Difference between revisions of "2014 AMC 8 Problems/Problem 14"
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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== | ==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>. |
==Solution 2== | ==Solution 2== |
Revision as of 18:25, 7 July 2020
Contents
Problem
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 5-12-13 triangle. So, the hypotenuse has to be .
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.