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 | AfterBoards
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IPMAT Indore 2024 (SA) PYQs

IPMAT Indore 2024

Geometry
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Triangles

Easy

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

Entered answer:

Correct Answer: 81
Let's place ABC\triangle ABC on a coordinate system with BB at the origin (0,0)(0,0).
Since angle BB is 90°90°, we can place AA at (0,18)(0,18) and CC at (18,0)(18,0).

Since B(0,0)B(0,0) is one vertex of the rectangle, the other vertices will be at (x,0)(x,0), (0,y)(0,y), and (x,y)(x,y) where (x,y)(x,y) lies on the hypotenuse ACAC.
The equation of line ACAC is y=18xy = 18 - x.

Let's find the area of the rectangle:
Area = x×y=x(18x)=18xx2x \times y = x(18 - x) = 18x - x^2
To maximize the area, we take the derivative and set it equal to zero:
dAdx=182x=0\dfrac{dA}{dx} = 18 - 2x = 0
x=9x = 9
When x=9x = 9, y=189=9y = 18 - 9 = 9

The rectangle has vertices at (0,0)(0,0), (9,0)(9,0), (0,9)(0,9), and (9,9)(9,9)
Maximum area = 9×9=819 \times 9 = 81 square units.

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