Chapter 0, Problem 45PO

To determine

The perimeter and area of the given rectangle.

Answer to Problem 45PO

  $\begin{array}{l}A=30{\text{ in}}^2\\ P=23\text{ in}\end{array}$

Explanation of Solution

Given:

Length $l=4$ in Width $w=7.5$ in

Formula used:

Area of rectangle

  $A=a\times b$

Perimeter of rectangle

  $P=2\left(l+w\right)$

Where,

  $l=$ Length

  $w=$ Width

Calculation:

Area of rectangle

  $\begin{array}{l}A=l\times w\\ A=4\times 7.5\\ A=30{\text{ in}}^2\end{array}$

Perimeter of the rectangle

  $\begin{array}{l}P=2\left(l+w\right)\\ P=2\left(4+7.5\right)\\ P=2\left(11.5\right)\\ P=23\text{ in}\end{array}$

Conclusion:

Hence, the perimeter of rectangle is $\text{23 in}$ and area of rectangleis ${\text{30 in}}^2$ .