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

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==Solution==
 
==Solution==
 
The area of <math>\bigtriangleup CDE</math> is <math>\frac{DC\cdot CE}{2}</math>. The area of <math>ABCD</math> is <math>AB\cdot AD=5\cdot 6=30</math>, which also must be equal to the area of <math>\bigtriangleup CDE</math>, which, since <math>DC=5</math>, must in turn equal <math>\frac{5\cdot CE}{2}</math>. Through transitivity, then, <math>\frac{5\cdot CE}{2}=30</math>, and <math>CE=12</math>. Then, using the Pythagorean Theorem, you should be able to figure out that <math>\bigtriangleup CDE</math> is a <math>5-12-13</math> triangle, so <math>DE=\boxed{13}</math>, or <math>\boxed{B}</math>.
 
The area of <math>\bigtriangleup CDE</math> is <math>\frac{DC\cdot CE}{2}</math>. The area of <math>ABCD</math> is <math>AB\cdot AD=5\cdot 6=30</math>, which also must be equal to the area of <math>\bigtriangleup CDE</math>, which, since <math>DC=5</math>, must in turn equal <math>\frac{5\cdot CE}{2}</math>. Through transitivity, then, <math>\frac{5\cdot CE}{2}=30</math>, and <math>CE=12</math>. Then, using the Pythagorean Theorem, you should be able to figure out that <math>\bigtriangleup CDE</math> is a <math>5-12-13</math> triangle, so <math>DE=\boxed{13}</math>, or <math>\boxed{B}</math>.
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==Solution 2==
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 +
The area of the rectangle is <math>5\times6=30.</math> Since the parallel line pairs are identical, <math>DC=5</math>. Let <math>CE</math> be <math>x</math>. <math>\dfrac{5x}{2}=30</math> is the area of the right triangle. Solving for <math>x</math>, we get <math>x=12.</math> According to the Pythagorean Theorem, we have a 5-12-13 triangle. So, the hypotenuse <math>DE</math> has to be <math>\boxed{B}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2014|num-b=13|num-a=15}}
 
{{AMC8 box|year=2014|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:06, 5 February 2018

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 $\boxed{B}$.

See Also

2014 AMC 8 (Problems • Answer Key • Resources)
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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