Question

A rhombus having each side 10 cm has one of its angles as ${120}^{\circ }$, then find the area of the rhombus.

Answer 1

Let ABCD be the rhombus in which $AB=AD=10cm$ $\angle B={120}^{\circ }$ as shown in figure

$\angle A={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

Draw DE perpendicular to AB

In $\Delta AED,\angle A={60}^{\circ },\angle AED={90}^{\circ },\angle ADE={30}^{\circ }$

The corresponding sides can be calculated as

$\Rightarrow \sin \left(30\right):\sin \left(60\right):\sin \left(90\right)$

$\Rightarrow \frac{1}{2}:\frac{\sqrt{3}}{2}:1$

$\Rightarrow 1:\sqrt{3}:2$

In a 30-60-90 right angled triangle, the sides are in the ratio 1 : $\sqrt{3}$ : 2

i.e AE : DE : AD = 1 : $\sqrt{3}$ : 2

Hence, let the sides be x , $\sqrt{3}$x , 2x

Given, 2x = 10 cm, $\Rightarrow$ x = 5 cm

Therefore DE = 5$\sqrt{3}$ cm

Area of rhombus = $AB\times DE$

$=10\times 5\sqrt{3}=50\sqrt{3}c{m}^2$