Chapter 4, Problem 21CT

To determine

To find : The original dimension of the rectangle.

Answer to Problem 21CT

The original dimensions of the rectangle is as follows:

Length is 9cm

Width is 5cm

Explanation of Solution

Given information:

A rectangle is $4\text{cm}$ longer than it is wide. $\text{I}$ t the length and width are both decreased by $2\text{cm},$ the area is decreased by $24{\text{cm}}^2$

Calculation:

Let the width of the rectangle be $x$ .

Then the Length of the rectangle is $x+4$ .

Write the expression as:

  $\begin{array}{c}\text{Area of the}\\ \text{original rectangle}\end{array}-24=\begin{array}{c}\text{Area of the}\\ \text{new rectangle}\end{array}$

  $x\cdot (x+4)-24=(x-2)(x+2)$

  $x^2+4x-24=x^2+2x-2x-4$

  $x^2+4x-24=x^2-4$

  $x^2-x^2+4x=-4+24$

  $4x=-4+24$

  $4x=20$

  $x=5$

Thus, the width of the rectangle is 5 and the length of the rectangle is $x+4=9$ .

Therefore, the original dimensions of the rectangle is length = 9cm and width = 5cm