A circle is inscribed in an equilateral triangle of side of length 12. If the area and perimeter of any square inscribed in this circle are and , respectively, then is equal to

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Solution

Q.
A circle is inscribed in an equilateral triangle of side of length 12. If the area and perimeter of any square inscribed in this circle are $m$ and $n$ , respectively, then $m + n^{2}$ is equal to [2024]

1 396

2 408

3 414

4 312

Ans.
(2)

Let radius of inscribed circle be $r .$

$r = \frac{1}{3} A D$

$= \frac{1}{3} \sqrt{a^{2} - \frac{a^{2}}{4}}, a is side of triangle.$

$= \frac{a}{2 \sqrt{3}} = \frac{12}{2 \sqrt{3}} = 2 \sqrt{3}$

Area of square DEFG, $m = \frac{1}{2} \times {(diagonal)}^{2}$

$= \frac{1}{2} \times {(2 r)}^{2} = \frac{1}{2} \times 4 \times 12 = 24 sq. units$

Side of square $= 2 \sqrt{6}$

Perimeter of square, $n = 4 \times 2 \sqrt{6} = 8 \sqrt{6}$

$∴ m + n^{2} = 24 + 384 = 408$

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