Difference between revisions of "2014 AMC 8 Problems/Problem 14"

(Solution 3)
(Solution 3)
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==Solution 3==
 
==Solution 3==
  
This problem can be solved with the Pythagorean Theorem (a^2 + b^2 = c^2.  We know AB = DC, so DC = 5.  CE is twice the length of AD, so CE = 12.  5^2 + 12^2 = c^2.  5^2 = 25.  12^2 = 144.  25 + 144 = 169.  169 has a square root of 13, so the hypotenuse or DE is 13.  The answer is <math>DE=\boxed{13}</math>, or <math>\boxed{(B)}</math>.
+
This problem can be solved with the Pythagorean Theorem (a^2 + b^2 = c^2.  We know AB = DC, so DC = 5.  CE is twice the length of AD, so CE = 12.  5^2 + 12^2 = c^2.  5^2 = 25.  12^2 = 144.  25 + 144 = 169.  169 has a square root of 13, so the hypotenuse or DE is 13.  The answer is <math>\boxed{(B)}</math>.
  
 
——MiracleMaths
 
——MiracleMaths

Revision as of 21:37, 20 December 2021

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

This problem can be solved with the Pythagorean Theorem (a^2 + b^2 = c^2. We know AB = DC, so DC = 5. CE is twice the length of AD, so CE = 12. 5^2 + 12^2 = c^2. 5^2 = 25. 12^2 = 144. 25 + 144 = 169. 169 has a square root of 13, so the hypotenuse or DE is 13. The answer is $\boxed{(B)}$.

——MiracleMaths

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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
All AJHSME/AMC 8 Problems and Solutions

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