Chapter 12.2, Problem 36STP

a.

To determine

To find: Translate the sentences into equations.

Answer to Problem 36STP

The sentences into equations are $3w=2l+3$ and $4l=P+12$ .

Explanation of Solution

Given information:

Three times the width of a certain rectangle exceeds twice its length by three inches, and four times its length is twelve more than its merimeters

Calculation:

Consider the first sentence:

Let the width and length of the rectangle be $w$ and $l$ respectively.

Then first sentence into an equation will be

  $3w=2l+3$

Consider the second sentence:

Let the perimeter of the rectangle be $P$ .

Then second sentence into an equation will be

  $4l=P+12$

Hence,

The sentences into equations are $3w=2l+3$ and $4l=P+12$ .

b.

To determine

To find: The dimensions of the rectangle.

Answer to Problem 36STP

The dimensions are $\text{length}=21\text{ inches}$ and $\text{width}=15\text{ inches}$ .

Explanation of Solution

Given information:

Three times the width of a certain rectangle exceeds twice its length by three inches, and four times its length is twelve more than its perimeters.

Calculation:

From (a):

  $\begin{array}{l}3w=2l+3\mathrm{......}\text{ (1)}\\ 4l=P+12\mathrm{......}\text{ (2)}\end{array}$

Also, it is known that $P=2l+2w$ , therefore equation (2) will become:

  $\begin{array}{l}4l=2l+2w+12\\ 2l-2w=12\\ l-w=6\\ l=w+6\mathrm{......}\text{ (3)}\end{array}$

Substitute value of $l$ from equation (3) in equation (1) and solve for $w$ :

  $\begin{array}{l}3w=2\left(w+6\right)+3\\ 3w=2w+12+3\\ 3w-2w=15\\ w=15\end{array}$

Substitute value of $w$ in equation (3) and solve for $l$ :

  $\begin{array}{l}l=15+6\\ l=21\end{array}$

Hence,

The dimensions are $\text{length}=21\text{ inches}$ and $\text{width}=15\text{ inches}$ .

c.

To determine

To find: the area of the rectangle.

Answer to Problem 36STP

The area of the rectangle is 315 ${\left(\text{inches}\right)}^2$ .

Explanation of Solution

Given information:

Three times the width of a certain rectangle exceeds twice its length by three inches, and four times its length is twelve more than its perimeters.

Calculation:

  $\text{length}=21\text{ inches}$ and $\text{width}=15\text{ inches}$ .

Area of the rectangle $A$ is:

  $\begin{array}{l}A=lw\\ A=\left(\text{21 inches}\right)\left(\text{15 inches}\right)\\ A=\text{315 }{\left(\text{inches}\right)}^2\end{array}$

Hence,

The area of the rectangle is 315 ${\left(\text{inches}\right)}^2$ .