Difference between revisions of "2014 AMC 8 Problems/Problem 14"
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+ | ==Problem 14== | ||
Rectangle <math>ABCD</math> and right triangle <math>DCE</math> have the same area. They are joined to form a trapezoid, as shown. What is <math>DE</math>? | Rectangle <math>ABCD</math> and right triangle <math>DCE</math> have the same area. They are joined to form a trapezoid, as shown. What is <math>DE</math>? | ||
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<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== | ||
+ | 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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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 <math>5-12-13</math> triangle. So, the hypotenuse <math>DE</math> has to be <math>\boxed{(B)}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | 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 | ||
+ | |||
+ | ==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== | ==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 12:53, 29 June 2024
Contents
Problem 14
Rectangle and right triangle
have the same area. They are joined to form a trapezoid, as shown. What is
?
Solution
The area of is
. The area of
is
, which also must be equal to the area of
, which, since
, must in turn equal
. Through transitivity, then,
, and
. Then, using the Pythagorean Theorem, you should be able to figure out that
is a
triangle, so
, or
.
Solution 2
The area of the rectangle is Since the parallel line pairs are identical,
. Let
be
.
is the area of the right triangle. Solving for
, we get
According to the Pythagorean Theorem, we have a
triangle. So, the hypotenuse
has to be
.
Solution 3
This problem can be solved with the Pythagorean Theorem (). We know
, so
.
is twice the length of
, so
.
.
.
.
.
has a square root of
, so the hypotenuse or
is
. The answer is
.
——MiracleMaths
Video Solution (CREATIVE THINKING)
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.