Difference between revisions of "2005 AMC 8 Problems/Problem 13"

m (Problem)
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==Problem==
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== Problem ==
 
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>?
 
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);
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 +
<asy>
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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);
 
draw(shift(0,-.25)*a--shift(.25,-.25)*a--shift(.25,0)*a);
 
draw(shift(0,-.25)*a--shift(.25,-.25)*a--shift(.25,0)*a);
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label("5", (0,6.5), W);
 
label("5", (0,6.5), W);
 
label("8", (4,9), N);
 
label("8", (4,9), N);
label("9", (8, 4.5), E);</asy>
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label("9", (8, 4.5), E);
 +
</asy>
  
 
<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 </math>
 
<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 </math>
  
==Solution==
+
== Solution ==
 
Notice that <math>AF + DE = BC</math>, so <math>DE=4</math>. Let <math>O</math> be the intersection of the extensions of <math>AF</math> and <math>DC</math>, which makes rectangle <math>ABCO</math>. The area of the polygon is the area of <math>FEDO</math> subtracted from the area of <math>ABCO</math>.
 
Notice that <math>AF + DE = BC</math>, so <math>DE=4</math>. Let <math>O</math> be the intersection of the extensions of <math>AF</math> and <math>DC</math>, which makes rectangle <math>ABCO</math>. The area of the polygon is the area of <math>FEDO</math> subtracted from the area of <math>ABCO</math>.
  
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Solving for the unknown, <math>EF=5</math>, therefore <math>DE+EF=4+5=\boxed{\textbf{(C)}\ 9}</math>.
 
Solving for the unknown, <math>EF=5</math>, therefore <math>DE+EF=4+5=\boxed{\textbf{(C)}\ 9}</math>.
  
==See Also==
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== See Also ==
 
{{AMC8 box|year=2005|num-b=12|num-a=14}}
 
{{AMC8 box|year=2005|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:34, 19 October 2020

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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