AMC10 2009 10A, Problem 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 $\triangle ABC$? [asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4;  pair B=(0,0), C=(sqrt(28),0), A=(0,sqrt(21)); pair D=foot(B,A,C); pair[] ps={B,C,A,D};  draw(A--B--C--cycle); draw(B--D); draw(rightanglemark(B,D,C));  dot(ps); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$3$",midpoint(A--D),NE); label("$4$",midpoint(D--C),NE); [/asy]

Solution: 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 $\frac{AC\cdot BD}2 = 7\sqrt3\Rightarrow\fbox{7\sqrt3}$. Proof: Consider the Pythagorean theorem for each of the triangles $ABC$, $ABD$, and $CBD$. We get: $AB^2 + BC^2 = AC^2 = (AD+DC)^2 = AD^2 + DC^2 + 2 \cdot AD \cdot DC$. $AB^2 = AD^2 + BD^2$ $BC^2 = BD^2 + CD^2$ Substituting equations 2 and 3 into the left hand side of equation 1, we get $BD^2 =  AD \cdot DC$. Alternatively, note that $\triangle ABD \sim \triangle BCD \Longrightarrow \frac{AD}{BD} = \frac{BD}{CD}$.

Answer: 7\sqrt3