Question

A rhombus of side 18 cm has one of its angles as ${135}^{\circ }$, then the area of the rhombus will be equal to

• A. $81c{m}^2$
• B. $162\sqrt{2}c{m}^2$
• C. $81\sqrt{2}c{m}^2$
• D. $162c{m}^2$

Answer 1

The correct option is B

$162\sqrt{2}c{m}^2$

In rhombus ABCD, draw $DE{\perp }^{r}AB$

Given ABCD is a rhombus having AB =CD =BC =AD =18 cm and $\angle B={135}^{\circ }$

$\angle DAB=({180}^{\circ }-{135}^{\circ })={45}^{\circ }\left(AD||CB\right)$

Angle of triangle AED are ${45}^{\circ },{45}^{\circ },{90}^{\circ }$ corresponding sides of triangle can be calculated as

$\Rightarrow \sin \left(45\right):\sin \left(45\right):\sin \left(90\right)$

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

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

$\begin{array}{ccccc}{45}^{\circ }& & {45}^{\circ }& & {90}^{\circ }\\ 1& :& 1& :& \sqrt{2}\\ x& :& x& :& x\sqrt{2}\\ AE& & DE& & AD\\ \downarrow & & \downarrow & & \downarrow \\ 9\sqrt{2}cm& & 9\sqrt{2}cm& & 18cm\end{array}$

($x\sqrt{2}=18\Rightarrow x=\frac{18}{\sqrt{2}}=9\sqrt{2}$)

So $DE=9\sqrt{2}cm$

Area of rhombus ABCD $=AB\times DE$

$=18\times 9\sqrt{2}=162\sqrt{2}c{m}^2$