JEE Mains · Maths · STD 11 - 9. straight line
A circle in inscribed in an equilateral triangle of side of length \(12\). If the area and perimeter of any square inscribed in this circle are \(\mathrm{m}\) and \(\mathrm{n}\), respectively, then \(m+n^2\) is equal to
- A \(396\)
- B \(408\)
- C \(312\)
- D \(414\)
Answer & Solution
Correct Answer
(B) \(408\)
Step-by-step Solution
Detailed explanation
\(\because r=\frac{\Delta}{s}=\frac{\sqrt{3} a^2}{4 \cdot \frac{3 a}{2}}=\frac{a}{2 \sqrt{3}}=\frac{12}{2 \sqrt{3}}=2 \sqrt{3}\) \( \therefore A=r \sqrt{2}=2 \sqrt{6} \) Area \(=m=A^2=24 \) Perimeter \(=n=4 A=8 \sqrt{6} \) \( \therefore m+n^2=24+384 \) \( =408\)
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