2003 AMC 8 Problems/Problem 6

Problem

Given the areas of the three squares in the figure, what is the area of the interior triangle?

$[asy] draw((0,0)--(-5,12)--(7,17)--(12,5)--(17,5)--(17,0)--(12,0)--(12,-12)--(0,-12)--(0,0)--(12,5)--(12,0)--cycle,linewidth(1)); label("$25$",(14.5,1),N); label("$144$",(6,-7.5),N); label("$169$",(3.5,7),N); [/asy]$

$\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 30 \qquad\mathrm{(C)}\ 60 \qquad\mathrm{(D)}\ 300 \qquad\mathrm{(E)}\ 1800$

Solution

The sides of the squares are $5, 12$ and $13$ for the square with area $25, 144$ and $169$ , respectively. The legs of the interior triangle are $5$ and $12$ , so the area is $\frac{5 \times 12}{2}=\boxed{\mathrm{(B)}\ 30}$

2003 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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2003 AMC 8 Problems/Problem 6

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