SOLUTION: The length of a rectangle is four times its width. If the perimeter of the rectangle is 50in, find its area.

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Question 621585: The length of a rectangle is four times its width.
If the perimeter of the rectangle is 50in, find its area.
Found 2 solutions by dragonwalker, Maths68:
Answer by dragonwalker(73) About Me  (Show Source):
You can put this solution on YOUR website!
So let us call the shorter side, i.e the width: 'x'
So each length is four times the width = 4x
The perimeter consists of two widths and two lengths and equals 50 inches.
So:
x+x+4x+4x=50
solve for x:
10x = 50
to find x divide both sides by 10:
10x/10 = 50/10
x=5
So width equals 5 in and length is four times this i.e. 4x5=20
To check: 5+5+20+20=50 !!!!

Answer by Maths68(1474) About Me  (Show Source):
You can put this solution on YOUR website!

Length = l
Width = w
Perimeter = p = 50
Given
------
The length of a rectangle is four times its width
l=4w



Perimeter of rectangle = 2(l+w)
50=2(4w+w)
50=2(5w)
50=10w
50/10=w
5=w
w=5



Area of rectangle = length * width
a=l*w
a=4w*w
a=4w^2............(1)
Put the value of w in (1)
a=4(5)^2
a=4(25)
a=100in^2


Area of rectangle is 100 square inches.