In Delta ABC,AB = AC. Let P_1 denote the inci | vedclass

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(D) Let $\alpha$ be half of the angle $\angle A$. Let $C_1$ and $C_2$ be the centers of circles $P_2$ and $P_1$ with radii $r_1 = 1$ and $r_2 = 2$ res

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$16\sqrt{2}$

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(D) Let $\alpha$ be half of the angle $\angle A$. Let $C_1$ and $C_2$ be the centers of circles $P_2$ and $P_1$ with radii $r_1 = 1$ and $r_2 = 2$ respectively.In the right triangle formed by the center $C_1$,the vertex $A$,and the point of tangency on $AB$,we have $\sin \alpha = \frac{r_1}{AC_1} = \frac{1}{AC_1}$.Since the circles are tangent,$AC_2 = AC_1 + r_1 + r_2 = AC_1 + 1 + 2 = AC_1 + 3$.Also,$\sin \alpha = \frac{r_2}{AC_2} = \frac{2}{AC_2}$.Equating the two expressions for $\sin \alpha$: $\frac{1}{AC_1} = \frac{2}{AC_1 + 3} \implies AC_1 + 3 = 2AC_1 \implies AC_1 = 3$.Thus,$\sin \alpha = \frac{1}{3}$.The height of the triangle $AD = AC_2 + r_2 = (AC_1 + 3) + 2 = 6 + 2 = 8$.In $\Delta ABD$,$	an \alpha = \frac{BD}{AD} = \frac{BD}{8}$.Since $\sin \alpha = \frac{1}{3}$,$\cos \alpha = \sqrt{1 - (\frac{1}{3})^2} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$.Then $	an \alpha = \frac{1/3}{2\sqrt{2}/3} = \frac{1}{2\sqrt{2}}$.So,$BD = 8 	imes \frac{1}{2\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.The base $BC = 2 	imes BD = 4\sqrt{2}$.Area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4\sqrt{2} 	imes 8 = 16\sqrt{2}$.

In $\Delta ABC$,$AB = AC$. Let $P_1$ denote the incircle of $\Delta ABC$. Circle $P_2$ is tangent to sides $AB$,$AC$ and to circle $P_1$. If the radii of circles $P_1$ and $P_2$ are $2$ and $1$ respectively,then the area of $\Delta ABC$ is:

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