If R = frac{65}{8},r_1 = frac{21}{2},and r_2 | vedclass

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(C) We know the formula for the exradii of a triangle: $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,where $\De

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$84$

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(C) We know the formula for the exradii of a triangle: $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,where $\Delta$ is the area and $s$ is the semi-perimeter.Also,$r_1 = 4R \sin(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2})$ is not needed here; we use $r_1 = s 	an(\frac{A}{2})$ and $R = \frac{abc}{4\Delta}$.Given $r_1 = \frac{21}{2}$ and $r_2 = 12$,we use the relation $\frac{1}{r_1} + \frac{1}{r_2} = \frac{s-a}{\Delta} + \frac{s-b}{\Delta} = \frac{2s-a-b}{\Delta} = \frac{c}{\Delta}$.Using the identity $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$,where $r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) = \frac{\Delta}{s}$.Alternatively,$\Delta = \sqrt{r r_1 r_2 r_3}$.Using the property $r_1 r_2 + r_2 r_3 + r_3 r_1 = s^2$,and $\Delta = rs$,we find the area $\Delta = 84$.

If $R = \frac{65}{8}$,$r_1 = \frac{21}{2}$,and $r_2 = 12$ are the circumradius and the radii of the excircles opposite to the vertices $A$ and $B$ of a triangle $ABC$ respectively,then the area of the triangle (in square units) is

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