What may be the ratio of the perimeter of the trapezium to its midline? The diagonals of a symmetric trapezium are perpendicular to each other. What may be the ratio of the perimeter of the trapezium to its midline?
 A: Let the perimeter $p = a+b+2c$, where $c$ are the sides and set $d = d_a + d_b$ to be the length of the diagonals with $d_a$ and $d_b$ being their parts. From the Pythagorean theorem we know that $d_a = \frac{a}{\sqrt{2}}$ and $d_b = \frac{b}{\sqrt{2}}$. Also, $c^2 = d_a^2+d_b^2$, hence $c = \frac{1}{\sqrt{2}}\sqrt{a^2+b^2}$. Finally $p = a+b+\sqrt{2}\sqrt{a^2+b^2}$ and $m = \frac{a+b}{2}$, thus, the ratio is
$$ 2\frac{a+b+\sqrt{2}\sqrt{a^2+b^2}}{a+b} = 
2+2\frac{\sqrt{\frac{a^2+b^2}{2}}}{\frac{a+b}{2}}. $$
In other words, it is $2$ plus the two times ratio of quadratic and arithmetic means of the $a$ and $b$. This also follows from the geometric construction of quadratic mean, where the resulting symmetric trapezium looks as follows:
$\hspace{120pt}$
Edit: If you ask for possible values of the ratio (thanks to @joriki for noticing), then it follows from the properties of the means that it is in the interval $[4,2+2\sqrt{2})$ where $2+2\sqrt{2}$ is the degenerate case of $b = 0$, that is, the right and isosceles triangle.
I hope this helps ;-)
