1955 AHSME Problems/Problem 9

Revision as of 22:01, 22 July 2020 by Duck master (talk | contribs) (created page w/ solution & categorization)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png