SOLUTION: A square poster has sides measuring 2 feet less than the sides of a square sign. If the difference between their areas is 44 aquare feet, find the lenghts of the sides of the poste
Algebra ->
Customizable Word Problem Solvers
-> Geometry
-> SOLUTION: A square poster has sides measuring 2 feet less than the sides of a square sign. If the difference between their areas is 44 aquare feet, find the lenghts of the sides of the poste
Log On
Question 472396: A square poster has sides measuring 2 feet less than the sides of a square sign. If the difference between their areas is 44 aquare feet, find the lenghts of the sides of the poster and sign. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A square poster has sides measuring 2 feet less than the sides of a square sign.
If the difference between their areas is 44 square feet, find the lengths of the sides of the poster and sign.
:
Let x = the length of the side of the sign
then
(x-2) = the length of the side of the poster
:
x^2 = the area of the sign
and
(x-2)^2 = the area of the poster
:
Sign area - poster area = 44 sq/ft
x^2 - (x-2)^2 = 44
:
FOIL (x-2)(x-2)
x^2 - (x^2 - 4x + 4) = 44
:
Removing the brackets changes the signs
x^2 - x^2 + 4x - 4 = 44
4x = 44 + 4
4x = 48
x =
x = 12 ft the side of the sign
and
12-2 = 10 ft the side of the poster
:
:
Check: 12^2 - 10^2 = 44
