The figure shows Delta ABC with AB = 3, AC = | vedclass

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(D) Let the radii of the circles centered at $A, B, C$ be $r_1, r_2, r_3$ respectively.Since the circles touch each other externally, we have:$r_1 + r

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$14$

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(D) Let the radii of the circles centered at $A, B, C$ be $r_1, r_2, r_3$ respectively.Since the circles touch each other externally, we have:$r_1 + r_2 = AB = 3$$r_1 + r_3 = AC = 4$$r_2 + r_3 = BC = 5$Adding these three equations: $2(r_1 + r_2 + r_3) = 3 + 4 + 5 = 12$, so $r_1 + r_2 + r_3 = 6$.Now, we find the individual radii:$r_3 = (r_1 + r_2 + r_3) - (r_1 + r_2) = 6 - 3 = 3$$r_2 = (r_1 + r_2 + r_3) - (r_1 + r_3) = 6 - 4 = 2$$r_1 = (r_1 + r_2 + r_3) - (r_2 + r_3) = 6 - 5 = 1$The sum of the areas of the three circles is $\pi r_1^2 + \pi r_2^2 + \pi r_3^2 = \pi(1^2 + 2^2 + 3^2) = \pi(1 + 4 + 9) = 14\pi$.

The figure shows $\Delta ABC$ with $AB = 3, AC = 4$ and $BC = 5$. Three circles $S_1, S_2$ and $S_3$ have their centers at $A, B$ and $C$ respectively and they touch each other externally. The sum of the areas of the three circles is: (in $\pi$)

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