Chapter 11.1, Problem 13WE

To determine

To Find: The area and perimeter of rectangle.

Answer to Problem 13WE

  $A=a^2-9$ , $P=4a$

Explanation of Solution

Given: $b=a+3,h=a-3$

Given: $b=a+3,h=a-3$

Area of rectangle

  $\begin{array}{l}A=bh\\ A=\left(a+3\right)\left(a-3\right)\\ A=a^2-3^2\\ A=a^2-9\end{array}$

Now

Perimeter of rectangle

  $\begin{array}{l}P=2(b+h)\\ P=2(a+3+a-3)\\ P=2(2a)\\ P=4a\end{array}$

Hence, $A=a^2-9$ , $P=4a$