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

Revision as of 10:09, 28 June 2023

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

Video Solution

https://youtu.be/-JsXX8WLASg ~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/j3QSD5eDpzU?t=88

~ pi_is_3.14

@@ Line 27: / Line 27: @@
 ==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>\boxed{(B)}</math>.
+This problem can be solved with the Pythagorean Theorem (<math>a^2 + b^2 = c^2</math>).  We know <math>AB = DC</math>, so <math>DC = 5</math>.  <math>CE</math> is twice the length of <math>AD</math>, so <math>CE = 12</math>.  <math>5^2 + 12^2 = c^2</math>.  <math>5^2 = 25</math>.  <math>12^2 = 144</math>.  <math>25 + 144 = 169</math>.  <math>169</math> has a square root of <math>13</math>, so the hypotenuse or <math>DE</math> is <math>13</math>.  The answer is <math>\boxed{(B)}</math>.
 ——MiracleMaths

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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