Chapter 11.1, Problem 1CYU

To determine

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

Answer to Problem 1CYU

The perimeter is $P=56in$ and the area of the given parallelogram is $A=180i{n}^2$ .

Explanation of Solution

Given information: From the given figure it can be observed that in the parallelogram $l=13in,b=15in,b_1=5in$ .

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

  $l=13in,b=15in,b_1=5in$

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

  $\begin{array}{c}{\text{(Height)}}^{\text{2}}\text{+}{\text{5}}^{\text{2}}\text{=}{\text{13}}^{\text{2}}\mathrm{.........}\text{(Pythagoras}\text{Theorem}\text{of}\text{Right}\text{Angle}\text{Δ)}\\ $ {(h)}^2={13}^2-5^2$h=\sqrt{\left(169-25\right)}\\ h=\sqrt{144}=12in\end{array}$

Then the perimeter of the parallelogram is

  $\begin{array}{c}P=2\times (l+b)\\ P=2\times (13+15)\\ =2\times (28)\\ P=56in\end{array}$

And the area of the parallelogram is

  $\begin{array}{c}A=b\times h\\ \text{=15×12}\\ A=180i{n}^2\end{array}$

Therefore the perimeter is $P=56in$ and the area of the given parallelogram is $A=180i{n}^2$ .