Question

The length of the rectangle is $ 5 $ more than the width. The perimeter is $ 22 $ feet. How do you find the length and the width?

Answer 1

Hint: In this question, we are given a rectangle with a certain perimeter and we are given a relation between the length of the rectangle and the width of the rectangle. Using the formula of perimeter, we can find the respective length and width of the rectangle.
Formula used:
Perimeter of the rectangle = $ 2 $ (length of the rectangle + width of the rectangle)

Complete step-by-step answer:
Firstly, seeing what is given in the question and what we have to calculate.
Let length of the rectangle be denoted by $ l $ and width by $ w $ . Now according to the question, the length is $ 5 $ more than the width. So, $ l = 5w $ and perimeter of the rectangle is given to be $ 22feet $ .
Formula of perimeter of the rectangle is = $ 2l + 2w $
Substituting the above values and form an equation,
 $
  22 = 2l + 2w \\
   \Rightarrow 22 = 2\left( {5w} \right) + 2w \\
  $
 $ \left( {\therefore l = 5w} \right) $
Solve for $ w, $
 $
  10w + 2w = 22 \\
  12w = 22 \\
   \Rightarrow w = \dfrac{{22}}{{12}} = \dfrac{{11}}{6}\;feet \;
  $
Now calculating the $ l $
 $ l = 5w = 5\left( {\dfrac{{11}}{6}} \right) = \dfrac{{55}}{6}feet $
Now, converting the feet into inches and we know $ 1feet = 12\;inches $
The width of the rectangle in inches is
  $
  w = \dfrac{{11}}{6}feet \\
   = \dfrac{{11}}{6} \times 12 = 22\;inches \;
  $
Now, calculating the same for length of the rectangle
 $
  l = \dfrac{{55}}{6}feet \\
   = \dfrac{{55}}{6} \times 12 = 110\;inches \;
  $
Hence, the length of the rectangle is $ 110\;inches $ and the width of the rectangle is $ 22\;inches $ .This is our required solution.

Note: We are not asked to convert the feet into inches. But we should know the formula for how to convert the feet into inches. The standard units of length based on the metric system is meter (m).
 $ 1meter = 3.28feet $ (approximate value)