Two sides of a triangle are equal to 10 and 12, and the median drawn to the third side is equal to 5. Find the area of the triangle.

#
On the extension of the median $A M$ beyond point $M$, lay off the segment $M D$, equal to $A M$.

## Solution

Let $A M$ be the median of triangle $A B C$, where $A M=5, A B=10, A C=12$. On the extension of the median $A M$ beyond point $M$, we lay off the segment $M D$, equal to $A M$. Then $A B D C$ is a parallelogram with diagonals $B C$ and $A D$, and the area of triangle $A B C$ is equal to the area of the isosceles triangle $A B D$, in which $A B=A D=10, B D=12$. We find the height $A H$ of triangle $A B D$ using the Pythagorean theorem in the right triangle $A B H$:

$$
A H=\sqrt{A B^{2}-B H^{2}}=\sqrt{100-36}=8.
$$

Therefore,

$$
S_{\triangle \mathrm{ABC}}=S_{\triangle \mathrm{ABD}}=\frac{1}{2} B D \cdot A H=\frac{1}{2} \cdot 12 \cdot 8=48
$$

## Answer

48.