Finding the area of a circle from the area of a tangent right triangle 
My younger brother asked me this question, and I didn't know how to answer it. I am not a math major, but I really wanted to know if the given information is sufficient to answer the question.
 A: The area of the circle is $400\pi\ \text{cm}^2$.
Since the area of the triangle is $60$, the other leg is $8$, so that the hypotenuse is $17$. Now, call the two non-right-angle vertices of the triangle $A$ and $B$, and let the three points of tangency be $P$, $Q$, and $R$, all right-to-left. Then $AP = AQ$ and $BQ = BR$. Now, if you finish drawing the square of which your picture produces parts of two sides, the side of that square is $15+8+AP+BR$, so the radius of the circle is $\frac{23+AP+BR}{2}$. But $AP+BR = AQ+BQ = AB = 17$, so the circle's radius is $20$ and its area is $400\pi$.
A: $\newcommand{\cm}{\mathrm{cm}}$Call $AOB$ the triangle, and label the tangency points $P,\,Q,\,R$ going from left to right.
You know $AO=\frac{120\cm^2}{15\cm}=8\cm$ and $AB=\sqrt{AO^2+BO^2}=\sqrt{289\cm^2}=17\cm$.
Moreover, you know that $PA=AQ$, that $QB=BR$, that $PA+AO=BR+BO$ and that $QB+PA=AB$. Putting the last two together: $$\begin{cases}PA+8\cm=BR+15\cm\\ PA+BR=17\cm\end{cases}$$
And this can be solved for $PA$ and $BR$. Since $RO\perp PO$, you have that $RO=BO+BR$ has the same length as the radius.
