Q45. - Algebra 1

Question

Find the perimeter and area of each figure.

Step-by-Step Solution

Answer

$\begin{array}{l}\text{Perimeter }P=18\text{ in}\\ \text{Area }A=13.5{\text{ in}}^2\end{array}$

Step 1. We have given a right angle triangle with all sides given.

$\begin{array}{l}\text{Height }h=6\text{ in}\\ \text{Base }b=4\frac{1}{2}\text{ in}\\ \text{Hypoteneous }l=7\frac{1}{2}\text{ in}\end{array}$

Step 2. Formula used.

$\begin{array}{l}\text{Perimeter of right angle triangle }P=h+b+l\\ \text{Area of right angle triangle }A=\frac{1}{2}bh\end{array}$

Step 3. Calculate the perimeter of the right-angle triangle.

$\begin{array}{l}\text{Parameter of right angle triangle }P=h+b+l\\ P=6+4\frac{1}{2}+7\frac{1}{2}\\ P=6+4.5+7.5\\ P=18\text{ in}\end{array}$

Step 4. Calculate the area of the right-angle triangle.

$\begin{array}{l}\text{Area of right angle triangle }A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(4\frac{1}{2}\right)\left(6\right)\\ A=13.5{\text{ in}}^2\end{array}$

Therefore,

$\begin{array}{l}\text{Perimeter }P=18\text{ in}\\ \text{Area }A=13.5{\text{ in}}^2\end{array}$