1955 AHSME Problems/Problem 9

A circle is inscribed in a triangle with sides $8, 15$, and $17$. The radius of the circle is:

$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 7$

Solution

We know that $A = sr$, where $A$ is the triangle's area, $s$ its semiperimeter, and $r$ its inradius. Since this particular triangle is a right triangle (which we can verify by the Pythagorean theorem), the area is half of $8*15 = 120$, and the semiperimeter is half of $8 + 15 + 17 = 40$. Therefore, the inradius is $\frac{120}{40} = 3$, so our answer is $\boxed{\textbf{(D)}}$ and we are done.


See also

1955 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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