# 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?

Dec 5, 2016

Let the length of the side of squared painting be $x f t$.Then the length of the side of the the squared mirror will be $\left(x - 2\right) f t$

By the given condition

${x}^{2} - {\left(x - 2\right)}^{2} = 32$

$\implies 4 x - 4 = 32$

$\implies x = \frac{36}{4} = 9 f t$

So the length of the side of squared painting is $9 f t$ and the length of the side of the the squared mirror is $\left(9 - 2\right) = 7 f t$

Dec 5, 2016

The side of the painting is $9 f t$ long and side of the mirror is $7 f t$ 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 ${\left(x - 2\right)}^{2}$.

The difference between the areas is $32 f {t}^{2}$

${x}^{2} - {\left(x - 2\right)}^{2} = 32$

${x}^{2} - \left[\left(x - 2\right) \left(x - 2\right)\right] = 32$

${x}^{2} - \left({x}^{2} - 2 x - 2 x + 4\right) = 32$

${x}^{2} - \left({x}^{2} - 4 x + 4\right) = 32$

${x}^{2} - {x}^{2} + 4 x - 4 = 32$

$4 x - 4 = \textcolor{w h i t e}{{a}^{2}} 32$
$\textcolor{w h i t e}{a a} + 4 = + 4$

$4 x = 36$

$\frac{4 x}{4} = \frac{36}{4}$

$x = 9 f t$ is the length of the side of the painting.

$x - 2 = 9 - 2 = 7 f t$ is the length of the side of the mirror.