IPMAT Indore 2026 · Quantitative Ability · Question 12
IPMAT Indore 2026 previous year question on Quadrilaterals, with the answer and a step-by-step solution.
Let ABCD be a cyclic quadrilateral with AB = 15, BC = 20, CD = 24 and AC = 25. Then AD equals_________
Answer & solution
Answer: 7
Step 1 — look at triangle ABC first. We know all three of its sides: AB=15, BC=20, AC=25. Check whether they form a right triangle:
{15}^{2}+{20}^{2}=225+400=625={25}^{2}
They do, so \angle ABC={90}^{∘}, with AC as the hypotenuse.
Step 2 — use the cyclic property. In a cyclic quadrilateral, opposite angles add to {180}^{∘}:
\angle ADC={180}^{∘}-\angle ABC={180}^{∘}-{90}^{∘}={90}^{∘}
Step 3 — now triangle ADC is right-angled at D, with the same hypotenuse AC=25:
A{D}^{2}+C{D}^{2}=A{C}^{2}\Rightarrow A{D}^{2}=625-{24}^{2}=625-576=49
AD=7
Notice 7–24–25 is the second Pythagorean triple hiding in this problem.
Answer: 7
Shortcut. Spot the triples. 15–20–25 is 5\times \left( 3,4,5 \right), so \angle B={90}^{∘}; the cyclic property instantly makes \angle D={90}^{∘}; and 24–25 demands the 7–24–25 triple. Answer in about ten seconds, with no algebra at all.
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