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Theorem of Pythagoras – Maharashtra Board Class 7 Solutions for Mathematics

Theorem of Pythagoras – Maharashtra Board Class 7 Solutions for Mathematics (English Medium)

MathematicsGeneral ScienceMaharashtra Board Solutions

Exercise 19:

Solution 1:

Theorem of Pythagoras - Maharashtra Board Class 7 Solutions for Mathematics-Ex-19-1

Exercise 20:

Solution 1:

By the Pythagoras Theorem,
(Hypotenuse)2 = (One Side)2 + (Other Side)2

  • (Hypotenuse)2 = (24)2 + (10)2 = 576 + 100 = 676
    (Hypotenuse)2 = (26)2
    Hypotenuse = 26
  • (Hypotenuse)2 = (9)2 + (12)2 = 81 + 144 = 225
    (Hypotenuse)2 = (15)2
    Hypotenuse = 15
  • (Hypotenuse)2 = (16)2 + (12)2 = 256 + 144 = 400
    (Hypotenuse)2 = (20)2
    Hypotenuse = 20
  • (Hypotenuse)2 = (18)2 + (24)2 = 324 + 576 = 900
    (Hypotenuse)2 = (30)2
    Hypotenuse = 30
  • Hypotenuse)2 = (20)2 + (21)2 = 400 + 441 = 841
    (Hypotenuse)2 = (29)2
    Hypotenuse = 29
  • (Hypotenuse)2 = (1.5)2 + (2.0)2 = 2.25 + 4.00 = 6.25
    (Hypotenuse)2 = (2.5)2
    Hypotenuse = 2.5
  • (Hypotenuse)2 = (2.7)2 + (3.6)2 = 7.29 + 12.96 = 20.25
    (Hypotenuse)2 = (4.5)2
    Hypotenuse = 4.5
  • (Hypotenuse)2 = (2.1)2 + (2.8)2 = 4.41 + 7.84 = 12.25
    (Hypotenuse)2 = (3.5)2
    Hypotenuse = 3.5

Solution 2:

Theorem of Pythagoras - Maharashtra Board Class 7 Solutions for Mathematics-Ex-20-2

Given, the diagonal of a rectangle divides it into two right angled triangles.
The sides of the right angled triangle forming the right angle are 15 m and 18 m. The diagonal of the rectangle is the hypotenuse of the right angled triangle.
∴ Diagonal = Hypotenuse
(Hypotenuse)2 = (One Side) 2 + (Other Side) 2
∴ (Hypotenuse)2 = (15) 2 + (8) 2 = 225 + 64 = 289
∴ (Hypotenuse)2 = (17) 2
∴ Hypotenuse = 17
∴ Diagonal = Hypotenuse = 17
Thus, the length of the diagonal of the rectangular piece of land is 17 m.

Solution 3:

From the figure,
In ΔLMN,
m∠M = 90°
Side LN is the hypotenuse.
By the Pythagoras Theorem,
[l(MN)]2 + [l(ML)]2 = [l(LN)]2
Thus, statement (2) is true.

Exercise 21:

Solution 1:

  • Given, Hypotenuse = 45 and One adjacent side = 27
    By the Pythagoras Theorem,
    (Hypotenuse)2 = (One  adjacent side)2 + (Second adjacent side)2
    (45)2 = (27)2 + (Second adjacent side)2
    (Second adjacent side)2 = (45)2 – (27)2 = 2025 – 729 = 1296
    (Second adjacent side)2 = (36)2
    Second adjacent side = 36
    Thus, the length of the second adjacent side is 36.
  • Given, Hypotenuse = 50 and Second adjacent side = 14
    By Pythagoras Theorem,
    (Hypotenuse)2 = (One  adjacent side)2 + (Second adjacent side)2
    (50)2 = (One  adjacent side)2 + (14)2
    (One adjacent side)2 = (50)2 – (14)2 = 2500 – 196 = 2304
    (One adjacent side)2 = (48)2
    One adjacent side = 48
    Thus, the length of one adjacent side is 48.
  • Given, Hypotenuse = 6.1 and One adjacent side = 6.0
    By the Pythagoras Theorem,
    (Hypotenuse)2 = (One  adjacent side)2 + (Second adjacent side)2
    (6.1)2 = (6.0)2 + (Second adjacent side)2
    (Second adjacent side)2 = (6.1)2 – (6.0)2 = 37.21 – 36.00 = 1.21
    (Second adjacent side)2 = (1.1)2
    Second adjacent side = 1.1
    Thus, the length of the second adjacent side is 1.1.
  • Given, Hypotenuse = 3.7 and Second adjacent side = 3.5,
    By the Pythagoras Theorem,
    (Hypotenuse)2 = (One  adjacent side)2 + (Second adjacent side)2
    (3.7)2 = (One  adjacent side)2 + (Second adjacent side)2
    (3.7)2 = (One  adjacent side)2 + (3.5)2
    (One adjacent side)2 = (3.7)2 – (3.5)2 = 13.69 – 12.25 = 1.44
    (One adjacent side)2 = (1.2)2
    One adjacent side = 1.2
    Thus, the length of one adjacent side is 1.2.
  • Given, Hypotenuse = 25 and One adjacent side = 7
    By the Pythagoras Theorem,
    (Hypotenuse)2 = (One  adjacent side)2 + (Second adjacent side)2
    (25)2 = (7)2 + (Second adjacent side)2
    (Second adjacent side)2 = (25)2 – (7)2 = 625 – 49 = 576
    (Second adjacent side)2 = (24)2
    Second adjacent side = 24
    Thus, the length of the second adjacent side is 24.
  • Given, Hypotenuse = 12.5 and Second adjacent side = 10
    By the Pythagoras Theorem,
    (Hypotenuse)2 = (One  adjacent side)2 + (Second adjacent side)2
    (12.5)2 = (One  adjacent side)2 + (10)2
    (One adjacent side)2 = (12.5)2 – (10)2 = 156.25 – 100 = 56.25
    (One adjacent side)2 = (7.5)2
    One adjacent Side = 7.5
    Thus, the length of one adjacent side is 7.5.
  • Given, Hypotenuse = 7.5 and One adjacent side = 6
    By the Pythagoras Theorem,
    (Hypotenuse)2 = (One adjacent side)2 + (Second adjacent side)2
    (7.5)2 = (6)2 + (Second adjacent side)2
    (Second adjacent side)2 = (7.5)2 – (6)2 = 56.25 – 36 = 20.25
    (Second adjacent side)2 = (4.5)2
    Second adjacent side = 4.5
    Thus, the length of the second adjacent side is 4.5.

Solution 2:

The length of the hypotenuse of the right angled triangle is same as the diagonal of the rectangle.

  1. Given, diagonal of the rectangle = Hypotenuse = 20 cm, One side = 16 cm.
    By the Pythagoras Theorem,
    (Hypotenuse)2 = (One side)2 + (Other side)2
    ∴ (20)2 = (16)2 + (Other side)2
    ∴ (Other side)2 = (20)2 – (16)2 = 400 – 256 = 144
    ∴ (Other side)2 = (12)2
    ∴ Other side = 12 cm
  2. Given, diagonal of the rectangle = Hypotenuse = 3.9 cm, One side = 15 cm.
    By the Pythagoras Theorem,
    (Hypotenuse)2 = (One side)2 + (Other side)2
    ∴ (3.9)2 = (1.5)2 + (Other side)2
    ∴ (Other side)2 = (3.9)2 – (1.5)2 = 15.21 – 2.25 = 12.96
    ∴ (Other side)2 = (3.6)2
    ∴ Other side = 3.6 cm
  3. Given, diagonal of the rectangle = Hypotenuse = 41 cm, One side = 40 cm.
    By the Pythagoras Theorem,
    (Hypotenuse)2 = (One side)2 + (Other side)2
    ∴ (41)2 = (40)2 + (Other side)2
    ∴ (Other side)2 = (41)2 – (40)2 = 1681 – 1600 = 81
    ∴ (Other side)2 = (9)2
    ∴ Other side = 9 cm

Exercise 22:

Solution 1:

  • Length of the longest side = 2.5 m
    (2.5)2 = 6.25
    Sum of squares of the other two sides are = (1.5)2 + (2)2 = 2.25 + 4 = 6.25
    Thus, the square of the longest side is equal to the sum of the squares of the other two sides.
    Hence, the given triangle is a right angled triangle.
  • Length of the longest side = 6 m
    (6)2 = 36
    Sum of squares of the other two sides are = (3)2 + (4)2 = 9 + 16 = 25
    Since the square of the longest side is not equal to the sum of the squares of the other two sides, the given triangle is not a right angled triangle.
  • Length of the longest side = 8 m
    (8)2 = 64
    Sum of squares of the other two sides are = (6)2 + (7)2 = 36 + 49 = 85
    Since the square of the longest side is not equal to the sum of the squares of the other two sides, the given triangle is not a right angled triangle.
  • Length of the longest side = 13 cm
    (13)2 = 169
    Sum of squares of the other two sides are = (5)2 + (12)2 = 25 + 144 = 169
    Thus, the square of the longest side is equal to the sum of the squares of the other two sides.
    Hence, the given triangle is a right angled triangle.
  • Length of the longest side = 41 cm
    (41)2 = 1681
    Sum of squares of the other two sides are = (9)2 + (40)2 = 81 + 1600 = 1681
    Thus, the square of the longest side is equal to the sum of the squares of the other two sides.
    Hence, the given triangle is a right angled triangle.
  • Length of the longest side = 1.7 cm
    (1.7)2 = 2.89
    Sum of squares of the other two sides are = (1.5)2 + (1.6)2 = 2.25 + 2.56 = 4.81
    Since the square of the longest side is not equal to the sum of the squares of the other two sides, the given triangle is not a right angled triangle.

Solution 2:

Given, l(KL) = 9 cm, l(LM) = 12 cm, l(KM) = 15 cm.
Length of the longest side = KM = 15 cm
(KM)2 = (15)2 = 225
Sum of the squares of the other two sides = (9)2 + (12)2 = 81 + 144 = 225
By the Pythagoras Theorem,
ΔKLM is a right angled triangle.
KM is the hypotenuse.
In ΔKLM , ∠L is a right angled triangle.
∴ m∠L = 90°

Solution 3:

We know that the sum of the measures of all angles of a triangle is 180°.
In ΔLMN,
m∠L + m∠M + m∠N = 180°
∴ m∠L + 40° + 50° = 180°
∴ m∠L = 180° – 40° – 50° = 90°
Thus, ΔLMN is a right angled triangle.
MN is the hypotenuse.
By the Pythagoras Theorem,
[l(MN)]2 = [l(LM)]2 + [l(LN)]2
Thus, statement 1 is true.

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