Chapter 11.1, Problem 5CYU

To determine

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

Answer to Problem 5CYU

The perimeter is $P=43.5in$ and the area of the given parallelogram is $A=20i{n}^2$ .

Explanation of Solution

Given information: From the given figure it can be observed that in the triangle $a=21.5in,b=5in,b_1=15in,h=8in$ .

Formula used:

The Perimeter and area of a parallelogram is given by

  $\begin{array}{l}P=\left(a+b+c\right)\\ A=\frac{1}{2}\times \left(b\times h\right)\\ b\text{is}\text{the}\text{base}\text{of}\text{the}\text{triangle}\text{.}\\ a\&care\text{the}\text{sides}\text{adjacent}\text{to}\text{the}\text{base}\left(b\right)\text{of}\text{the}\text{triangle}\text{.}\\ h\text{is}\text{the}\text{height}\text{of}\text{the}\text{triangle}\text{.}\end{array}$

Calculation:

The given information is

  $a=21.5in,b=5in,b_1=15in,h=8in$

Also to find the adjacent side of the triangle, $\text{Consider}\text{the}\text{unshaded}\text{Δ}\text{.}$

  $\begin{array}{l}{h}^{\text{2}}+b_1{}^2=c^2\mathrm{...............}\text{By}\text{Pythagoras}\text{Theoram}\\ c^2={\left(8\right)}^2+{\left(15\right)}^2\\ c=\sqrt{64+225}=\sqrt{289}\\ c=17in\end{array}$

Then the perimeter of the parallelogram is

  $\begin{array}{c}P=\left(a+b+c\right)\\ =\left(21.5+5+17\right)\\ P=43.5in\end{array}$

And the area of the parallelogram is

  $\begin{array}{c}A=\frac{1}{2}\times \left(b\times h\right)\\ \text{=}\frac{1}{2}\times \left(5\times 8\right)=\frac{1}{2}\times 40\\ A=20i{n}^2\end{array}$

Therefore the perimeter is $P=43.5in$ and the area of the given parallelogram is $A=20i{n}^2$ .