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Difference between revisions of "2002 AMC 10A Problems/Problem 25"

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It shouldn't be hard to use [[trigonometry]] to bash this and find the height, but there is a much easier way. Extend <math>\overline{AD}</math> and <math>\overline{BC}</math> to meet at point <math>E</math>:
 
It shouldn't be hard to use [[trigonometry]] to bash this and find the height, but there is a much easier way. Extend <math>\overline{AD}</math> and <math>\overline{BC}</math> to meet at point <math>E</math>:
  
−
<asy>
+
<center><asy>
 
size(250);
 
size(250);
 
defaultpen(0.8);
 
defaultpen(0.8);
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label("5",(A+D)/2,E);
 
label("5",(A+D)/2,E);
 
label("12",(B+C)/2,WSW);
 
label("12",(B+C)/2,WSW);
−
</asy>
+
</asy></center>
  
 
Since <math>\overline{AB} || \overline{CD}</math> we have <math>\triangle AEB \sim \triangle DEC</math>, with the ratio of [[proportion]]ality being <math>\frac {39}{52} = \frac {3}{4}</math>. Thus
 
Since <math>\overline{AB} || \overline{CD}</math> we have <math>\triangle AEB \sim \triangle DEC</math>, with the ratio of [[proportion]]ality being <math>\frac {39}{52} = \frac {3}{4}</math>. Thus

Revision as of 13:49, 11 February 2008

Problem

In trapezoid $ABCD$ with bases $AB$ and $CD$, we have $AB = 52$, $BC = 12$, $CD = 39$, and $DA = 5$. The area of $ABCD$ is

$\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$

Solution

It shouldn't be hard to use trigonometry to bash this and find the height, but there is a much easier way. Extend $\overline{AD}$ and $\overline{BC}$ to meet at point $E$:

[asy] size(250); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), F=(100/13,240/13); draw(A--B--C--D--cycle); draw(D--F--C,dashed); label("\(A\)",A,S); label("\(B\)",B,S); label("\(C\)",C,NE); label("\(D\)",D,W); label("\(E\)",F,N); label("39",(C+D)/2,N); label("52",(A+B)/2,S); label("5",(A+D)/2,E); label("12",(B+C)/2,WSW); [/asy]

Since $\overline{AB} || \overline{CD}$ we have $\triangle AEB \sim \triangle DEC$, with the ratio of proportionality being $\frac {39}{52} = \frac {3}{4}$. Thus \begin{align*} \frac {CE}{CE + 12} = \frac {3}{4} & \Longrightarrow CE = 36 \\ \frac {DE}{DE + 5} = \frac {3}{4} & \Longrightarrow DE = 15 \end{align*} So the sides of $\triangle CDE$ are $15,36,39$, which we recognize to be a $5 - 12 - 13$ right triangle. Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared), \[[ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{210} \Rightarrow \mathrm{(C)}.\]

See also

2002 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions
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