Difference between revisions of "2014 AMC 8 Problems/Problem 14"
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==Solution 2== | ==Solution 2== | ||
− | The area of the | + | The area of the ectangle 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== | ==Solution 3== |
Revision as of 00:15, 6 October 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 ectangle 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.