SOLUTION: The length of a rectangle is twice its width. If the perimeter of the rectangle is 30IN , find its area.
so far i know, l=2w 2w(w)=30 2w^2=30 w^2=15 i know i need to get rid o
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-> SOLUTION: The length of a rectangle is twice its width. If the perimeter of the rectangle is 30IN , find its area.
so far i know, l=2w 2w(w)=30 2w^2=30 w^2=15 i know i need to get rid o
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Question 253526: The length of a rectangle is twice its width. If the perimeter of the rectangle is 30IN , find its area.
so far i know, l=2w 2w(w)=30 2w^2=30 w^2=15 i know i need to get rid of the square root
You can put this solution on YOUR website! You are close but the perimeter of a rectangle is all the sides added together so you would have length + length + width + width = 30
you can use L for length and w for width
So the two equations you will have are:
L = 2W
L+L+W+W=30 or 2L + 2W = 30
use substitution:
2(2W) + 2W = 30
4W + 2W = 30
6W = 30 divide both sides by 6
W = 5
Sub back into L=2(w) so L = 2(5)
L = 10
Hope this helps
