2011 AMC 8 Problem 20

Attempt Problem 20 of the 2011 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 2011 AMC 8 solutions, or check the answer key.

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

20.

Quadrilateral ABCDABCD is a trapezoid, AD=15,AD = 15, AB=50,AB = 50, BC=20,BC = 20, and the altitude is 12.12. What is the area of the trapezoid?
600600
650650
700700
750750
800800
Answer: D
Concepts:trapezoidPythagorean Triple
Small hint:
Use 99-1212-1515 and 1212-1616-2020 right triangles.
Big hint:
Drop perpendiculars from the top vertices to the base.
Solution:
We can drop the following altitudes to more easily find the area.
We can use the Pythagorean Theorem to get that DE=152−122=9 DE = \sqrt{15^2 - 12^2} = 9 and FC=202−122=16. FC = \sqrt{20^2 - 12^2} = 16.
We also know that EF=AB=50, EF = AB = 50, so DC=DE+EF+FC=75. DC = DE + EF + FC = 75.
Then the area of ABCDABCD is 12⋅(DC+50)⋅12=6⋅125 \dfrac{1}{2} \cdot (DC + 50) \cdot 12 = 6 \cdot 125 =750. = 750.
Thus, D is the correct answer.
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