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A rectangle is inscribed in an equilateral triangle of side length 2a units. The maximum area of this rectangle can be

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\sqrt{3} a^{2}

\frac{\sqrt{3} a^{2}}{4}

\frac{\sqrt{3} a^{2}}{2}

a^{2}

answer is C.

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Detailed Solution

Given,
Side length =2a Suppose length of rectangle

= l

Breadth of rectangle

= b

△ DBF

ta n 60^{\circ} = \frac{2 b}{2 a - l} (since tan 60^{\circ} = \sqrt{3})

\sqrt{3} = \frac{2 b}{2 a - l}

b = \frac{\sqrt{3} (2 a - l)}{2}

Area of

DEFG

= l \times b

A = 1 \times \frac{\sqrt{3}}{2} (2 a - l)

Now, for maximum area differentiate the above area with respect to length and equate it to zero,

\frac{dA}{dl} = \frac{\sqrt{3}}{2} (2 a - 2 l)

\frac{dA}{dl} = 0

\frac{dA}{dl} = \frac{\sqrt{3}}{2} (2 a - 2 l)

2 a - 2 l

a = l

Therefore, maximum Area of rectangle
(A)

= l \times b

A = a \times \frac{\sqrt{3}}{2} (2 a - a)

A = \frac{\sqrt{3}}{2} a^{2}

Hence, the maximum area of rectangle is

\frac{\sqrt{3} a^{2}}{2}

Correct option is 3.

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