SOLUTION: The length of a rectangle is twice it's width. If the perimeter of the rectangle is 48 cm, find it's area. I came up with 24 cm. could some review and let me know if I did this

Algebra ->  Linear-equations -> SOLUTION: The length of a rectangle is twice it's width. If the perimeter of the rectangle is 48 cm, find it's area. I came up with 24 cm. could some review and let me know if I did this       Log On


   



Question 156578: The length of a rectangle is twice it's width. If the perimeter of the rectangle is 48 cm, find it's area. I came up with 24 cm. could some review and let me know if I did this right..thanks upfront..
Found 3 solutions by gonzo, checkley77, Earlsdon:
Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
length = 2 * w.
if w = width then l = length = 2w.
p = perimeter = 48 cm.
perimeter = 2*l + 2*w.
since l = 2w, then perimeter = 2*(2w) + 2*w.
4w + 2w = 6w = 48, so w = 48/6 = 8 cm.
if w = 8 cm, then l = 2w = 16 cm.
area of a rectangle is length * width = 8 * 16 = 128 cm.



Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
L=2W
2*2W+2*W=48
4W+2W=48
6W=48
W=48/6
W=8 ANSWER FOR THE WIDTH.
L=2*8=16 ANSWER FOR THE LENGTH.
L*W=AREA
8*16=128 AREA.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let's start by finding the length, L, and the width, W.
We are told that:
L = 2W "The length is twice its (no apstrophe here, please) width" and we are also told that:
P = 48 "The perimeter is 48cm.
The perimeter is found by:
P = 2(L+W) Replace the L here with the L from the first equation (L = 2W) and the P with 48cm.
48 = 2(2W+W) Simplify.
48 = 6W Divide both sides by 6.
W = 8cm and L = 2(8) = 16cm.
Now you can find the area because:
A = L*W
A = 16*8
A = 128 sq.cm