Question

Equilateral $\triangle ABC$ has side length $2,$ $M$ is the midpoint of $\overline{AC},$ and $C$ is the midpoint of $\overline{BD}.$ What is the area of $\triangle CDM?$

$\dfrac{\sqrt{2}}{2}$

$\dfrac{3}{4}$

$\dfrac{\sqrt{3}}{2}$

$1$

$\sqrt{2}$

Solution:

Take $\overline{CD}$ as the base. Since $C$ is the midpoint of $\overline{BD}$ and $BC = 2,$ we have $CD = 2.$

The height of $\triangle CDM$ is the distance from $M$ to line $BD.$ Because $M$ is the midpoint of $\overline{AC},$ this distance is half the height of $\triangle ABC,$ which is $\dfrac12 \cdot \sqrt3 = \dfrac{\sqrt3}{2}.$

The area is $\dfrac12 \cdot 2 \cdot \dfrac{\sqrt3}{2} = \dfrac{\sqrt3}{2}.$

Thus, C is the correct answer.

2005 AMC 10B Problem 14

Solution:

Problem 14 in Other Years