Question

Rhombus of a side of $10$ cm has two angles of ${60}^{\circ }$ each.

Find a if the area a$\sqrt{3}c{m}^2$

Answer 1

$\Rightarrow$ In given figure ABCD is a rhombus.

$\Rightarrow$ AC and BD are diagonals of rhombus.

$\Rightarrow$ AD = CD = CB = BA = 10cm --- (Given)

$\Rightarrow$ $\angle$ADC =60${}^{\circ }$ -- (given)

$\Rightarrow$ $\angle$ADO = 30${}^{\circ }$ --- (Diagonals of rhombus bisect angle equally.)

$\Rightarrow$ $\angle$AOD = 90${}^{\circ }$ ---- (Diagonals of rhombus perpendicularly bisect each other)

$\Rightarrow$ In $△$AOD

$\Rightarrow$ sin30${}^{\circ }$ = $\frac{AO}{AD}$

$\Rightarrow$ $\frac{1}{2}$ = $\frac{AO}{10}$

$\Rightarrow$ AO = 5cm.

$\therefore$ Diagonal AC = 2$\times$AO = 10cm. -- (1)

Now,

$\Rightarrow$ cos30${}^{\circ }$ = $\frac{OD}{AD}$

$\Rightarrow$ $\frac{\sqrt{3}}{2}$ = $\frac{OD}{10}$

$\Rightarrow$ OD = 5$\sqrt{3}$cm

$\therefore$ Diagonal BD = 2$\times$OD = 10$\sqrt{3}$cm ---- (2)

Area of rhombus ABCD = $\frac{AC\times BD}{2}$

From (1) and (2) we get,

$\Rightarrow$ $\frac{10\times 10\sqrt{3}}{2}$ = 50$\sqrt{3}c{m}^2$

$\therefore$ Area of rhombus ABCD is 50$\sqrt{3}c{m}^2$.

So a must be 50