2009 AMC 10A Problem 10

The altitude from the right angle splits $\triangle ABC$ into two triangles, $\triangle ABD$ and $\triangle BCD$, each similar to the original. In particular $\triangle ABD \sim \triangle BCD$ (both right-angled, and $\angle BAD = \angle CBD$ since each is complementary to $\angle ABD$). Matching legs,
$$\frac{AD}{BD} = \frac{BD}{DC} \quad\Longrightarrow\quad BD^2 = AD \cdot DC = 3 \cdot 4 = 12,$$
so $BD = 2\sqrt{3}$.

The hypotenuse is $AC = 3 + 4 = 7$ and $BD$ is the altitude to it, so
$$[ABC] = \frac{1}{2} \cdot 7 \cdot 2\sqrt{3} = 7\sqrt{3}.$$

The answer is $\boxed{\textbf{(B)}\ 7\sqrt3}$.

In a right triangle the altitude to the hypotenuse is the geometric mean of the two segments of the hypotenuse; this comes straight from the similarity of the two small triangles. Whenever a right triangle is given with the hypotenuse already split by the altitude, compute that altitude first; the area then needs no side lengths at all.

Use the leg relations $AB^2 = AD \cdot AC = 21$ and $BC^2 = DC \cdot AC = 28$. Then $[ABC] = \frac12 \sqrt{21}\sqrt{28} = \frac12 \sqrt{588} = \frac12 \cdot 14\sqrt{3} = 7\sqrt{3}$.

• Treating 3 and 4 as the legs of a 3-4-5 triangle (area 6), or dropping the 1/2 and answering 14 sqrt 3.
• Using $BD = \frac{3 + 4}{2}$ or $BD = \sqrt{3^2 + 4^2}$ instead of the geometric mean $\sqrt{3 \cdot 4}$.