Chapter 11.1, Problem 15WE

To determine

To Find: The Perimeter and height of rectangle.

Answer to Problem 15WE

  $h=x-3$ , $P=4x-6$

Explanation of Solution

Given: $b=x,A=x^2-3x$

Given: $b=x,A=x^2-3x$

Area of rectangle

  $\begin{array}{l}A=bh\\ x^2-3x=\left(x\right)\left(h\right)\\ h=\frac{x^2-3x}{x}\\ h=x-3\end{array}$

The height of rectangle is $h=x-3$

Now,

Perimeter of rectangle

  $\begin{array}{l}P=2(b+h)\\ P=2(x+x-3)\\ P=2(2x-3)\\ P=4x-6\end{array}$

Hence, $h=x-3$ , $P=4x-6$