2009 AMC 10A Problem 10

The area of a triangle is half of base times height. Take $AC$ as the base: $AC = 3 + 4 = 7$. The height to that base is $BD$. So we only need $BD$. Call it $h$.

1. Triangle $ABD$ has a right angle at $D$, with legs $3$ and $h$. By the Pythagorean theorem,
$$AB^2 = 3^2 + h^2 = 9 + h^2.$$
2. Triangle $CBD$ also has a right angle at $D$, with legs $4$ and $h$:
$$BC^2 = 4^2 + h^2 = 16 + h^2.$$
3. The big triangle $ABC$ has a right angle at $B$, and its long side is $AC = 7$:
$$AB^2 + BC^2 = 7^2 = 49.$$
4. Put steps 1 and 2 into step 3:
$$(9 + h^2) + (16 + h^2) = 49, \qquad 25 + 2h^2 = 49, \qquad h^2 = 12.$$
5. So $h = \sqrt{12} = 2\sqrt{3}$.
6. Area:
$$\frac{1}{2} \times 7 \times 2\sqrt{3} = 7\sqrt{3}.$$

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

Drawing the height from the right angle splits the triangle into two smaller right triangles, and the Pythagorean theorem works in all three. Writing all three equations and adding makes the unknown sides drop out. The result is a fact worth remembering: the height to the long side, squared, equals the product of the two pieces it cuts ($3 \times 4 = 12$).

Use similar triangles. The small triangles $ABD$ and $BCD$ both have a right angle at $D$, and $\angle BAD = \angle CBD$ (each one plus $\angle ABD$ makes $90^\circ$). So they are similar, and matching sides gives
$$\frac{AD}{BD} = \frac{BD}{DC}, \qquad BD^2 = 3 \times 4 = 12.$$
Then $BD = 2\sqrt3$ and the area is $\frac12 \times 7 \times 2\sqrt3 = 7\sqrt3$.