Let A be the area of the in-circle and A_1, A | vedclass

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(D) The area of a circle is given by $\pi r^2$. Given the areas of the ex-circles $A_1, A_2, A_3$ as $4, 9, 16$ respectively, we have:$\pi r_1^2 = 4 \

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$\frac{144}{169}$

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(D) The area of a circle is given by $\pi r^2$. Given the areas of the ex-circles $A_1, A_2, A_3$ as $4, 9, 16$ respectively, we have:$\pi r_1^2 = 4 \Rightarrow r_1 = \frac{2}{\sqrt{\pi}}$$\pi r_2^2 = 9 \Rightarrow r_2 = \frac{3}{\sqrt{\pi}}$$\pi r_3^2 = 16 \Rightarrow r_3 = \frac{4}{\sqrt{\pi}}$The relationship between the inradius $r$ and the exradii $r_1, r_2, r_3$ is given by $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$.Substituting the values:$\frac{1}{r} = \frac{\sqrt{\pi}}{2} + \frac{\sqrt{\pi}}{3} + \frac{\sqrt{\pi}}{4} = \sqrt{\pi} \left( \frac{6+4+3}{12} \right) = \frac{13\sqrt{\pi}}{12}$.Thus, $r = \frac{12}{13\sqrt{\pi}}$.The area of the in-circle $A$ is $\pi r^2 = \pi \left( \frac{12}{13\sqrt{\pi}} \right)^2 = \pi \left( \frac{144}{169\pi} \right) = \frac{144}{169}$.

Let $A$ be the area of the in-circle and $A_1, A_2, A_3$ be the areas of the ex-circles of a triangle. If $A_1=4, A_2=9, A_3=16$, then $A=$

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