Question

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?

Answer 1

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

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.