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max2010maxim [7]
3 years ago
9

The perimeter of a rectangle is 42 inches, and its area is 108 square inches. find the length and width of the rectangle.

Mathematics
1 answer:
Anit [1.1K]3 years ago
8 0
p = 2w + 2l              p = 42 
a = l * w                   a = 108

You want to make either the length or width variable (I chose the width variable to be by itself. It doesn't matter) by itself by:

Divide both sides by 2.
(p) ÷ 2 = (2w + 2l) ÷ 2 
               2's  cancel  out
\frac{p}{2} = w + l

Subtract both sides by either length or width depending on what variable you chose to make by itself (in this case I subtracted the length variable because I wanted the width variable by itself).
\frac{p}{2} - l = w + l - l
                 l's  cancel  out 
p/2 - l = w

a = l * w

You replace w with the equation above p/2 - l = w 
a = l( \frac{p}{2} - l)

Multiply out the L.
a = \frac{p}{2}l - l^{2}

Plug in all your numbers a = 108 and p = 42
108 = \frac{42}{2}l - l^{2}
108 = 21l - l^{2

Move everything to the left side so l^{2} would be positive (makes the equation easier when l^{2} is positive).
l^{2} - 21l + 108 = 0

Factor.
(l -9)(l-12) = 0

Make each parenthesis set equal to 0.
l-9=0     l-12=0

Add.
l = 9  l = 12

By doing this you solve for both the width and length so the answer is:
w = 12    L = 9
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