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 $B$ is $90°$ , we can place $A$ at $(0,18)$ and $C$ at $(18,0)$ .

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 $AC$ is $y = 18 - x$ .

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.

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IPMAT Indore Past Year Questions (Topic-Wise)

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: