2014 AMC 8 Problems/Problem 14

Problem

Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$ ?

$[asy] size(250); defaultpen(linewidth(0.8)); pair A=(0,5),B=origin,C=(6,0),D=(6,5),E=(18,0); draw(A--B--E--D--cycle^^C--D); draw(rightanglemark(D,C,E,30)); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,N); label("$E$",E,S); label("$5$",A/2,W); label("$6$",(A+D)/2,N); [/asy]$

$\textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16$

Solution

The area of $\bigtriangleup CDE$ is $\frac{DC\cdot CE}{2}$ . The area of $ABCD$ is $AB\cdot AD=5\cdot 6=30$ , which also must be equal to the area of $\bigtriangleup CDE$ , which, since $DC=5$ , must in turn equal $\frac{5\cdot CE}{2}$ . Through transitivity, then, $\frac{5\cdot CE}{2}=30$ , and $CE=12$ . Then, using the Pythagorean Theorem, you should be able to figure out that $\bigtriangleup CDE$ is a $5-12-13$ triangle, so $DE=\boxed{13}$ , or $\boxed{(B)}$ .

Solution 2

The area of the rectangle is $5\times6=30.$ Since the parallel line pairs are identical, $DC=5$ . Let $CE$ be $x$ . $\dfrac{5x}{2}=30$ is the area of the right triangle. Solving for $x$ , we get $x=12.$ According to the Pythagorean Theorem, we have a $5-12-13$ triangle. So, the hypotenuse $DE$ has to be $\boxed{(B)}$ .

Solution 3

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AJHSME/AMC 8 Problems and Solutions

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2014 AMC 8 Problems/Problem 14

Contents

Problem

Solution

Solution 2

Solution 3

See Also