2014 AMC 8 Problem 14

Attempt Problem 14 of the 2014 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

Rectangle ABCDABCD and right triangle DCEDCE have the same area. They are joined to form a trapezoid, as shown. What is DE?DE?
12 12
13 13
14 14
15 15
16 16
Answer: B
Concepts:areaPythagorean Theorem
Small hint:
Set the triangle area equal to the rectangle area.
Big hint:
After finding CECE, use the 55-1212-1313 right triangle.
Solution:
The area of ABCDABCD is 5⋅6=30.5 \cdot 6 = 30.
The area of △DCE=12DC⋅CE=30\begin{align*} \triangle DCE &= \frac{1}{2}DC \cdot CE \\ &= 30 \end{align*} Therefore: 12⋅5⋅CE=30CE=12.\begin{align*} \dfrac{1}{2} \cdot 5 \cdot CE &= 30 \\ CE &= 12. \end{align*} Then using the Pythagorean theorem we get that DE=52+122=169=13.\begin{align*} DE &= \sqrt{5^2 + 12^2} \\ &= \sqrt{169} \\ &= 13. \end{align*}
Thus, B is the correct answer.
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