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 = 90^{0}
By Pythagoras’ Theorem,
(Hypotenuse)^{ 2} = (base)^{ 2} + (height)^{ 2}
Hypotenuse is the largest side opposite to 90^{0}
LN is Hypotenuse.
LM is Base.
MN is Height.
LN^{2} = LM^{2} + MN^{2}
LN^{2} = 7^{2} + 24^{2}
LN^{2} = 49 + 576
LN^{2} = 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 = 90^{0}
By Pythagoras’ Theorem,
(Hypotenuse)^{ 2} = (base)^{ 2} + (height)^{ 2}
Hypotenuse is the largest side opposite to 90^{0}
PR is Hypotenuse.
PQ is Base.
QR is Height.
PR^{2} = PQ^{2} + QR^{2}
41^{2} = 9^{2} + QR^{2}
1681 – 81 =QR^{2}
QR^{2} = 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 = 90^{0}
By Pythagoras’ Theorem,
(Hypotenuse)^{ 2} = (base)^{ 2} + (height)^{ 2}
Hypotenuse is the largest side opposite to 90^{0}
EF is Hypotenuse.
DF is Base.
DE is Height.
EF^{2} = DF^{2} + DE^{2}
17^{2} = 8^{2} + QR^{2}
289 – 64 = QR^{2}
QR^{2} = 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 90^{0}
QR is Hypotenuse.
PR is Base.
PQ is Height.
QR^{2} = PR^{2} + PQ^{2}
QR^{2} = 10^{2} + 24^{2}
QR^{2} = 100 + 576
QR^{2} = 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 90^{0}
LN is Hypotenuse.
LM is Base.
MN is Height.
LN^{2} = LM^{2} + MN^{2}
20^{2} = 12^{2} + MN^{2}
MN^{2 }= 400 – 144
MN^{2 }= 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 90^{0}
15^{2} = (base)^{ 2} +9^{2}
(Base)^{ 2} = 15^{2} – 9^{2}
(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
5^{2 }= 3^{2} + 4^{2}
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
5^{2 }= 2^{2} + 4^{2}
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
6^{2 }= 4^{2} + 5^{2}
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
7^{2 }= 2^{2} + 6^{2}
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
41^{2 }= 40^{2} + 9^{2}
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
8^{2 }= 7^{2} + 4^{2}
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
17^{2 }= 8^{2} + 15^{2}
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
15^{2 }= 11^{2} + 12^{2}
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
61^{2 }= 11^{2} + 60^{2}
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
1.7^{2 }= 1.5^{2} + 1.6^{2}
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
40^{2 }= 20^{2} + 30^{2}
1600 = 400 + 900
1600 ≠ 1300
L.H.S ≠ R.H.S
This not follows Pythagoras’ Theorem.
This triangle is not right-angled triangle.