IPMAT Indore 2022 · Quantitative Ability · Question 20

Step 1 — identify the midsegment. D is the midpoint of AB and E the midpoint of AC, so DE joins the midpoints of two sides of triangle ABC. By the midpoint theorem, DE is parallel to the third side BC and half its length:

DE=\frac{BC}{2}

Step 2 — solve for BC.

6=\frac{BC}{2}\Rightarrow BC=12

Step 3 — sanity-check against the right angle. The triangle is right-angled with hypotenuse AC=13, so

AB=\sqrt{{13}^{2}-{12}^{2}}=\sqrt{169-144}=\sqrt{25}=5

giving the familiar 5–12–13 triple ✔ Everything is consistent.

Answer: Option 1 — 12 cm

Shortcut. The midpoint theorem answers this in one step: BC=2\times DE=12. The hypotenuse length is not even needed — it is there to let you confirm the 5–12–13 triple, and to tempt you into Pythagoras first.

Option (b) 5 cm is the trap: it is the other leg, which you get if you use the right angle instead of the midsegment.

Recognise the three small Pythagorean triples on sight — 3–4–5, 5–12–13, 8–15–17. Seeing 13 with a leg of 12 should immediately suggest 5, which is the consistency check that confirms the answer.