2008 AMC 10A Problem 20

Below is the professionally curated solution for Problem 20 of the 2008 AMC 10A, 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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Concepts:trapezoid similarity area ratio

Difficulty rating: 1710

20.

Trapezoid $ABCD$ has bases $AB$ and $CD$ and diagonals intersecting at $K.$ Suppose that $AB = 9,$ $DC = 12,$ and the area of $\triangle AKD$ is $24.$ What is the area of trapezoid $ABCD?$

$92$

$94$

$96$

$98$

$100$

Solution:

Triangles $AKB$ and $CKD$ are similar with ratio $\dfrac{9}{12} = \dfrac{3}{4}.$

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

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

Thus, the correct answer is D.

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