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.

#### Question 1:

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

(i)

In the right angled triangle LMN, ∠M = 90. Hence, side LN is the hypotenuse.
According to Pythagoras' theorem,
l(LN)2l(LM)2l(MN)2
⇒(x)2 = (7)2 + (24)2
x2 = 49 + 576
x2 = 625
x2 = (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 = x2 + 81
x2 = 1681 − 81
x2 = 1600
x2 = (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 = x2 + 64
x2 = 289 − 64
x2 = 225
x2 = (15)2
x = 15
∴ the value of x is 15.

#### 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.

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.

#### 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.

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.

#### 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 ?

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.

#### 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

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.
52 = 25; 42 = 16; 32 = 9
Now, 16 + 9 = 25
∴ 42 + 32 = 52
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.
52 = 25; 42 = 16; 22 = 4
Now, 16 + 4 = 20 ≠ 25
∴ 42 + 22 ≠  52
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.
62 = 36; 52 = 25; 42 = 16
Now, 25 + 16 = 41 ≠ 36
∴ 52 + 42 ≠  62
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.
72 = 49; 62 = 36; 22 = 4
Now, 4 + 36 = 40 ≠ 49
∴ 22 + 62 ≠  72
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.
92 = 81; 402 = 1600; 412 = 1681
Now, 81 + 1600 = 1681
∴ 92 + 402 = 412
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.
82 = 64; 72 = 49; 42 = 16
Now, 16 + 49 = 65 ≠ 64
∴ 42 + 72 ≠  82
Thus, (4, 7, 8) does not form a Pythagorean triplet.

#### 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

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