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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