Find the dimensions of a rectangle whose perimeter is 22meters and whose area is 24 square meters.

Question

Find the dimensions of a rectangle whose perimeter is 2222 meters and whose area is 2424 square meters.

Answer 1

Perimeter = 2 x Length + 2 x width.

So P = 2L + 2w

 

 

Area = length x width.

So A = L * w

 

22 = 2L + 2w = 2(L + w) 

11 = L + w <--- divides both sides by 2

L = 11 - w  <--- solves for L

 

24 = L * w = (11 - w)* w <--- substitutes L = 11 - w 

 

24 = 11w - w^2

 

w^2 - 11w + 24 = 0  <--- everybody to the left side so leading coefficient stays positive

 

(  w   -  8 )( w    -    3   ) = 0

 

w - 8 = 0 ---> w= 8  ---> L = 3   

 

w - 3 = 0  ----> w = 3 ---> L = 8

 

The rectangle is 8 x 3.

The length is 8.

The width is 3.