Question : The length of the diagonal of a square is $9\sqrt{2}$ cm. The square is reshaped to form an equilateral triangle. What is the area (in cm2) of the largest incircle that can be formed in that triangle?
Option 1: $6\pi$
Option 2: $9\pi$
Option 3: $12\pi$
Option 4: $15\pi$
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Correct Answer: $12\pi$
Solution : Diagonal of square = $9\sqrt2$ Diagonal = side$\sqrt2$ $9\sqrt2=side\sqrt2$ Side = 9 cm Perimeter of the square = 9 × 4 = 36 $\therefore$ Perimeter of the triangle = 36 Side = $\frac{36}{3}=12$ We know that Radius of the incircle = $\frac{\text{Side of the triangle}}{2\sqrt3}$ = $\frac{12}{2\sqrt3}$ = $\frac{6}{\sqrt3}$ Area of the circle = $\pi r^2$ = $\pi (\frac{6}{\sqrt3})^2$ = $12 \pi$ Hence, the correct answer is $12\pi$.
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