imc

The area of a triangle can be given by $\frac{1}{2}ab\sin C$. $MC=1$ because it is the midpoint of a side, and $CD=2$ because it is the same length as $BC$. Each angle of an equilateral triangle is $60^{\circ }$ so $\angle MCD=120^{\circ }$. The area is $\frac{1}{2}(1)(2)\sin 120^{\circ }=\boxed{\frac{\sqrt{3}}{2}}$. Note: Even if you don't know the value of $\sin 120^{\circ }$, you can use the fact that $\sin x=\sin (180^{\circ }-x)$, so $\sin 120^{\circ }=\sin 60^{\circ }$. You can easily calculate $\sin 60^{\circ }$ to be $\frac{\sqrt{3}}{2}$ using equilateral triangles.