IPMAT Indore 2024 · Quantitative Ability · Question 4

Step 1 — set up coordinates. Put B at the origin with the legs along the axes: B\left( 0,0 \right), A\left( 0,18 \right), C\left( 18,0 \right). The hypotenuse AC is the line

x+y=18

Step 2 — describe the rectangle. With B as one vertex and the sides along the legs, the rectangle has corners \left( 0,0 \right), \left( x,0 \right), \left( x,y \right), \left( 0,y \right). For it to fit inside, the far corner \left( x,y \right) must lie on or below the hypotenuse:

x+y\le 18

Step 3 — maximise the area. Area =xy subject to x+y\le 18. The maximum sits on the boundary x+y=18, and for a fixed sum the product is largest when the two are equal (AM–GM):

\frac{x+y}{2}\ge \sqrt{xy}\Rightarrow xy\le{\left( \frac{18}{2}\right)}^{2}=81

with equality at x=y=9.

Answer: 81

Shortcut. Memorise the result. The largest rectangle inscribed in a triangle with a vertex at the right angle is the one whose far corner is the midpoint of the hypotenuse, and its area is exactly half the triangle’s. Here the triangle is 1/2\left( 18 \right)\left( 18 \right)=162, so the rectangle is 81 — no calculus, no AM–GM, one halving.