What is the ratio of empty to filled volume of the glass? The base diameter of a glass is $20$% smaller than the diameter at the rim. The glass is filled to half of the height. Then what is the ratio of empty to filled volume of the glass ?

I know it is an old question. But I think my way of solving the problem is different from the others and an easy to understand.
Solution: Let radius of the rim be $10$ unit. 
Now since the base diameter of a glass is $20~\%$ smaller than the diameter at the rim, so the radius. Hence the radius of the base is $10-\left(10\times \frac {20}{100}\right)=8$ unit.
Let the radius of the circular layer of the water in the glass be $x$ unit and the height of the glass is $2h_1$ unit.
Therefore as per the given condition the height of the water is $h_1$ unit and the height of the remaining part is also $h_1$ unit.
If you try to understand the problem graphically, it seems like the following

From the figure, using the formula for the similar triangles, we have
$$\dfrac{2h_1+h}{h}=\dfrac{10}{8}\implies 16h_1+8h=10h\implies h=8h_1$$
Also by the same approach,
$$\dfrac{9h_1}{8h_1}=\dfrac{x}{8}\implies x=9$$
Therefore ratio of empty to filled volume of the glass is
$$\dfrac{\pi/3\left[10^2\times 10h_1-9^2\times 9h_1\right]}{\pi/3\left[9^2\times 9h_1-8^2\times 8h_1\right]}=\dfrac{10^3-9^3}{9^3-8^3}~.$$
