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JEE Main 20246 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2024 · 6 April · Morning Shift · Mathematics · Circle

A circle is inscribed in an equilateral triangle of side of length...

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 mm and nn, respectively, then m+n2m+n^2 is equal to

Options

  1. A408
  2. B414
  3. C396
  4. D312

Correct answer

A. 408

Step-by-step solution

∵r=Δs=3a24⋅3a2=a23=1223=23\because \mathrm{r}=\frac{\Delta}{\mathrm{s}}=\frac{\sqrt{3} \mathrm{a}^2}{4 \cdot \frac{3 \mathrm{a}}{2}}=\frac{\mathrm{a}}{2 \sqrt{3}}=\frac{12}{2 \sqrt{3}}=2 \sqrt{3}∴A=r2=26 Area =m=A2=24 Perimeter =n=4A=86∴m+n2=24+384=408\begin{aligned} & \therefore A=\mathrm{r} \sqrt{2}=2 \sqrt{6} \\ & \text{ Area }=\mathrm{m}=\mathrm{A}^2=24 \\ & \text{ Perimeter }=\mathrm{n}=4 \mathrm{ A}=8 \sqrt{6} \\ & \therefore \mathrm{m}+\mathrm{n}^2=24+384 \\ & =408\end{aligned}$$
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