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Difference between revisions of "2005 AMC 8 Problems/Problem 13"
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==Problem==
 
==Problem==
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The area of polygon <math> ABCDEF</math> is 52 with <math> AB\equal{}8</math>, <math> BC\equal{}9</math> and <math> FA\equal{}5</math>. What is <math> DE\plus{}EF</math>?
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The area of polygon <math> ABCDEF</math> is 52 with <math> AB=8</math>, <math>BC=9</math> and <math>FA=5</math>. What is <math> DE+EF</math>?
 
<asy>pair a=(0,9), b=(8,9), c=(8,0), d=(4,0), e=(4,4), f=(0,4);
 
<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(a--b--c--d--e--f--cycle);

Revision as of 14:16, 23 August 2015

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 (Problems • Answer Key • Resources)
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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