Chapter 11.1, Problem 30PPE

To determine

Find the simplified expression for the width of the rectangle B.

Answer to Problem 30PPE

Width of Rectangle B is $6(x+1)$

Explanation of Solution

Given:

Rectangle A has length 2x + 6 and width 3x. Rectangle B has length (x + 2) and an area 12 square units greater than Rectangle A`s area. What is a simplified expression for the width of Rectangle B?

Concept Used:

Area of Rectangle: $\text{A}\text{=}\text{length}·\text{width}$

Condition 1: Rectangle A has $\text{length}=2x+6\text{and}\text{width}\text{=}\text{3x}$

Condition 2: Rectangle B has length (x + 2) and an area 12 square units greater than Rectangle A`s area.

Calculation:

Rectangle A has $\text{length}=2x+6\text{and}\text{width}\text{=}\text{3x}$

Area of Rectangle: $\text{A}\text{=}\text{length}·\text{width}=(2x+6)·3x$

Rectangle B has length (x + 2) and an area 12 square units greater than Rectangle A`s area.

Area of Rectangle B = Area of Rectangle A + 12

Let the width of Rectangle is w .

Area of Rectangle B: $\text{A}\text{=}\text{length}·\text{width}=(x+2)·w$

Equation: Area of Rectangle B = Area of Rectangle A + 12

  $\begin{array}{l}(x+2)·w=(2x+6)·3x+12\\ \frac{(x+2)·w}{(x+2)}=\frac{6x^2+18x+12}{(x+2)}\\ w=\frac{6(x^2+3x+2)}{(x+2)}=\frac{6(x+2)(x+1)}{(x+2)}=6(x+1)\\ w=6(x+1)\end{array}$ Width of Rectangle B is $6(x+1)$

Thus, width of Rectangle B is $6(x+1)$