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

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2009 AMC 10A Problem 10

Attempt Problem 10 of the 2009 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 2009 AMC 10A solutions, or check the answer key.

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

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 $\triangle ABC?$

$4\sqrt{3}$

$7\sqrt{3}$

$21$

$14\sqrt{3}$

$42$

Answer: B

Concepts:right triangle altitude triangle area

Difficulty rating: 1240

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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