SOLUTION: Each side of a rectangle increased by l%. The perimeter of the rectangle is increased by a 4% b 2% c 1% d 0.25% e 0.5%

Algebra ->  Rectangles -> SOLUTION: Each side of a rectangle increased by l%. The perimeter of the rectangle is increased by a 4% b 2% c 1% d 0.25% e 0.5%      Log On


   

Question 319627: Each side of a rectangle increased by l%. The perimeter of the rectangle is
increased by
a 4% b 2% c 1% d 0.25% e 0.5%
Found 2 solutions by nerdybill, Alan3354:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Each side of a rectangle increased by l%. The perimeter of the rectangle is
increased by
.
Let x=width
and y=length
.
Original perimeter is
2(x+y)
.
Increased by 1%
1.1x = width
1.1y = lenth
perimeter is:
2(1.1x + 1.1y)
2*1.1(x+y)
2.2(x+y)
.
amount increased:
2.2(x+y) - 2(x+y)
(x+y)(2.2-2)
(x+y)(0.2)
.2(x+y)
.
percent increased is then
[.2(x+y)]/[2(x+y)] * 100
[.2]/[2] * 100
.1 * 100
10%

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Each side of a rectangle increased by l%. The perimeter of the rectangle is
increased by
.
Let x=width
and y=length
.
Original perimeter is
2(x+y)
.
Increased by 1%
1.01x = width
1.01y = lenth
perimeter is:
2(1.01x + 1.01y)
2*1.01(x+y)
2.02(x+y)
.
amount increased:
2.02(x+y) - 2(x+y)
(x+y)(2.02-2)
(x+y)(0.02)
0.02(x+y)
.
percent increased is then
[.02(x+y)]/[2(x+y)] * 100
[.02]/[2] * 100
.01 * 100
1%
----------------
I copied and pasted Nerdy Bill's solution, and changed it from 10% to the requested 1%.
---------------
Nerdy Bill used 10% increase (inadvertently) and got a 10% increase in perimeter. I used 1%, and got a 1% increase in perimeter.
The perimeter is linearly related to the side lengths. s*x --> p*x