Chapter 5.7, Problem 5PSA

a.

To determine

To calculate:Thearea of rectangle ABCD.

Answer to Problem 5PSA

Theareais $18x^2-45x$ .

Explanation of Solution

Given information:

Side AB = 9x,

Side BC = 2x-5.

Formula used:

Area of rectangle: $A=l\times w$

Where “l”is length of rectangle and “w” is width of rectangle

Calculation:

  

Area of rectangle:

  $\begin{array}{l}A=l\times w\\ A=AB\times BC\\ A=9x\times \left(2x-5\right)\\ A=18x^2-45x\end{array}$

b.

To determine

To calculate: The perimeter of rectangle ABCD.

Answer to Problem 5PSA

The perimeter is $22x-10$ .

Explanation of Solution

Given information:

Side AB = 9x,

Side BC = 2x-5.

Formula used:

Perimeter is sum of all sides.

Calculation:

  

Perimeter is sum of all sides.

  $\begin{array}{l}P=AB+BC+CA+DA\\ P=9x+(2x-5)+9x+(2x-5)\\ P=9x+2x-5+9x+2x-5\\ P=22x-10\end{array}$

c.

To determine

To find: The area and perimeter of rectangle ABCDif x is $4.2$.

Answer to Problem 5PSA

The area is $128.52$ and perimeter is $82.4$ .

Explanation of Solution

Given information:

Side AB = 9x,

Side BC = 2x-5.

Formula used:

Area of rectangle: $A=l\times w$

Where “l”is length of rectangle and “w” is width of rectangle

Perimeter is sum of all sides.

Calculation:

  

Area of rectangle:

  $\begin{array}{l}A=l\times w\\ A=AB\times BC\\ A=9x\times \left(2x-5\right)\\ A=18x^2-45x\\ x=4.2\\ A=\left(18\times {\left(4.2\right)}^2\right)-\left(45\left(4.2\right)\right)\\ A=\left(18\times 17.64\right)-\left(45\left(4.2\right)\right)\\ A=317.52-189\\ A=128.52\end{array}$

Perimeter is sum of all sides.

  $\begin{array}{l}P=AB+BC+CA+DA\\ P=9x+(2x-5)+9x+(2x-5)\\ P=9x+2x-5+9x+2x-5\\ P=22x-10\\ x=4.2\\ P=\left(22\times 4.2\right)-10\\ P=92.4-10\\ P=82.4\end{array}$