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.

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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
Difficulty rating: 1410
Solution:

We can drop the following altitudes to more easily find the area.

We can use the Pythagorean Theorem to get that DE=152122=9 DE = \sqrt{15^2 - 12^2} = 9 and FC=202122=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=6125 \dfrac{1}{2} \cdot (DC + 50) \cdot 12 = 6 \cdot 125 =750. = 750.

Thus, D is the correct answer.

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