Question

A rhombus whose side measures 40 cm has one of its angles as ${60}^{\circ },$, then the length of its diagonals are

• A. 40 cm, 80 cm
• B. $40cm,40\sqrt{3}cm$
• C. $40\sqrt{3}cm,80cm$
• D. $80cm,80\sqrt{3}cm$

Answer 1

The correct option is B

$40cm,40\sqrt{3}cm$

Let ABCD be the given rhombus with $AB=AD=40cm$ and $\angle A={60}^{\circ }$

$\angle CAB=CAD={30}^{\circ }$ (Diagonal of a rhombus bisects the angle)

In $\Delta AOB$

$\angle AOB={90}^{\circ },$ (diagonal bisect each other at ${90}^{\circ }$)

$\angle OAB={30}^{\circ },\angle OBA={60}^{\circ }$

Hence, angle of triangle AOB are ${30}^{\circ },{60}^{\circ },{90}^{\circ }$

So, the corresponding sides can be calculated as

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

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

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

$\begin{array}{ccccc}{30}^{\circ }& & {60}^{\circ }& & {90}^{\circ }\\ x& :& x\sqrt{3}& :& 2x\\ \downarrow & & \downarrow & & \downarrow \\ OB& & OA& & AB\\ \downarrow & & \downarrow & & \downarrow \\ 20cm& & 20\sqrt{3}cm& & 40cm\end{array}$

$OA=20\sqrt{3,}$ hence $AC=40\sqrt{3}cm$ (O is midpoint of AC as diagonals of a rhombus bisect each other)

Similarly $OB=20cm,$ hence $BD=40cm$