The perimeter of a rectangle is 40 cm and the area of the same rectangle is 20 cm^2. find the dimensions of the rectangle to the nearest hundredth.

Question

I need help with this problem you have to solve each application of the problem and show work But I have no Idea how to do this problem please help me!!!!

Answer 1

Let w = the width

Let l = the length

P = 2l + 2w

40 = 2l + 2w

40 - 2l = 2w

20 - l = w

A = lw

20 = lw

20 = l(20 - l)

20 = 20l - l²

l² - 20l + 20 = 0

l = 20 ± √[(20² - 4(1)(20)]

            2(1)

l = 20 ± √320

            2

l = 20 ± √64√5

           2

 l = 20 ± 8√5

          2

l = 20 ± 17.889

           2

l = 18.94 cm      or    l = 1.06 cm

w = 20 - l                 w = 20 - l

w = 1.06 cm             w = 18.94 cm