## RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4A.

Other Exercises

Question 1.
Solution:
(i)Angle : When two rays OA and OB meet at a point o, then ∠AOB is called an angle.

(ii) Interior of angle : The interior an angle is a set of all points in its plane which lie on the same side of OA as B and also on the same side of OB as A.
(iii) Obtuse angle : An angle greater than 90° but less than 180° is called an obtuse angle.
(iv) Reflex angle : An angle more than 180° but less than 360° is called a reflex angle.
(v) Complementary angles : Two angles are said to be complementary angles if their sum is 90°.
(vi) Supplementary angles : Two angles are said to be supplementary angles if their sum is 180°.

Question 2.
Solution:
∠A = 36°27’46”
∠B = 28° 43’39”
Adding, ∠A + ∠B = 64° 70′ 85″
We know that 60″ = 1′ and 60′ = 1°
∠A+ ∠B = 65° 11′ 25″ Ans.

Question 3.
Solution:
36° – 24° 28′ 30″
= 35° 59’60” – 25° 28’30”

= 10° 31′ 30″ Ans.

Question 4.
Solution:
We know that two angles are complementary of their sum is 90°. Each of these two angles is complement to the other, therefore.
(i) Complement of 58° = 90° – 58° = 32°
(ii) Complement of 16° = 90° – 16° = 74°
(iii) Complement of $$\frac { 2 }{ 3 }$$ of a right angle i.e.
of $$\frac { 2 }{ 3 }$$ x 90° or 60° = 90° – 60° = 30°
= $$\frac { 2 }{ 3 }$$ of right angle,
(iv) Complement of 46° 30′
= 90° – 46° 30′
= 43° 30′
(v) Complement of 52° 43′ 20°= 90° – 52° 43′ 20″
= 37° 16′ 40″
(vi) Complement of 68° 35′ 45″
= 90° – 68° 35′ 45″
= 21° 24′ 15″ Ans.

Question 5.
Solution:
We know that two angles are said to be supplement to each other of their sum is 180° therefore
(i) Supplement of 68° = 180° – 68° =112°
(ii) Supplement of 138° = 180° – 138° = 42°
(iii) Supplement of $$\frac { 3 }{ 5 }$$ of a right angle or $$\frac { 3 }{ 5 }$$ x 90° or 54°
= 180° – 54° = 126°
(iv) Supplement of 75° 36′ = 180° – 75° 36′ = 104° 24′
(v) Supplement of 124° 20′ 40″
= 180° – 124° 20′ 40″
= 55° 39′ 20″
(vi) Supplement of 108° 48′ 32″ = 180° – 108″ 48′ 32″ = 71° 11′ 28″ Ans.

Question 6.
Solution:
(i) Let the measure of required angle = x ,
their its complement = 90° – x
According to the condition,
x = 90° – x => 2x = 90°
=>x = $$\frac { { 90 }^{ o } }{ 2 }$$ = 45°
Required angle = 45°
(ii) Let the measure of required angle = x then its supplement = 180° – x
According to the condition,
x = 180° – x => 2x = 180° = 90°
=>x = $$\frac {{ 180 }^{ o }}{ 2 }$$ = 90°
Hence required angle = 90° Ans.

Question 7.
Solution:
Let required angle = x
then its complement = 90° – x
According to the condition,
x – (90° – x) = 36°
=> x – 90° + x = 36°
=> 2x = 36° + 90° = 126°
= $$\frac { { 126 }^{ o } }{ 2 }$$ = 63°
Required angle = 63° Ans.

Question 8.
Solution:
Let the required angle = x
then its supplement = 180° – x
According to the condition,
(180° – x) – x = 25°
=> 180° – x – x = 25°
=> – 2x = 25° – 180°
=> – 2x = – 155°
=> x = $$\frac { { – 155 }^{ o } }{ – 2 }$$
= 77.5°
Hence required angle = 77.5° Ans.

Question 9.
Solution:
Let required angle = x
Then its complement = 90° – x
According to the condition,
x = 4 (90° – x) => x = 360° – 4x
=> x + 4x = 360° => 5x = 360°
x = $$\frac { { 360 }^{ o } }{ 5 }$$ = 72°
Required angle = 72° Ans.

Question 10.
Solution:
Let required angle = x
Then its supplement = 180° – x
According to the condition,
x = 5 (180° – x)
=> x = 900° – 5x
=> x + 5x = 900°
=> 6x = 900°
=> x = $$\frac { { 900 }^{ o } }{ 6 }$$ = 150°
Hence, required angle = 150° Ans

Question 11.
Solution:
Let required angle = x
then its supplement = 180° – x
and complement = 90° – x
According to the condition,
180° – x = 4 (90°- x)
=> 180° – x = 360° – 4x
=> – x + 4x — 360° – 180°
=> 3x= 180°
=> x = $$\frac { { 180 }^{ o } }{ 3 }$$ = 60°
Required angle = 60° Ans.

Question 12.
Solution:
Let required angle = x
Then, its complement = 90° – x
and its supplement = 180° – x
According to the condition,
90° – x = $$\frac { 1 }{ 3 }$$ (180° – x)
=> 90° – x = 60° – $$\frac { 1 }{ 3 }$$ x
=> 90° – 60° = x – $$\frac { 1 }{ 3 }$$ x
=> $$\frac { 2 }{ 3 }$$ x = 30° =>x = $$\frac { { 30 }^{ o }X3\quad \quad }{ 2 }$$ => x = 45° Ans.

Question 13.
Solution:
Let one angle = x
Then, its supplement = 180° – x
According to the condition,
x : (180° – x) = 3:2
=> $$\frac { x }{ { 180 }^{ o }-x } =\frac { 3 }{ 2 }$$
=>2x = 3(180°- x)
=> 2x = 540° – 3x
=> 2x + 3x = 540°
=> 5x = 540° => x = $$\frac { { 540 }^{ o } }{ 5 }$$ = 108°
Angle = 108° and its supplement = 180° – 108° = 72°
Hence, angles are 108° and 72° Ans.

Question 14.
Solution:
Let angle = x
Then, its complementary angle = 90° – x
According to the condition,
x : (90° – x) = 4 : 5
=> $$\frac { x }{ { 90 }^{ o }-x } =\frac { 4 }{ 5 }$$
=> 5x = 4 (90° – x)
=> 5x = 360° – 4x
=> 5x + 4x = 360°
=> 9x = 360°
=> x = $$\frac { { 360 }^{ o } }{ 9 } ={ 40 }^{ o }$$
and its complement = 90° – 40° = 50°
Hence, angles are 40° and 50° Ans.

Question 15.
Solution:
Let the required angle = x
.’. its complement = 90° – x
and supplement = 180° – x
According to the condition,
7(90° – x) = 3(180° – x) – 10°
=> 630° – 7x = 3 (180° – x) – 10°
=> 630° – 7x = 540° – 3x – 10°
=> – 7x + 3x = 540° – 10° – 630°
– 4x = – 100°
x = $$\frac { { -100 }^{ o } }{ -4 } ={ 25 }^{ o }$$
Hence required angle = 25° Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4A are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14D

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14D.

Other Exercises

Question 1.
Solution:
(i) First eight natural numbers are 1, 2, 3, 4, 5, 6, 7, 8
∴ Mean = $$\frac { 1+2+3+4+5+6+7+8 }{ 8 }$$ = $$\frac { 36 }{ 8 }$$ = $$\frac { 9 }{ 2 }$$ = 4.5
(ii) First ten odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
∴ Mean = $$\frac { 1+3+5+7+9+11+13+15+17+179 }{ 10 }$$ = $$\frac { 100 }{ 10 }$$ = 10
(iii) First five prime numbers are 2, 3, 5, 7, 11.
∴ Mean = $$\frac { 2+3+5+7+11 }{ 5 }$$ = $$\frac { 28 }{ 5 }$$ = 5.6
(iv) First six even numbers are 2, 4, 6, 8, 10, 12
∴ Mean = $$\frac { 2+4+6+8+10+12 }{ 10 }$$ = $$\frac { 42 }{ 6 }$$ = 7
(v) First seven multiples of 5 are 5, 10, 15, 20, 25, 30, 35
∴ Mean = $$\frac { 5+10+15+20+25+30+35 }{ 7 }$$ = $$\frac { 140 }{ 7 }$$ = 20
(vi) All the factors of 20 are, 1, 2, 4, 5, 10, 20
∴ Mean = $$\frac { 1+2+4+5+10+20 }{ 6 }$$ = $$\frac { 42 }{ 6 }$$ = 7

Question 2.
Solution:
No. of families (n) = 10
Sum of children (∑x1) = 2 + 4 + 3 + 4 + 2 + 0 + 3 + 5 + 1 + 6 = 30
∴ Mean $$\left( \overline { x } \right) =\frac { \sum { x1 } }{ n } =\frac { 30 }{ 10 } =3$$

Question 3.
Solution:
Here number of days (n) = 7
Number of books (∑x1) = 105 + 216 + 322 + 167 + 273 + 405 + 346 = 1834
∴ Mean $$\left( \overline { x } \right) =\frac { \sum { x1 } }{ n } =\frac { 1864 }{ 7 } =262 books$$

Question 4.
Solution:
Number of days (n) = 6
Sum of temperature (∑x) = 35.5 + 30.8 + 27.3 + 32.1 + 23.8 + 29.9 = 179.4°F
∴ Mean temperature = $$\frac { \sum { x } }{ n } =\frac { 179.4 }{ 6 }$$=29.9°F

Question 5.
Solution:
Number of students (n) = 12
Sum of marks (∑x) = 64 + 36 + 47 + 23 + 0 + 19 + 81 + 93 + 72 + 35 + 3 + 1 = 474
= $$\frac { \sum { x1 } }{ n } =\frac { 474 }{ 12 }$$ = 39.5

Question 6.
Solution:
Here, n = 6
and arithmetic mean =13
Total sum = 13 x 6 = 78.
But sum of 7 + 9+ 11 + 13 + 21=61
Value of x = 78 – 61 = 17
Hence x = 17 Ans.

Question 7.
Solution:
Let x1, x2, x3, … x24 be the 24 numbers
$$\frac { x1+x2+x3+…..+x24 }{ 24 } =35$$
=> x1 + x2 + x3 +….+ x24 = 35 x 24

Question 8.
Solution:
Let x1 + x2 + x3……..x20 be the 20 numbers

Question 9.
Solution:
Let x1, x2, x3 … x15 be the numbers
Mean = $$\frac { x1+x2+x3+…..+x15 }{ 15 } = 27$$

Question 10.
Solution:
Let x1, x2, x3 … x12 be the numbers
Mean = $$\frac { x1+x2+x3+…..+x12 }{ 12 }$$ = 40
=> x1+x2+x3+….+x12 = 40 X 12 =480
Now,new numbers are $$\frac { x1 }{ 8 } ,\frac { x2 }{ 8 } ,\frac { x3 }{ 8 } ,..\frac { x12 }{ 8 }$$
Mean = $$\frac { \frac { x1 }{ 8 } +\frac { x2 }{ 8 } +\frac { x3 }{ 8 } +…..+\frac { x12 }{ 8 } }{ 12 }$$
= $$\frac { 1 }{ 8 } \frac { \left( x1+x2+x3+….x12 \right) }{ 12 }$$
= $$\frac { 480 }{ 8X12 }$$ = 5
Mean of new numbers =5

Question 11.
Solution:
Let x1, x2, x3,…..x20 are the numbers
Mean = $$\frac { x1+x2+x3+…x20 }{ 20 }$$ = 18
=> x1 + x2 + x3 +….+ x20 = 18 X 20 =360

Question 12.
Solution:
Mean weight of 6 boys = 48 kg
Their total weight = 48 x 6 = 288 kg
Weights of 5 boys among them, are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg
Sum of weight of 5 boys = (51 + 45 + 49 + 46 + 44) kg = 235 kg
Weight of 6th boy = 288 kg – 235 kg = 53 kg Ans.

Question 13.
Solution:
Mean of 50 students = 39
Total score = 39 x 50 = 1950
Now correct sum of scores = 1950 – Wrong item + Correct item= 1950 – 23 + 43
= 1950 + 20 = 1970
Correct mean = $$\frac { 1970 }{ 50 }$$ = 39.4 Ans.

Question 14.
Solution:
Mean of 100 items = 64
The sum of 100 items = 64 x 100 = 6400
New sum of 100 items = 6400 + 36 + 90 – 26 – 9 = 6526 – 35 = 6491,
Correct mean = $$\frac { 6491 }{ 100 }$$ =64.91 = 64.91 Ans.

Question 15.
Solution:
Mean of 6 numbers = 23
Sum of 6 numbers = 23 x 6 = 138
Excluding one number, the mean of remaining 5 numbers = 20
Total of 5 numbers = 20 x 5 = 100
Excluded number = 138 – 100 = 38 Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14D are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14E

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14E.

Other Exercises

Question 1.
Solution:
Mean marks of 7 students = 226
∴Total marks of 7 students = 226 x 1 = 1582
Marks obtained by 6 of them are 340, 180, 260, 56, 275 and 307
∴ Sum of marks of these 6 students = 340 + 180 + 260 + 56 + 275 + 307 = 1418
∴ Marks obtained by seventh student = 1582 – 1418 = 164 Ans.

Question 2.
Solution:
Mean weight of 34 students = 46.5 kg.
∴ Total weight of 34 students = 46.5 x 34 kg = 1581 kg
By including the weight of teacher, the mean weight of 35 persons = 500 g + 46.5 kg
= 46.5 + 0.5 = 47.0 kg
∴ Total weight = 47.0 x 35 = 1645 kg
∴ Weight of the teacher = (1645 – 1581) kg = 64 kg Ans.

Question 3.
Solution:
Mean weight of 36 students = 41 kg
∴ Total weight of 36 students = 41 x 36 kg = 1476 kg
After leaving one student, number of students = 36 – 1 =35
Their new mean = 41.0 – 200g = (41.0 – 0.2) kg = 40.8 kg.
∴ Total weight of 35 students = 40.8 x 35 = 1428 kg
∴ Weight of leaving student = (1476 – 1428) kg = 48 kg Ans.

Question 4.
Solution:
Average weight of 39 students = 40 kg
∴ Total weight of 39 students = 40 x 39 = 1560 kg
By admitting of a new student, no. of students = 39 + 1 =40
and new mean = 40 kg – 200 g = 40 kg – 0.2 kg = 39.8 kg
∴Weight of 40 students = 39.8 x 40 kg = 1592 kg
∴Weight of new student = 1592 – 1560 kg = 32 kg Ans.

Question 5.
Solution:
Average salary of 20 workers = Rs. 7650
∴Their total salary = Rs. 7650 x 20 = Rs. 153000
By adding the salary of the manager, their mean salary = Rs. 8200
∴Their total salary = Rs. 8200 x 21 = Rs. 172200
∴Salary of the manager = Rs. 172200 – Rs. 153000 = Rs. 19200 Ans.

Question 6.
Solution:
Average wage of 10 persons = Rs. 9000
Their total wage = Rs. 9000 x 10 = Rs. 90000
Wage of one person among them = Rs. 8100
and wage of new member = Rs. 7200
∴ New total wage = Rs. (90000 – 8100 + 7200) = Rs. 89100
Their new mean wage = Rs.$$\frac { 89100 }{ 10 }$$ = Rs. 8910

Question 7.
Solution:
Mean consumption of petrol for 7 months of a year = 330 litres
Mean consumption of petrol for next 5 months = 270 litres
∴Total consumption for first 7 months = 7 x 330 = 2320 l
and total consumption for next 5 months = 5 x 270 = 1350 l
Total consumption for 7 + 5 = 12 months = 2310 + 1350 = 3660 l
Average consumption = $$\frac { 3660 }{ 12 }$$ = 305 liters per month.

Question 8.
Solution:
Total numbers of numbers = 25
Mean of 15 numbers = 18
∴Total of 15 numbers = 18 x 15 = 270
Mean of remaining 10 numbers = 13
∴Total = 13 x 10 = 130
and total of 25 numbers = 270 + 130 = 400
∴Mean = $$\frac { 400 }{ 25 }$$ = 16 Ans.

Question 9.
Solution:
Mean weight of 60 students = 52.75 kg.
Total weight = 52.75 x 60 = 3165 kg
Mean of 25 out of them = 51 kg.
∴Their total weight = 51 x 25 = 1275 kg
∴ Total weight of remaining 60 – 25 = 35 students = 3165 – 1275 = 1890 kg 1890
Mean weight = $$\frac { 1890 }{ 35 }$$ = 54 kg Ans.

Question 10.
Solution:
Average increase of 10 oarsman = 1.5 kg.
∴ Total increased weight =1.5 x 10 = 15kg
Weight of out going oarsman = 58 kg .
∴ Weight of new oarsman = 58 + 15 = 73 kg Ans.

Question 11.
Solution:
Mean of 8 numbers = 35
∴Total of 8 numbers = 35 x 8 = 280
After excluding one number, mean of 7 numbers = 35 – 3 = 32
Total of 7 numbers = 32 x 7 = 224
Hence excluded number = 280 – 224 = 56 Ans.

Question 12.
Solution:
Mean of 150 items = 60
Total of 150 items = 60 x 150 = 9000
New total = 9000 + 152 + 88 – 52 – 8 = 9000 + 240 – 60 = 9180
∴New mean = $$\frac { 9180 }{ 150 }$$ = 61.2 Ans.

Question 13.
Solution:
Mean of 31 results = 60
∴Total of 31 results = 60 x 31 = 1860
Mean of first 16 results = 58
∴Total of first 16 results = 58 x 16 = 928
and mean of last 16 results = 62
∴Total of last 16 results = 62 x 16 = 992
∴16th result = (928 + 992) – 1860 = 1920 – 1860 = 60 Ans.

Question 14.
Solution:
Mean of 11 numbers = 42
∴Total of 11 numbers = 42 x 11 = 462
Mean of first 6 numbers = 37
∴Total of first 6 numbers = 37 x 6 = 222
Mean of last 6 numbers = 46
∴Total of last 6 numbers = 46 x 6 = 276
∴6th number = (222 + 276) – 462 = 498 – 462 = 36 Ans.

Question 15.
Solution:
Mean weight of 25 students = 52 kg
∴ Total weight of 25 students = 52 x 25 = 1300 kg
Mean weight of first 13 students = 48 kg
∴ Total weight of first 13 students = 48 x 13 kg = 624 kg
Mean of last 13 students = 55 kg
∴Total of last 13 students = 55 x 13 kg = 715 kg
∴Weight of 13th student = (624 + 715) – 1300 = 1339 – 1300 = 39 kg Ans.

Question 16.
Solution:
Mean of 25 observations = 80
Total of 25 observations = 80 x 25 = 2000
Mean of another 55 observations = 60
∴ Total of these 55 observations = 60 x 55 = 3300
Total number of observations = 25 + 55 = 80
and total of 80 numbers = 2000 + 3300 = 5300
Mean of 80 observations = $$\frac { 5300 }{ 80 }$$ = 66.25 Ans.

Question 17.
Solution:
Marks in English = 36
Marks in Hindi = 44
Marks in Mathematics = 75
Marks in Science = x
∴Total number of marks in 4 subjects = 36 + 44 + 75 + x = 155 + x
Average marks in 4 subjects = 50
∴Total marks = 50 x 4 = 200
∴155 + x = 200
=> x = 200 – 155
=> x = 45
Hence, marks in Science = 45 Ans.

Question 18.
Solution:
Mean of monthly salary of 75 workers = Rs. 5680
Total salary of 75 workers = Rs. 5680 x 75 = Rs. 426000
Mean salary of 25 among them = Rs. 5400
Total of 25 workers = Rs. 5400 x 25 = Rs. 135000
Mean salary of 30 among them = Rs. 5700
∴Total of 30 among them = Rs. 5700 x 30 = Rs. 171000
∴ Total salary of 25 + 30 = 55 workers = Rs. 135000 + 171000 = Rs. 306000
∴Total salary of remaining 75 – 55 = 20 workers = Rs. 426000 – 306000 = Rs. 120000
∴ Mean of remaining 20 workers = Rs. $$\frac { 120000 }{ 20 }$$ = Rs. 6000 Ans.

Question 19.
Solution:
Let distance between two places = 60 km
∴Time taken at the speed of 15 km/h = $$\frac { 60 }{ 15 }$$ = 4 hours
and time taken at speed of 10 km/h for coming back = $$\frac { 60 }{ 10 }$$ = 6 hours
Total the taken = 4 + 6 = 10
hours and distance covered = 60 + 60 = 120 km
∴Average speed = $$\frac { 120 }{ 10 }$$ =12 km/hr.

Question 20.
Solution:
No. of total students = 50
No. of boys = 40
∴ No. of girls = 50 – 40 = 10
Average weight of class = 44kg
∴Total weight of 50 students = 44 x 50 = 2200 kg
Average weight of 10 girls = 40 kg
∴Total weight = 40 x 10 = 400 kg
Total weight of 40 boys = 2200 – 400 = 1800 kg
∴Average weight of 40 boys = $$\frac { 1800 }{ 40 }$$kg = 45 kg Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14E 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.

## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H.

Other Exercises

Question 1.
Solution:
Arranging the given data in ascending order :
0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6
We see that 6 occurs in maximum times.
Mode = 6 Ans.

Question 2.
Solution:
Arranging in ascending order, we get:
15, 20, 22, 23, 25, 25, 25, 27, 40
We see that 25 occurs in maximum times.
Mode = 25 Ans.

Question 3.
Solution:
Arranging in ascending order we get:
1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9
Here, we see that 9 occurs in maximum times.
Mode = 9 Ans.

Question 4.
Solution:
Arranging in ascending order, we get:
9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60
Here, we see that 50 occurs in maximum times.
Modal score = 50 scores Ans.

Question 5.
Solution:
Arranging in ascending order, we get:
10, 10, 11, 11, 12, 12, 13, 14,15, 17
Here, number of terms is 10, which is even
∴ Median = $$\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (5th term + 6th term)

Question 6.
Solution:

Question 7.
Solution:
Writing its cumulative frequency table

Question 8.
Solution:
Writing its cumulative frequency table

Here, number of items is 40 which is even.
∴ Median = $$\frac { 1 }{ 2 } \left[ \frac { 40 }{ 2 } th\quad term+\left( \frac { 40 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (20th term + 21th term)
= $$\frac { 1 }{ 2 }$$ (30 + 30) = $$\frac { 1 }{ 2 }$$ x 60 = 30
Mean= $$\frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } }$$ = $$\frac { 1161 }{ 40 }$$ = 29.025
∴Mode = 3 median – 2 mean = 3 x 30 – 2 x 29.025 = 90 – 58.05 = 31.95

Question 9.
Solution:
Preparing its cumulative frequency table we get:

Here number of terms is 50, which is even

Question 10.
Solution:
Preparing its cumulative frequency table :

Question 11.
Solution:
Preparing its cumulative frequency table we have,

Question 12.
Solution:
Preparing its cumulative frequency table we have,

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H 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.

## RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A.

Question 1.
Solution:
Number of trials = 500 times
Let E be the no. of events in each case, then
∴No. of heads (E1) = 285 times
and no. of tails (E2) = 215 times
∴ Probability in each case will be
∴(i)P(E1) = $$\frac { 285 }{ 500 }$$ = $$\frac { 57 }{ 100 }$$ = 0.57
(ii) P(E2) = $$\frac { 215 }{ 500 }$$ = $$\frac { 43 }{ 100 }$$ = 0.43

Question 2.
Solution:
No. of trials = 400
Let E be the no. of events in each case, then
No. of 2 heads (E1) = 112
No. of one head (E2) = 160 times
and no. of O. head (E3) = 128 times
∴ Probability in each case will be:
∴ (i)P(E1) = $$\frac { 112 }{ 400 }$$ = $$\frac { 28 }{ 100 }$$ = 0.28
(ii)P(E2) = $$\frac { 160 }{ 400 }$$ = $$\frac { 40 }{ 100 }$$= 0.40
(iii) P(E3) = $$\frac { 128 }{ 400 }$$ = $$\frac { 32 }{ 100 }$$ = 0.32 Ans.

Question 3.
Solution:
Number of total trials = 200
Let E be the no. of events in each case, then
No. of three heads (E1) = 39 times
No. of two heads (E2) = 58 times
No. of one head (E3) = 67 times
and no. of no head (E4) = 36 times
∴ Probability in each case will be .
(i) P(E1) = $$\frac { 39 }{ 200 }$$ = 0.195
(ii) P(E3) = $$\frac { 67 }{ 200 }$$ = 0.335
(iii) P(E4) = $$\frac { 36 }{ 200 }$$ = $$\frac { 18 }{ 100 }$$ = 0.18
(iv) P(E2) = $$\frac { 58 }{ 200 }$$ = $$\frac { 29 }{ 100 }$$ = 0.29

Question 4.
Solution:
Solution No. of trials = 300 times
Let E be the no. of events in each case, then
No. of outcome of 1(E1) = 60
No. of outcome of 2(E2) = 72
No. of outcome of 3(E3) = 54
No. of outcome of 4(E4) 42
No. of outcome of 5(E5) = 39
No. of outcome of 6(E6) = 33
The probability of
(i) P(E3) = $$\frac { 54 }{ 300 }$$ = $$\frac { 18 }{ 100 }$$ = 0.18
(ii) P(E6) = $$\frac { 33 }{ 100 }$$ = $$\frac { 11 }{ 100 }$$= 0.11
(iii) P(E5) = $$\frac { 39 }{ 300 }$$ = $$\frac { 13 }{ 100 }$$ = 0.13
(iv) P(E1) = $$\frac { 60 }{ 300 }$$ = $$\frac { 20 }{ 100 }$$= 0.20 Ans.

Question 5.
Solution:
Let E be the no. of events in each case.
No. of ladies who like coffee (E1) = 142
No. of ladies who like coffee (E2) = 58
Probability of
(1) P(E1) = $$\frac { 142 }{ 200 }$$ = $$\frac { 71 }{ 100 }$$ = 0.71
(ii) P(E2) = $$\frac { 58 }{ 200 }$$ = $$\frac { 29 }{ 100 }$$ = 0.29 Ans.

Question 6.
Solution:
Total number of tests = 6
No. of test in which the students get more than 60% mark = 2
Probability will he
P(E) = $$\frac { 2 }{ 6 }$$ = $$\frac { 1 }{ 3 }$$Ans.

Question 7.
Solution:
No. of vehicles of various types = 240
No. of vehicles of two wheelers = 64.
Probability will be P(E) = $$\frac { 84 }{ 240 }$$ = $$\frac { 7 }{ 20 }$$ = 0.35 Ans.

Question 8.
Solution:
No. of phone numbers are one page = 200
Let E be the number of events in each case,
Then (i) P(E5) = $$\frac { 24 }{ 200 }$$ = $$\frac { 12 }{ 100 }$$ = 0.12
(ii) P(E8) = $$\frac { 16 }{ 200 }$$ = $$\frac { 8 }{ 100 }$$ = 0.08 Ans.

Question 9.
Solution:
No. of students whose blood group is checked = 40
Let E be the no. of events in each case,
Then (i) P(E0) = $$\frac { 14 }{ 40 }$$ = $$\frac { 7 }{ 20 }$$ = 0.35
(ii) P(EAB) = $$\frac { 6 }{ 40 }$$ = $$\frac { 3 }{ 20 }$$ = 0.15 Ans.

Question 10.
Solution:
No. of total students = 30.
Let E be the number of elements, this probability will be of interval 21 – 30
P(E) = $$\frac { 6 }{ 30 }$$ = $$\frac { 1 }{ 5 }$$ = 0.2 Ans.

Question 11.
Solution:
Total number of patients of various age group getting medical treatment = 360
Let E be the number of events, then
(i) No. of patient which are 30 years or more but less than 40 years = 60.
P(E) = $$\frac { 60 }{ 360 }$$ = $$\frac { 1 }{ 6 }$$
(ii) 50 years or more but less than 70 years = 50 + 30 = 80
P(E) = $$\frac { 80 }{ 360 }$$ = $$\frac { 2 }{ 9 }$$
(iii) Less than 10 years = zero
P(E) = $$\frac { 0 }{ 360 }$$ = 0
(iv) 10 years or more 90 + 50 + 60 + 80 + 50 + 30 = 360

Hope given RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14F.

Other Exercises

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Mean = 8

Question 6.
Solution:
Mean = 28.25

Question 7.
Solution:
Mean = 16.6

Question 8.
Solution:
Mean = 50

Question 9.
Solution:

Question 10.
Solution:
Let assumed mean = 67

Question 11.
Solution:
Here h = 1, Let assumed mean (A) = 21

Question 12.
Solution:
Here h = 400 and let assumed mean (A) = 1000

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14F are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G.

Other Exercises

Question 1.
Solution:
Arranging in ascending order, we get:
2,2,3,5,7,9,9,10,11
Here, number of terms is 9 which is odd.
∴ Median = $$\frac { n+1 }{ 2 }$$ th term = $$\frac { 9+1 }{ 2 }$$ th term = 5th term = 7 Ans.
(ii) Arranging in ascending order, we get: 6, 8, 9, 15, 16, 18, 21, 22, 25
Here, number of terms is 9 which is odd.
∴ Median = $$\frac { n+1 }{ 2 }$$ th term = $$\frac { 9+1 }{ 2 }$$ th term = 5th term = 16 Ans.
(iii) Arranging in ascending order, we get: 6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25
Here, number of terms is 11 which is odd.
∴ Median = $$\frac { 11+1 }{ 2 }$$ th term = $$\frac { 12 }{ 2 }$$ th term = 6th term = 16 Ans.
(iv) Arranging in ascending order, we get:
0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10
Here, number of terms is 13, which is odd.
Median = $$\frac { 13+1 }{ 2 }$$ th term = $$\frac { 14 }{ 2 }$$ th term = 7th term = 4 Ans.

Question 2.
Solution:
Arranging in ascending order, we get 9, 10, 17, 19, 21, 22, 32, 35
Here, number of terms is 8 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 8 }{ 2 } th\quad term+\left( \frac { 8 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ [4th term + 5th term] = $$\frac { 1 }{ 2 }$$ (19 + 21) = $$\frac { 1 }{ 2 }$$ x 40 = 20
(ii) Arranging in ascending order, we get:
29, 35, 51, 55, 60, 63, 72, 82, 85, 91
Here number of terms is 10 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (60 + 63) = $$\frac { 1 }{ 2 }$$ x 123 = 61.5 Ans.
(iii) Arranging in ascending order we get
3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81
Here number of terms is 12 which is even.
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 12 }{ 2 } th\quad term+\left( \frac { 12 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (6th term + 7th term) = $$\frac { 1 }{ 2 }$$ (15 + 17)= $$\frac { 1 }{ 2 }$$ x 32
= 16 Ans.

Question 3.
Solution:
Arranging the given data in ascending order, we get :
17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40
∴ Median = $$\frac { 15+1 }{ 2 }$$ th term = $$\frac { 16 }{ 2 }$$ th term = 8th term = 23
∴ Median score = 23 Ans.

Question 4.
Solution:
Arranging in ascending order, we get:
143.7, 144.2, 145, 146.5, 147.3, 148.5, 149.6, 150, 152.1
Here, number of terms is 9 which is odd.
Median = $$\frac { 9+1 }{ 2 }$$ th term = $$\frac { 10 }{ 2 }$$ th term = 5th term = 147.3 cm
Hence, median height = 147.3 cm Ans.

Question 5.
Solution:
Arranging in ascending order, we get:
9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2
Here number of terms is 8 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 8 }{ 2 } th\quad term+\left( \frac { 8 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$[4th term + 5th term]
= $$\frac { 1 }{ 2 }$$ (13.4 + 14.3) = $$\frac { 1 }{ 2 }$$ (27.7) = 13.85
∴ Median weight = 13.85 kg. Ans.

Question 6.
Solution:
Arranging in ascending order, we get:
32, 34, 36, 37, 40, 44, 47, 50, 53, 54
Here, number of terms is 10 which is even.
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ [5th term + 6th term ] = $$\frac { 1 }{ 2 }$$ (40 + 44) = $$\frac { 1 }{ 2 }$$ x 84 = 42 .
∴ Median age = 42 years.

Question 7.
Solution:
The given ten observations are 10, 13, 15, 18, x + 1, x + 3, 30, 32, 35, 41
These are even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ [5th term + 6th term ] = $$\frac { 1 }{ 2 }$$(x + 1 + x + 3) = $$\frac { 1 }{ 2 }$$(2x + 4)
= x + 2
But median is given = 24
∴ x + 2 = 24 => x = 24 – 2 = 22
Hence x = 22.

Question 8.
Solution:
Preparing the cumulative frequency table, we have:

Here, number of terms (n) = 41, which is odd,
Median = $$\frac { 41+1 }{ 2 }$$ th term = $$\frac { 42 }{ 2 }$$ th term = 21st term = 50 (∵ 20th to 28th term = 50)
Hence median weight = 50 kg Ans.

Question 9.
Solution:
Arranging first in ascending order, we get:

Now preparing its cumulative frequency table

Here, number of terms is 37 which is odd.
Median = $$\frac { 37+1 }{ 2 }$$ th term = $$\frac { 38 }{ 2 }$$ th term = 19 th term = 22 (∵18th to 21st = 22)
Hence median – 22 Ans.

Question 10.
Solution:
first arranging in ascending order we get

Now preparing its cumulative frequency table,we find:

Here, number of terms is 43, which if odd.
Median = $$\frac { 43+1 }{ 2 }$$ th term = $$\frac { 44 }{ 2 }$$ th term = 22nd term = 25 25 (∵ 11th to 26th = 25)

Question 11.
Solution:
Arranging in ascending order,we get

Now preparing its cumulative frequency table, we find :

Here, number of terms = 50 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 50 }{ 2 } th\quad term+\left( \frac { 50 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (154 + 155) = $$\frac { 1 }{ 2 }$$ (309) = 154.5 (∵ 22nd to 25th = 154, 26th to 34th= 155)

Question 12.
Solution:
Arranging in ascending order, we get:

Now, preparing its cumulative frequency table.

Here, number of terms is 60 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 60 }{ 2 } th\quad term+\left( \frac { 60 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (30th term + 31st term)
= $$\frac { 1 }{ 2 }$$ (20 + 23) = $$\frac { 1 }{ 2 }$$ x 43 = 21.5 (∵ 18th to 30th term = 20, 31st term to 34th = 23)
Hence median = 21.5 Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11A.

Other Exercises

Question 1.
Solution:
Let AB be a chord of a circle with centre O. OC⊥AB and OA be the radius of the circle, then
AB = 16cm, OA = 10cm .

OC ⊥ AB.
OC bisects AB at C
AC = $$\frac { 1 }{ 2 }$$ AB = $$\frac { 1 }{ 2 }$$ x 16 = 8cm

Question 2.
Solution:
Let AB be the chord of the circle with centre O and OC ⊥ AB, OA be the radius of the circle,
then OC = 3cm, OA = 5cm

Now in right ∆ OAC,
OA² = AC² = OC² (Pythagoras Theorem)

Question 3.
Solution:
Let AB be the chord, OA be the radius of
the circle OC ⊥ AB
then AB = 30 cm, OC = 8cm

Question 4.
Solution:
AB and CD are parallel chords of a circle with centre O.

Question 5.
Solution:
Let AB and CD be two chords of a circle with centre O.
OA and OC are the radii of the circle. OL⊥AB and OM⊥CD.

Question 6.
Solution:
In the figure, a circle with centre O, CD is its diameter AB is a chord such that CD⊥AB.
AB = 12cm, CE = 3cm.
Join OA.
∵ COD⊥AB which intersects AB at E

Question 7.
Solution:
A circle with centre O, AB is diameter which bisects chord CD at E
i.e. CE = ED = 8cm and EB = 4cm
Join OC.
Let radius of the circle = r

Question 8.
Solution:
Given : O is the centre of a circle AB is a chord and BOC is the diameter. OD⊥AB
To prove : AC || OD and AC = 20D
Proof : OD⊥AB
∵ D is midpoint of AB

Question 9.
Solution:
Given : O is the centre of the circle two
chords AB and CD intersect each other at P inside the circle. PO bisects ∠BPD.
To prove : AB = CD.

Question 10.
Solution:
Given : PQ is the diameter of the circle with centre O which is perpendicular to one chord AB and chord AB || CD.
PQ intersects AB and CD at E and F respectively
To prove : PQ⊥CD and PQ bisects CD.

Question 11.
Solution:
Two circles with centre O and O’ intersect each other.

To prove : The two circles cannot intersect each other at more than two points.
Proof : Let if opposite, the two circles intersect each other at three points P, Q and R.
Then these three points are non-collinear. But, we know that through three non- collinear points, one and only one circle can be drawn.
∵ Our supposition is wrong
Hence two circle can not intersect each other at not more than two points.
Hence proved

Question 12.
Solution:
Given : Two circles with centres O and O’ intersect each other at A and B. AB is a common chord. OO’ is joined.
AO and AO is joined.

Question 13.
Solution:
Given : Two equal circles intersect each other at P and Q.
A straight line drawn through
P, is drawn which meets the circles at A and B respectively
To prove : QA = QB

Question 14.
Solution:
Given : A circle with centre 0. AB and CD are two chords and diameter PQ bisects AB and CD at L and M
To Prove : AB || CD.

Question 15.
Solution:
Given : Two circles with centres A and B touch each other at C internally. A, B arc joined. PQ is the perpendicular bisector of AB intersecting it at L and meeting the bigger circle at P and Q respectively and radii of the circles are 5cm and 3cm. i.e. AC = 5cm,BC = 3cm
Required : To find the lenght of PQ

Question 16.
Solution:
Given : AB is a chord of a circle with centre O. AB is produced to C such that BC = OB, CO is joined and produced to meet the circle at D.
∠ ACD = y°, ∠ AOD = x°
To prove : x = 3y

Question 17.
Solution:
Given : O is the centre of a circle AB and AC are two chords such that AB = AC
OP⊥AB and OQ⊥AC.
which intersect AB and AC at M and N
respectively. PB and QC are joined.
To prove : PB = QC.

Question 18.
Solution:
Given : In a circle with centre O, BC is its diameter. AB and CD are two chords such that AB || CD.
To prove : AB = CD
Const. Draw OL⊥AB
OM⊥CD.
Proof : In ∆ OLB and ∆ OCM,
OB = OC (radii of the same circle)
∠ OLB = ∠ OMC (each 90°)
∠ OBL = ∠ OCM (alternate angles)

Question 19.
Solution:
Equilateral ∆ ABC in inscribed in a circle in which
AB = BC = CA = 9cm.

Question 20.

Solution:
Given : AB and AC are two equal chords of a circle with centre O
To Prove : O lies on the bisector of ∠ BAC

Question 21.
Solution:
Given : OPQR is a square with centre O, a circle is drawn which intersects the square at X and Y.
To Prove : Q = QY
Const. Join OX and OY

Hope given RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11A are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C.

Other Exercises

Question 1.
Solution:
Radius of base (r) = 35cm
and height (h) = 84cm.

Question 2.
Solution:
Height of cone (h) = 6cm
Slant height (l) = 10cm.

Question 3.
Solution:
Volume of right circular cone = (100 π) cm3
Height (h) = 12cm.
Let r be the radius of the cone

Question 4.
Solution:
Circumference of the base = 44cm

Question 5.
Solution:
Slant height of the cone (l) = 25cm
Curved surface area = 550 cm2
πrl = curved surface area

Question 6.
Solution:
Slant height (l) = 37cm.
We know that

Question 7.
Solution:
Curved surface area = 4070 cm2
Diameter of the base = 70cm

Question 8.
Solution:
Radius of the conical tent = 7m
and height = 24 m.

Question 9.
Solution:
Radius of the first cone (r) = 1.6 cm.
and height (h) = 3.6 cm.

Question 10.
Solution:
Ratio in their heights =1:3
and ratio in their radii = 3:1
Let h1,h2 he their height and r1,r2 be their radii, then

The ratio between their volumes is 3:1
hence proved

Question 11.
Solution:
Diameter of the tent = 105m

Question 12.
Solution:
No. of persons to be s accommodated =11
Area to be required for each person = 4m2

Question 13.
Solution:
Height of the cylindrical bucket (h) = 32cm
Volume of sand filled in it = πr2h
= π x 18 x 18 x 32 cm3
= 10368π cm3
Volume of conical sand = 10368 π cm3
Height of cone = 24 cm

Question 14.
Solution:
Let h be the height and r be the radius of the cylinder and cone.
Curved surface area of cylinder = 2πrh
and curved surface area of cone = πrl

Question 15.
Solution:
Diameter of the pillar = 20cm
Radius (r) = $$\frac { 20 }{ 2 }$$ = 10cm

Question 16.
Solution:
Height of the bigger cone (H) = 30cm
By cutting a small cone from it, then volume of smaller cone = $$\frac { 1 }{ 27 }$$ of volume of big cone

Let radius and height of the smaller cone be r and h
and radius and height of the bigger cone be R and H.

Hence at the height of 20cm from the base it was cut off. Ans.

Question 17.
Solution:
Height of the cylinder (h) = 10cm.
Height of the cone = 10cm

Question 18.
Solution:
Diameter of conical vessel = 40cm
Radius (r) = $$\frac { 40 }{ 2 }$$ = 20cm
and depth (h) = 24cm.
.’. Volume = $$\frac { 1 }{ 3 }$$ πr2h

Hope given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11C.

Other Exercises

Question 1.
Solution:
In cyclic quad. ABCD, ∠ DBC = 60° and ∠BAC = 40°
∴∠ CAD and ∠ CBD are in the same segment of the circle.
∴∠ CAD = ∠ CBD or ∠ DBC
= 40° + 60° = 100°
(Sum of opposite angles)
=> 100° + ∠BCD = 180°
=> ∠BCD = 180° – 100°
∴ ∠ BCD = 80°
Hence (i) ∠BCD = 80° and

Question 2.
Solution:
In the figure, POQ is diameter, PQRS is a cyclic quad, and ∠ PSR =150° In cyclic quad. PQRS.
∠ PSR + ∠PQR = 180°
(Sum of opposite angles)
=> 150° + ∠PQR = 180°
=> ∠PQR = 180°- 150° = 30°
=> ∠PQR =180° – 150° = 30°
Now in ∆ PQR,
∴∠ PRQ = 90° (Angle in a semicircle)
∴∠ RPQ + ∠PQR = 90°
=> ∠RPQ + 30° = 90°
=> ∠RPQ = 90° – 30° = 60° Ans.

Question 3.
Solution:
AB || DC and ∠BAD = 100°
(co-interior angles)

=> ∠ ADC + 100° = 180°
=> ∠ADC = 180° – 100° = 80°
∴ ABCD is a cyclic quadrilateral.
∴ ∠BAD + ∠BCD = 180°
=> 100° + ∠ BCD = 180°
=> ∠BCD = 180° – 100°
=> ∠BCD = 80°
Similarly ∠ABC + ∠ADC = 180°
=> ∠ABC + 80° = 180°
=> ∠ABC = 180° – 80° = 100°
Hence (i) ∠BCD = 80° (ii) ∠ADC = 80° and (iii) ∠ABC = 100° Ans.

Question 4.
Solution:
O is the centre of the circle and arc ABC subtends an angle of 130° at the centre i.e. ∠AOC = 130°. AB is produced to P
Reflex ∠AOC = 360° – 130° = 230°
Now, arc AC subtends reflex ∠ AOC at the centre and ∠ ABC at the remaining out of the circle.

Question 5.
Solution:
In the figure, ABCD is a cyclic quadrilateral in which BA is produced to F and AE is drawn parallel to CD.
∠ABC = 92° and ∠FAE = 20°

Question 6.
Solution:
In the figure, BD = DC and ∠CBD = 30°
In ∆ BCD,
BD = DC (given)
∠ BCD = ∠ CBD
(Angles opposite to equal sides)
= 30°
But ∠BCD + ∠CBD + ∠BDC = 180° (Angles of a triangle)
=> 30°+ 30°+ ∠BDC = 180°
=> 60°+ ∠BDC = 180°
=> ∠ BDC =180° – 60° = 120°
But ABDC is a cyclic quadrilateral
∠BAC + ∠BDC = 180°
=> ∠BAC + 120°= 180°
=> ∠ BAC = 180° – 120° = 60°
Hence ∠ BAC = 60° Ans.

Question 7.
Solution:
(i) Arc ABC subtends ∠ AOC at the centre , and ∠ ADC at the remaining part of the circle.
∠ AOC = 2 ∠ ADC

Question 8.
Solution:
In the figure, ABC is an equilateral triangle inscribed is a circle
Each angle is of 60°.
∠ BAC = ∠ BDC
(Angles in the same segment)
∠BDC = 60°
∠BDC + ∠BEC = 180°
=> 60°+ ∠BEC = 180°
=> ∠BEC = 180° – 60°= 120°
Hence ∠BDC = 60° and ∠BEC = 120° Ans.

Question 9.
Solution:
(opposite angles of a cyclic quad.)
so ∠BAD = 180° – 100° = 80°
Now in ∆ ABD,

=> 80° + 50° + ∠ADB = 180°
=> ∠ADB = 180° – 130° = 50°

Question 10.
Solution:
Arc BAD subtends ∠ BOD at the centre and ∠BCD at the remaining part of the circle.

Question 11.
Solution:
In ∆ OAB,
OA = OB (radii of the same circle)
∠OAB = ∠OBA = 50°
and Ext ∠BOD = ∠OAB + ∠OBA
=>x° = 50° + 50° – 100°
(opposite angles of a cyclic quad.)
=> 50°+ y° = 180°
=> y° = 180° – 50° = 130°
Hence x = 100° and y = 130° Ans.

Question 12.
Solution:
Sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively.
∠CBF = 130°, ∠CDE = x.
∠CBF + ∠CBA = 180° (Linear pair)
=> 130°+ ∠CBA = 180°
=> ∠CBA = 180° – 130° = 50°
But Ext. ∠ CDE = Interior opp. ∠ CBA (In cyclic quad. ABCD)
=> x = 50° Ans.

Question 13.
Solution:
In a circle with centre O AB is its diameter and DO || CB is drawn. ∠BCD = 120°
To Find : (i) ∠BAD (ii) ABD
(v) Show that ∆ AOD is an equilateral triangle.
(i) ABCD is a cyclic quadrilateral.

Question 14.
Solution:
AB = 6cm, BP = 2cm, DP = 2.5cm
Let CD = xcm
Two chords AB and CD

Question 15.
Solution:
O is the centre of the circle
∠ AOD = 140° and ∠CAB = 50°
BD is joined.
(i) ABDC is a cyclic quadrilateral.

Question 16.
Solution:
Given : ABCD is a cyclic quadrilateral whose sides AB and DC are produced to meet each other at E.
To Prove : ∆ EBC ~ ∆ EDA
Proof : In ∆ EBC and ∆ EDA
∠ E = ∠ E (common)
{Exterior angle of a cyclic quad, is equal to its interior opposite angle}
and ∠ EBC = ∠EDA
∆ EBC ~ ∆ EDA (AAS axiom)
Hence proved

Question 17.
Solution:
Solution Given : In an isosceles ∆ ABC, AB = AC
A circle is drawn x in such a way that it passes through B and C and intersects AB and AC at D and E respectively.
DE is joined.
To Prove : DE || BC
Proof : In ∆ ABC,
AB = AC (given)
∠ B = ∠ C (angles opposite to equal sides)
But ∠ ADE = ∠ C (Ext. angle of a cyclic quad, is equal E to its interior opposite angle)
But, these are corresponding angles
DE || BC.
Hence proved.

Question 18.
Solution:
Given : ∆ ABC is an isosceles triangle in which AB = AC.
D and E are midpoints of AB and AC respectively.
DE is joined.
To Prove : D, B, C, E are concyclic.
Proof: D and E are midpoints of sides AB and AC respectively.
DE || BC
In ∆ ABC, AB = AC
∠B = ∠C
But ∠ ADE = ∠ B (alternate angles)
But ∠ADE is exterior angle of quad. DBCE which is equal to its interior opposite angle C.
Hence D, B, C, E are con cyclic.
Hence proved.

Question 19.
Solution:
Given : ABCD is a cyclic quadrilateral whose perpendicular bisectors l, m, n, p of the side are drawn

To prove : l, m, n and p are concurrent.
Proof : The sides AB, BC, CD and DA are the chords of the circle passing through the vertices’s of quad. A, B, C and D. and perpendicular bisectors of a chord always passes through the centre of the circle.
l,m, n and p which are the perpendicular bisectors of the sides of cyclic quadrilateral will pass through O, the same point Hence, l, m, n and p are concurrent.
Hence proved.

Question 20.
Solution:
Given : ABCD is a rhombus and four circles are drawn on the sides AB, BC, CD and DA as diameters. Diagonal AC and BD intersect each other at O.

Question 21.
Solution:
Given: ABCD is a rectangle whose diagonals AC and BD intersect each other at O.
To prove : O is the centre of the circle passing through A, B, C and D

Question 22.
Solution:
Construction.
(i) Let A, B and C are three points
(ii) With A as centre and BC as radius draw an arc
(iii) With centre C, and radius AB, draw another arc which intersects the first arc at D.
D is the required point.
Join BD and CD, AC and BA and CB

BC = BC (common)
AC = BD (const.)
AB = DC
∴ ∆ ABC ≅ ∆ DBC (SSS axiom)
∴ ∠BAC ≅ ∠BDC (c.p.c.t.)
But these are angles on the same sides of BC
Hence these are angles in the same segment of a circle
A, B, C, D are concyclic Hence D lies on the circle passing througtvA, B and C.
Hence proved.

Question 23.
Solution:
Given : ABCD is a cylic quadrilateral (∠B – ∠D) = 60°
To prove : The small angle of the quad, is 60°

Question 24.

Solution:
To prove : A, B, C and D lie on a circle

Question 25.
Solution:
Given : In the figure, two circles intersect each other at D and C
∠BAD = 75°, ∠DCF = x° and ∠DEF = y°

Question 26.
Solution:
Given : ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect at O at right angle.

Question 27.
Solution:
In a circle, two chords AB and CD intersect each other at E when produced.

Question 28.
Solution:
Given : Two parallel chords AB and CD of a circle BD and AC are joined and produced to meet at E.

Question 29.
Solution:
Given : In a circle with centre O, AB is its diameter. ADE and CBE are lines meeting at E such that ∠BAD = 35° and ∠BED = 25°.
To Find : (i) ∠DBC (ii) ∠DCB (iii) ∠BDC
Solution. Join BD and AC,

Hope given RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11C are helpful to complete your math homework.

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## RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A.

Other Exercises

Question 1.
Solution:
We know that sum of angles of a quadrilateral is 360°
Now, sum of three angles = 56° + 115° + 84° = 255°
Fourth angle = 360° – 255° = 105° Ans.

Question 2.
Solution:
Sum of angles of a quadrilateral = 360°
Their ratio = 2 : 4 : 5 : 7
Let first angle = 2x
then second angle = 4x
third angle = 5x
and fourth angle = 7x
∴ 2x + 4x + 5x + 7x = 360°
=> 18x = 360°
=> x = $$\frac { { 360 }^{ o } }{ 18 }$$ = 20°
Hence, first angle = 2x = 2 x 20° = 40°
Second angle = 4x = 4 x 20° = 80°
Third angle = 5x = 5 x 20° = 100°
and fourth angle = 7x = 7 x 20° = 140°Ans.

Question 3.
Solution:
In the trapezium ABCD
DC || AB
∴ ∠ A + ∠ D = 180° (Co-intericr angles)
∴ 55°+ ∠D = 180°
∠D = 180° – 55°
∴ ∠D = 125°
Similarly, ∠B + ∠C = 180°
(Co-interior angles)
=> 70° + ∠C = 180°
=> ∠C = 180° – 70°
∠C = 110°
Hence ∠C = 110° and ∠D = 125° Ans.

Question 4.
Solution:
Given : In the figure, ABCD is a square and ∆ EDC is an equilateral triangles on DC. AE and BE are joined.
To Prove : (i) AE = BE
(ii) ∠DAE = 15°

Question 5.
Solution:
Given : In the figure,
BM ⊥ AC, DN ⊥ AC.
BM = DN

Question 6.
Solution:
AB = AD and BC = DC
AC is joined.
To Prove : (i) AC bisects ∠ A and ∠ C
(ii) BE = DE
Const. Join BD.
Proof : (i) In ∆ ABC and ∆ ADB

Question 7.
Solution:
Given : In square ABCD,
∠ PQR = 90°
PB = QC = DR

Question 8.
Solution:
Given : In quadrilateral ABCD, O is any point inside it. OA, OB, OC and OD are joined.
To Prove : OA + OB + OC + OD > AC + BD

Question 9.
Solution:
Given : In quadrilateral ABCD, AC is its one diagonal.

Question 10.
Solution: