angle of 110° and cut off QR = 5 cm.
(iv) At R, draw a ray RZ making an angle of 70° which intersects PX at S.
RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5 6
Then PQRS is the required quadrilateral.

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