The square with vertices $(-a, -a), (a, -a), (-a, a), (a, a)$ is cut by the line $y = x/2$ into congruent quadrilaterals. The perimeter of one of these congruent quadrilaterals divided by $a$ equals what? Express your answer in simplified radical form.
y = x/2
This line will intersect the square at x values -a and a
So....the associated y values are
y = (-a)/2 = -a/2 and (a)/2
So....the perimeter of one of the quadrilaterals =
2a + [(a) - (-a/2)] + [(a) - a/2 ] + sqrt [ ( a - (-a)) ^2 + ( a/2 - (-a/2)^2] =
2a + 3a/2 + a/2 + sqrt [ 4a^2 + a^2 ] =
4a + a√ 5 dividing this by a produces =
4 + √ 5