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Class 8 Maths: The Baudhayana-Pythagoras Theorem, 8 practice questions with answers

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These are 8 practice questions on Chapter 2, The Baudhayana-Pythagoras Theorem, of the new NCERT Class 8 Maths textbook, Ganita Prakash, Part 2. Each has a worked answer and the textbook page it comes from. They cover a chain of squares built on diagonals, why A-size paper is √2 times as long as it is wide, the area of an isosceles right triangle from its hypotenuse, how many decimal places the square of a decimal has, the longest rod in a box, why adding the same number never keeps a triple, why hypotenuse plus longer leg is a square, and two screens with the same diagonal. Try each one on paper before you open the answer.

These questions are our own, drafted with AI help. Every answer was re-worked by a test script and each question was compared with the chapter's examples, activities and exercises, but no practising teacher has reviewed them yet. If you find a mistake, please tell us.

What the chapter says you need

Facts from the NCERT textbook (Ganita Prakash, Grade 8, Part 2, Chapter 2, printed pages 33 to 54):

Squares, halves and square roots

Question 1. Start with a square of side 3 cm. Build a new square on its diagonal, then another square on the diagonal of that one, and so on, four times in all. (a) What is the area of each square in the chain? (b) What is the side of the fourth new square? (c) Why does the side double after every two steps, although the area doubles at every step?

Answer: (a) 9, 18, 36, 72 and 144 cm²; (b) 12 cm; (c) two doublings of the area make it 4 times as big, and a square with 4 times the area has twice the side. (a) The square built on the diagonal of a square has double its area, so the areas go 9, 18, 36, 72 and 144 cm². (b) The last square has area 144 cm², so its side is 12 cm, because 12 × 12 = 144. (c) After two steps the area is 2 × 2 = 4 times as large, and a square with 4 times the area has a side twice as long (9 to 36 cm² takes the side from 3 to 6 cm; 36 to 144 cm² takes it from 6 to 12 cm). After one step only, the side is not a whole number of times longer. (A chain of squares, each double the one before, page 35.)

Question 2. A sheet of A-size paper has the useful property that folding it in half across its longer side gives a smaller sheet of exactly the same shape. (a) If the sides are l (long) and s (short), show that l² = 2s², so l ÷ s = √2. (b) An A4 sheet is 21 cm by 29.7 cm. How close is 29.7 ÷ 21 to √2?

Answer: (a) the half sheet is s by l/2, and the same shape needs l ÷ s = s ÷ (l/2), which gives l² = 2s²; (b) 29.7 ÷ 21 ≈ 1.414, very close to √2. (a) Folding across the long side gives a sheet whose sides are s and l/2, with s now the longer side. The same shape means the long side divided by the short side is the same for both: l ÷ s = s ÷ (l/2) = 2s ÷ l. Multiplying both sides by ls gives l² = 2s², so (l ÷ s)² = 2 and l ÷ s = √2. (b) 29.7 ÷ 21 = 1.4142…, and √2 lies between 1.414 and 1.415, so the A4 sheet matches √2 to three decimal places. (The paper sizes are from outside the chapter; √2 and its bounds, pages 37 and 38.)

Question 3. The two equal sides of an isosceles right triangle are not given, but its area is 162 cm². (a) How long are the equal sides? (b) Show that for ANY isosceles right triangle, the area is the square of the hypotenuse divided by 4, and check it with this triangle.

Answer: (a) 18 cm; (b) area = a²/2 and c² = 2a², so area = c²/4; here 648 ÷ 4 = 162. (a) The triangle is half of a square whose side is one of the equal sides, so that square has area 2 × 162 = 324 cm², and the equal sides are 18 cm (18 × 18 = 324). (b) With equal sides a and hypotenuse c, the area is a × a ÷ 2 = a²/2, and the chapter shows c² = 2a², so a² = c²/2 and the area is c²/4. Check: c² = 2 × 324 = 648, and 648 ÷ 4 = 162. (The hypotenuse of an isosceles right triangle, c² = 2a², page 40.)

Question 4. (a) A square's side is a decimal with exactly 2 digits after the point, the last of them not 0. How many digits after the point does its area have, and why is the last one never 0? (b) Can a square whose area is written with exactly 3 digits after the point (the last not 0) have a side that is a terminating decimal? Explain.

Answer: (a) exactly 4, and the last is never 0; (b) no. (a) A side with 2 digits after the point is a whole number of hundredths, so its square is a whole number of ten-thousandths: 4 places. The last of them comes from the side's last digit times itself, and the squares of 1 to 9 end in 1, 4, 9, 6, 5, 6, 9, 4 and 1, never 0. For example, 1.25 × 1.25 = 1.5625. (b) In the same way, a side with k digits after the point (the last not 0) gives an area with exactly 2k digits after the point, an even number. An area with exactly 3 digits after the point would need 2k = 3, which no whole number k gives, so its side cannot be a terminating decimal. (Squares of decimals and their last digits, pages 38 and 39.)

Right triangles and triples

Question 5. What is the longest straight rod that fits inside a closed box 12 cm long, 9 cm wide and 8 cm tall, from one bottom corner to the opposite top corner?

Answer: 17 cm. First, the diagonal of the 12 cm by 9 cm floor: it is the hypotenuse of a right triangle with sides 12 and 9, so its square is 144 + 81 = 225, and the floor diagonal is 15 cm. The rod, that floor diagonal and the 8 cm upright edge make a second right triangle (the edge stands at right angles to the floor), so the rod's square is 15² + 8² = 225 + 64 = 289, and the rod is 17 cm. (The Baudhayana-Pythagoras theorem, pages 46 and 47, used twice.)

Question 6. Ria adds 1 to each number of the triple (20, 21, 29) and gets (21, 22, 30). It is not a triple, because 21² + 22² = 925 but 30² = 900. Show that adding the same positive number d to all three numbers of ANY Baudhayana triple can never give a triple.

Answer: the two smaller squares always add up to more than the largest square. Let (a, b, c) be a triple, so a² + b² = c². Then (a + d)² + (b + d)² = a² + b² + 2d(a + b) + 2d², while (c + d)² = c² + 2dc + d². Taking the second from the first leaves 2d(a + b − c) + d². In a triangle the two shorter sides add up to more than the longest, so a + b − c is positive, and d² is positive too. So the sum of the two smaller squares is always bigger than the largest square, and the new numbers are never a triple. (Working with triples algebraically, as the chapter does with (ka)², page 49.)

Question 7. The triple (13, 84, 85) comes from the chapter's odd-square method, and 85 + 84 = 169 = 13². Explain why, in every triple made by this method, the hypotenuse plus the longer leg equals the square of the shorter leg.

Answer: the method's triple is (m, n − 1, n) with m² = 2n − 1, and n + (n − 1) = 2n − 1. The method starts from (n − 1)² + (2n − 1) = n², where the odd number 2n − 1 is a square, say m². Then m² + (n − 1)² = n², so the triple is (m, n − 1, n): the hypotenuse is n and the longer leg is n − 1. Their sum is n + (n − 1) = 2n − 1 = m², the square of the shorter leg. For (13, 84, 85): n = 85, and 2 × 85 − 1 = 169 = 13². (Triples from odd squares, pages 49 and 50.)

Question 8. Two screens both have a diagonal of 65 inches. One is 4 : 3 in shape (width to height) and the other 12 : 5. Find the width, height and area of each.

Answer: 52 by 39 inches (2,028 square inches) and 60 by 25 inches (1,500 square inches). A 4 : 3 rectangle has sides 4k and 3k, and its diagonal is 5k, because (4k)² + (3k)² = 25k². So 5k = 65, k = 13, and the screen is 52 by 39, area 52 × 39 = 2,028. A 12 : 5 rectangle has sides 12k and 5k and diagonal 13k, because 144k² + 25k² = 169k². So 13k = 65, k = 5, and the screen is 60 by 25, area 1,500. The same diagonal can hold very different areas. (The theorem, page 46; scaled triples, page 49.)

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Sources

All opened and read on 30 October 2026. Page numbers are the printed page numbers of the textbook; printed page = PDF page + 32.

The rules are from the textbook pages above. The questions, worked answers and checks are our own, and every number was re-computed by script. This guide is general information; follow your teacher and the textbook.

Practise with the worksheet: Baudhayana-Pythagoras Theorem, Class 8 worksheet. Printable, in English, Hindi and Marathi, with an answer key.