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

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Now the area of the original trapezoid is <math>\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{210}</math>.
 
Now the area of the original trapezoid is <math>\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{210}</math>.
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=== Solution 3 ===
 
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Draw altitudes from points <math>C</math> and <math>D</math>:
 
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<center><asy>
 
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unitsize(0.2cm);
 
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defaultpen(0.8);
 
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pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0);
 
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draw(A--B--C--D--cycle);
 
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draw(C--E,dashed);
 
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draw(D--F,dashed);
 
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label("\(A\)",A,SW);
 
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label("\(B\)",B,S);
 
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label("\(C\)",C,NE);
 
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label("\(D\)",D,N);
 
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label("\(D'\)",F,SSE);
 
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label("\(C'\)",E,S);
 
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label("39",(C+D)/2,N);
 
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label("52",(A+B)/2,S);
 
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label("5",(A+D)/2,W);
 
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label("12",(B+C)/2,ENE);
 
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</asy></center>
 
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Call the length of <math>AD'</math> to be <math>y</math>, the length of <math>BC'</math> to be <math>z</math>, and the height of the trapezoid to be <math>x</math>.
 
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By the Pythagorean Theorem, we have:
 
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<cmath>z^2 + z^2 = 144</cmath>
 
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<cmath>y^2 + x^2 = 25</cmath>
 
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Subtracting these two equation yields:
 
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<cmath>z^2-y^2=119 \implies (z+y)(z-y)=119</cmath>
 
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We also have: <math>z+y=52-39=13</math>.
 
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We can substitute the value of <math>z+y</math> into the equation we just obtained, so we now have:
 
  
 
== See also ==
 
== See also ==

Revision as of 00:56, 26 January 2016

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

Solution 1

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), \[[ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{\mathrm{(C)}\ 210}\]

Solution 2

Draw altitudes from points $C$ and $D$:

[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label("\(A\)",A,SW); label("\(B\)",B,S); label("\(C\)",C,NE); label("\(D\)",D,N); label("\(D'\)",F,SSE); label("\(C'\)",E,S); label("39",(C+D)/2,N); label("52",(A+B)/2,S); label("5",(A+D)/2,W); label("12",(B+C)/2,ENE); [/asy]

Translate the triangle $ADD'$ so that $DD'$ coincides with $CC'$. We get the following triangle:

[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0); draw(A--B--C--cycle); draw(C--F,dashed); label("\(A'\)",A,SW); label("\(B\)",B,S); label("\(C\)",C,N); label("\(C'\)",F,SE); label("5",(A+C)/2,W); label("12",(B+C)/2,ENE); [/asy]

The length of $A'B$ in this triangle is equal to the length of the original $AB$, minus the length of $CD$. Thus $A'B = 52 - 39 = 13$.

Therefore $A'BC$ is a well-known $(5,12,13)$ right triangle. Its area is $[A'BC]=\frac{A'C\cdot BC}2 = \frac{5\cdot 12}2 = 30$, and therefore its altitude $CC'$ is $\frac{[A'BC]}{A'B} = \frac{60}{13}$.

Now the area of the original trapezoid is $\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{210}$.

See also

2002 AMC 10A (Problems • Answer Key • Resources)
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