Answer:
6 in
Step-by-step explanation:
Let x = the side length of the original square.
They removed 3 in from each side of the original square, so the side lengths of the remaining square are x - 3 in.
The area of the smaller square is (x - 3)².
The area of the original square is x²
I assume the area of the smaller square is ¼ that of the original square. Then
1. Solve for x
[tex]\begin{array}{rcl}\frac{1}{4}x^{2} & = & (x - 3)^{2}\\x^{2} & = & 4(x - 3)^{2}\\& = & 4(x^{2} - 6x + 9)\\x^{2}& = & 4x^{2} - 24x + 36\\3x^{2} - 24x + 36 & = & 0\\x^{2} - 8x + 12 & = & 0\\(x - 2)(x - 6) & = & 0\\x - 2 = 0& \qquad &x - 6 = 0\\x = 2& \qquad &x = 6\\\end{array}[/tex]
2. Calculate the side length of the smaller square
(a) x = 2
Side length = x - 3 = 2 - 3 = -1 in.
IMPOSSIBLE. You can't have a negative side length.
(b) x = 6
Side length of smaller square = 6 - 3 = 3 in.
Side length of original square = x = 6 in
Check:
[tex]\begin{array}{rcl}\frac{1}{4}(6)^{2} & = & (6 - 3)^{2}\\\frac{1}{4}\times 36 & = & 3^{2}\\9 & = & 9\\\end{array}[/tex]
OK.
