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UPSC NDA I 2025 Mathematics Paper

Question Numbers: 73-74

Consider the following for the two (02) items that follow: Let ABC be a triangle right-angled at B and AB+A

C = 3

units.

Question : 74 of 120

Marks: +1, -0

What is the maximum area of the triangle?

Solution:

Concept:

For a right triangle with legs

AB

and

BC

and hypotenuse

AC

, area is

\frac12 \times AB \times BC

. Given

AB+AC=3

, we express area in one variable and maximize it using differentiation.

Explanation:

Let

AB = x

. Then

AC = 3 - x

. Since

AC

is the hypotenuse,

BC = \sqrt{AC^2 - AB^2} = \sqrt{(3-x)^2 - x^2} = \sqrt{9 - 6x}

Area

A = \frac12 \times x \times \sqrt{9-6x}

To maximize, differentiate

f(x)=x\sqrt{9-6x}

and set

f'(x)=0

f'(x) = \sqrt{9-6x} - \frac{3x}{\sqrt{9-6x}} = 0

\Rightarrow

9-6x = 3x

\Rightarrow

9=9x

\Rightarrow

x=1

Then

BC = \sqrt{9-6} = \sqrt{3}

. Maximum area

A_{\text{max}} = \frac12 \times 1 \times \sqrt{3} = \frac{\sqrt{3}}{2}

Answer:

\frac{\sqrt{3}}{2}

square unit (Option A).

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