Class 8 Mathematics

Chapter 9 Solutions — The Baudhayana-Pythagoras Theorem

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Solutions

Overview

Step-by-step NCERT solutions for The Baudhayana-Pythagoras Theorem (Chapter 9, NCERT Class 8 Mathematics) — every question and answer worked out in full, not just the final result. You can also read the The Baudhayana-Pythagoras Theorem textbook chapter.

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What these solutions cover

All 19 questions in The Baudhayana-Pythagoras Theorem are solved in the PDF. Here's what's inside, exercise by exercise:

Hypotenuse of an Isosceles Right Triangle

Earlier, we saw a method to create a square with double the area of a given square paper. There is another method to do this in which two identical square papers are cut diagonally into two triangles each (pieces 1, 2, 3, 4). Can you arrange these pieces to create a square with double the area of either square?
2.
The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse. Find bounds on the length of the hypotenuse such that they have at least one digit after the decimal point.
- (i) 3
- (ii) 4
- (iii) 6
- (iv) 8
- (v) 9
The hypotenuse of an isosceles right triangle is 10. What are its other two sidelengths? [Hint: Find the area of the square composed of two such right triangles.]

Using Baudhayana's Theorem

If a right-angled triangle has shorter sides of lengths 5 cm and 12 cm, then what is the length of its hypotenuse? First draw the right-angled triangle with these sidelengths and measure the hypotenuse, then check your answer using Baudhayana's Theorem.
If a right-angled triangle has a short side of length 8 cm and hypotenuse of length 17 cm, what is the length of the third side? Again, try drawing the triangle and measuring, and then check your answer using Baudhayana's Theorem.
Using the constructions you have now seen, how would you construct a square whose area is triple the area of a given square? Five times the area of a given square? (Baudhayana's Sulba-Sutra, Verse 1.10)
4.
Let a, b and c denote the length of the sides of a right triangle, with c being the length of the hypotenuse. Find the missing sidelength in each of the following cases:
- (i) a = 5, b = 7
- (ii) a = 8, b = 12
- (iii) a = 9, c = 15
- (iv) a = 7, b = 12
- (v) a = 1.5, b = 3.5

Right-Triangles Having Integer Sidelengths

Find 5 more Baudhayana triples using this idea. [The method: if the nth odd number (2n - 1) is a perfect square, then (n - 1)² + (2n - 1) = n² gives a Baudhayana triple.]
Does this method yield non-primitive Baudhayana triples? [Hint: Observe that among the triples generated, one of the smaller sidelengths is one less than the hypotenuse.]
Are there primitive triples that cannot be obtained through this method? If yes, give examples.

Further Applications of the Baudhayana-Pythagoras Theorem

Find the diagonal of a square with sidelength 5 cm.
2.
Find the missing sidelengths in the following right triangles (six triangles shown in figure):
- (a) legs 7 and 9, hypotenuse missing
- (b) legs 4 and 10 (right angle between them), hypotenuse missing
- (c) leg 40 and unknown leg, hypotenuse 41
- (d) leg 10 and unknown leg, hypotenuse √200
- (e) legs 10 and √150 (right angle between them), hypotenuse missing
- (f) leg 27 and unknown leg, hypotenuse 45
Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units.
Is the hypotenuse the longest side of a right triangle? Justify your answer.
True or False — Every Baudhayana triple is either a primitive triple or a scaled version of a primitive triple.
Give 5 examples of rectangles whose sidelengths and diagonals are all integers.
Construct a square whose area is equal to the difference of the areas of squares of sidelengths 5 units and 7 units.
8.
Using the dots of a grid as the vertices, can you create a square that has an area of
- (a) 2 sq. units,
- (b) 3 sq. units,
- (c) 4 sq. units, and
- (d) 5 sq. units?
- (ii) Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?
Find the area of an equilateral triangle with sidelength 6 units. [Hint: Show that an altitude bisects the opposite side. Use this to find the height.]

Sample

Worked examples

3 questions from The Baudhayana-Pythagoras Theorem, solved step by step.

Hypotenuse of an Isosceles Right Triangle

Earlier, we saw a method to create a square with double the area of a given square paper. There is another method to do this in which two identical square papers are cut diagonally into two triangles each (pieces 1, 2, 3, 4). Can you arrange these pieces to create a square with double the area of either square?

Each square is cut along one diagonal, giving two right triangles per square — four triangles in total.

Arrange the four triangles so that their hypotenuses (the diagonals of the original squares) form the sides of a new square.

Place the four triangles with their right-angle vertices pointing outward and hypotenuses facing inward. The four hypotenuses together form a square whose side equals the diagonal of the original square.

Area of new square = 4 × (area of one triangle) = 4 × (1/2 × area of original square) = 2 × area of original square.

Thus the new arrangement forms a square with double the area of either original square.

Answer: Arrange the four triangles with their hypotenuses forming the sides of a new tilted square; this square has double the area of either original square.

Using Baudhayana's Theorem

If a right-angled triangle has a short side of length 8 cm and hypotenuse of length 17 cm, what is the length of the third side? Again, try drawing the triangle and measuring, and then check your answer using Baudhayana's Theorem.

Let the known short side be a = 8 cm, the hypotenuse c = 17 cm, and the unknown side be b.

By Baudhayana's Theorem: a² + b² = c²

8² + b² = 17²

64 + b² = 289

b² = 289 - 64 = 225

b = √225 = 15 cm

Answer: The third side is 15 cm.

Right-Triangles Having Integer Sidelengths

Does this method yield non-primitive Baudhayana triples? [Hint: Observe that among the triples generated, one of the smaller sidelengths is one less than the hypotenuse.]

In every triple generated by this method, the two largest numbers are n-1 and n (consecutive integers).

For example: (3,4,5), (5,12,13), (7,24,25), (9,40,41), ...

In each case, the hypotenuse c = n and the larger leg b = n - 1 = c - 1.

Since b and c = b + 1 are consecutive integers, gcd(b, c) = 1.

Also, the smallest side is an odd number (the square root of the odd square used), and gcd(odd square root, n-1) needs checking.

For (3,4,5): gcd(3,4,5) = 1 — primitive.

For (5,12,13): gcd(5,12,13) = 1 — primitive.

For (7,24,25): gcd(7,24,25) = 1 — primitive.

In general, since c and c-1 are consecutive, they share no common factor. The smallest side is always odd, and one can verify it shares no common factor with the others.

Therefore, this method always yields primitive Baudhayana triples — it never produces non-primitive ones.

Answer: No. This method always produces primitive Baudhayana triples; the two larger numbers in each triple are consecutive integers and therefore share no common factor.

All 19 questions in The Baudhayana-Pythagoras Theorem are solved step by step in the Solutions PDF above.

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Chapter 9 Solutions — The Baudhayana-Pythagoras Theorem

Overview

What these solutions cover

Hypotenuse of an Isosceles Right Triangle

Using Baudhayana's Theorem

Right-Triangles Having Integer Sidelengths

Further Applications of the Baudhayana-Pythagoras Theorem

Worked examples

If a right-angled triangle has a short side of length 8 cm and hypotenuse of length 17 cm, what is the length of the third side? Again, try drawing the triangle and measuring, and then check your answer using Baudhayana's Theorem.

Does this method yield non-primitive Baudhayana triples? [Hint: Observe that among the triples generated, one of the smaller sidelengths is one less than the hypotenuse.]

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