Question 267174
Let x = length of the sides of the original square.
Then x-3 = the length of the side of the square after the sides have been decreased by 3<br>
The area of any square is {{{s^2}}} where s is the length of the sides. For the original square, since the length of the sides of the original square is x, the area would be {{{x^2}}}. The area of the reduced square would be {{{(x-3)^2}}}.<br>
We are told that the area of the reduced square is 45 square meters less than the area of the original square. In other words, the area of the reduced square, {{{(x-3)^2}}}, plus 45 equals the area of the original square. So
{{{(x-3)^2 + 45 = x^2}}}
We now have an equation we can solve. We start by simplifying. (Remember to use FOIL or the pattern for (a-b)^2 on the left side. {{{(x-3)^2}}} <b>is not</b> {{{x^2 - 9}}}!)
{{{x^2 -6x + 9 + 45 = x^2}}}
{{{x^2 -6x + 54 = x^2}}}
Subtract {{{x^2}}} form each side:
{{{-6x + 54 = 0}}}
Subtract 54 from each side:
{{{-6x = -54}}}
Divide both sides by -6:
{{{x = (-54)/(-6) = 9}}}<br>
Since x is the length of the side of the original square and since that is what the problem asks us to find, then the answer to the problem is 9.<br>
If we want to check our answer we can find the side of the reduced square which is x-3 or 6. Then we can find the areas of the two squares:
{{{x^2 = (9)^2 = 81}}} square meters
{{{(x-3)^2 = (6)^2 = 36}}} square meters
The difference in these two areas is
{{{81 - 36 = 45}}} square meters. Check!