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

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