Chapter 8.5, Problem 58PFA

a.

To determine

To express: The areas of rectangle and the right triangle.

Answer to Problem 58PFA

Area of rectangle is $\left(24x^2+28x\right)\text{sq}\text{. units}$

Area of right triangle is $\left(4x^2-2x\right)\text{sq}\text{. units}$

Explanation of Solution

Given:

Rectangle has length of $4x$ units and a width of $\left(6x+7\right)$ units.

Right triangle has legs of lengths $x$ and $\left(8x-4\right)$ units.

Calculation:

Area of rectangle is the product of length and the width. Hence, the area of rectangle is given by

  $\begin{array}{l}4x\left(6x+7\right)\\ =\left(24x^2+28x\right)\text{sq}\text{. units}\end{array}$

And the area of the right triangle is given by

  $\begin{array}{l}\frac{1}{2}\cdot x\cdot \left(8x-4\right)\\ =\frac{1}{2}x\cdot 4\left(2x-1\right)\\ =2x\left(2x-1\right)\\ =\left(4x^2-2x\right)\text{sq}\text{. units}\end{array}$

b.

To determine

To find: The factored form of the difference of areas of rectangle and the right triangle.

Answer to Problem 58PFA

  $10x\left(2x+3\right)$

Explanation of Solution

Given:

Rectangle has length of $4x$ units and a width of

  $\left(6x+7\right)$ units.

Right triangle has legs of lengths $x$ and $\left(8x-4\right)$ units.

Calculation:

The difference of the areas of rectangle and the right triangle is

  $\begin{array}{l}24x^2+28x-\left(4x^2-2x\right)\\ =24x^2+28x-4x^2+2x\\ =20x^2+30x\end{array}$

Factored out the GCF

  $10x\left(2x+3\right)$