imc

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 $B{D}^2=AD\cdot CD$. (See below for a proof.) Then $BD=\sqrt{3\cdot 4}=2\sqrt{3}$, and the area of the triangle $ABC$ is $\frac{AC\cdot BD}{2}=7\sqrt{3}\Rightarrow \boxed{7\sqrt{3}}$. Proof: Consider the Pythagorean theorem for each of the triangles $ABC$, $ABD$, and $CBD$. We get: 1. $A{B}^2+B{C}^2=A{C}^2=(AD+DC{)}^2=A{D}^2+D{C}^2+2\cdot AD\cdot DC$. 2. $A{B}^2=A{D}^2+B{D}^2$ 3. $B{C}^2=B{D}^2+C{D}^2$ Substituting equations 2 and 3 into the left hand side of equation 1, we get $B{D}^2=AD\cdot DC$. Alternatively, note that $△ABD\sim △BCD⟹\frac{AD}{BD}=\frac{BD}{CD}$.