The square with vertices (-a, -a), (a, -a), (-a, a), (a, a) is cut by the line y = x/3 into congruent quadrilaterals. The perimeter of one of these congruent quadrilaterals divided by a equals what? Express your answer in simplified radical form.
+2666 Note that the line will leave the bottom side intact, and split the two vertical sides into parts that sum to 2a.
This means that the perimeter of the shape is 4a + the length of the line.
The length of the line is the hypotenuse of a right triangle with legs of 2a and \({4 \over 3}a\)
Using the Pythagorean theorem, we find that the length of the hypotenuse is \(\sqrt{2^2 + {2 \over 3}^2} = \sqrt{40 \over 9} = {2 \sqrt {10} \over 3}a\).
This means the perimeter of the shape is \(4a + {2\sqrt{10} \over 3}a\), which can be factored into \(a(4 + {2\sqrt{10} \over 3})\).
Can you take it from here?
