Extra Questions for Class 8 Maths Chapter 9: The Baudhayana-Pythagoras Theorem – HOTS

Original HOTS-level (Higher Order Thinking Skills) practice questions for Class 8 Maths Chapter 9 (The Baudhayana-Pythagoras Theorem), genuinely harder than the textbook exercise, using fresh numbers not found in the NCERT book. Full worked solutions included. These Class 8 Mathematics Chapter 9 important questions are handy for last-minute exam practice.

Q1 (Ladder problem). A 15 m ladder rests against a wall, its foot 9 m from the wall’s base. How high up the wall does it reach?
Solution: height² = 15² − 9² = 225 − 81 = 144 ⇒ height = 12 m.

Q2 (Generating a triple, Euclid-style). Using m = 3, n = 2, generate a Pythagorean triple via a = m²−n², b = 2mn, c = m²+n².
Solution: a = 9−4 = 5, b = 12, c = 13 ⇒ (5, 12, 13). Check: 25+144 = 169 = 13². ✓

Q3 (Assertion–Reason). Assertion (A): (9, 12, 15) is a primitive Pythagorean triple. Reason (R): A triple is primitive if the HCF of the three numbers is 1.
Solution: HCF(9,12,15) = 3 ≠ 1, so (9,12,15) is not primitive — it’s 3×(3,4,5). Assertion is false; Reason is true (R correctly defines primitivity, it’s just that the triple in A fails that test).

Q4 (Reverse problem). A rectangle has diagonal 25 cm and one side 24 cm. Find its perimeter.
Solution: Other side² = 25² − 24² = 625 − 576 = 49 ⇒ other side = 7 cm. Perimeter = 2(24+7) = 62 cm.

Q5 (Multi-step, sliding ladder). A 17 m ladder leans against a wall reaching 15 m up. If the foot of the ladder is pulled out by 7 m, how far down the wall does the top slide?
Solution: Initial foot distance = √(17²−15²) = √64 = 8 m. New foot distance = 8+7 = 15 m. New height = √(17²−15²) = √64 = 8 m. The top slides down from 15 m to 8 m — a drop of 7 m.

Q6 (Spot the error). A student claims that doubling all three sides of a right triangle also doubles its area. Check this claim.
Solution: Original area = ½ab. Doubled sides: ½(2a)(2b) = 4 × (½ab), i.e. 4 times the original area, not 2 times. The student’s claim is false.

Q7 (Lattice grid extension). Can a square of area 10 square units be drawn on a lattice/dot grid? If yes, give the vector.
Solution: 10 = 1²+3² = 1+9. Yes — use the vector (1,3).

Q8 (Proof, similar triangles). Prove that dropping an altitude from the right angle to the hypotenuse of a right triangle divides it into two smaller triangles, each similar to the original.
Solution: Let ▵ABC be right-angled at A, with altitude AD drawn to hypotenuse BC. In ▵ABD and ▵CBA: ∠ADB = ∠BAC = 90°, and ∠B is common to both, so ▵ABD ∼ ▵CBA (AA similarity). By the same reasoning with ∠C common, ▵ACD ∼ ▵BCA. Hence all three triangles — ▵ABD, ▵ACD and ▵ABC — are similar to one another.

Q9 (Integer hypotenuse). A right triangle has integer hypotenuse 25 units. List every possible pair of positive integer legs.
Solution: Need a²+b² = 625 with positive integers a ≤ b. 7²+24² = 49+576 = 625 ✓. 15²+20² = 225+400 = 625 ✓. So the pairs are (7, 24) — a primitive triple — and (15, 20), which is 5×(3,4,5).

Q10 (Real-world application). A rectangular TV screen advertised as “50 inches” (measured along its diagonal) has a width-to-height ratio of 16:9. Find its actual width and height, to 1 decimal place.
Solution: Let width = 16k, height = 9k. Diagonal² = (16k)²+(9k)² = 337k² = 50² = 2500 ⇒ k² = 2500/337 ≈ 7.418 ⇒ k ≈ 2.724. Width = 16k ≈ 43.6 in; Height = 9k ≈ 24.5 in.

More on This Chapter

See also: NCERT Solutions for Class 8 Maths Chapter 9, Class 8 Maths NCERT Book and the Class 8 Maths Formulas Handbook.

Revision Notes: Revision Notes for Class 8 Maths Chapter 9

Written by Satish

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