Class 8 Mathematics

Chapter 9 Solutions — The Baudhayana-Pythagoras Theorem

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

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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.]

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