Std 6 Mathematics · Chapter 16

16. Baudhayana-Pythagoras Theorem

Solutions to Practice Set 16.1, Practice Set 16.2 and Exercise 16. Maharashtra State Board (Balbharati), Grade Six Mathematics.

About the chapter

This chapter is about right angled triangles. The side opposite the right angle is the hypotenuse. It is the longest side. The Baudhayana-Pythagoras theorem gives a relation between the three sides. We use it to find an unknown side and to check whether a triangle is right angled.

Key points

In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
The Indian mathematician Baudhayana stated this rule in his Shulba Sutra.
Three numbers in which the square of the largest number is equal to the sum of the squares of the other two numbers form a triplet.
If the sides of a triangle form a triplet, the triangle is right angled.
If every number of a triplet is multiplied by the same number, we get another triplet.

Formulas and terms

Hypotenuse – the side opposite the right angle.
(Hypotenuse)² = (side 1)² + (side 2)²
(Side)² = (Hypotenuse)² − (other side)²
Baudhayana-Pythagorean triplet – like (3, 4, 5), (5, 12, 13), (8, 15, 17).

Practice Set 16.1

1Look at the figures below and find the length of the unknown side.

i) ∠M is the right angle, so LN is the hypotenuse.

ii) ∠B is the right angle, so DF is the hypotenuse.

i) In △LMN, ∠M = 90°, l(LM) = 9, l(MN) = 12. Find l(LN).

By the Baudhayana-Pythagoras theorem,
[l(LN)]² = [l(LM)]² + [l(MN)]²
∴ [l(LN)]² = 9² + 12²
∴ [l(LN)]² = 81 + 144
∴ [l(LN)]² = 225
∴ l(LN) = 15 … (15² = 225)

The length of side LN is 15.

ii) In △BDF, ∠B = 90°, l(BD) = 8, l(BF) = 15. Find l(DF).

[l(DF)]² = [l(BD)]² + [l(BF)]²
∴ [l(DF)]² = 8² + 15²
∴ [l(DF)]² = 64 + 225
∴ [l(DF)]² = 289
∴ l(DF) = 17 … (17² = 289)

The length of side DF is 17.

2A person walks 12 m. east from his house. There he turns north at a right angle and stops after walking 9 m. How far is the person from his original place?

Walking east, then turning north at a right angle.

The two paths make a right angle. The distance from the house is the hypotenuse.

(hypotenuse)² = (side 1)² + (side 2)²
= 12² + 9²
= 144 + 81
= 225
∴ hypotenuse = 15 m … (15² = 225)

The person is 15 metres away from his house.

3A ladder 35 ft. long reaches a window 21 ft. above the ground. Find the distance between the base of the wall and the lower end of the ladder.

The ladder is the hypotenuse of a right angled triangle.

The wall and the ground make a right angle. The ladder is the hypotenuse.

(35)² = (21)² + (distance)²
∴ 1225 = 441 + (distance)²
∴ (distance)² = 1225 − 441
∴ (distance)² = 784
∴ distance = 28 ft … (28² = 784)

The lower end of the ladder is 28 feet from the base of the wall.

Practice Set 16.2

1Determine which are the Baudhayana-Pythagoras triplets in the number set given below. i) 4, 5, 6 ii) 2, 6, 7 iii) 15, 20, 25

In a triplet, the square of the largest number is equal to the sum of the squares of the other two numbers.

i) 4, 5, 6

Sum of squares of the smaller numbers = 4² + 5² = 16 + 25 = 41
Square of the largest number = 6² = 36
41 ≠ 36

(4, 5, 6) is not a Baudhayana-Pythagorean triplet.

ii) 2, 6, 7

2² + 6² = 4 + 36 = 40
7² = 49
40 ≠ 49

(2, 6, 7) is not a Baudhayana-Pythagorean triplet.

iii) 15, 20, 25

15² + 20² = 225 + 400 = 625
25² = 625
625 = 625

(15, 20, 25) is a Baudhayana-Pythagorean triplet.

2The sides of some triangles are given below. Find out which ones are right-angled triangles? i) 11, 60, 61 ii) 40, 20, 30 iii) 9, 40, 41

i) 11, 60, 61

11² + 60² = 121 + 3600 = 3721
61² = 3721
3721 = 3721

The sides form a triplet. So, it is a right angled triangle.

ii) 40, 20, 30

The largest side is 40.
20² + 30² = 400 + 900 = 1300
40² = 1600
1300 ≠ 1600

The sides do not form a triplet. So, it is not a right angled triangle.

iii) 9, 40, 41

9² + 40² = 81 + 1600 = 1681
41² = 1681
1681 = 1681

The sides form a triplet. So, it is a right angled triangle.

Exercise 16

1Choose the correct option.

i) The hypotenuse of a right-angled triangle is 41 cm and one side forming the right angle is 40 cm. What is the length of the other side? — c) 9 cm

(41)² = (40)² + (other side)²
∴ 1681 = 1600 + (other side)²
∴ (other side)² = 81
∴ other side = 9 cm

ii) A ladder 17 m long is placed against a wall at a distance of 8 m from it. At what height from the ground will the top of the ladder be? — a) 15 m

(17)² = (8)² + (height)²
∴ 289 = 64 + (height)²
∴ (height)² = 225
∴ height = 15 m

iii) Two pillars of height 20 m and 27 m are standing on the ground 24 m apart. What is the distance between their upper ends? — c) 25 m

Difference in heights = 27 − 20 = 7 m
Distance between the pillars = 24 m
(distance between the tops)² = 24² + 7²
= 576 + 49 = 625
∴ distance between the tops = 25 m

2Observe the adjacent figure and find the length of the unknown side.

∠Q is the right angle, so PR is the hypotenuse.

In △PQR, ∠Q = 90°, so PR is the hypotenuse.
[l(PR)]² = [l(PQ)]² + [l(QR)]²
∴ [l(PR)]² = (30)² + (40)²
∴ [l(PR)]² = 900 + 1600
∴ [l(PR)]² = 2500
∴ l(PR) = 50 cm … (50² = 2500)

The length of side PR is 50 cm.

3Determine whether the set of numbers (6, 8, 10) is a Baudhayana-Pythagorean triplet or not.

Square of the largest number = 10² = 100
Sum of squares of the smaller numbers = 6² + 8² = 36 + 64 = 100
100 = 100

So, (6, 8, 10) is a Baudhayana-Pythagorean triplet.

4Find Baudhayana-Pythagorean triplets.

Some Baudhayana-Pythagorean triplets are:

(3, 4, 5) – 9 + 16 = 25
(5, 12, 13) – 25 + 144 = 169
(8, 15, 17) – 64 + 225 = 289
(7, 24, 25) – 49 + 576 = 625
(20, 21, 29) – 400 + 441 = 841
(9, 40, 41) – 81 + 1600 = 1681

Multiply each number of (3, 4, 5) by 2, 3, 4, … to get more triplets: (6, 8, 10), (9, 12, 15), (12, 16, 20).

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