Chapter 11.1, Problem 4CYU

To determine

To calculate: The perimeter and area of the given parallelogram.

Answer to Problem 4CYU

The perimeter is $P=60.1m$ and the area of the given parallelogram is $A=115m^2$ .

Explanation of Solution

Given information: From the given figure it can be observed that in the parallelogram $b=23m,b_1=5m,\theta ={45}^{\circ }$ .

Formula used:

The Perimeter and area of a parallelogram is given by

  $\begin{array}{l}P=2\left(l+b\right)\\ A=b\times h\\ b\text{is}\text{the}\text{base}\text{of}\text{the}\text{parallelogram}\text{.}\\ l\text{is}\text{the}\text{side}\text{adjacent}\text{to}\text{the}\text{base}\left(b\right)\text{of}\text{the}\text{parallelogram}\text{.}\\ h\text{is}\text{the}\text{height}\text{of}\text{the}\text{parallelogram}\text{.}\end{array}$

Calculation:

The given information is

  $b=23m,b_1=5m,\theta ={45}^{\circ }$

Also to find the height and adjacent side of the parallelogram, $\text{Consider}\text{the}\text{Δ}\text{inscribed}\text{in}\text{Parallelogram}\text{.}$

  $\begin{array}{l}\tan \theta =\frac{h}{b_1}\\ \tan {45}^{\circ }=\frac{h}{5}\\ h=5\times \left(1\right)=5m\\ \\ \cos \theta =\frac{b_1}{l}\\ \cos {45}^{\circ }=\frac{5}{l}\\ l=\frac{5}{\left(\frac{1}{\sqrt{2}}\right)}=5\sqrt{2}=7.07m\end{array}$

Then the perimeter of the parallelogram is

  $\begin{array}{c}P=2\times (l+b)\\ P=2\times (7.07+23)\\ =2\times (30.07)\\ P=60.14m\\ P=60.1m\end{array}$

And the area of the parallelogram is

  $\begin{array}{c}A=b\times h\\ \text{=23}\times 5\\ A=115m^2\end{array}$

Therefore the perimeter is $P=60.1m$ and the area of the given parallelogram is $A=115m^2$ .