Chapter 11.1, Problem 1BCYP

To determine

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

Answer to Problem 1BCYP

The perimeter is $P=96ft$ and the area of the given parallelogram is $A=552f{t}^2$ .

Explanation of Solution

Given information: From the given figure it can be observed that in the parallelogram $h=24ft,b=23ft,b_1=7ft$ .

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

  $h=24ft,b=23ft,b_1=7ft$

Also to find the other side of the parallelogram, $\text{Consider}\text{the}\text{Δ}\text{formed}\text{outside}\text{the}\text{Parallelogram}\text{.}$

  $\begin{array}{c}$ {\text{(}h\text{)}}^{\text{2}}\text{+}{\text{7}}^{\text{2}}\text{=}{\left(l\right)}^{\text{2}}\mathrm{.........}\text{(Pythagoras}\text{Theorem}\text{of}\text{Right}\text{Angle}\text{Δ)}$$ {(l)}^2={24}^2+7^2$l=\sqrt{\left(576+49\right)}\\ l=\sqrt{625}=25ft\end{array}$

Then the perimeter of the parallelogram is

  $\begin{array}{c}P=2\times (l+b)\\ P=2\times (25+23)\\ =2\times (48)\\ P=96ft\end{array}$

And the area of the parallelogram is

  $\begin{array}{c}A=b\times h\\ \text{=23×24}\\ A=552f{t}^2\end{array}$

Therefore the perimeter is $P=96ft$ and the area of the given parallelogram is $A=552f{t}^2$ .