RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS

RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions MCQS

Other Exercises

Choose the correct answer in each of the following questions.
Question 1.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 1

Question 2.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 1

Question 3.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 3

Question 4.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 4

Question 5.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 5

Question 6.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 6

Question 7.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 7

Question 8.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 8
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 9

Question 9.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 10

Question 10.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 11

Question 11.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 12

Question 12.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 13
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 14

Question 13.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 15

Question 14.
Solution:
Sum of first n odd natural numbers = (n)²
Sum of first 20 odd natural numbers = (20)² = 400 (c)

Question 15.
Solution:
First 40 positive integers divisible by 6 are 6, 12, 18, 24, … to 40 terms
Here, a = 6, d = 6, n = 40
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 16
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 17

Question 16.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 18

Question 17.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 19

Question 18.
Solution:
In an AP,
a18 – a14 = 32
Let a be the first term and d be the common difference, then
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 20

Question 19.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 21

Question 20.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 22

Question 21.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 23
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 24

Question 22.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 25

Question 23.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 26

Question 24.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 27
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 28

Question 25.
Solution:
Sum of first 16 terms of AP 10, 6, 2, …
Here, a = 10, d = 6 – 10 = -4, n = 16
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 29

Question 26.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 30
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 31

Question 27.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 32

Question 28.
Solution:
In an AP, T17 = T10 + 21
Let a be the first term and d be the common difference, then
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 33

Question 29.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 34

Question 30.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 35

Hope given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions MCQS 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 10 Solutions Chapter 10 Quadratic Equations Ex 10C

RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10C.

Other Exercises

Find the discriminant of each of the following equations:
Question 1.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 1
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 2
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 3

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:
Question 2.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 4

Question 3.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 5
x = 3 + √5 or x = 3 – √5

Question 4.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 6

Question 5.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 7

Question 6.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 8
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 9

Question 7.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 10

Question 8.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 11

Question 9.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 12

Question 10.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 13
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 14

Question 11.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 15

Question 12.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 16
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 17

Question 13.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 18

Question 14.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 19
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 20

Question 15.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 21

Question 16.
Solution:
x² + x + 2 = 0
Comparing it with ax² + bx + c = 0
a = 1, b = 1, c = 2
D = b² – 4ac = (1)² – 4 x 1 x 2 = 1 – 8 = -7
D < 0
There are no real roots.

Question 17.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 22

Question 18.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 23
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 24

Question 19.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 25

Question 20.
Solution:
3x² – 2x + 2 = 0
Comparing it with ax² + bx + c = 0
a = 3, b = -2, c = 2
D = b² – 4ac = (-2 )² – 4 x 3 x 2 = 4 – 24 = -20
D < 0
Roots are not real.

Question 21.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 26

Question 22.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 27

Question 23.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 28

Question 24.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 29
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 30

Question 25.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 31

Question 26.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 32

Question 27.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 33
x = (a + 2b) and x = (a – 2b)

Question 28.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 34

Question 29.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 35

Question 30.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 36

Question 31.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 37

Question 32.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 38
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 39

Question 33.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 40

Question 34.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 41
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 42

Question 35.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 43
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 44

Question 36.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 45
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10C 46

Hope given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10C 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 10 Solutions Chapter 10 Quadratic Equations Ex 10E

RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10E.

Other Exercises

PROBLEMS ON NUMBERS
Question 1.
Solution:
Let natural number be = x
According to the condition,
x² + x = 156
⇒ x² + x – 156 = 0
⇒ x² + 13x – 12x – 156 = 0
⇒ x (x + 13) – 12 (x + 13) = 0
⇒ (x + 13) (x – 12) = 0
Either, x + 13 = 0, then x = -13 but it is not a natural number
or x – 12 = 0, then x = 12
Hence, required number = 12

Question 2.
Solution:
Let the natural number be x,
Then according to the condition,
x + √x = 132
Let √x = y, then
y² + y = 132
⇒ y² + y – 132 = 0
⇒ y² + 12y – 11y – 132 = 0
⇒ y (y + 12) – 11 (y + 12) = 0
⇒ (y + 12) (y – 11) = 0
Either, y + 12 = 0, then y = -12 but it is not a natural number.
or y – 11 = 0 ⇒ y = 11
√x = 11 ⇒ x = 121
Hence, required number =121

Question 3.
Solution:
Sum of two natural numbers = 28
Let first number = x
Then second number = 28 – x
According to the condition,
x (28 – x) = 192
⇒ 28x – x² = 192
⇒ x² – 28x + 192 = 0
⇒ x² – 16x – 12x + 192 = 0
⇒ x (x – 16) – 12 (x – 16) = 0
⇒ (x – 16)(x – 12) = 0
Either, x – 16 = 0, then x = 16
or x – 12 = 0, then x = 12
Required natural numbers are 16 and 12

Question 4.
Solution:
Let first integer = x
Then second = x + 1
According to the condition,
(x)² + (x + 1)² = 365
⇒ x² + x² + 2x + 1 = 365
⇒ 2x² + 2x + 1 – 365 = 0
⇒ 2x² + 2x – 364 = 0
⇒ x² + x – 182 = 0
⇒ x² + 14x – 13x – 182 = 0
⇒ x(x + 14) – 13(x + 14) = 0
⇒ (x + 14)(x – 13) = 0
Either, x + 14 = 0, then x = -14 but it is negative.
or x – 13 = 0, then x = 13
First number = 13
and second number = 13 + 1 = 14
Hence, the numbers are 13, 14

Question 5.
Solution:
Let first odd number = x
Then second consecutive odd number = x + 2
According to the condition,
x² + (x + 2)² = 514
⇒x² + x² + 4x + 4 = 514
⇒ 2x² + 4x + 4 – 514 = 0
⇒ 2x² + 4x – 510 = 0
⇒ x² + 2x – 255 = 0
⇒ x² + 17x – 15x – 255 = 0
⇒ x (x + 17) – 15 (x + 17) = 0
⇒ (x + 17) (x – 15) = 0
Either, x + 17 = 0, then x = -17 but it is not positive.
or x – 13 = 0, then x = 13
First odd number = 13
and second number = 13 + 2 = 15
Hence, the number are 15, 17

Question 6.
Solution:
Let first positive even number = x
Then second even number = x + 2
According to the condition, x² + (x + 2)² = 452
⇒ x² + x² + 4x + 4 = 452
⇒ 2x² + 4x + 4 – 452 = 0
⇒ 2x² + 4x – 448 = 0
⇒ x² + 2x – 224 = 0
⇒ x² + 16x – 14x – 224 = 0
⇒ x (x + 16) – 14 (x + 16) = 0
⇒ (x + 16) (x – 14) = 0
Either, x + 16 = 0, then x = 16 which is not positive.
or x – 14 = 0, then x = 14
First even number = 14
and second number = 14 + 2 = 16
Hence, the numbers are 14, 16

Question 7.
Solution:
Let first positive integer = x
Then second integer = x + 1
According to the condition,
x (x + 1) = 306
⇒ x² + x – 306 = 0
⇒ x² + 18x – 17x – 306 = 0
⇒ x (x + 18) – 17 (x + 18) = 0
⇒ (x + 18)(x – 17) = 0
Either, x + 18 = 0, then x = -18 which is not positive.
or x – 17 = 0, then x = 17
First number =17
and second integer = 17 + 1 = 18
Hence, the numbers are 17, 18

Question 8.
Solution:
Let first natural number = x
Then second number = x + 3
According to the condition,
x (x + 3) = 504
⇒ x² + 3x – 504 = 0
⇒ x² + 24x – 21x – 504 = 0
⇒ x(x + 24) – 21(x + 24) = 0
⇒ (x + 24)(x – 21) = 0
Either x + 24 = 0, then x = -24 but it is not positive.
or x – 21 =0, then x = 21
First natural number = 21
and second number = 21 + 3 = 24
Hence, two required numbers are 21, 24

Question 9.
Solution:
Let first multiple of 3 = 3x
and second = 3x + 3
According to the condition,
3x (3x + 3) = 648
⇒ 9x² + 9x – 648 = 0
⇒ x² + x – 72 = 0
⇒ x² + 9x – 8x – 72 = 0
⇒ x (x + 9) – 8 (x + 9) = 0
⇒ (x + 9)(x – 8) = 0
Either, x + 9 = 0, then x = -9 which is negative.
or x – 8 = 0, then x = 8
First multiple of 3 = 3 x 8 = 24
and second = 24 + 3 = 27
Hence, numbers are 24, 27

Question 10.
Solution:
Let first positive odd integer = x
Then second = x + 2
According to the condition,
x (x + 2) = 483
⇒ x² + 2x – 483 = 0
⇒ x² + 23x – 21x – 483 = 0
⇒ x (x + 23) – 21(x + 23) = 0
⇒ (x + 23)(x – 21) = 0
Either, x + 23 = 0, then x = -23 which is negative.
or x – 21 =0, then x = 21
First odd number = 21
and second = 21 + 2 = 23
Hence, numbers are 21, 23

Question 11.
Solution:
Let first positive even integer = 2x
Then second even integer = 2x + 2
According to the condition,
2x (2x + 2) = 288
⇒ 4x² + 4x – 288 = 0
⇒ x² + x – 12 = 0
⇒ x² + 9x – 8x – 72 = 0
⇒ x (x + 9) – 8 (x + 9) = 0
⇒ (x + 9) (x – 8) = 0
Either, x + 9 = 0, then x = -9 which is negative.
or x – 8 = 0, then x = 8
First even integer = 2 x 8 = 16
and second even integer = 16 + 2 = 18
Hence, required integers are 16, 18

Question 12.
Solution:
Sum of two natural numbers = 9
Let first number = x
Then second number = 9 – x
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 1

Question 13.
Solution:
Sum of two natural numbers = 15
Let first number = x
Then second number = 15 – x
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 2

Question 14.
Solution:
Difference of two natural numbers = 3
Let first number = x
Then second number = x + 3
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 3
Either, x + 7 = 0, then x = -7 which is negative.
or x – 4 = 0, then x = 4.
First number = 4
and second number = 4 + 3 = 7
Hence, the numbers are 4, 7

Question 15.
Solution:
Difference of two natural numbers = 5
Let first number = x
Then second number = x – 5
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 4
⇒ x (x – 7) + 2 (x – 7) = 0
⇒ (x – 7) (x – 2) = 0
Either, x – 7 = 0, then x = 7
or x + 2 = 0, then x = -2 which is negative
First number = 7
and second number = 7 – 5 = 2
Hence, numbers are 7, 2

Question 16.
Solution:
Let first number = 7x
Then second number = 7x + 7
According to the condition,
(7x)² + (7x + 7)² = 1225
⇒ 49x² + 49x² + 49 + 98x = 1225
⇒ 98x² + 98x + 49 – 1225 = 0
⇒ 98x² + 98x – 1176 = 0
⇒ x² + x – 12 = 0 (Dividing by 98)
⇒ x² + 4x – 3x – 12 = 0
⇒ x (x + 4) – 3 (x + 4) = 0
⇒ (x + 4) (x – 3) = 0
Either, x + 4 = 0, then x = -4 which is negative.
or x – 3 = 0, then x = 3
First number 3 x 7 = 21
and second number = 21 + 7 = 28
Hence, numbers are 21, 28

Question 17.
Solution:
Let the natural number = x
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 5
⇒ 8x² – 64x – x + 8 = 0
⇒ 8x (x – 8) – 1 (x – 8) = 0
⇒ (x – 8) (8x – 1) = 0
Either, x – 8 = 0, then x = 8
or 8x – 1 = 0, then 8x = 1 ⇒ x = \(\frac { 1 }{ 8 }\) which is not a natural number.
Required natural number = 8

Question 18.
Solution:
Let first part = x
Then second part = 57 – x
According to the condition,
x (57 – x) = 680
⇒ 57x – x² = 680
⇒ x² – 57x + 680 = 0
⇒ x² – 40x – 17x + 680 = 0
⇒ x (x – 40) – 17 (x – 40) = 0
⇒ (x – 40) (x – 17) = 0
Either, x – 40 = 0, then x = 40
or x – 17 = 0, then x = 17
If first part is 40, then second part = 57 – 40 = 17
Parts are 40, 17 or 17, 40

Question 19.
Solution:
Let first part = x
Then second part = 27 – x
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 6
⇒ 81x – 3x² = 540
⇒ 3x² – 81x + 540 = 0
⇒ x² – 27x + 180 = 0
⇒ x² – 12x – 15x + 180 = 0
⇒ x(x – 12) – 15(x – 12) = 0
⇒ (x – 12)(x – 15) = 0
Either, x – 12 = 0, then x = 12
or x – 15 = 0, then x = 15
If first part is 12, then
second part = 27 – 12 = 15
Parts are 12, 15 or 15, 12

Question 20.
Solution:
Let first part (larger) = x
Then second part = 16 – x
According to the condition,
2(x²) – (16 – x)² = 164
⇒ 2x² – (256 + x² – 32x) = 164
⇒ 2x² – 256 – x² + 32x – 164 = 0
⇒ x² + 32x – 420 = 0
⇒ x² + 42x – 10x – 420 = 0
⇒ x (x + 42) – 10 (x + 42) = 0
⇒ (x + 42) (x – 10) = 0
either x + 42 = 0, then x = -42 which is negative.
or x – 10 = 0, then x = 10
First part =10
and second part =16 – 10 = 6
Hence, parts are 10 and 6

Question 21.
Solution:
Let a and b be the two natural numbers.
According to the condition,
a² + b² = 25(a + b) and a² + b² = 50(a – b)
25(a + b) = 50(a – b)
⇒ a + b = 2(a – b)
⇒ a + b = 2a – 2b
⇒ 2a – a = 2b + b
⇒ a = 3b
Now, a² + b² = 25(a + b)
⇒ (3b)² + b² = 25(3b + b)
⇒ 9b² + b² = 25 x 4b
⇒ 10b² = 100b
⇒ b² = 10b
⇒ b² – 10b = 0
⇒ b(b – 10) = 0
Either, b = 0 which is not possible
or b – 10 = 0, then b = 10
a = 3b = 3 x 10 = 30
Numbers are 30 and 10

Question 22.
Solution:
Let the two natural numbers be a and b.
According to the condition,
a² – b² = 45 and b² = 4a
⇒ a² – 4a = 45
⇒ a² – 4a – 45 = 0
⇒ a² – 9a + 5a – 45 = 0
⇒ a(a – 9) + 5 (a – 9) = 0
⇒ (a – 9) (a + 5) = 0
Either, a – 9 = 0, then a = 9
or a + 5 = 0, then a = -5 which is not a natural number.
a = 9
and b² = 4a = 4 x 9 = 36 = (6)²
b = 6
Hence, numbers are 9, 6

Question 23.
Solution:
Let three consecutive positive integers be x, x + 1 and x + 2
According to the condition,
x² + (x + 1) (x + 2) = 46
⇒ x² + x² + 3x + 2 – 46 = 0
⇒ 2x² + 3x – 44 = 0
⇒ 2x² – 8x + 11x – 44 = 0
⇒ 2x (x – 4) + 11 (x – 4) = 0
⇒ (x – 4) (2x + 11) = 0
Either, x – 4 = 0, then x = 4
or 2x + 11 = 0, then 2x = -11 ⇒ x = \(\frac { -11 }{ 2 }\)
But it is not a natural number.
x = 4
Hence, three consecutive numbers are 4, 5, 6.

Question 24.
Solution:
Let the digits of a two digit number be x and y
Let one’s digit = x
and ten’s digit = y
Number = x + 10y
According to the condition,
x + 10y = 4(x + y)
and x + 10y = 2xy
Now, x + 10y = 4x + 4y
⇒ 10y – 4y = 4x – x
⇒ 3x = 6y
⇒ x = 2y …(i)
x + 10y = 2xy
⇒ 2y + 10y = 2 x 2y x y
⇒ 12y = 4y²
⇒ 3y = y²
⇒ y² – 3y = 0
⇒ y (y – 3) = 0
Either, y = 0, which is not possible,
or y – 3 = 0, then y = 3
x = 2y = 2 x 3 = 6
and number = x + 10y = 6 + 10 x 3 = 6 + 30 = 36

Question 25.
Solution:
Let the ones digit of a two digit number = x
and ten’s digit = y
Number = x + 10y
By reversing the digits,
the One’s digit = y
and ten’s digit = x
Number = y + 10x
According to the condition,
x + 10y + 45 = y + 10x
⇒ 10y – y + x – 10x = -45
9y – 9x = -45
⇒ x – y = 5
⇒ x = y + 5
and xy = 14
⇒ (y + 5) x y = 14
⇒ y² + 5y – 14 = 0
⇒ y² + 7y – 2y – 14 = 0
⇒ y (y + 7) – 2(y + 7) = 0
⇒ (y + 7)(y – 2) = 0
Either, y + 7 = 0, then x = -7 which is negative, so not possible.
or y – 2 = 0, then y = 2
x = y + 5 = 2 + 5 = 7
Number = x + 10y = 7 + 10 x 2 = 7 + 20 = 27

Question 26.
Solution:
Let numerator = x
Then denominator = x + 3
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 7
⇒ x² + 5x – 2x – 10 = 0
⇒ x (x + 5) – 2 (x + 5) = 0
⇒ (x + 5)(x – 2) = 0
Either x + 5 = 0, then x = -5 which not possible being negative,
or x – 2 = 0, then x = 2
Fraction = \(\frac { x }{ x + 3 }\) = \(\frac { 2 }{ 2 + 3 }\) = \(\frac { 2 }{ 5 }\)

Question 27.
Solution:
Let the denominator of a fraction = x
Then its numerator = x – 3
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 8

Question 28.
Solution:
Let the number = x
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 9

SOME GENERAL PROBLEMS
Question 29.
Solution:
Let number of students = x
According to the given condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 10
x² = (x + 25)(x – 24)
x² = x² + (25 – 24) x + 25 x (-24)
x² = x² + x – 600
x = 600
Number of students = 600

Question 30.
Solution:
Total number of apples = 300
Let number of students = x
Then according to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 11

Question 31.
Solution:
Total number of marks = 40
Let marks in Mathematics = x
Then marks in English = 40 – x
According to the given condition,
(x + 3)(40 – x – 4) = 360
(x + 3)(36 – x) = 360
⇒ 36x – x² + 108 – 3x = 360
⇒ -x² + 33x + 108 – 360 = 0
⇒ -x² + 33x – 252 = 0
⇒ x² – 33x + 252 = 0
⇒ x² – 21x – 12x + 252 = 0
⇒ x(x – 21) – 12(x – 21) = 0
⇒ (x – 21)(x – 12) = 0
Either, x – 21 =0, then x = 21
or x – 11 = 0, then x = 11
Marks obtain in Mathematic and English = (21 and 19) or (12 or 28)

Question 32.
Solution:
Total budget = ₹ 2000
Let number of students = x
5 students failed to attend
Remaining students = x – 5
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 12
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 13

Question 33.
Solution:
Let original price of each book = ₹ x
and total amount = ₹ 600
Commission in price = ₹ 5 per book
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 14
⇒ x² – 30x + 25x – 750 = 0
⇒ x(x – 30) + 25(x – 30) = 0
⇒ (x – 30)(x + 25) = 0
Either, x – 30 = 0, then x = 30
or x + 25 = 0, then x = -25 which is not possible being negative.
Original price of each book = ₹ 30
and number of books = \(\frac { 600 }{ 30 }\) = 20

Question 34.
Solution:
Total expenses on tour = ₹ 10800
Let number of days = x
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 15
⇒ x(x + 24) – 20(x + 24) = 0
⇒ (x + 24)(x – 20) = 0
Either, x + 24 = 0, then x = -24 which is not possible being negative,
or x – 20 = 0, then x = 20
Number of days of tour = 20 days

Question 35.
Solution:
Total marks obtained = 28
Let marks in Mathematics = x
Then marks in Science = 28 – x
According to the condition,
(x + 3)(28 – x – 4) = 180
⇒ (x + 3)(24 – x) = 180
⇒ 24x – x² + 72 – 3x = 180
⇒ -x² + 21x – 180 + 72 = 0
⇒ x² – 21x + 108 = 0
⇒ x² – 12x – 9x + 108 = 0
⇒ x(x – 12) – 9(x – 12) = 0
⇒ (x – 12)(x – 9) = 0
Either, x – 12 = 0, then x = 12
or x – 9 = 0, then x = 9
Marks obtained in Mathematics and Science = (12, 16) or (9, 19)

Question 36.
Solution:
Amount = ₹ 180
Let number of pens purchased = x
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 16
⇒ x(x + 15) – 12(x + 15) = 0
⇒ (x + 15) (x – 12) = 0
Either, x + 15 = 0, then x = -15 which is not possible being negative,
or x – 12 = 0, then x = 12
Number of pens purchased = 12

Question 37.
Solution:
S.P. of an article = ₹ 75
Gain % = C.P. of the article
Let gain % = x %
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 17
⇒ x² + 150x – 50x – 7500 = 0
⇒ x(x + 150) – 50(x + 150) = 0
⇒ (x + 150) (x – 50) = 0
Either, x + 150 = 0, then x = -150 which is not possible being negative,
or x – 50 = 0, then x = 50
Cost price of the article = ₹ 50

PROBLEMS ON AGES
Question 38.
Solution:
(i) Let present age of son = x years
His age 1 year ago = (x – 1) years
His father’s age = (x – 1) x 8
According to the condition,
(x – 1) x 8 + 1 = x2
⇒ 8x – 8 + 1 = x2
⇒ x² – 8x + 7 = 0
⇒ x² – x – 7x + 7 = 0
⇒ (x – 1)(x – 7) = 0
Either, x – 1 = 0, then x = 1 which is not possible.
or x – 7 = 0, then x = 7
Son’s age = 7 years
and his father’s age = (7)² = 49 years
(ii) Let the age of the son = x years
and the age of father = 3.5x years
According to the condition,
(x)² + (3.5x)² = 1325
⇒ x² + 12.25x² = 1325
⇒ 13.25x² = 1325
⇒ x² = 100
⇒ x = 10
Hence, the age of the son = 10 years
and the age of father 3.5 x 10 = 35 years

Question 39.
Solution:
Present age of Meena = x years
3 years ago her age = (x – 3) years
and 5 years hence, her age = (x + 5) years
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 18
(x – 7) (x + 3) = 0
Either, x – 7 = 0, then x = 7
or x + 3 = 0, then x = -3 which is not possible being negative.
Her age = 7 years

Question 40.
Solution:
Sum of ages of a boy and his brother = 25 years
Let age of the boy = x years
Then age of his brother = (25 – x) years
According to the condition,
x (25 – x) = 126
⇒ 25x – x² – 126 = 0
⇒ x² – 25x + 126 = 0
⇒ x² – 18x – 7x + 126 = 0
⇒ x (x – 18) – 7 (x – 18) = 0
⇒ (x – 18) (x – 7) = 0
Either, x – 18 = 0, then x = 18
or x – 7 = 0, then x = 7
Age of boy = 7 years or 18 years
and age of his brother = 18 years or 7 years.

Question 41.
Solution:
Let present age of Tanvy = x years
Then 5 years ago her age = x – 5 years
8 years hence, her age = x + 8 years
According to the condition,
(x – 5) (x + 8) = 30
⇒ x² + 3x – 40 = 30
⇒ x² + 3x – 40 – 30 = 0
⇒ x² – 10x + 7x – 70 = 0
⇒ x (x – 10) + 7 (x – 10) = 0
⇒ (x – 10) (x + 7) = 0
Either, x – 10 = 0, then x = 10
or x + 7 = 0, then x = -7 which is not possible being negative.
Her present age = 10 years

Question 42.
Solution:
2 years ago,
Let son’s age = x years
and his father’s age = 3x²
Present age of son = (x + 2) years
and his father’s age = 3x² + 2
3 years hence, son’s age will be = x + 2 + 3 = x + 5 years
and age of his father = 3x² + 2 + 3 = (3x² + 5) years
According to the condition,
3x² + 5 = 4(x + 5)
⇒ 3x² + 5 = 4x + 20
⇒ 3x² – 4x + 5 – 20 = 0
⇒ 3x² – 4x – 15 = 0
⇒ 3x² – 9x + 5x – 15 = 0
⇒ 3x (x – 37) + 5 (x – 3) = 0
⇒ (x – 3)(3x + 5) = 0
Either, x – 3 = 0, then x = 3
or 3x + 5 = 0, then 3x = -5 ⇒ x = \(\frac { -5 }{ 3 }\) which is not possible being negative.
x = 3
Son’s present age = 3 + 2 = 5 years
and his father’s age = 3x² + 2
= 3 x (3)² + 2
= 27 + 2 = 29 years

PROBLEMS ON TIME AND DISTANCE
Question 43.
Solution:
Distance travelled = 150 km
Let the average speed of the truck = x km/hr
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 19
⇒ x (x – 60) + 10 (x – 60) 0
⇒ (x – 60) (x + 10) = 0
Either, x – 60 = 0, then x = 60
or x + 10 = 0, then x = -10 which is not possible being negative.
Original speed of the truck = 60 km/h

Question 44.
Solution:
Distance covered = 1500 km
Let original speed of the plane = x km/h
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 20
⇒ x² + 600x – 500x – 300000 = 0
⇒ x(x + 600) – 500(x + 600) = 0
⇒ (x + 600) (x – 500) = 0
Either, x + 600 = 0, then x = -600 which is not possible being negative.
or x – 500 = 0, then x = 500
Original speed of the plane = 500 km/hr
Yes, his promptness was appreciable.

Question 45.
Solution:
Distance covered by a train = 480 km
Let uniform speed = x km/h
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 21
⇒ x² – 40x + 32x – 1280 = 0
⇒ x(x – 40) + 32(x – 40) = 0
⇒ (x – 40) (x + 32) = 0
Either, x – 40 = 0, then x = 40
or x + 32 = 0, then x = -32 which is not possible being negative.
Speed of the train = 400 km/hr

Question 46.
Solution:
Distance covered = 54 km and then 63 km
Let first average speed = x km/hr
and next speed = (x + 6) km/hr
Total time taken = 3 hours
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 22
⇒ x² – 33x – 108 = 0
⇒ x² – 36x – 3x – 108 = 0
⇒ x (x – 36) + 3 (x – 36) = 0
⇒ (x – 36) (x + 3) = 0
Either, x – 36 = 0, then x = 36
or x + 3 = 0, then x – 3 which is not possible being negative.
Speed of the train = 36 km/hr.

Question 47.
Solution:
Distance travelled by a train =180 km
Let the uniform speed = x km/hr
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 23
⇒ x² + 45x – 36x – 1620 = 0
⇒ x (x + 45) – 36 (x + 45) = 0
⇒ (x + 45) (x – 36) = 0
Either, x + 45 = 0, then x = -45 which is not possible being negative,
or (x – 36) = 0, then x = 36
Uniform speed of the train = 36 km/hr.

Question 48.
Solution:
Distance travelled = 300 km
Let the original speed of train be x km
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 24
⇒ 300x + 1500 – 300x = 2x² + 10x
⇒ 2x² + 10x – 1500 = 0
⇒ x² + 5x – 750 = 0 (Dividing by 2)
⇒ x² + 30x – 25x – 750 = 0
⇒ x (x + 30) – 25 (x + 30) = 0
⇒ (x + 30) (x – 25) = 0
Either, x + 30 = 0 ⇒ x = -30 (Rejected ∵ speed cannot be negative)
or x – 25 = 0 ⇒ x = 25
Speed of the train = 25 km/hr

Question 49.
Solution:
Distance = 300 km
Let the speed of the passenger = x km/hr
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 25
⇒ x (x + 30) – 25 (x + 30) = 0
⇒ (x + 30) (x – 25) = 0
Either, x + 30 = 0, then x = -30 which is not possible being negative,
or x – 25 = 0, then x = 25
Speed of the train = 25 km/hr.

Question 50.
Solution:
Distance between Mumbai and Pune = 192 km
Let the speed of Deccan Queen = x km/hr
and speed of another train = (x – 20) km/hr
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 26
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 27

Question 51.
Solution:
Distance travelled = 24 km up and down
Speed of motor boat in still water = 18 km/hr
Let speed of stream = x km/hr
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 28
⇒ x (x + 54) – 6 (x + 54) = 0
⇒ (x + 54) (x – 6) = 0
Either, x + 54 = 0, then x = -54 which is not possible being negative,
or x – 6 = 0, then x = 6
Speed of stream = 6 km/hr

Question 52.
Solution:
Speed of a boat in still water = 8 km/hr
Let speed of water in stream = x km/hr
Total time taken = 5 hours
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 29

Question 53.
Solution:
Speed of motorboat in still water = 9 km/hr
Let speed of water = x km/hr
Total time taken = 3 hours
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 30
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 31

PROBLEMS ON TIME AND WORK AND PIPES AND CISTERN
Question 54.
Solution:
Let B finishes the work in = x days
Then A will finish it in = (x – 10) days
According to the condition,
(A + B)’s one day’s work
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 32
⇒ x² – 30x – 4x + 120 = 0
⇒ x (x – 30) – 4 (x – 30) = 0
⇒ (x – 30)(x – 4) = 0
Either, x – 30 = 0, then x = 30
or x – 4 = 0, then x = 4
But x = 4 is not possible
x = 30
B can finish the work in 30 days
and A can in 30 – 10 = 20 days.

Question 55.
Solution:
2 pipes can fill a cistern in 3\(\frac { 1 }{ 13 }\) = \(\frac { 40 }{ 13 }\) minutes
Let one pipe can fill it in = x minute
Then second will fill it in = (x + 3) minutes
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 33
⇒ 13x (x – 5) + 24 (x – 5) = 0
⇒ (x – 5) (13x + 24) = 0
Either, x – 5 = 0, then x = 5
or 13x + 24 = 0
⇒ 13x = -24
⇒ x = \(\frac { -24 }{ 13 }\) which is not possible being negative.
x = 5
First pipe can fill the tank in 5 minutes
and second pipe will fill it in 5 + 3 = 8 minutes.

Question 56.
Solution:
Two pipes can fill a tank together in = 11\(\frac { 1 }{ 9 }\) = \(\frac { 100 }{ 9 }\) minutes
Let first pipe can fill it in = x minute
Then second pipe will fill it in = (x + 5) min
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 34
⇒ 9x² – 155x – 500 = 0
⇒ 9x² – 180x + 25x – 500 = 0
⇒ 9x (x – 20) + 25(x – 20) = 0
⇒ (x – 20) (9x + 25) = 0
Either, x – 20 = 0, then x = 20
or 9x + 25 = 0, then 9x = -25 ⇒ x = \(\frac { -25 }{ 9 }\)
But it is not possible being negative.
First pipe will fill it in 20 minutes
and second will fill in = 20 + 5 = 25 minutes-

Question 57.
Solution:
Two water taps together can fill a tank in = 6 hours
Let first tap of larger diameter can fill it in x hour
Then second tap will fill it in (x – 9) hour
According to the condition,
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 35

PROBLEMS ON AREA AND GEOMETRY
Question 58.
Solution:
Let breadth of a rectangle (b) = x cm
Then length (l) = 2x cm
Area = l x b = 2x x x
2x x x = 288
⇒ 2x² = 288
⇒ x² = 144 = (12)²
x = 12
Length = 2x = 2 x 12 = 24 cm
and breadth = x = 12 cm

Question 59.
Solution:
Let breadth of a rectangular field (b) = x m
Then its length (l) = 3x m
Area = l x b = 3x x x = 3x²
3x² = 147
⇒ x² = 49 = (7)²
x = 7
Length = 3x = 3 x 7 = 21 m
and breadth = x = 7 m

Question 60.
Solution:
Let breadth of a hall (b) = x m
Then its length (l) = (x + 3) m
Area = l x b = (x + 3) x x = x² + 3x
x² + 3x = 238
⇒ x² + 3x – 238 = 0
⇒ x² + 11x – 14x – 238 = 0
⇒ x (x + 17) – 14 (x – 17) = 0
⇒ (x + 17) (x – 14) = 0
Either, x + 17 = 0, then x = -17 which is not possible being negative.
or x – 14 = 0, then x = 14
Breadth = 14 m
and length = 14 + 3 = 17 m

Question 61.
Solution:
Perimeter of a rectangular plot = 62 m
and area = 228 m²
Let length of plot = x m
Then width = \(\frac { 62 }{ 2 }\) – x = 31 – x
Area = l x b = x (31 – x)
x (31 – x) = 228
⇒ 31x – x² = 228
⇒ x² – 31x + 228 = 0
⇒ x² – 19x – 12x + 228 = 0
⇒ x (x – 19) – 12 (x – 19) = 0
⇒ (x – 19) (x – 12) = 0
Either, x – 19 = 0, then x = 19
or x – 12 = 0, then x = 12
If length is 19 m, then breadth = 31 – 19 = 12 m
and if length is 12 m then breadth = 31 – 12 = 19 m
which is not possible as length > breadth
Dimensions are (19, 12) m

Question 62.
Solution:
Area of path around it = 120 m²
Length of a rectangular field (l) = 16 m
and breadth (b) = 10 m
Let a path of x m wide be made around it, then its
Outer length (L) = (16 + 2x) m
and breadth (B) = (10 + 2x) m
Area of path = L x B – l x b
= (16 + 2x) (10 + 2x) – 16 x 10
⇒ 160 + 52x + 4x² – 160 = 120
⇒ 4x² + 52x – 120 = 0
⇒ x² + 13x – 30 = 0 (Dividing by 4)
⇒ x² + 15x – 2x – 30 = 0
⇒ x (x + 15) – 2 (x + 15) = 0
⇒ (x + 15) (x – 2) = 0
Either, x + 15 = 0, then x = -15 which is not possible being negative,
or x – 2 = 0, then x = 2
Width of path = 2 m

Question 63.
Solution:
The sum of areas of two squares = 640 m²
and difference between their perimeters = 64 m
Let 4 sides of first square = x m
Then perimeter = 4x
Perimeter of second square = 4x – 64
and side = \(\frac { 4x – 64 }{ 4 }\) = (x – 16) m
Now sum of their areas,
x² + (x – 16)² = 640
⇒ x² + x² – 32x + 256 = 640
⇒ 2x² – 32x + 256 – 640 = 0
⇒ 2x² – 32x – 384 = 0
⇒ x² – 16x – 192 = 0
⇒ x² – 24x + 8x – 192 = 0
⇒ x (x – 24) + 8 (x – 24) = 0
⇒ (x – 24) (x + 8) = 0
Either, x – 24 = 0, then x = 24
or x + 8 = 0, then x = -8 which not possible being negative,
x = 24
Side of first square = 24 m
and side of second square = 24 – 16 = 8 m

Question 64.
Solution:
Let side of a square = x m
Then length of rectangle = 3x m
and width = (x – 4) m
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 36
Now area of square = x² m²
and area of rectangle = l x b = (3x) x (x – 4) m²
According to the condition,
x² = 3x (x – 4)
⇒ x² = 3x² – 12x
⇒ 3x² – 12x – x2 = 0
⇒ 2x² – 12x = 0
⇒ 2x (x – 6) = 0
⇒ x – 6 = 0
⇒ x = 6m
Side of square = 6 m
and length of rectangle = 3 x 6 m = 18m
and width = x – 4 = 6 – 4 = 2m

Question 65.
Solution:
Area of a rectangular garden = 180 sq m
Let length of garden = x m
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 37
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 38

Question 66.
Solution:
Area of a right triangle = 600 cm²
Let altitude = x cm
Then base = (x + 10) cm
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 39
Area = \(\frac { 1 }{ 2 }\) base x altitude
600 = \(\frac { 1 }{ 2 }\) x (x + 10) x
x² + 10x = 1200
⇒ x² + 10x – 1200 = 0
⇒ x² + 40x – 30x – 1200 = 0
⇒ x (x + 40) – 30 (x + 40) = 0
⇒ (x + 40) (x – 30) = 0
Either, x + 40 = 0, then x = -40 which is not possible being negative.
or x – 30 = 0, then x = 30
Altitude = 30 cm
Base = 30 + 10 = 40 cm
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 40

Question 67.
Solution:
Area of right triangle = 96 sq. m
Let altitude of the triangle = x m
Then base = 3x m
Area = \(\frac { 1 }{ 2 }\) x base x altitude
96 = \(\frac { 1 }{ 2 }\) x 3x x x
⇒ 3x² = 96 x 2
x² = 64 = (8)²
x = 8 m
Base = 3x = 3 x 8 = 24m
and altitude = x = 8 m

Question 68.
Solution:
Area of right triangle = 165 sq m
Let altitude = x m
Then base = – (x + 7) m
Area = \(\frac { 1 }{ 2 }\) x base x altitude
165 = \(\frac { 1 }{ 2 }\) x x x (x + 7)
330 = x² + 7x
⇒ x² + 7x – 330 = 0
⇒ x² + 22x – 15x – 330 = 0
⇒ x (x + 22) – 15 (x + 22) = 0
⇒ (x + 22) (x – 15) = 0
Either, x + 22 = 0, then x = -22 which is not possible being negative,
or x – 15 = 0, then x = 15
Altitude = x + 7 = 15 + 7 = 22m
and base = x = 15 m

Question 69.
Solution:
In a right triangle,
Hypotenuse = 20 m
Difference of other two sides = 4 m
Let first side = x m
Then second side = (x + 4) m
According to the condition,
(x)² + (x + 4)² = (20)²
⇒ x² + x² + 8x + 16 = 400
⇒ 2x² + 8x + 16 – 400 = 0
⇒ 2x² + 8x – 384 = 0
⇒ x² + 4x – 192 = 0
⇒ x² + 16x – 12x – 192 = 0
⇒ x (x + 16) – 12 (x + 16) = 0
⇒ (x + 16) (x – 12) = 0
Either, x + 16 = 0, then x = -16 which is not possible being negative,
or x – 12 = 0, then x = 12
First side = 12 m
Second side = x + 4= 12 + 4 = 16 m

Question 70.
Solution:
In a right triangle, let base = x m
Length of hypotenuse = (x + 2) m
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10E 41

Question 71.
Solution:
Let shorter side of a right triangle = x m
Then hypotenuse = (2x – 1) m
and longer side = (x + 1) m
According to the condition,
(2x – 1)² = x² + (x + 1)² (Use of Pythagoras Theorem)
⇒ 4x² – 4x + 1 = x² + x² + 2x + 1
⇒ 4x² – 4x + 1 – x² – x² – 2x – 1 = 0
⇒ 2x² – 6x = 0
⇒ 2x (x – 3) = 0
⇒ x – 3 = 0
⇒ x = 3
Shorter side = 3 m
Hypotenuse = 2x – 1 = 3 x 2 – 1 = 6 – 1 = 5 m
and longer side = x + 1 = 3 + 1 = 4m
Sides 3 m, 4 m, 5 m

Hope given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10E are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D

RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10D.

Other Exercises

Question 1.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 1
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 2
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 3

Question 2.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 4

Question 3.
Solution:
x² + px – q² = 0
Comparing it with ax² + bx + c = 0
a = 1, b = p, c = -q²
Discriminant (D) = b² – 4ac
= (p)² – 4 x 1 x (-q²)
= p² + 4 q²
p and q are of two powers
p² + 4q² is always greater than 0
The roots are real for all real values of p and q.

Question 4.
Solution:
The quadratic equation is 3x² + 2kx + 27 = 0
Comparing it with ax² + bx + c = 0
a = 3, b = 2k, c = 27
Discriminant (D) = b² – 4ac
= (2k)² – 4 x 3 x 27
= (2k)² – 324
Roots are real and equal
(2k)² – 324 = 0
⇒ (2k)² – (18)² = 0
⇒ (k)² – (9)² = 0
⇒ (k + 9) (k – 9) = 0
Either k + 9 = 0, then k = -9
or k – 9 = 0, then k = 9
Hence, k = 9, -9

Question 5.
Solution:
The quadratic equation is
kx (x – 2√5) x + 10 = 0
kx² – 2√5 kx + 10 = 0
Comparing it with ax² + bx + c = 0
a = k, b = -2√5 k, c = 10
D = b² – 4ac = (-2 k)² – 4 x k x 10 = 20k² – 40k
Roots are real and equal.
D = 0
20k² – 40k = 0
⇒ k² – 2k = 0
⇒ k (k – 2) = 0
Either, k = 0 or k – 2 = 0, then k = 2
k = 0, k = 2

Question 6.
Solution:
The quadratic equation is 4x² + px + 3 = 0
Comparing it with ax² + bx + c = 0
a = 4, b = p, c = 3
D = b² – 4ac = p² – 4 x 4 x 3 = p²- 48
Roots are real and equal.
D = 0
⇒ p² – 48 = 0
⇒ p² = 48 = (±4√3)²
⇒ P = ± 4√3
P = 4√3, p = -4√3

Question 7.
Solution:
The quadratic equation is 9x² – 3kx + k = 0
Comparing it with ax? + bx + c = 0
a = 9, b = -3k, c = k
D = b² – 4ac = (-3k)² – 4 x 9 x k = 9k² – 36k
Roots are real and equal.
D = 0
9k² – 36k = 0
⇒ 9k (k – 4) = 0
Either, k = 0 or k – 4 = 0, then k = 4
The value of k is non-zero.
k = 4

Question 8.
Solution:
(i) The equation is (3k + 1) x² + 2(k + 1) x + 1 = 0
Comparing it with ax² + bx + c = 0
a = (3k + 1), b = 2(k + 1), c = 1
D = b² – 4 ac
= [2(k + 1)]² – 4(3k + 1) x 1
= 4k² + 4 + 8k – 12k – 4
= 4k² – 4k
= 4k (k – 1)
Roots are real and equal.
Either, k = 0 or k – 1 = 0, then k = 1
k = 0, k = 1
(ii) x² + k(2x + k – 1) + 2 = 0
⇒ x² + 2kx + (k² – k + 2) = 0
Here, a = 1, b = 2k, c = (k² – k + 2)
Discriminant (D) = b² – 4ac = (2k)² – 4 x 1 x (k² – k + 2)
= 4k² – 4k² + 4k – 8
= 4k – 8
Roots are real and equal.
D = 0
⇒ 4k – 8 = 0
⇒ k = 2
Hence, k = 2

Question 9.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 5

Question 10.
Solution:
The given quadratic equation is
(p + 1) x² – 6(p + 1) x + 3(p + 9) = 0, p ≠ -1
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 6

Question 11.
Solution:
-5 is a root of 2x² + px – 15 = 0
x = -5 will satisfy it
Now, substituting the value of x = -5
⇒ 2(-5)² + p(-5) – 15 = 0
⇒ 50 – 5p – 15 = 0
⇒ 35 – 5p = 0
⇒ 5p = 35
⇒ P = 7
In quadratic equation p(x² + x) + k = 0
⇒ 7 (x² + x) + k = 0 (p = 7)
⇒ 7x² + 7x + k = 0
Comparing it with ax² + bx + c = 0
a = 7, b = 7, c = k
D = b² – 4ac = (7)² – 4 x 7 x k
= 49 – 28k
Roots are real and equal.
49 – 28k = 0
⇒ 28k = 49
k = \(\frac { 49 }{ 28 }\) = \(\frac { 7 }{ 4 }\)

Question 12.
Solution:
3 is a root of equation x² – x + k = 0
It will satisfy it
Now, substituting the value of x = 3 in it
(3)² – (3) + k = 0
⇒ 9 – 3 + k = 0
⇒ 6 + k = 0
⇒ k = -6
Now in the equation, x² + k (2x + k + 2) + p = 0
x² + (-6)(2x – 6 + 2) + p = 0
⇒ x² – 12x + 36 – 12 + p = 0
⇒ x² – 12x + (24 + p) = 0
Comparing it with ax² + bx + c = 0
a = 1, b = -12, c = 24 + p
D = b² – 4ac
= (-12)² – 4 x 1 x (24 + p)
= 144 – 96 – 4p = 48 – 4p
Roots are real and equal.
D = 0
48 – 4p = 0
⇒ 4p = 48
⇒ p = 12
Hence, p = 12

Question 13.
Solution:
-4 is a root of the equation x² + 2x + 4p = 0
Then it will satisfy the equation
Now, substituting the value of x = -4
(-4)² + 2(-4) + 4p = 0
16 – 8 + 4p = 0
⇒ 8 + 4p = 0
⇒ 4p = -8
⇒ p = -2
In the quadratic equation x² + px (1 + 3k) + 7(3 + 2k) = 0
⇒ x² – 2x (1 + 3k) + 7(3 + 2k) = 0
Comparing it with ax² + bx + c = 0
a = 1, b = -2 (1 + 3k), c = 7 (3 + 2k)
D = b² – 4ac
= [-2(1 + 3k)]² – 4 x 1 x 7(3 + 2k)
= 4(1 + 9k² + 6k) – 28(3 + 2k)
= 4 + 36k² + 24k – 84 – 56k
= 36k² – 32k – 80
Roots are equal.
D = 0
⇒ 36k² – 32k – 80 = 0
⇒ 9k² – 8k – 20 = 0
⇒ 9k² – 18k + 10k – 20 = 0
⇒ 9k (k – 2) + 10(k – 2) = 0
⇒ (k – 2) (9k + 10) = 0
Either, k – 2 = 0, then k = 2
or 9k + 10 = 0, then 9k = -10 ⇒ k = \(\frac { -10 }{ 9 }\)
k = 2, k = \(\frac { -10 }{ 9 }\)

Question 14.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 7

Question 15.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 8

Question 16.
Solution:
The quadratic equation is 2x² + px + 8 = 0
Comparing it with ax² + bx + c = 0
a = 2, b = p, c = 8
D = b2 – 4ac = p² – 4 x 2 x 8 = p² – 64
Roots are real.
D ≥ 0
p² – 64 ≥ 0
⇒ p² ≥ 64 ≥ (±8)²
p ≥ 8 or p ≤ -8

Question 17.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 9
Roots are equal
D = 0
⇒ 4(α – 12) (α – 14) = 0
⇒ α – 14 = 0 {(α – 12) ≠ 0}
⇒ α = 14
Hence, α = 14

Question 18.
Solution:
9x² + 8kx + 16 = 0
Comparing it with ax2 + bx + c = 0
a = 9, b = 8k, c = 16
D = b² – 4ac
= (8k)² – 4 x 9 x 16 = 64k² – 576
Roots are real and equal.
D = 0
64k² – 576 = 0
64k² = 576
⇒ k² = 9 = (±3)²
k = 3, k = -3

Question 19.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 10
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 11
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 12

Question 20.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 13
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 14

Question 21.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 15
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 16
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 17

Question 22.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10D 18

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RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B

RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 10 Quadratic Equations Ex 10B.

Other Exercises

Solve each of the following equations by using the method of completing the square:
Question 1.
Solution:
x² – 6x + 3 = 0
=> x² – 2 x 3 x x = -3
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 1

Question 2.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 2

Question 3.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 3

Question 4.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 4
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 5

Question 5.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 6
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 7

Question 6.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 8
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 9

Question 7.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 10
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 11

Question 8.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 12
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 13

Question 9.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 14

Question 10.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 15
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 16

Question 11.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 17
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 18
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 19

Question 12.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 20
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 21

Question 13.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 22
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 23

Question 14.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 24
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 25

Question 15.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 26
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 27

Question 16.
Solution:
RS Aggarwal Class 10 Solutions Chapter 10 Quadratic Equations Ex 10B 28

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RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D

RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4C.

Other Exercises

Question 1.
Solution:
For the given triangle to be right-angled, the sum of the squares of the two smaller sides must be equal to the square of the largest side.
(i) 9 cm, 16 cm, 18 cm
Longest side = 18
Now (18)² = 324
and (9)² + (16)² = 81 + 256 = 337
324 ≠ 337
It is not a right triangle.
(ii) 1 cm, 24 cm, 25 cm
Here longest side = 25 cm
(25)² = 625
and (7)² x (24)² = 49 + 576 = 625
625 = 625
It is a right triangle
(iii) 1.4 cm, 4.8 cm, 5 cm
Here longest side = 5 cm
(5)² = 25
and (1.4)² + (4.8)² = 1.96 + 23.04 = 25.00 = 25
25 = 25
It is a right triangle
(iv) 1.6 cm, 3.8 cm, 4 cm
Here longest side = 4 cm
(4 )² = 16
and (1.6)² + (3.8)² = 2.56 + 14.44 = 17.00 = 17
16 ≠ 17
It is not a right triangle
(v) (a- 1) cm, 2√a cm, (a + 1) cm
Here longest side = (a + 1) cm
(a + 1)² = a² + 2a + 1
and (a – 1)² + (2 √a )² = a² – 2a + 1 + 4a = a² + 2a + 1
a² + 2a + 1 = a² + 2a + 1
It is a right triangle.

Question 2.
Solution:
A man goes 80 m from O to east side and reaches A, then he goes 150 m due north from A and reaches B.
Join OB.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 1
In right ∆OAB,
OB² = OA²+² (Pythagoras Theorem) = (80)² + (150)² = 6400 + 22500 = 28900
⇒ OB = √28900 = 170
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 2
He is 170 m away from the starting point.

Question 3.
Solution:
A man goes 10 m due south from O and reaches A and then 24 m due west from A and reaches B.
Join OB.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 3

Question 4.
Solution:
Length of a ladder = 13 m
Height of the window = 12 m
Distance between the foot of the ladder and building.
In the figures,
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 4
AB is ladder, A is window of building AC
AB² = AC² + BC² (Pythagoras Theorem)
⇒ (13)² = (12)² + x²
⇒ 169 = 144 + x²
⇒ x² = 169 – 144 = 25 = (5)²
x = 5
Hence, distance between foot of ladder and building = 5 m.

Question 5.
Solution:
Let length of ladder AB = x m
Height of window AC = 20 m
and distance between the foot of the ladder and the building (BC) = 15 m
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 5
AB² = AC² + BC² (Pythagoras Theorem)
⇒ x² = 20² + 15² = 400 + 225 = 625 = (25)²
x = 25
Length of ladder = 25 m

Question 6.
Solution:
Height of first pole AB = 9 m
and of second pole CD = 14 m
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 6
Let distance between their tops CA = x m
From A, draw AE || BD meeting CD at E.
Then EA = DB = 12 m CE = CD – ED = CD – AB = 14-9 = 5 m
In right ∆AEC,
AC² = AE² + CE² = 122 + 52 = 144 + 25 = 169 = (13)²
AC = 13
Distance between their tops = 13 m

Question 7.
Solution:
Height of the pole AB = 18 m
and length of wire AC = 24 m
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 7
Distance between the base of the pole and other end of the wire
BC = x m (suppose)
In right ∆ABC,
AC² = AB² + BC² (Pythagoras Theorem)
(24)² = (18)² + x²
⇒ 576 = 324 + x²
⇒ x² = 576 – 324 = 252
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 8

Question 8.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 9
Solution:
In ∆PQR, O is a point in it such that
OP = 6 cm, OR = 8 cm and ∠POR = 90°
PQ = 24 cm, QR = 26 cm
To prove : ∆PQR is a right angled.
In ∆POR, ∠O = 90°
PR² = PO² + OR² = (6)² + (8)² = 36 + 64 = 100 = (10)²
PR = 10
Greatest side QR is 26 cm
QR² = (26)² = 676
and PQ² + PR² = (24)² + (10)² = 576 + 100 = 676
676 = 676
QR² = PQ² + PR²
∆PQR is a right angled triangle and right angle at P.

Question 9.
Solution:
In isosceles ∆ABC, AB = AC = 13 cm
AL is altitude from A to BC
and AL = 5 cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 10
Now, in right ∆ALB
AB² = AL² + BL²
(13)² = (5)² + BL²
⇒ 169 = 25 + BL²
⇒ BL² = 169 – 25 = 144 = (12)²
BL = 12 cm
L is mid point of BC
BC = 2 x BC = 2 x 12 = 24 cm

Question 10.
Solution:
In an isosceles ∆ABC in which
AB = AC = 2a units, BC = a units
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 11
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 12

Question 11.
Solution:
∆ABC is an equilateral triangle
and AB = BC = CA = 2a
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 13
AD ⊥ BC
D is mid point of BC
BD = DC = \(\frac { 1 }{ 2 }\) BC
= \(\frac { 1 }{ 2 }\) x 2a = a
Now, in right ∆ADB,
AB² = AD² + BD² (Pythagoras Theorem)
(2a)² = AD² + a²
⇒ 4a² – a² = AD²
⇒ AD² = 3a² = (√3 a)²
AD = √3 a = a√3 units

Question 12.
Solution:
∆ABC is an equilateral triangle in which
AB = BC = CA = 12 cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 14
AD ⊥ BC which bisects BC at D
BD = DC = \(\frac { 1 }{ 2 }\) BC = \(\frac { 1 }{ 2 }\) x 12 = 6cm
Now, in right ∆ADB,
AB² = AD² + BD²
⇒ (12)² = AD² + (6)²
⇒ 144 = AD² + 36
AD² = 144 – 36 = 108
AD = √108 = √(36 x 3) = 6√3 cm

Question 13.
Solution:
Let ABCD is a rectangle in which adjacent sides.
AB = 30 cm and BC = 16 cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 15
AC is its diagonal.
In right ∆ABC,
AC² = AB² + BC² (Pythagoras Theorem)
= (30)² + (16)² = 900 + 256 = 1156 = (34)²
Diagonal AC = 34 cm

Question 14.
Solution:
ABCD is a rhombus
Its diagonals AC and BD bisect each other at O.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 16
AO = OC = \(\frac { 1 }{ 2 }\) AC.
and BO = OD = \(\frac { 1 }{ 2 }\) BD
BD = 24 cm and AC = 10 cm
BO = \(\frac { 1 }{ 2 }\) x BD = \(\frac { 1 }{ 2 }\) x 24 = 12 cm
AO = \(\frac { 1 }{ 2 }\) x AC = \(\frac { 1 }{ 2 }\) x 10 = 5 cm
Now, in right ∆AOB,
AB² = AO² + BO² = (5)² + (12)² = 144 + 25 = 169 = (13)²
AB = 13
Hence, each side of rhombus = 13 cm

Question 15.
Solution:
Given : In ∆ABC, AC > AB.
D is the mid point of BC and AE ⊥ BC.
AD is joined.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 17
To prove: AB² = AD² – BC x DE + \(\frac { 1 }{ 4 }\) BC2
Proof: In ∆AEB, ∠E = 90°
AB² = AE² + BE² …..(i) (Pythagoras Theorem)
In ∆AED, ∠E = 90°
AD² = AE² + DE²
⇒ AE² = AD² – DE² …..(ii)
Now, substitute eq. (ii) in eq. (i)
AB² = AE² + BE²
AB² = AD² – DE² + BE² [from (ii)]
AB² = (AD² – DE²) + (BD – DE)² [BE = BD – DE²]
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 18

Question 16.
Solution:
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 19

Question 17.
Solution:
Given : In ∆ABC, D is the mid point of BC, AE ⊥ BC,
BC = a, AC = b, AB = c, ED = x, AD =p and AE =
AD is joined.
To prove :
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 20
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 21
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 22

Question 18.
Solution:
Given : In ∆ABC, AB =AC
BC is produced to D and AD is joined.
To prove : (AD² – AC²) = BD x CD
Construction : Draw AE ⊥ BC.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 23
Proof: In ∆ABC,
AB = AC and AE ⊥ BC
BE = EC
Now, in right ∆AED, ∠E = 90°
AD² = AE² + ED² …..(i)
and in right ∆AEC, ∠E = 90°
AC² = AE² + EC² …..(ii)
Now, subtracting (i) and (ii),
AD² – AC² = (AE² + ED²) – (AE² + EC²)
= AE² + ED² – AE² – EC²
= ED² – EC²
= (ED + EC) (ED – EC) (BE = EC proved above)
= BD x CD = BD x CD
AD² – AC² = BD x CD
Hence proved.

Question 19.
Solution:
Given : In ∆ABC, AB = BC and ∠ABC = 90°
∆ACD and ∆ABE are similar to each other.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 24
To prove : Ratio between area ∆ABE and area ∆ACD.
Proof: Let AB = BC = x
Now, in right ∆ABC,
⇒ AC² = AB² + BC² = AB² + AB² = 2AB² = 2x²
∆ABE and ∆ACD are similar
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 25
Ratio between the areas of ∆ABE and ∆ACD = 1 : 2

Question 20.
Solution:
An aeroplane flies from airport to north at the speed of 1000 km/hr.
Another aeroplane flies from the airport to west at the speed of 1200 km/hr.
Period = 1\(\frac { 1 }{ 2 }\) hours
Distance covered by the first plane in 1\(\frac { 1 }{ 2 }\) hours = 1000 x \(\frac { 3 }{ 2 }\) km = 1500 km
and distance covered by another plane in 1\(\frac { 1 }{ 2 }\) hours = 1200 x \(\frac { 3 }{ 2 }\) km = 1800 km
At present, the distance between them
AB² = (BO)² + (AO)²
= (1800)² + (1500)²
= 3240000 + 2250000
= 5490000
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 26

Question 21.
Solution:
Given : In ∆ABC,
D is the mid point of BC and AL ⊥ BC
AD is joined.
To prove:
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 27
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 28
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 29
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 30

Question 22.
Solution:
AM is rod and BC is string out of rod.
In ∆BMC,
BC² = BM² + CM² = (1.8)² + (2.4)²
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 31
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4D 32

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RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS

RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQS.

Other Exercises

Choose the correct answer in each of the following questions.
Question 1.
Solution:
(c) A man goes from O to 24 m due west at A and then 10 m due north at B.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 1
Now, AB² = OA² + OB²
= (24)² + (10)² = 576 + 100 = 676 = (26)²
AB = 26 m

Question 2.
Solution:
(b) Two poles AB and CD are standing on the plane ground 8 m apart.
AB = 13 m and CD = 7 m, CE || DB
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 2
In right ∆ACE,
AC² = CE² + AE²
= (8)² + (6)² = 64 + 36 = 100 = (10)²
AC = 10 m
Distance between the tops of poles = 10 m

Question 3.
Solution:
(c) A vertical stick AB = 1.8 m
and its shadow = 45 cm = 0.45 m
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 3
At the same time, let x cm be the shadow of 6 m long pole.
∆ABC ~ ∆DEF
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 4

Question 4.
Solution:
(d) Shadow of a vertical pole 6 m long is 3.6 m on the ground and shadow of a tower at the same, is 18 m.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 5

Question 5.
Solution:
(d) Shadow of 5 m long stick = 2 m
Let shadow of 12.5 m high tree at the same time = x
∆ABC ~ ∆DEF
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 6

Question 6.
Solution:
(a) Length of ladder AB = 25 m .
Height above the ground = 24 m
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 7
Let its foot is x m away from the foot of building.
In right ∆ABC,
AB² = AC² + BC² (Pythagoras Theorem)
(25)² = (24)² + x²
⇒ 625 = 576 + x²
⇒ x² = 625 – 576 = 49 = (7)²
x = 7
Distance = 7 m

Question 7.
Solution:
(b) O is a point inside ∆MNP such that
MOP = 90°, OM = 16 cm, OP = 12 cm.
If MN = 21 cm ∠NMP = 90°, then NP = ?
Let MP = x Now, in right ∆MOP,
∠O = 90°
MP² = OM² + OP² (Pythagoras Theorem)
= (16)² + (12)² = 256 + 144 = 400 = (20)²
MP = 20 cm
Now, in right ∆MNP, ∠M = 90°
NP² = MN² + MP²
= (21)² + (20)² = 441 + 400 = 841 = (29)²
NP = 29 cm

Question 8.
Solution:
(b) Let ∆ABC is a right angled triangle with ∠B = 90°
AC = 25 cm
Let one side AB of the other two sides = x cm
then second side BC = (x + 5) cm
According to the Pythagoras Theorem,
AC² = AB² + BC²
(25)² = x² + (x + 5)²
625 = x² + x² + 10x + 25
⇒ 2x² + 10x + 25 – 625 = 0
⇒ 2x² + 10x – 600 = 0
⇒ x² + 5x – 300 = 0
⇒ x² + 20x – 15x – 300 = 0
⇒ x (x + 20) – 15 (x + 20) = 0
⇒ (x + 20)(x – 15) = 0
Either x + 20 = 0, then x = -20 which is not possible being negative,
or x – 15 = 0, then x = 15
First side = 15 cm
and second side = 15 + 5 = 20 cm

Question 9.
Solution:
(b) Side of an equilateral triangle = 12 cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 8

Question 10.
Solution:
(d) In isosceles ∆ABC,
AB = AC = 13 cm
Length of altitude AB, (from A to BC) = 5 cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 9
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 10

Question 11.
Solution:
(a) In the given figure,
AB = 6 cm, AC = 8 cm
AD is the bisector of ∠A which meets BC at D.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 11

Question 12.
Solution:
(d) In the given figure,
AD is the internal bisector of ∠A
BD = 4 cm, DC = 5 cm, AB = 6 cm
Let AC = x cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 12

Question 13.
Solution:
(b) In the given figure,
AD is the bisector of ∠A of ∆ABC.
AB = 10 cm, AC = 14 cm and BC = 6 cm
Let CD = x cm
Then BD = (6 – x) cm
Now, AD is the bisector of ∠A
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 13

Question 14.
Solution:
(b) In a ∆ABC, AD ⊥ BC and BD = DC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 14
In a ∆ABC, AD = \(\frac { 1 }{ 2 }\) BC and BD = DC.
In right ∆ABD and ∆ACD
AD = AD (common)
∠ABD = ∠ADC (each 90°)
BD = DC (given)
∆ABD = ∆ACD (SAS axiom)
AB = AC
∆ABC is an isosceles triangle.

Question 15.
Solution:
(c) In equilateral ∆ABC, AD ⊥ BC
Then BD = DC = \(\frac { 1 }{ 2 }\) BC
Now, in right ∆ABD,
AB² = BD² + AD² (Pythagoras Theorem)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 15

Question 16.
Solution:
(c) In a rhombus, each side = 10 cm and one diagonal = 12 cm
AB = BC = CD = DA = 10 cm BD = 12 cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 16
The diagonals of a rhombus bisect each other at right angles.
In ∆AOB,
AB² = AO² + BO²
⇒ (10)² = AO² + (6)²
⇒ AO² = (10)² – (6)² = 100 – 36 = 64 = 8²
AO = 8 cm
Diagonals AC = 2 x AO = 2 x 8 = 16 cm

Question 17.
Solution:
(b) Length of diagonals of a rhombus are 24 cm and 10 cm.
The diagonals of a rhombus bisect each other at right angles.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 17
In rhombus ABCD
AO = OC, BO = OD
Let AO = OC = \(\frac { 24 }{ 2 }\) = 12 cm
BO = OD = \(\frac { 10 }{ 2 }\) = 5 cm
Now, in right ∆AOB,
AB² = AO² + BO² (Pythagoras Theorem)
= (12)² + (5)² = 144 + 25 = 169 = (13)²
AB = 13
Each side of rhombus = 13 cm

Question 18.
Solution:
(b) Diagonals of e. quadrilateral divides each other proportionally, then it is
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 18
In quadrilateral ABCD, diagonals AC and BD intersect each-other at O and \(\frac { AO }{ OC }\) = \(\frac { BO }{ OD }\)
Then, quadrilateral ABCD is a trapezium.

Question 19.
Solution:
(a) In the given figure,
ABCD is a trape∠ium and its diagonals AC
and BD intersect at O.
and OA = (3x – 1) cm OB = (2x + 1) cm, OC and OD = (6x – 5) cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 19

Question 20.
Solution:
(a) The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram

Question 21.
Solution:
(c) If the bisector of angle of a triangle bisects the opposite side of a triangle.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 20

Question 22.
Solution:
(a) In ∆ABC,
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 21
∠B = 70° and ∠C = 50°
But ∠A + ∠B + ∠C = 180° (Angles of a triangle)
∠A = 180° – (∠B + ∠C)
= 180° – (70° + 50°)
= 180° – 120° = 60°
\(\frac { AB }{ AC }\) = \(\frac { BD }{ DC }\)
AD is the bisector of ∠A
∠BAD = \(\frac { 60 }{ 2 }\) = 30°

Question 23.
Solution:
(b) In ∆ABC, DE || BC
AD = 2.4 cm, AE = 3.2 cm, EC = 4.8 cm
Let AD = x cm
DE || BC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 22

Question 24.
Solution:
(b) In ∆ABC, DE || BC
AB = 7.2 cm, AC = 6.4 cm, AD = 4.5 cm
Let AE = x cm
DE || BC
∆ADE ~ ∆ABC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 23

Question 25.
Solution:
(c) In ∆ABC, DE || BC
AD = (7x – 4) cm, AE = (5x – 2) cm DB = (3x + 4) cm and EC = 3x cm
In ∆ABC, DE || BC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 24
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 25

Question 26.
Solution:
(d) In ∆ABC, DE || BC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 26

Question 27.
Solution:
(b) ∆ABC ~ ∆DEF
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 27
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 28

Question 28.
Solution:
(a) ∆ABC ~ ∆DEF
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 29

Question 29.
Solution:
(d) ∆DEF ~ ∆ABC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 30
Perimeter of ∆DEF = DE + EF + DF
= 12 + 8 + 10 = 30 cm

Question 30.
Solution:
(d) ABC and BDE are two equilateral triangles such that D is the midpoint of BC.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 31

Question 31.
Solution:
(b) ∆ABC ~ ∆DFE.
∠A = 30°, ∠C = 50°, AB = 5cm, AC = 8 cm and DF = 7.5 cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 32

Question 32.
Solution:
(c) In ∆ABC, ∠A = 90°
AD ⊥ BC
In ∆ABD and ∆ADC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 33

Question 33.
Solution:
(c) In ∆ABC, AB = 6 cm, AC = 12 cm and BC = 6 cm.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 34
Longest side (AC)2 = (12)2 = 144
AB2 + BC2 = (6√3)2 + (6)2 = 108 + 36 = 144
AC2 = AB2 + BC2 (Converse of Pythagoras Theorem)
∠B = 90°

Question 34.
Solution:
(c) In ∆ABC and ∆DEF, \(\frac { AB }{ DE }\) = \(\frac { BC }{ FD }\)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 34
For similarity,
Here, included angles must be equal and these
are ∠B = ∠D.

Question 35.
Solution:
(b) In ∆DEF and ∆PQR,
∠D = ∠Q and ∠R = ∠E
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 35

Question 36.
Solution:
(c) ∆ABC ~ ∆EDF
∠A = ∠E, ∠B = ∠D, ∠C = ∠F
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 36

Question 37.
Solution:
(b) In ∆ABC and ∆DEF,
∠B = ∠E, ∠F = ∠C and AB = 3DE
The triangles are similar as two angles are equal but including sides are not proportional.

Question 38.
Solution:
(a)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 37

Question 39.
Solution:
(d) In the given figure, two line segments AC and BD intersect each other at P such that
PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°
In ∆ABP and ∆CPD,
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 37

Question 40.
Solution:
(d) Corresponding sides of two similar triangles = 4:9
The areas of there triangle will be in the ratio
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 39

Question 41.
Solution:
(d)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 40
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 41

Question 42.
Solution:
(b) In the given figure,
∆ABC is an equilateral triangle.
D is midpoint of AB and E is the midpoint of AC.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 42

Question 43.
Solution:
(b)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 43
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 44

Question 44.
Solution:
(b) ∆ABC ~ ∆DEF
ar (∆ABC) = 36 cm² and ar (∆DEF) = 49 cm²
i.e. areas are in the ratio 36 : 49
Ratio in their corresponding sides = √36 : √49 = 6 : 7

Question 45.
Solution:
(c) Two isosceles triangles have their corresponding angles equal and ratio in their areas is 25 : 36.
The ratio in their corresponding altitude
(heights) = √25 : √36 = 5 : 6 (∆s are similar)

Question 46.
Solution:
(b) The line segments joining the midpoints of a triangle form 4 triangles which are similar to the given (original) triangle.

Question 47.
Solution:
(b) ∆ABC ~ ∆QRP
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 45

Question 48.
Solution:
(c) In the given figure, O is the point of intersection of two chords AB and CD.
OB = OD and ∠AOC = 45°
∠B = ∠D (Angles opposite to equal sides)
∠A = ∠D, ∠C = ∠B (Angles in the same segment)
and ∠AOC = ∠BOD = 45° each
∆OAC ~ ∆ODB (AAA axiom)
OA = OC (Sides opposite to equal angles)
∆OAC and ∆ODB are isosceles and similar.

Question 49.
Solution:
(d) In an isosceles ∆ABC,
AC = BC
⇒ AB² = 2 AC²
⇒ AB² = AC² + AC²
⇒ AB² = AC² + BC² (AC = BC)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 46
Converse of the Pythagoras Theorem,
∆ABC is a right triangle and angle opposite to AB = 90°
∠C = 90°

Question 50.
Solution:
(b) In ∆ABC,
AB = 16 cm, BC = 12 cm and AC = 20 cm
(Longest side)2 = 20² = 400
Sum of square on other sides = AB² + BC²
= 162 + 122 = 256 + 144 = 400
AC² = AB² + BC²
∆ABC is a right triangle.

True/False type
Question 51.
Solution:
(c) (a) False. Not always congruent.
(b) False. Two similar figures are similar if they have same shape, not size in every case.
(c) True.
(d) False. Not in each case.

Question 52.
Solution:
(a) True
(b) False, as ratio of the areas of the two similar triangles is equal to the ratio of the square of their corresponding sides.
(c) True
(d) True

Question 53.
Matching of columns : (2 marks)
Solution:
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 47
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 48
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 49
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 50
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 51

Question 54.
Solution:
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 52
RS Aggarwal Class 10 Solutions Chapter 4 Triangles MCQS 53
correct answer is
(a) → (r)
(b) → (q)
(c) → (p)
(d) → (s)

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQS are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A

RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4A.

Other Exercises

Question 1.
Solution:
In ∆ABC, points D and E are on the sides AB and AC respectively such that DE || BC.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 1
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 2
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 3
⇒ 7x = 28 ⇒ x = 4

Question 2.
Solution:
D and E are the points on the sides AB and AC of ∆ABC respectively and DE || BC,
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 4
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 5
By cross multiplication,
⇒ (7x – 4) (3x) = (5x – 2) (3x + 4)
⇒ 21x² – 12x = 15x² + 20x – 6x – 8
⇒ 21x² – 12x – 15x² – 20x + 6x = -8
⇒ 6x² – 26x + 8 = 0
⇒ 3x² – 13x + 4 = 0
⇒ 3x² – 12x – x + 4 = 0
⇒ 3x (x – 4) – 1 (x – 4) = 0
⇒ (x – 4) (3x – 1) = 0
Either x – 4 = 0, then x = 4
or 3x – 1 = 0 then 3x = 1 ⇒ x = \(\frac { 1 }{ 3 }\) which is not possible
x = 4

Question 3.
Solution:
D and E are the points on the sides AB and AC of ∆ABC
(i) AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm and EC = 8 cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 6
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 7
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 8

Question 4.
Solution:
In ∆ABC, AD is the bisector of ∠A
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 9
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 10
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 11
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 12

Question 5.
Solution:
Given : M is a point on the side BC of a parallelogram ABCD.
DM when produced meets AB produced at N.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 13

Question 6.
Solution:
Given : In trapezium AB || DC.
M and N are the mid points of sides AD and BC respectively.
MN is joined.
To prove : MN || AB or DC.
Construction: Produce AD and BC to meet at P.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 14

Question 7.
Solution:
ABCD is a trapezium and its diagonals AC and BD intersect each other at O.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 15
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 16

Question 8.
Solution:
Given : In ∆ABC, M and N are points on the sides AB and AC respectively such that
BM = CN and if ∠B = ∠C.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 17
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 18

Question 9.
Solution:
Given : ∆ABC and ∆DBC are on the same side of BC.
P is any point on BC.
PQ || AB and PR || BD are drawn, which meet AC at Q and CD at R.
AD is joined.
To prove : QR || AD.
Proof:
In ∆ABC, PQ||AB
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 19
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 20
Hence proved.

Question 10.
Solution:
Given : In the given figure, D is the mid point of BC. AD is joined. O is any point on AD. BO and CO are produced to meet AC at F and AB at E respectively. AD is produced to X such that D is the mid point of OX.
To prove: AO : AX = AF : AB and EF || BC
Construction. Join BX and CX.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 21
D is the midpoint of BC.
Proof: BD = DC and OD = DX. (given)
OBXC is a parallelogram. (Diagonals of a ||gm bisect each other)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 22

Question 11.
Solution:
Given : In ||gm, ABCD, P is mid point of DC and Q is the mid point of AC. Such that
CQ = \(\frac { 1 }{ 4 }\) AC.
PQ is produced to meet BC at R.
To prove : R is the mid point of BC.
Construction : Join BD which intersects AC at O.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 23
Proof: Diagonals of a parallelogram bisect each other
CO = \(\frac { 1 }{ 2 }\) AC.
But CQ = \(\frac { 1 }{ 2 }\) CO (given)
Q is the mid point of CO.
PQ || DO
⇒ PR || DB and P is mid point of CD
R is the midpoint BC.
Hence proved.

Question 12.
Solution:
Given : In ∆ABC, AB = AC
D and E are the points on AB and AC such that
AD = AE
To prove : BCED is a cyclic quadrilateral.
Proof: In ∆ABC,
AB = AC,
D and E are the mid points of AB and AC respectively.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 24
But these are opposite angles of quad. BCED
BCED is a cyclic quadrilateral
Hence, B, C, E and D are concyclic.

Question 13.
Solution:
Given : In ∆ABC, BD is the bisector of AB which meets AC at D.
A line PQ || AC which meets AB, BC, BD at P, Q, R respectively.
To prove : PR x BQ = QR x BP
Proof: In ∆ABC, PQ || AC
and BD is the bisector of ∠B
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4A 25

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4A 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 10 Solutions Chapter 4 Triangles Ex 4B

RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4B.

Other Exercises

Question 1.
Solution:
We know that two triangles are similarity of their corresponding angles are equal and corresponding sides are proportional.
(i) In ∆ABC and ∆PQR
∠A = ∠Q = 50°
∠B = ∠P = 60° and ∠C = ∠R = 70°
∆ABC ~ ∆QPR (AAA axiom)
(ii) In ∆ABC and ∆DEF
In ∆ABC,
AB = 3 cm, BC = 4.5
and in ∆DEF
DF = 6 cm, DE = 9 cm
∆ABC is not similar to ∆DEF
As in ∆ABC, ∠A is not included of two sides AB and BC.
(iii) In ∆ABC and ∆PQR
In ∆ABC,
AC = 8 cm BC = 6 cm
Included ∠C = 80°
In ∆PQR,
PQ = 4.5 cm, QR = 6 cm
and included ∠Q = 80°
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 1
(v) In ∆ABC,
∠A = 80°, ∠C = 70°
and third angle
∠B = 180° – (80° + 70°)
⇒ ∠B = 180° – 150° = 30°
In ∆MNR,
∠M = 80°, ∠N = 30°, and ∠R = [180° – (80° + 30°)]
∠R = 180° – 110° = 70°
Now, in ∆ABC
∠A = ∠M – 80°, ∠B = ∠N = 30°
and ∠C = ∠R = 70°
∆ABC ~ ∆MNR (AAA or AA axiom)

Question 2.
Solution:
In the given figure, ∆ODC ~ ∆OBA and ∠BOC =115°, ∠CDO = 70°
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 2
To find :
(i) ∠DOC
(ii) ∠DCO
(iii) ∠OAB
(iv) ∠OBA
∆ODC ~ ∆OBA
∠D = ∠B = 70°
∠C = ∠A
∠COD = ∠AOB
(i) But ∠DOC + ∠BOC = 180° (Linear pair)
⇒ ∠DOC + 115°= 180°
⇒ ∠DOC = 180° – 115° = 65°
(ii) ∠DOC + ∠CDO + ∠DCO = 180° (Angles of a triangle)
⇒ 65° + 70° + ∠DCO = 180°
⇒ 135° + ∠DCO = 180°
⇒ ∠DCO = 180° – 135°
∠DCO = 45°
(iii) ∠AOB = ∠DOC = 65° (vertically opposite angles)
∠OAB = ∠DCO = 45° (∆ODC ~ ∆OBA)
(iv) ∠OBA = ∠CDO = 70° (∆ODC ~ ∆OBA)

Question 3.
Solution:
In the given figure, ∆OAB ~ ∆OCD
AB = 8 cm, BO = 6.4 cm OC = 3.5 cm, CD = 5 cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 3
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 4

Question 4.
Solution:
Given : In the given figure,
∠ADE = ∠B
To prove:
(i) ∆ADE ~ ∆ABC
(ii) If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm and BC = 4.2 cm, find DE
Proof: (i) In ∆ADE and ∆ABC
∠ADE = ∠B (given)
∠A = ∠A (common)
∆ADE ~ ∆ABC (AA axiom)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 5

Question 5.
Solution:
∆ABC ~ ∆PQR,
PQ = 12 cm
To find AB.
Perimeter of ∆ABC = AB + BC + CA = 32 cm
Perimeter of ∆PQR = PQ + QR + RP = 24 cm
Now,
∆ABC ~ ∆PQR
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 6

Question 6.
Solution:
∆ABC ~ ∆DEF
BC = 9.1 cm, EF = 6.5 cm
Perimeter of ∆DEF = 25 cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 7
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 8

Question 7.
Solution:
Given : In the given figure,
∠CAB = 90° and AD ⊥ BC
To prove : ∆BDA ~ ∆BAC
If AC = 75 cm, AB = 1 m or 100 cm,
BC = 1.25 m or 125 cm
Find AD.
∆BDA ~ ∆BAC (corresponding sides and proportional)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 9

Question 8.
Solution:
In the given figure,
∠ABC = 90°, BD ⊥ AC.
AB = 5.7 cm, BD = 3.8 cm, CD = 5.4 cm
To find BC,
In ∆ABC and ∆BDC,
∠ABC = ∠BDC (each 90°)
∠BCA = ∠BCD (common)
∆ABC ~ ∆BDC (AA axiom)
Corresponding sides are proportional
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 10

Question 9.
Solution:
In the given figure, ”
∠ABC = 90°, BD ⊥ AC
BD = 8 cm, AD = 4cm
To find CD,
Let CD = x
Now in ∆DBC and ∆BDA
∠BDC = ∠BDA (each 90°)
∠C = ∠ABD
∆DBC ~ ∆BDA
Sides are proportional
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 11

Question 10.
Solution:
In ∆ABC, P and Q are points on the sides AB and AC respectively such that
AP = 2 cm, PB = 4 cm, AQ = 3 cm and QC = 6 cm.
To prove : BC = 3 PQ
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 12
BC = 3 PQ
Hence proved.

Question 11.
Solution:
Given: ABCD is a parallelogram.
E is a point on BC and diagonal BD intersects AE at F.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 13
To prove : AF x FB = EF x FD
Proof: In ∆AFD and ∆BFE
∠AFD = ∠BFE (vertically opposite angles)
∠ADF = ∠FBE (alternate angles)
∆AFD ~ ∆BFE (AA axiom)
\(\frac { AF }{ EF }\) = \(\frac { FD }{ FB }\)
By cross multiplication,
⇒ AF x FB = EF x FD
Hence proved.

Question 12.
Solution:
Given : In the given figure,
DB ⊥ BC, DE ⊥ AB and AC ⊥ BC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 14

Question 13.
Solution:
In ∆ABC and ∆DEF
AC is stick and BC is its shadow.
DF is the tower and EF is its shadow
AC = 7.5 m, BC = 5 m EF = 24 m,
let DF = x m
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 15
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 16

Question 14.
Solution:
Given : In isosceles ∆ABC,
CA = CB, base AB and BA are produced in P and Q such that
AP x BQ = AC²
To prove : ∆ACP ~ ∆BCQ
Proof: In ∆ABC,
CA = CB
∠CAB = ∠CBA (Angles opposite to equal sides)
⇒ 180° – ∠CAB = 180° – ∠CBA (Subtracting each from 180°)
⇒ ∠CAP = ∠CBQ
AP x BQ = AC² (given)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 17

Question 15.
Solution:
Given : In the given figure, ∠1 = ∠2
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 18

Question 16.
Solution:
Given : In quadrilateral ABCD,
AD = BC
P, Q, R and S are midpoints of AB, AC, CD and BD respectively.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 19
To proof : PQRS is a rhombus.
Proof: In ∆ABC,
P and Q are the midpoints of sides AB and AC respectively.
PQ || BC ……(i)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 20

Question 17.
Solution:
In the given circle, two chords AB and CD intersect each other at P inside the circle.
To prove :
(a) ∆PAC ~ ∆PDB
(b) PA . PB = PC . PD
Proof:
(a) In ∆PAC and ∆PDB
∠APC = ∠BPD (Vertically opposite angles)
∠A = ∠D (Angles in the same segment)
∆PAC ~ ∆PDB (AA axiom)
\(\frac { PA }{ PD }\) = \(\frac { PC }{ PB }\)
⇒ PA x PB = PC x PD
Hence proved.

Question 18.
Solution:
In a circle, two chords AB and CD intersect each other at the point P outside the circle.
AC and BD are joined.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 21
To prove :
(a) ∆PAC ~ ∆PDB
(b) PA . PB = PC . PD
Proof: In the circle, quadrilateral ABDC is a cyclic.
Ext. ∠PAC = ∠D
Now, in ∆PDB and ∆PAC
∠P = ∠P (common)
∠PAC = ∠D (proved)
∆PDB ~ ∆PAC (AA axiom)
or ∆PAC ~ ∆PBD
\(\frac { PA }{ PD }\) = \(\frac { PC }{ PB }\)
⇒ PA . PB = PC . PD
Hence proved

Question 19.
Solution:
Given : In right ∆ABC, ∠B = 90°
D is a point on hypotenuse AC such that
BD ⊥ AC and DP ⊥ AB and DQ ⊥ BC.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4B 22
To prove :
(a) DQ² = DP . QC
(b) DP² = DQ . AP
Proof: AB ⊥ BC and DQ ⊥ BC
AB || DQ
DP ⊥ AB
DP || BC
Now, AB or PB || DQ and BC or BQ || DP BQDP is a rectangle
BQ = DP and BP = DQ
Now, in right ∆BQD
∠1 + ∠2 = 90° …..(i)
Similarly in rt. ∆DQC,
∠3 + ∠4 = 90° (DQ ⊥ BC) …(ii)
and in right ∆BDC,
∠2 + ∠3 = 90° …(iii)
∠BDC = 90° (BD ⊥ AC)
From (i) and (ii),
∠1 = ∠3
and from (ii) and (iii),
∠2 = ∠4
Now, in ∆BQD and ∆DQC
∠1 = ∠3
∠2 = ∠4 (proved)
∆BQD ~ ∆DQC (AA axiom)
\(\frac { BQ }{ DQ }\) = \(\frac { DQ }{ QC }\)
⇒ DQ² = BQ x QC
⇒ DQ² = DP x QC
(b) Similarly, we can prove that
∆PDA ~ ∆PBD
\(\frac { PD }{ PB }\) = \(\frac { AP }{ DP }\)
⇒ DP² = BP x AP
⇒ DP² = DQ . AP (BP = DQ)
Hence proved.

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4B are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself

RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Test Yourself.

Other Exercises

MCQ
Question 1.
Solution:
(b) ∆ABC ~ ∆DEF
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 1

Question 2.
Solution:
(a) In the given figure,
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 2
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 3

Question 3.
Solution:
(b) Length of pole AB = 6 m
and CD = 11 m
and distance between the BD = 12 m
Draw AE || BD, then
ED = AB = 6 m, AE = BD = 12 m
CE = CD – ED = 11 – 6 = 5 m
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 4
Now, in right ∆ACE,
AC² = AE² + CE² (Pythagoras Theorem)
= (12)² + (5)² = 144 + 25 = 169 = 13²
AC = 13
Distance between their tops = 13 m

Question 4.
Solution:
(c) Area of two similar triangles ABC and DEF are 25 cm² and 36 cm².
Altitude of first ∆ABC is AL = 3.5 m.
Let DM be the altitude of second triangle.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 5

Short-Answer Questions
Question 5.
Solution:
∆ABC ~ ∆DEF
and 2AB = DE
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 6

Question 6.
Solution:
In the given figure,
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 7
DE || BC, AL ⊥ BC
AD = x cm,
DB = (3x + 4) cm AE = (x + 3) cm, EC = (3x + 19) cm
In ∆ABC, DE || BC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 8

Question 7.
Solution:
Let AB be the ladder, and A is the window.
Length of ladder AB = 10 m and AC = 8 m
Distance between the foot of the ladder from the house = x m
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 9
In right ∆ABC,
AB² = AC² + BC²
⇒ (10)² = (8)² + (x)²
⇒x² = (10)² – (8)²
⇒ x² = 100 – 64 = 36
⇒ x² = (6)²
⇒ x = 6
Distance between the foot of ladder and foot of the house = 6m.

Question 8.
Solution:
In ∆ABC, AB = BC = CA = 2a cm
AD ⊥ BC which bisects the base BC at D
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 10
BD = DC = \(\frac { 2a }{ 2 }\) = a cm
Now, in right ∆ABD
AB² = AD² + BD²
(2a)² = AD² + a²
⇒ AD² = 4a² – a² = 3a²
AD = √3a² = √3 a
Height of altitude = √3 a cm

Question 9.
Solution:
Let EF = x cm
∆ABC ~ ∆DEF
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 11

Question 10.
Solution:
ABCD is a trape∠ium in which AB || DC
AB = 2CD
Diagonals AC and BD intersect each other at O.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 12
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 13

Question 11.
Solution:
Let corresponding sides of two similar triangles ABC and DEF are in the ratio 2 : 3.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 14

Question 12.
Solution:
In the given figure,
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 15

Question 13.
Solution:
Given : In ∆ABC, AD is the bisector of ∠A which meets BC at D.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 16
Construction : Produce BA and draw CE || DA meeting BA produced at E.
DA || CE
∠3 = ∠1 (corresponding angles)
∠4 = ∠2 (alternate angles)
But ∠3 = ∠4 (AD is the bisector of ∠A)
∠1 = ∠2
AC = AE (Sides opposite to equal angles)
Now, in ∆AEC,
AD || EC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 17

Question 14.
Solution:
Given : In ∆ABC,
AB = BC = CA = a cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 18
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 19

Question 15.
Solution:
In rhombus ABCD, diagonals AC = 24 cm and BD = 10 cm.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 20
The diagonals of a rhombus bisect each other at right angles.
AO = OC = \(\frac { 24 }{ 2 }\) = 12 cm
and BO = OD = \(\frac { 10 }{ 2 }\) = 5 cm
Now, in right ∆AOB,
AB² = AO² + BO² = 12² + 5² = 144 + 25 = 169 = (13)²
AB = 13
Each side of a rhombus = 13 cm

Question 16.
Solution:
Given : ∆ABC ~ ∆DEF
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 21
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 22
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 23

Long-Answer Questions
Question 17.
Solution:
Given: In the given figure, ∆ABC and ∆DBC are on the same base BC but in opposite sides.
AD and BC intersect at O.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 24
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 25
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 26

Question 18.
Solution:
In the given figure,
XY || AC and XY divides
∆ABC into two regions equal in area
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 27
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 28
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 29

Question 19.
Solution:
In the given figure, ∆ABC is an obtuse triangle, obtuse angle at B.
AD ⊥ CB produced.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 30
To prove : AC² = AB² + BC² + 2BC x BD
Proof: In ∆ADB, ∠D = 90°
AB² = AD² + DB² (Pythagoras Theorem)
⇒ AD² = AB² – DB² ……(i)
Similarly, in right ∆ADC,
AC² = AD² + DC²
= AB² – DB² + (DB + BC)² [From (i)]
= AB² – DB² + DB² + BC² + 2BC x BD
= AB² + BC² + 2 BC x BD

Question 20.
Solution:
In the given figure,
PA, QB and RC is perpendicular to AC.
AP = x, QB = z, RC =y, AB = a and BC = b.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 31
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself 32

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Test Yourself 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 10 Solutions Chapter 4 Triangles Ex 4C

RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4C.

Other Exercises

Question 1.
Solution:
Given : Area of ∆ABC = 64 cm²
and area of ∆DEF =121 cm²
EF = 15.4 cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 1
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 2

Question 2.
Solution:
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 3

Question 3.
Solution:
∆ABC ~ ∆PQR
ar (∆ABC) = 4ar (∆PQR),
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 4
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 5

Question 4.
Solution:
Areas of two similar triangles are 169 cm² and 121 cm²
Longest side of largest triangle = 26 cm
Let longest side of smallest triangle = x
∆s are similar
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 6

Question 5.
Solution:
Area of ∆ABC = 100 cm²
and area of ∆DEF = 49 cm²
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 7

Question 6.
Solution:
Given : Corresponding altitudes of two similar triangles are 6 cm and 9 cm
We know that the areas of two similar triangles are in the ratio of the squares of their corresponding altitudes.
Ratio in the areas of two similar triangles = (6)² : (9)² = 36 : 81 = 4 : 9 (Dividing by 9)

Question 7.
Solution:
The areas of two similar triangles are 81 cm² and 49 cm²
Altitude of the first triangle = 6.3 cm
Let altitude of second triangle = x cm
The areas of two similar triangles are in the ratio of the squares on their corresponding altitude,
Let area of ∆ABC = 81 cm²
and area of ∆DEF = 49cm²
Altitude AL = 6 – 3 cm
Let altitude DM = x cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 8

Question 8.
Solution:
Areas of two similar triangles are 100 cm² and 64 cm²
Let area of ∆ABC = 100 cm²
and area of ∆DEF = 64 cm²
Median DM of ∆DEF = 5.6 cm
Let median AL of ∆ABC = x
The areas of two similar triangles is proportional to the squares of their corresponding median.
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 9
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 10

Question 9.
Solution:
Given : In ∆ABC, PQ is a line which meets AB in P and AC in Q.
AP = 1 cm, PB = 3 cm, AQ = 1.5 cm QC = 4.5 cm
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 11
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 12

Question 10.
Solution:
In ∆ABC,
DE || BC
DE = 3 cm, BC = 6 cm
area (∆ADE) = 15 cm²
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 13

Question 11.
Solution:
Given : In right ∆ABC, ∠A = 90°
AD ⊥ BC
BC = 13 cm, AC = 5 cm
To find : Ratio in area of ∆ABC and ∆ADC
In ∆ABC and ∆ADC
∠C = ∠C (common)
∠BAC = ∠ADC (each 90°)
∆ABC ~ ∆ADC (AA axiom)
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 14

Question 12.
Solution:
In the given figure ∆ABC,
DE || BC and DE : BC = 3 : 5.
In ∆ABC and ∆ADE,
DE || BC
∆ABC ~ ∆ADE
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 15

Question 13.
Solution:
In ∆ABC, D and E are the midpoints of sides AB and AC respectively.
DE || BC and DE = \(\frac { 1 }{ 2 }\) BC
∆ADE ~ ∆ABC
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 16
RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C 17

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4C 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.