Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2014 AMC 8 Problems/Problem 14: Difference between revisions

← Older edit
Mathsmagic123 (talk | contribs)
editor
34 edits
Trigoeinstein (talk | contribs)
editor
9 edits
 
(3 intermediate revisions by 2 users not shown)
Line 19: Line 19:
<math> \textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16 </math>
<math> \textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16 </math>


==Solution==
==Solution 1==
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>.


Line 31: Line 31:


——MiracleMaths
——MiracleMaths
Note: Another way to find out that CE is 12 is by using logic. Since DC = 5, we can reason that the triangle is a 5 - 12 - 13 triangle, because the only Pythagorean Triple with 5 as a leg(not hypotenuse, as that would be the case for a 3 - 4 - 5 right triangle), is 5 - 12 - 13. We know DE is the hypotenuse, so it can’t be 12, which means CE has to be 12, which, in turn, means that DE has to be: 13 <math>\boxed{(B)}</math>.


==Video Solution (CREATIVE THINKING)==
==Video Solution (CREATIVE THINKING)==

Latest revision as of 11:08, 7 January 2025

Problem 14

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 1

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

Note: Another way to find out that CE is 12 is by using logic. Since DC = 5, we can reason that the triangle is a 5 - 12 - 13 triangle, because the only Pythagorean Triple with 5 as a leg(not hypotenuse, as that would be the case for a 3 - 4 - 5 right triangle), is 5 - 12 - 13. We know DE is the hypotenuse, so it can’t be 12, which means CE has to be 12, which, in turn, means that DE has to be: 13 $\boxed{(B)}$.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_14&oldid=239262"