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2002 AMC 10A Problems/Problem 25

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Contents

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See also

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 $E$ :

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--...

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 $\begin{align*} \frac {CE}{CE + 12} = \frac {3}{4} & \Longrightarrow CE = 36 \\\frac {DE}{DE + 5} = \frac {3}{4} & \Lo...$ 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 $D$ :

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);...

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

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)...

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 • Resources)
Preceded by Problem 24	Followed by Last problem
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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2002_AMC_10A_Problems/Problem_25"

Category: Introductory Geometry Problems

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