Question 515421: If the length and width of a rectangle are doubled, how do the perimeters of the original and new rectangles compare?
Found 2 solutions by richard1234, oberobic:
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! The perimeter of the new rectangle is double. To show this algebraically, we let L and W be the length, width of the original rectangle. The perimeter is 2L + 2W. The new rectangle would have dimensions 2L and 2W, and the perimeter would be 2(2L ) + 2(2W) = 4L + 4W, or double the original perimeter.
Answer by oberobic(2304)  (Show Source):
You can put this solution on YOUR website! P = perimeter = 2(L+W) = 2L + 2W
A = area = L*W
.
If you double L and W, you have 2L and 2W
so
P = 2(2L + 2W) = 4L + 4W
which means the perimeter has doubled from 2L + 2W to 4L + 4W.
.
A = 2L*2W
A = 4L*W
.
However, the area has quadrupled.
Why?
2*2 = 4
.
As a check, consider a rectangle with sides = 4 ft.
(Yes, it's a square.)
.
P = 2(4+4) = 2(8) = 16 ft
A = 4*4 = 16 sq ft
.
Now double the sides = 8 ft
.
P = 2(8+8) = 2*16 = 32
A = 8*8 = 64
.
The perimeter doubled from 16 to 32 ft.
The area quadrupled from 16 to 64.
.
Done.
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