Question

A parallelogram containing a ${60}^{\circ }$ angle has a perimeter of length $p$ and its longer diagonal is of length $d$. Find its area.

• A. $\frac{\sqrt{3}}{8}(p^2-4d^2)$
• B. $\frac{\sqrt{3}}{4}(p^2-4d^2)$
• C. $\frac{\sqrt{3}}{8}(p^2-2d^2)$
• D. $\frac{\sqrt{3}}{4}(p^2-2d^2)$

Answer 1

The correct option is A $\frac{\sqrt{3}}{8}(p^2-4d^2)$

Perimeter$=4a$

$P=4a$

a=$\frac{P}{4}$

d is the longer diagnol.

$x^2$+$\frac{d^2}{4}$=$a^2$, $x^2=(a^2-\frac{d^2}{4})$

Area of para||elogram = $2\times \frac{\sqrt{3}}{4}{a}^2\Rightarrow 2\times \frac{\sqrt{3}}{4}(a^2-\frac{d^2}{4})$

$\Rightarrow 2\times \frac{\sqrt{3}}{4}(\frac{P^2}{16}-\frac{d^2}{4})$

$\Rightarrow \frac{\sqrt{3}}{8}(P^2-4d^2)$

Hence, $\frac{\sqrt{3}}{8}(P^2-4d^2)$ is the correct answer.