2008 AMC 10A Problem 20

Attempt Problem 20 of the 2008 AMC 10A 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 2008 AMC 10A solutions, or check the answer key.

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20.

Trapezoid ABCDABCD has bases ABAB and CDCD and diagonals intersecting at K.K. Suppose that AB=9,AB = 9, DC=12,DC = 12, and the area of AKD\triangle AKD is 24.24. What is the area of trapezoid ABCD?ABCD?

9292

9494

9696

9898

100100

Answer: D
Concepts:trapezoidsimilarityarea ratio
Difficulty rating: 1710
Solution:

Triangles AKBAKB and CKDCKD are similar with ratio 912=34.\dfrac{9}{12} = \dfrac{3}{4}.

Since AKD\triangle AKD and KCD\triangle KCD share the base and have collinear vertices, [KCD][AKD]=KCAK=43,\dfrac{[KCD]}{[AKD]} = \dfrac{KC}{AK} = \dfrac{4}{3}, so [KCD]=32.[KCD] = 32. Similarly [AKB]=18.[AKB] = 18.

Also [BKC]=[AKD]=24.[BKC] = [AKD] = 24. The total is 24+32+18+24=98.24 + 32 + 18 + 24 = 98.

Thus, the correct answer is D.

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