Chapter 6.6, Problem 40STP

To determine

To find: the height and width of the rectangle

Answer to Problem 40STP

The length of rectangle is 17.5 and width of rectangle is 8.5

Explanation of Solution

Given:

  $x-y=9$

Perimeter 52

Calculation:

Objective: To find the height and width of the rectangle.

Let $x$ be the length of the rectangle and $y$ be the width of the rectangle.

Then, $x-y=9$ First equation

Perimeter of rectangle: $2x+2y=52$ Second equation

Multiply first equation by 2,

  $2(x-y)=2(9)$

  $2x-2y=18$ Third equation

Add second equation and third equation,

  $2x-2y+(2x+2y)=18+52$

Simplify $2x-2y+2x+2y=70$

  $4x=70$

Multiply both sides by $\frac{1}{4},$

  $\frac{1}{4}\times 4x=\frac{1}{4}\times 70$

Therefore, $x=17.5$

Substitute $x=17.5$ in first equation,

  $17.5-y=9$

Subtract 17.5 from both sides,

  $17.5-y-17.5=9-17.5$

  $y=8.5$

Hence length of rectangle is 17.5 and width of rectangle is 8.5

Conclusion:

Hence length of rectangle is 17.5 and width of rectangle is 8.5