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 $\frac{5 \times 12}{2}=\boxed{\mathrm{(B)}\ 30}$

(note: the ab/2 area method only works because the converse Pythagorean theorem holds for the triple [5, 12, 13]. Therefore, we can find the solution this way because we know the triangle is right ~megaboy6679)

See Also

2003 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AJHSME/AMC 8 Problems and Solutions

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