Posted by Gabby on Wednesday, February 13, 2013 at 7:24pm.
Jim has designed a rectangle with an area of 100 square feet and a perimeter of 401 feet.
What are the dimensions of Jim's rectangle?please show me how u got the answer.
- Math ms sue please help again - Ms. Sue, Wednesday, February 13, 2013 at 7:49pm
A = LW
Let's see which factors of 100 would add up to 401.
10 * 10
2 * 50
4 * 25
5 * 20
P = 2L + 2W
None of those come close to a perimeter of 401.
Are you sure you copied the problem correctly?
- Math ms sue please help again - Gabby, Wednesday, February 13, 2013 at 7:53pm
Yes and there is only one way
- Math ms sue please help again - Reiny, Wednesday, February 13, 2013 at 8:53pm
lw = 100
and 2l + 2w = 401
from the 1st .... l = 100/w
into the 2nd:
2(100/w) + 2w = 401
200 + 2w^2 = 401w
2w^2 - 401w + 200 = 0
by the formula:
w = (401 ± √159201)/4
= (401 ± 399)/4
= 200 or .5
or , it factors to
(w-200)(2w - 1) = 0
w = 200 or w = 1/2
if w=200, the l = 100/200 = 1/2
if w = 1/2, the l = 100/(1/2) = 200
The rectangle is 200 feet by 1/2 ft
area = (1/2)(200) = 100 , checks!
perimeter = 200 + 200 + 1/2 + 1/2 = 401, checks!
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