2002 AMC 10A Problems/Problem 25

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 $\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{\mathrm{(C)}\ 210}$

Solution 3

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'$",E,S); label("39",(C+D)/2,N); label("52",(A+B)/2,S); label("5",(A+D)/2,W); label("12",(B+C)/2,ENE); [/asy]$

Call the length of $AD'$ to be $y$ , the length of $BC'$ to be $z$ , and the height of the trapezoid to be $x$ . By the Pythagorean Theorem, we have: $z^2 + x^2 = 144$ $y^2 + x^2 = 25$

Subtracting these two equation yields: $z^2-y^2=119 \implies (z+y)(z-y)=119$

We also have: $z+y=52-39=13$ .

We can substitute the value of $z+y$ into the equation we just obtained, so we now have:

$(13) (z-y)=119 \implies z-y=\frac{119}{13}$ .

We can add the $z+y$ and the $z-y$ equation to find the value of $z$ , which simplifies down to be $\frac{144}{13}$ . Finally, we can plug in $z$ and use the Pythagorean theorem to find the height of the trapezoid.

$\frac{12^4}{13^2} + x^2 = 12^2 \implies x^2 = \frac{(12^2)(13^2)}{13^2} -\frac{12^4}{13^2} \implies x^2 = \frac{(12 \cdot 13)^2 - (144)^2}{13^2} \implies x^2 = \frac{(156+144)(156-144)}{13^2} \implies x = \sqrt{\frac{3600}{169}} = \frac{60}{13}$

Now that we have the height of the trapezoid, we can multiply this by the median to find our answer.

The median of the trapezoid is $\frac{39+52}{2} = \frac{91}{2}$ , and multiplying this and the height of the trapezoid gets us:

$\frac{60 \cdot 91}{13 \cdot 2} = \boxed{\mathrm{(C)}\ 210}$

Solution 4

We construct a line segment parallel to $\overline{AD}$ from point $C$ to line $\overline{AB},$ and label the intersection of this segment with line $\overline{AB}$ as point $E.$ Then quadrilateral $AECD$ is a parallelogram, so $CE=5, AE=39,$ and $EB=13.$ Triangle $EBC$ is therefore a right triangle, with area $\frac12 \cdot 5 \cdot 12 = 30.$

By continuing to split $\overline{AB}$ and $\overline{CD}$ into segments of length $13,$ we can connect these vertices in a "zig-zag," creating seven congruent right triangles, each with sides $5,12,$ and $13,$ and each with area $30.$ The total area is therefore $7 \cdot 30 = \boxed{\textbf{(C)} 210}.$

Solution 2 but quicker

From Solution $2$ we know that the the altitude of the trapezoid is $\frac{60}{13}$ and the triangle's area is $30$ . Note that once we remove the triangle we get a rectangle with length $39$ and height $\frac{60}{13}$ . The numbers multiply nicely to get $180+30=\boxed{(C) 210}$ -harsha12345

Quick Time Trouble Solution 5

First note how the answer choices are all integers. The area of the trapezoid is $\frac{39+52}{2} * h = \frac{91}{2} h$ . So h divides 2. Let $x$ be $2h$ . The area is now $91x$ . Trying x=1 and x=2 can easily be seen to not work. Those make the only integers possible so now you know x is a fraction. Since the area is an integer the denominator of x must divide either 13 or 7 since $91 = 13*7$ . Seeing how $39 = 3*13$ and $52 = 4*13$ assume that the denominator divides 13. Letting $y = \frac{x}{13}$ the area is now $7y$ . Note that (A) and (C) are the only multiples of 7. We know that A doesn't work because that would mean h is 4 which we ruled out. So the answer is $\boxed{\textbf{(C)} 210}$ . - megateleportingrobots

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by -(Last question)
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All AMC 10 Problems and Solutions

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