Chapter 11.4, Problem 8PSB

To determine

To find: The area of rhombus if perimeter is 40 and one angle is 60

Answer to Problem 8PSB

Area of rhombus = $50\sqrt{3}{\text{ unit}}^2$

Explanation of Solution

Given information :

Perimeter of rhombus ( P ) = 40 units

One angle = 60

Formula used :

Area of rhombus = $\frac{1}{2}$ (product of diagonals)

Calculation :

  

The perimeter of the rhombus is given as

  $\begin{array}{l}P=4a\text{ [a is the side of the rhombus]}\\ 40=4a\\ a=10\text{ units}\end{array}$

We know that

  $\mathrm{Sin}\theta =\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

In triangle AOD, we have

  $\begin{array}{l}\mathrm{Sin}\theta =\frac{OA}{AD}\\ \mathrm{Sin}{60}^o=\frac{OA}{10}\text{ [Where }AD=a=10\text{]}\\ \frac{\sqrt{3}}{2}=\frac{OA}{10}\\ OA=5\sqrt{3}\text{ units}\end{array}$

Therefore,

  $\begin{array}{l}AC=2\times OA\\ AC=2\times 5\sqrt{3}\\ AC=10\sqrt{3}\text{ units}\end{array}$

Similarly,in triangle AOD, we have

  $\begin{array}{l}\mathrm{Cos}\theta =\frac{\text{Base}}{\text{Hypotenuse}}\\ \mathrm{Cos}{60}^o=\frac{OD}{AD}\\ \mathrm{Cos}{60}^o=\frac{OD}{10}\text{ [Where }AD=a=10\text{]}\\ \frac{1}{2}=\frac{OD}{10}\\ OD=5\text{ units}\end{array}$

Therefore,

  $\begin{array}{l}BD\text{ = 2}\times \text{OD}\\ BD=2\times 5\\ BD=\text{10 units}\end{array}$

Now, Area of rhombus = $\frac{1}{2}$ (product of diagonals)

  $\Rightarrow$ Area of rhombus

= $\begin{array}{l}=\frac{1}{2}\times \text{AC}\times \text{BD}\\ \text{=}\frac{1}{2}\times 10\sqrt{3}\times 10\\ =50\sqrt{3}{\text{ unit}}^2\end{array}$