Question

The length of the diagonal AC of a rhombus ABCD is 20 cm and one of its angle measures ${120}^{\circ },$ then the length of the other diagonal BD is

• A. $10\sqrt{3}cm$
• B. 10 cm
• C. 20 cm
• D. $\frac{20}{\sqrt{3}}cm$

Answer 1

The correct option is D

$\frac{20}{\sqrt{3}}cm$

ABCD is a rhombus with $AC=20cm$

$\angle A={60}^{\circ },\angle B={120}^{\circ }$

Here $AO=\frac{1}{2}AC=\frac{1}{2}\times 20=10cm$ (diagonals of a rhombus bisect each other at right angles)

$\angle AOB={90}^{\circ }$

In triangle AOB, the angles are ${30}^{\circ },{60}^{\circ },{90}^{\circ }$

So, the corresponding sides can be calculated as

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

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

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

$\begin{array}{ccccc}{30}^{\circ }& & {60}^{\circ }& & {90}^{\circ }\\ x& :& x\sqrt{3}& :& 2x\\ \downarrow & & \downarrow & & \downarrow \\ BO& & AO& & AB\\ \downarrow & & \downarrow & & \downarrow \\ \frac{10}{\sqrt{3}}cm& & 10cm& & \frac{20}{\sqrt{3}}cm\end{array}$

$BO=\frac{10}{\sqrt{3}}cm$

$BD=2\times BO$

$BD=2\times \frac{10}{\sqrt{3}}=\frac{20}{\sqrt{3}}cm$