Question

In the rhombus ABCD, Side AD = 12 cm and $\angle B$ is ${135}^{\circ }$ as shown in figure. The height (DE) and the area of the rhombus will be equal to

• A. $6\sqrt{2}cm,36\sqrt{2}c{m}^2$
• B. $6\sqrt{2}cm,72\sqrt{2}c{m}^2$
• C. $12\sqrt{2}cm,36c{m}^2$
• D. $12\sqrt{2}cm,72c{m}^2$

Answer 1

The correct option is B

$6\sqrt{2}cm,72\sqrt{2}c{m}^2$

Given ABCD is a rhombus

$AB=CD=AD=CB$ , $\angle B={135}^{\circ }$ and $\angle A={45}^{\circ }$

then $\angle ADE=({180}^{\circ }-{135}^{\circ })={45}^{\circ }$

Angle of $\Delta ADE$ are ${45}^{\circ },{45}^{\circ },{90}^{\circ }$

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

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

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

So, the sides AE, ED, AD will be in the ratio $1:1:\sqrt{2}$

The corresponding sides will be calculated as

$\begin{array}{ccccc}{45}^{\circ }& & {45}^{\circ }& & {90}^{\circ }\\ 1& :& 1& :& \sqrt{2}\\ x& :& x& :& x\sqrt{2}\\ AE& & ED& & AD\\ \downarrow & & \downarrow & & \downarrow \\ 6\sqrt{2}cm& & 6\sqrt{2}cm& & 12cm\end{array}$

($x\sqrt{2}=12\Rightarrow x=\frac{12}{\sqrt{2}}=6\sqrt{2}$)

The distance between parallel lines is $ED=6\sqrt{2}cm$

Area of rhombus will be equal to $AB\times DE=12\times 6\sqrt{2}$

$=72\sqrt{2}c{m}^2$