2014 AMC 8 Problems/Problem 14
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 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 | |
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All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.