2009 AMC 10A Problem 10 — Solution | LIVE by Po-Shen Loh

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2009 AMC 10A Problem 10

Below is the professionally curated solution for Problem 10 of the 2009 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:right triangle altitude triangle area

Difficulty rating: 1240

10.

Triangle $ABC$ has a right angle at $B.$ Point $D$ is the foot of the altitude from $B,$ $AD = 3,$ and $DC = 4.$ What is the area of $\triangle ABC?$

$4\sqrt{3}$

$7\sqrt{3}$

$21$

$14\sqrt{3}$

$42$

Solution:

For the altitude from the right angle to the hypotenuse, $BD^2 = AD \cdot DC = 3 \cdot 4 = 12,$ so $BD = 2\sqrt{3}.$

The hypotenuse is $AC = 3 + 4 = 7,$ so the area is $\dfrac12 \cdot 7 \cdot 2\sqrt3 = 7\sqrt3.$

Thus, the correct answer is B.

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