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

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==Solution 3==
 
==Solution 3==
  
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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.
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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>.
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——MiracleMaths
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==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/ToM-f4WMWjQ
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
 
 +
==Video Solution==
 +
https://youtu.be/-JsXX8WLASg ~savannahsolver
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 +
==Video Solution ==
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https://youtu.be/j3QSD5eDpzU?t=88
 +
 
 +
~ pi_is_3.14
  
 
==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}}

Latest revision as of 11:11, 2 July 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 (CREATIVE THINKING)

https://youtu.be/ToM-f4WMWjQ

~Education, the Study of Everything


Video Solution

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

Video Solution

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

~ pi_is_3.14

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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