IPMAT Indore 2024 (SA) - Let ABC be a triangle right-angled at B with AB = BC = 18. The area of the largest rectangle that can be inscribed in this triangle and has B as one of the vertices is: | PYQs + Solutions

IPMAT Indore 2024

Geometry

Triangles

Easy

Let $\triangle ABC$ be a triangle right-angled at $B$ with $AB = BC = 18$ . The area of the largest rectangle that can be inscribed in this triangle and has $B$ as one of the vertices is:

Entered answer:

✅ Correct Answer: 81

Let's place

\triangle ABC

on a coordinate system with

B

at the origin

(0,0)

Since angle

B

90°

, we can place

A

(0,18)

and

C

(18,0)

Since

B(0,0)

is one vertex of the rectangle, the other vertices will be at

(x,0)

(0,y)

, and

(x,y)

where

(x,y)

lies on the hypotenuse

AC

The equation of line

AC

y = 18 - x

Let's find the area of the rectangle:

Area =

x \times y = x(18 - x) = 18x - x^2

To maximize the area, we take the derivative and set it equal to zero:

\dfrac{dA}{dx} = 18 - 2x = 0

x = 9

When

x = 9

y = 18 - 9 = 9

The rectangle has vertices at

(0,0)

(9,0)

(0,9)

, and

(9,9)

Maximum area =

9 \times 9 = 81

square units.