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

MathematicsGeneral ScienceMaharashtra Board Solutions

**Exercise 19:**

**Solution 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:**

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.

- 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 - 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 - 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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