A Square Mirror has sides measuring 2ft. less than the sides of a square painting. if the difference between their areas is 32ft^2, how do you find the lengths of the sides of the mirror and the painting?

2 Answers
Dec 5, 2016

Let the length of the side of squared painting be #xft#.Then the length of the side of the the squared mirror will be #(x-2)ft#

By the given condition

#x^2-(x-2)^2=32#

#=>4x-4=32#

#=>x=36/4=9ft#

So the length of the side of squared painting is #9ft# and the length of the side of the the squared mirror is #(9-2)=7ft#

Dec 5, 2016

Answer:

The side of the painting is #9 ft# long and side of the mirror is #7 ft# long.

Explanation:

Let #x=# the side of the painting. The area is then #x^2#.

#x-2=# the side of the mirror. The area is #(x-2)^2#.

The difference between the areas is #32 ft^2#

#x^2-(x-2)^2=32#

#x^2-[(x-2)(x-2)]=32#

#x^2-(x^2-2x-2x+4)=32#

#x^2-(x^2-4x+4)=32#

#x^2 -x^2+4x-4=32#

#4x-4=color(white)(a^2)32#
#color(white)(aa)+4=+4#

#4x=36#

#(4x)/4=36/4#

#x=9 ft# is the length of the side of the painting.

#x-2=9-2=7ft# is the length of the side of the mirror.