2005 AMC 8 Problems/Problem 13

Problem

The area of polygon $ABCDEF$ is 52 with $AB=8$, $BC=9$ and $FA=5$. What is $DE+EF$?

[asy] pair a=(0,9), b=(8,9), c=(8,0), d=(4,0), e=(4,4), f=(0,4); draw(a--b--c--d--e--f--cycle); draw(shift(0,-.25)*a--shift(.25,-.25)*a--shift(.25,0)*a); draw(shift(-.25,0)*b--shift(-.25,-.25)*b--shift(0,-.25)*b); draw(shift(-.25,0)*c--shift(-.25,.25)*c--shift(0,.25)*c); draw(shift(.25,0)*d--shift(.25,.25)*d--shift(0,.25)*d); draw(shift(.25,0)*f--shift(.25,.25)*f--shift(0,.25)*f); label("$A$", a, NW); label("$B$", b, NE); label("$C$", c, SE); label("$D$", d, SW); label("$E$", e, SW); label("$F$", f, SW); label("5", (0,6.5), W); label("8", (4,9), N); label("9", (8, 4.5), E); [/asy]

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution

Notice that $AF + DE = BC$, so $DE=4$. Let $O$ be the intersection of the extensions of $AF$ and $DC$, which makes rectangle $ABCO$. The area of the polygon is the area of $FEDO$ subtracted from the area of $ABCO$.

\[\text{Area} = 52 = 8 \cdot 9- EF \cdot 4\]

Solving for the unknown, $EF=5$, therefore $DE+EF=4+5=\boxed{\textbf{(C)}\ 9}$.

See Also

2005 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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
All AJHSME/AMC 8 Problems and Solutions

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