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September 16, 2014

September 16, 2014

Posted by **Gabby** on Wednesday, February 13, 2013 at 7:24pm.

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 7:49pmA = 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 7:53pmYes and there is only one way

- Math ms sue please help again -
**Reiny**, Wednesday, February 13, 2013 at 8:53pmlw = 100

and 2l + 2w = 401

from the 1st .... l = 100/w

into the 2nd:

2(100/w) + 2w = 401

times w

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

check:

area = (1/2)(200) = 100 , checks!

perimeter = 200 + 200 + 1/2 + 1/2 = 401, checks!

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