2009 AMC 10A Problems/Problem 10

Problem

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 $\triangle ABC$ ?

It is a well-known fact that in any right triangle $ABC$ with the right angle at $B$ and $D$ the foot of the altitude from $B$ onto $AC$ we have $BD^2 = AD\cdot CD$ . (See below for a proof.) Then $BD = \sqrt{ 3\cdot 4 } = 2\sqrt 3$ , and the area of the triangle $ABC$ is $\frac{AC\cdot BD}2 = \boxed{7\sqrt 3}$ .

Proof: Consider the Pythagorean theorem for each of the triangles $ABC$ , $ABD$ , and $CBD$ . We get:

Substituting equations 2 and 3 into the left hand side of equation 1, we get $BD^2 = AD \cdot DC$ .