Chapter 11.1, Problem 6CYU

To determine

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

Answer to Problem 6CYU

The perimeter is $P=80mm$ and the area of the given parallelogram is $A=240m{m}^2$ .

Explanation of Solution

Given information: From the given figure it can be observed that in the triangle $b=30mm,b_1=12mm,a_1=20mm$ .

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

  $b=30mm,b_1=12mm,a_1=20mm$

To find the height of the triangle, $\text{Consider}\text{the}\text{unshaded}\text{Δ}\text{.}$

  $\begin{array}{l}{h}^{\text{2}}+b_1{}^2=a_1{}^2\mathrm{...............}\text{By}\text{Pythagoras}\text{Theoram}\\ h^{\text{2}}+{12}^2={20}^2\\ h^2=400-144\\ h=\sqrt{400-144}=\sqrt{256}\\ h=16mm=a\end{array}$

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

  $\begin{array}{l}{a}^{\text{2}}+b^2=c^2\mathrm{...............}\text{By}\text{Pythagoras}\text{Theoram}\\ h^{\text{2}}+b^2=c^2\left(\because a=h\right)\\ c^2={\left(30\right)}^2+{\left(16\right)}^2\\ c=\sqrt{900+256}=\sqrt{1156}\\ c=34mm\end{array}$

Then the perimeter of the parallelogram is

  $\begin{array}{c}P=\left(a+b+c\right)\\ =\left(16+30+34\right)\\ P=80mm\end{array}$

And the area of the parallelogram is

  $\begin{array}{c}A=\frac{1}{2}\times \left(b\times h\right)\\ \text{=}\frac{1}{2}\times \left(30\times 16\right)=\frac{1}{2}\times 480\\ A=240m{m}^2\end{array}$

Therefore the perimeter is $P=80mm$ and the area of the given parallelogram is $A=240m{m}^2$ .