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1999 AHSME Problems/Problem 16

Problem

What is the radius of a circle inscribed in a rhombus with diagonals of length $10$ and $24$?

$\mathrm{(A) \ }4 \qquad \mathrm{(B) \ }\frac {58}{13} \qquad \mathrm{(C) \ }\frac{60}{13} \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ }6$

Solution

Let $d_1=10$ and $d_2=24$ be the lengths of the diagonals, $a$ the side, and $r$ the radius of the inscribed circle.

Using Pythagorean theorem we can compute $a=\sqrt{ (d_1/2)^2 + (d_2/2)^2 }=13$.

We can now express the area of the rhombus in two different ways: as $d_1 d_2 / 2$, and as $2ar$. Solving $d_1 d_2 / 2 = 2ar$ for $r$ we get $r=\boxed{\frac{60}{13}}$.

(The first formula computes the area as one half of the circumscribed rectangle whose sides are parallel to the diagonals. The second one comes from the fact that we can divide the rhombus into $4$ equal triangles, and in those the height on the side $a$ is equal to $r$. See pictures below.)

[asy] unitsize(1cm); pair A=(-3,0), B=(0,-1.25), C=(3,0), D=(0,1.25); pair E=(-3,-1.25), F=(-3,1.25), G=(3,1.25), H=(3,-1.25); fill ( E -- F -- G -- H -- cycle, lightgray ); draw ( E -- F -- G -- H -- cycle ); draw ( A -- B -- C -- D -- cycle ); label("$d_1$",(E+F)*0.5,W); label("$d_2$",(F+G)*0.5,N); [/asy]
[asy] unitsize(1cm); pair A=(-3,0), B=(0,-1.25), C=(3,0), D=(0,1.25); pair P = intersectionpoint( A--B--C--D--cycle, circle( (0,0), 15/13 ) ); fill ( A -- B -- (0,0) -- cycle, lightgray ); draw ( A -- B -- (0,0) -- cycle ); draw ( A -- B -- C -- D -- cycle ); draw ( circle( (0,0), 15/13 ) ); draw ( (0,0) -- P ); label("$a$",(A+B)*0.5,SW); label("$r$",P*0.5,0.5*NW); [/asy]

See also

1999 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AHSME Problems and Solutions

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