Question

ABC is an equilateral triangle with sides equal to 2 cm. BC is extended its own length to D, and E is the midpoint of AB. ED meets AC at F. Find the area of the quadrilateral BEFC in square centimeters in simplest radical form.

Answer 1

Answer:

$\frac{2\sqrt3}{3}$ $cm^2$

Step-by-step explanation:

Since $\triangle ABC$ is equilateral triangle, then $\angle A = \pi /3$

So,

$S_{\triangle ABC} = \frac{2 * 2 * sin(\pi /3)}{2} = \frac{2 * 2 * \sqrt{3}/2}{2} = \sqrt{3}$ $cm^2$

Then we need to find $S_{\triangle AEF}$ which can be computed by finding length_AF.

Let's call x = length_AF.

By Menelao's Theorem,

$\frac{BE*x*CD}{AE*(2-x)*BD} = \frac{1*x*2}{1*(2-x)*4}  = 1$

⇒ x = 4/3 cm

Thus,

$S_{\triangle AEF} = \frac{1 * x * sin(\pi /3)}{2} = \frac{1 * 4/3 * \sqrt{3}/2}{2} = 1/\sqrt{3}$ $cm^2$

To find the area of quadrilateral $BEFC$, we have to subtract $S_{\triangle AEF}$ from $S_{\triangle ABC}$

Hence,

$S_{BEFC} = S_{\triangle ABC} - S_{\triangle AEF} = \sqrt3 -1/\sqrt3 = \frac{2\sqrt3}{3}$ $cm^2$