Mathematics Solutions Solutions for Class 7 Maths Chapter 14 Pythagoras’ Theorem are provided here with simple step-by-step explanations. These solutions for Pythagoras’ Theorem are extremely popular among Class 7 students for Maths Pythagoras’ Theorem Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 7 Maths Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

#### Page No 90:

#### Question 1:

In the figures below, find the value of '*x*'.

#### Answer:

(i)

In the right angled triangle LMN, ∠M = 90^{∘}. Hence, side LN is the hypotenuse.

According to Pythagoras' theorem,

*l*(LN)^{2} = *l*(LM)^{2} + *l*(MN)^{2}

⇒(*x*)^{2} = (7)^{2} + (24)^{2}

⇒*x*^{2} = 49 + 576

⇒*x*^{2} = 625

⇒*x*^{2} = (25)^{2}

⇒*x* = 25

∴ the value of *x* is 25.

(ii)

In the right angled triangle PQR, ∠Q = 90^{∘}. Hence, side PR is the hypotenuse.

According to Pythagoras' theorem,

*l*(PR)^{2} = *l*(QR)^{2} + *l*(PQ)^{2}

⇒(41)^{2} = (*x*)^{2} + (9)^{2}

⇒1681 = *x*^{2} + 81

⇒*x*^{2} = 1681 − 81

⇒*x*^{2} = 1600

⇒*x*^{2} = (40)^{2}

⇒*x* = 40

∴ the value of *x* is 40.

(iii)

In the right angled triangle EDF, ∠D = 90^{∘}. Hence, side EF is the hypotenuse.

According to Pythagoras' theorem,

*l*(EF)^{2} = *l*(ED)^{2} + *l*(DF)^{2}

⇒(17)^{2} = (*x*)^{2} + (8)^{2}

⇒289 = *x*^{2} + 64

⇒*x*^{2} = 289 − 64

⇒*x*^{2} = 225

⇒*x*^{2} = (15)^{2}

⇒*x* = 15

∴ the value of *x* is 15.

#### Page No 90:

#### Question 2:

In the right-angled ∆PQR, ∠ P = 90°. If *l*(PQ) = 24 cm and *l*(PR) = 10 cm, find the length of seg QR.

#### Answer:

In the right angled triangle PQR, ∠P = 90^{∘}. Hence, side QR is the hypotenuse.

According to Pythagoras' theorem,

*l*(QR)^{2} = *l*(PQ)^{2} + *l*(PR)^{2}

⇒*l*(QR)^{2} = (24)^{2} + (10)^{2}

⇒*l*(QR)^{2} = 576 + 100

⇒*l*(QR)^{2} = 676

⇒*l*(QR)^{2} = (26)^{2}

⇒*l*(QR) = 26

∴ Length of seg QR = 26 cm.

#### Page No 90:

#### Question 3:

In the right-angled ∆LMN, ∠ M = 90°. If* l*(LM) = 12 cm and *l*(LN) = 20 cm, find the length of seg MN.

#### Answer:

In the right angled triangle LMN, ∠M = 90^{∘}. Hence, side LN is the hypotenuse.

According to Pythagoras' theorem,

*l*(LN)^{2} = *l*(MN)^{2} + *l*(LM)^{2}

⇒(20)^{2} = *l*(MN)^{2} + (12)^{2}

⇒400 = *l*(MN)^{2} + 144

⇒*l*(MN)^{2} = 400 − 144

⇒*l*(MN)^{2} = 256

⇒*l*(MN)^{2} = (16)^{2}

⇒*l*(MN) = 16

∴ Length of seg MN = 16 cm.

#### Page No 90:

#### Question 4:

The top of a ladder of length 15 m reaches a window 9 m above the ground. What is the distance between the base of the wall and that of the ladder ?

#### Answer:

Let LN be ladder of length 15 *m* that is resting against a wall. Let M be the base of the wall and L be the position of the window.

The window is 9 m above the ground. Now, MN is the distance between base of the wall and that of the ladder.

In the right angled triangle LMN, ∠M = 90^{∘}. Hence, side LN is the hypotenuse.

According to Pythagoras' theorem,

*l*(LN)^{2} = *l*(MN)^{2} + *l*(LM)^{2}

⇒(15)^{2} = *l*(MN)^{2} + (9)^{2}

⇒225 = *l*(MN)^{2} + 81

⇒*l*(MN)^{2} = 225 − 81

⇒*l*(MN)^{2} = 144

⇒*l*(MN)^{2} = (12)^{2}

⇒*l*(MN) = 12

∴ Length of seg MN = 16 *m*.

Hence, the distance between base of the wall and that of the ladder is 12 *m*.

#### Page No 90:

#### Question 1:

Find the Pythagorean triplets from among the following sets of numbers.

(i) 3, 4, 5

(ii) 2, 4, 5

(iii) 4, 5, 6

(iv) 2, 6, 7

(v) 9, 40, 41

(vi) 4, 7, 8

#### Answer:

It is known that, if in a triplet of natural numbers, the square of the biggest number is equal to sum of the squares of the other two numbers, then the three numbers form a Pythgorean triplet.

(i) The given set of numbers is (3, 4, 5).

The biggest number among the given set is 5.

5^{2} = 25; 4^{2} = 16; 3^{2} = 9

Now, 16 + 9 = 25

∴ 4^{2} + 3^{2} = 5^{2}

Thus, (3, 4, 5) forms a Pythagorean triplet.

(ii) The given set of numbers is (2, 4, 5).

The biggest number among the given set is 5.

5^{2} = 25; 4^{2} = 16; 2^{2} = 4

Now, 16 + 4 = 20 ≠ 25

∴ 4^{2} + 2^{2} ≠ 5^{2}

Thus, (2, 4, 5) does not form a Pythagorean triplet.

(iii) The given set of numbers is (4, 5, 6).

The biggest number among the given set is 6.

6^{2} = 36; 5^{2} = 25; 4^{2} = 16

Now, 25 + 16 = 41 ≠ 36

∴ 5^{2} + 4^{2} ≠ 6^{2}

Thus, (4, 5, 6) does not form a Pythagorean triplet.

(iv) The given set of numbers is (2, 6, 7).

The biggest number among the given set is 7.

7^{2} = 49; 6^{2} = 36; 2^{2} = 4

Now, 4 + 36 = 40 ≠ 49

∴ 2^{2} + 6^{2} ≠ 7^{2}

Thus, (2, 6, 7) does not form a Pythagorean triplet.

(v) The given set of numbers is (9, 40, 41).

The biggest number among the given set is 41.

9^{2} = 81; 40^{2} = 1600; 41^{2} = 1681

Now, 81 + 1600 = 1681

∴ 9^{2} + 40^{2} = 41^{2}

Thus, (9, 40, 41) forms a Pythagorean triplet.

(vi) The given set of numbers is (4, 7, 8).

The biggest number among the given set is 8.

8^{2} = 64; 7^{2} = 49; 4^{2} = 16

Now, 16 + 49 = 65 ≠ 64

∴ 4^{2} + 7^{2} ≠ 8^{2}

Thus, (4, 7, 8) does not form a Pythagorean triplet.

#### Page No 90:

#### Question 2:

The sides of some triangles are given below. Find out which ones are right-angled triangles?

(i) 8, 15, 17

(ii) 11, 12, 15

(iii) 11, 60, 61

(iv) 1.5, 1.6, 1.7

(v) 40, 20, 30

#### Answer:

It is known that, if in a triplet of natural numbers, the square of the biggest number is equal to sum of the squares of the other two numbers, then the three numbers form a Pythgorean triplet. If the lengths of the sides of a triangle form such a triplet, then the triangle is right angled triangle.

(i) The sides of the given triangle is 8, 15 and 17.

Let us check whether the given set (8, 15, 17) forms a Pythagorean triplet or not.

The biggest number among the given set is 17.

(17)^{2} = 289; (15)^{2} = 225; (8)^{2} = 64

Now, 225 + 64 = 289

∴ (15)^{2} + (8)^{2} = (17)^{2}

Thus, (8, 15, 17) forms a Pythagorean triplet.

Hence, the given triangle with sides 8, 15 and 17 is a right-angled triangle.

(ii) The sides of the given triangle is 11, 12 and 15.

Let us check whether the given set (11, 12, 15) forms a Pythagorean triplet or not.

The biggest number among the given set is 15.

(15)^{2} = 225; (11)^{2} = 121; (12)^{2} = 144

Now, 121 + 144 = 265 ≠ 225

∴ (11)^{2} + (12)^{2} ≠ (15)^{2}

Thus, (11, 12, 15) does not form a Pythagorean triplet.

Hence, the given triangle with sides 8, 15 and 17 is not a right-angled triangle.

(iii) The sides of the given triangle is 11, 60 and 61.

Let us check whether the given set (11, 60, 61) forms a Pythagorean triplet or not.

The biggest number among the given set is 61.

(61)^{2} = 3721; (11)^{2} = 121; (60)^{2} = 3600

Now, 121 + 3600 = 3721

∴ (11)^{2} + (60)^{2} = (61)^{2}

Thus, (11, 60, 61) forms a Pythagorean triplet.

Hence, the given triangle with sides 11, 60 and 61 is a right-angled triangle.

(iv) The sides of the given triangle is 1.5, 1.6 and 1.7.

Let us check whether the given set (1.5, 1.6, 1.7) forms a Pythagorean triplet or not.

The biggest number among the given set is 1.7.

(1.7)^{2} = 2.89; (1.5)^{2} = 2.25; (1.6)^{2} = 2.56

Now, 2.25 + 2.56 = 4.81 ≠ 2.89

∴ (1.5)^{2} + (1.6)^{2} ≠ (1.7)^{2}

Thus, (1.5, 1.6, 1.7) does not form a Pythagorean triplet.

Hence, the given triangle with sides 1.5, 1.6 and 1.7 is not a right-angled triangle.

(v) The sides of the given triangle is 40, 20 and 30.

Let us check whether the given set (40, 20, 30) forms a Pythagorean triplet or not.

The biggest number among the given set is 40.

(40)^{2} = 1600; (20)^{2} = 400; (30)^{2} = 900

Now, 400 + 900 = 1300 ≠ 1600

∴ (20)^{2} + (30)^{2} ≠ (40)^{2}

Thus, (40, 20, 30) does not form a Pythagorean triplet.

Hence, the given triangle with sides 40, 20 and 30 is not a right-angled triangle.

View NCERT Solutions for all chapters of Class 7