Question 519099: The length of a rectangle is 3 yds longer than its width.
If the perimeter of the rectangle is 42 yds , find its area.
Answer by mamiya(56)   (Show Source):
You can put this solution on YOUR website! let's p be the perimeter, a the area, l the length and w the width .
since we are talking about a rectangle, p=2(l+w) and a=lw
l is 3 yds longer than w, means l=w+3
p=2(l+w)
=2(w+3+w)
=4w +6
from this we have w= (p-6)/4
At this level, there are two possibilities.
the first one, find the value of l and w and then a
w=(p-6)/4= (42-6)/4 = 9 yds
l= w+3= 9+3 =12 yds
then a= lw= 12*9= 108 yds^2
the second one, express w as a function of p and and we change the l into w+3 and we put both of them in the formula of the area
a=lw
=(w+3)w we know p=4w+6 , so w= (p-6)/4
= [(p-6)/4 +3] (p-6)/4
= [ (42-6)/4 +3] (42-6)/4
= 108 yds^2
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