$$ABC$$ is an equilateral triangle of side $$a$$ units. Midpoint of the side $$BC$$ is $$D$$. $$D$$ is also the midpoint of the side $$PQ$$ of a rectangle $$PQRS$$ having its side $$PQ$$ along $$BC$$. Also $$P$$ is the midpoint of the line segment $$BD$$. If the vertex $$A$$ of the triangle lies on the side $$RS$$ of the rectangle, the area of the rectangle is __________

PLACEHOLDER ANSWER - the answer is an expression in $$a$$, not a number. The required area is $$\frac{\sqrt{3}}{4}a^2$$. Since $$BD = \frac{a}{2}$$ and $$P$$ is its midpoint, $$PD = \frac{a}{4}$$, and as $$D$$ is the midpoint of $$PQ$$, $$PQ = \frac{a}{2}$$. $$A$$ lies on $$RS$$, so the width of the rectangle is the altitude $$\frac{\sqrt{3}}{2}a$$. Area $$= \frac{a}{2} \times \frac{\sqrt{3}}{2}a = \frac{\sqrt{3}}{4}a^2$$.