Chapter 8.6, Problem 60PFA

a.

To determine

To find: The perimeter of the rectangle.

Answer to Problem 60PFA

  $4x+8$

Explanation of Solution

Given:

The perimeter of rectangle is same as the perimeter of square.

The area of rectangle is $x^2+4x-21$ and side lengths are integers.

Calculation:

Factor the polynomial $x^2+4x-21$

  $\begin{array}{l}{x}^2+7x-3x-21\\ =x\left(x+7\right)-3\left(x+7\right)\\ =\left(x+7\right)\left(x-3\right)\end{array}$

Hence, the dimensions of the rectangle are $\left(x+7\right)\left(x-3\right)$

Therefore, the perimeter of the rectangle is given by

  $\begin{array}{l}2\left(x+7\right)+2\left(x-3\right)\\ =2x+14+2x-6\\ =4x+8\end{array}$

b.

To determine

To find: The area of the square

Answer to Problem 60PFA

  $x^2+4x+4$

Explanation of Solution

Given:

The perimeter of rectangle is same as the perimeter of square.

The area of rectangle is $x^2+4x-21$ and side lengths are integers.

Calculation:

From part (a), the perimeter of the rectangle is given by $4x+8$

Since, perimeter of square and the rectangle are equal. Hence, perimeter of square is $4x+8$

This, the side length of the square is given by

  $\begin{array}{l}\frac{1}{4}\left(4x+8\right)\\ =\frac{1}{4}\cdot 4\cdot \left(x+2\right)\\ =\left(x+2\right)\end{array}$

Therefore, the area of the square is given by

  $\begin{array}{l}{\left(x+2\right)}^2\\ =x^2+4x+4\end{array}$