2009 AMC 10A Problem 10

Question

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

The altitude to the hypotenuse satisfies

BD^2 = AD \cdot DC

The hypotenuse is

AC = AD + DC = 7

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.

Problem 10 in other years