Chapter SR17, Problem 4P

Solution

Find the perimeter and area of the given figure.

Perimeter of rhombus $=24m$

Area of rhombus $=30m^2$

Given:

The given figure.

  

Concept Used:

• Formula of Perimeter of rhombus $=4a$ .
• Formula of Area of rhombus $=\frac{(diagonal1)\cdot (diagonal2)}{2}$

Calculation:

The given figure is -

  

As the given figure has all sides are equal and opposite sides are parallel so it is a Rhombus.

The Perimeter of rhombus $=4a$

  $\begin{array}{l}=4\cdot 6\\ =24\end{array}$

So, Perimeter of rhombus $=24m$

The Area of rhombus $=\frac{(diagonal1)\cdot (diagonal2)}{2}$

  $=\frac{AC\cdot BD}{2}$

In $\Delta ABE$ -

  $\begin{array}{l}A{B}^2=A{E}^2+B{E}^2\\ 6^2=5^2+B{E}^2\\ B{E}^2=36-25\\ B{E}^2=11\\ BE=\sqrt{11}\end{array}$

So $\begin{array}{c}CE=BC-BE\\ =6-\sqrt{11}\end{array}$

In $\Delta ACE$ -

  $\begin{array}{l}A{C}^2=A{E}^2+C{E}^2\\ $ A{C}^2=5^2+{(6-\sqrt{11})}^2$A{C}^2=25+36+11-2(6)(\sqrt{11})\\ A{C}^2=72-12\sqrt{11}\\ AC=\sqrt{72-12\sqrt{11}}\\ AC=\sqrt{4(18-3\sqrt{11})}\\ AC=2\sqrt{18-3\sqrt{11}}\end{array}$

  $\begin{array}{c}AO=\frac{1}{2}AC\\ =\frac{1}{2}(2\sqrt{18-3\sqrt{11}})\\ =\sqrt{18-3\sqrt{11}}\end{array}$

In $\Delta AOB$ -

  $\begin{array}{c}A{B}^2=A{O}^2+O{B}^2\\ $ 6^2={(\sqrt{18-3\sqrt{11}})}^2+O{B}^2$36=18-3\sqrt{11}+O{B}^2\\ 36-18+3\sqrt{11}=O{B}^2\\ O{B}^2=18+3\sqrt{11}\\ OB=\sqrt{18+3\sqrt{11}}\end{array}$

  $\begin{array}{c}BD=2OB\\ =2\sqrt{18+3\sqrt{11}}\end{array}$

So, Area of rhombus $=\frac{AC\cdot BD}{2}$

  $\begin{array}{l}=\frac{\left(2\sqrt{18-3\sqrt{11}}\right)\cdot \left(2\sqrt{18+3\sqrt{11}}\right)}{2}\\ =\frac{4\left(\sqrt{18-3\sqrt{11}}\right)\cdot \left(\sqrt{18+3\sqrt{11}}\right)}{2}\\ =2\sqrt{(18-3\sqrt{11})\cdot (18+3\sqrt{11})}\\ =2\sqrt{{(18)}^2-{(3\sqrt{11})}^2}\\ =2\sqrt{324-99}\\ =2\sqrt{225}\\ =2(15)\\ =30\end{array}$

So, Area of rhombus $=30m^2$