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?

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Answer 1 · 2016-12-05T00:23:41

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 $(9-2)=7ft$

Answer 2 · 2016-12-05T00:24:20

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

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.