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

Revision as of 13:27, 24 January 2009

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

@@ Line 5: / Line 5: @@
 == 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 <math>\overline{AD}</math> and <math>\overline{BC}</math> to meet at point <math>E</math>:
@@ Line 32: / Line 35: @@
 <cmath>
 [ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{\mathrm{(C)}\ 210}</cmath>
+=== Solution 2 ===
+Draw altitudes from points <math>C</math> and <math>D</math>:
+<center><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></center>
+Translate the triangle <math>ADD'</math> so that <math>DD'</math> coincides with <math>CC'</math>. We get the following triangle:
+<center><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></center>
+The length of <math>A'B</math> in this triangle is equal to the length of the original <math>AB</math>, minus the length of <math>CD</math>.
+Thus <math>A'B = 52 - 39 = 13</math>.
+Therefore <math>A'BC</math> is a well-known <math>(5,12,13)</math> right triangle. Its area is <math>[A'BC]=\frac{A'C\cdot BC}2 = \frac{5\cdot 12}2 = 30</math>, and therefore its altitude <math>CC'</math> is <math>\frac{[A'BC]}{A'B} = \frac{60}{13}</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>.
 == See also ==

2002 AMC 10A (Problems • Answer Key • 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
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