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 (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 .
——MiracleMaths
Video Solution
https://youtu.be/-JsXX8WLASg ~savannahsolver
Video Solution by OmegaLearn
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.