# Maharashtra Board Class 7 Math Chapter 13 Pythagoras Theorem Solution

## Maharashtra Board Class 7 Math Solution Chapter 13 – Pythagoras Theorem

Balbharati Maharashtra Board Class 7 Math Solution Chapter 13: Pythagoras Theorem. Marathi or English Medium Students of Class 7 get here Pythagoras Theorem full Exercise Solution.

 Std Maharashtra Class 7 Subject Math Solution Chapter Pythagoras Theorem

1.) In the figures below, find the value of ‘x’.

ANS:

(i)

Here given that,

l (LM) = 7 and l(MN) = 24

We have to find l (LN) =?

Here, M = 900

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (base) 2 + (height) 2

Hypotenuse is the largest side opposite to 900

LN is Hypotenuse.

LM is Base.

MN is Height.

LN2 = LM2 + MN2

LN2 = 72 + 242

LN2 = 49 + 576

LN2 = 625

LN = 25

Value of x = 25

(ii)

ANS:

Here given that,

l (PQ) = 9 and l(PR) = 41

We have to find l (QR) =?

Here, Q = 900

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (base) 2 + (height) 2

Hypotenuse is the largest side opposite to 900

PR is Hypotenuse.

PQ is Base.

QR is Height.

PR2 = PQ2 + QR2

412 = 92 + QR2

1681 – 81 =QR2

QR2 = 1600

QR = 40

Value of x = 40

(iii)

ANS:

Here given that,

l (DF) = 8 and l(EF) = 17

We have to find l (DE) =?

Here, D = 900

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (base) 2 + (height) 2

Hypotenuse is the largest side opposite to 900

EF is Hypotenuse.

DF is Base.

DE is Height.

EF2 = DF2 + DE2

172 = 82 + QR2

289 – 64 = QR2

QR2 = 225

QR = 15

Value of x = 15

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

ANS:

Given that,

The right-angled ∆PQR, ∠P = 90°.

l (PQ) = 24 cm and l(PR) = 10 cm.

We have to find Length of QR.

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (base) 2 + (height) 2

Hypotenuse is the largest side opposite to 900

QR is Hypotenuse.

PR is Base.

PQ is Height.

QR2 = PR2 + PQ2

QR2 = 102 + 242

QR2 = 100 + 576

QR2 = 676

QR = 26

Value of QR is 26 cm.

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

ANS:

Given that,

The right-angled ∆LMN, ∠M = 90°

l (LM) = 12 cm and l(LN) = 20 cm

We have to find length of seg MN.

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (base) 2 + (height) 2

Hypotenuse is the largest side opposite to 900

LN is Hypotenuse.

LM is Base.

MN is Height.

LN2 = LM2 + MN2

202 = 122 + MN2

MN2 = 400 – 144

MN2 = 256

MN = 16

Value of segment MN is 16 cm.

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?

ANS:

Given that,

The top of a ladder of length 15 m reaches a window 9 m above the ground.

This is a Right angled figure.

The top of a ladder of length 15 m. is the largest side which is called as Hypotaneous.

Hypotaneous of a figure = 15 m.

The top of a ladder of length 15 m reaches a window 9 m above the ground.

Height of a figure = 9 m.

Now,

We have to find the base of the figure.

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (base) 2 + (height) 2

Hypotenuse is the largest side opposite to 900

152 = (base) 2 +92

(Base) 2 = 152 – 92

(Base) 2 = 225 – 81

(Base) 2 = 144

Base = 12 m.

The distance between the base of the wall and that of the ladder is 12m.

### Practice Set 49

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

(i) 3, 4, 5

ANS:

We know,

The Largest side is called Hypotaneous.

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (side 1)2 + (side 2)2

Hypotenuse = 5

Side 1 = 3

Side 2 = 4

52 = 32 + 42

25 = 9 + 16

25 = 25

L.H.S = R.H.S

This is a Pythagorean triplets.

(ii) 2, 4, 5

ANS:

We know,

The Largest side is called Hypotaneous.

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (side 1)2 + (side 2)2

Hypotenuse = 5

Side 1 = 2

Side 2 = 4

52 = 22 + 42

25 = 4 + 16

25 ≠ 20

L.H.S ≠ R.H.S

This is a not Pythagorean triplets.

(iii) 4, 5, 6

ANS:

We know,

The Largest side is called Hypotaneous.

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (side 1)2 + (side 2)2

Hypotenuse = 6

Side 1 = 4

Side 2 = 5

62 = 42 + 52

36 = 16 + 25

36 ≠ 41

L.H.S ≠ R.H.S

This is a not Pythagorean triplets.

(iv) 2, 6, 7

ANS:

We know,

The Largest side is called Hypotaneous.

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (side 1)2 + (side 2)2

Hypotenuse = 7

Side 1 = 2

Side 2 = 6

72 = 22 + 62

49 = 4 + 36

49 ≠ 40

L.H.S ≠ R.H.S

This is a not Pythagorean triplets.

(v) 9, 40, 41

ANS:

We know,

The Largest side is called Hypotaneous.

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (side 1)2 + (side 2)2

Hypotenuse = 41

Side 1 = 40

Side 2 = 9

412 = 402 + 92

1681 = 1600 + 81

1681 = 1681

L.H.S = R.H.S

This is a Pythagorean triplets.

(vi) 4, 7, 8

ANS:

We know,

The Largest side is called Hypotaneous.

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (side 1)2 + (side 2)2

Hypotenuse = 8

Side 1 = 7

Side 2 = 4

82 = 72 + 42

64 = 49 + 16

64 ≠ 65

L.H.S ≠ R.H.S

This is a not Pythagorean triplets.

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

(i) 8, 15, 17

ANS:

We know that,

Any triangle is right-angled triangles only when it follows Pythagoras’ Theorem.

When it follows Pythagoras’ Theorem then the triangle is right-angled triangles.

We know,

The Largest side is called Hypotaneous.

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (side 1)2 + (side 2)2

Hypotenuse = 17

Side 1 = 8

Side 2 = 15

172 = 82 + 152

289 = 64 + 225

289 = 289

L.H.S = R.H.S

This follows Pythagoras’ Theorem

This triangle is right-angled triangle.

(ii) 11, 12, 15

ANS:

We know that,

Any triangle is right-angled triangles only when it follows Pythagoras’ Theorem.

When it follows Pythagoras’ Theorem then the triangle is right-angled triangles.

We know,

The Largest side is called Hypotaneous.

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (side 1)2 + (side 2)2

Hypotenuse = 15

Side 1 = 11

Side 2 = 12

152 = 112 + 122

225 = 121 + 144

225 ≠ 265

L.H.S ≠ R.H.S

This not follows Pythagoras’ Theorem.

This triangle is not right-angled triangle.

(iii) 11, 60, 61

ANS:

We know that,

Any triangle is right-angled triangles only when it follows Pythagoras’ Theorem.

When it follows Pythagoras’ Theorem then the triangle is right-angled triangles.

We know,

The Largest side is called Hypotaneous.

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (side 1)2 + (side 2)2

Hypotenuse = 61

Side 1 = 11

Side 2 = 60

612 = 112 + 602

3721 = 121 + 3600

3721 = 3721

L.H.S = R.H.S

This follows Pythagoras’ Theorem

This triangle is right-angled triangle.

(iv) 1.5, 1.6, 1.7

ANS:

We know that,

Any triangle is right-angled triangles only when it follows Pythagoras’ Theorem.

When it follows Pythagoras’ Theorem then the triangle is right-angled triangles.

We know,

The Largest side is called Hypotaneous.

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (side 1)2 + (side 2)2

Hypotenuse = 1.7

Side 1 = 1.5

Side 2 = 1.6

1.72 = 1.52 + 1.62

2.89 = 2.25 + 2.56

2.89 ≠ 4.81

L.H.S ≠ R.H.S

This not follows Pythagoras’ Theorem.

This triangle is not right-angled triangle.

(v) 40, 20, 30

ANS:

We know that,

Any triangle is right-angled triangles only when it follows Pythagoras’ Theorem.

When it follows Pythagoras’ Theorem then the triangle is right-angled triangles.

We know,

The Largest side is called Hypotaneous.

By Pythagoras’ Theorem,

(Hypotenuse) 2 = (side 1)2 + (side 2)2

Hypotenuse = 40

Side 1 = 20

Side 2 = 30

402 = 202 + 302

1600 = 400 + 900

1600 ≠ 1300

L.H.S ≠ R.H.S

This not follows Pythagoras’ Theorem.

This triangle is not right-angled triangle.

Updated: August 4, 2021 — 6:03 pm