SOLUTION: Find the perimeter and area of a rectangle if its length is 8x-2, and its width is 4x.

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Question 11404: Find the perimeter and area of a rectangle if its length is 8x-2, and its width is 4x.
Found 2 solutions by tishy, Earlsdon:
Answer by tishy(4) About Me  (Show Source):
You can put this solution on YOUR website!
Find the perimeter and area of a rectangle if its length is 8x-2, and its width is 4x.

8x-2= -16(length)
therefore perimeter is when you add all the sides together so therefore
perimeter = -16+-16+4+4
= -32 + 8
= -24
area is L (length) times W (width)
area = LxW
-16 x 4
= -64
answer :-)

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
The perimeter of a rectangle (or of any polygon) is the sum of the lengths of the sides.
P+=+2%28L%2BW%29
P+=+2%28%288x-2%29+%2B+4x%29
P+=+2%2812x+-+2%29
P+=+24x+-+4 This is the perimeter.
The area of a rectangle is length (L) times width (W):
A+=+L%2AW
A+=+%288x-2%29%284x%29
A+=+32x%5E2+-+8x This is the area.