A cone is inverted and filled with water to 3/4 of its height. What percent of the cone's volume is filled with water? Express your answer as a decimal to the nearest ten-thousandth.
Let's call the height of the cone " h " and the area of its base " a ".
volume of cone = (1/3)(a)(h)
The water will also be in the shape of a cone, with a height that is 3/4 that of the container.
volume of water = (1/3)(a)(3/4 * h)
volume of water = (3/4) * (1/3)(a)(h)
volume of water = (3/4) * volume of cone
volume of water = 75% of the volume of cone
75% of the cone's volume is filled with water.
Hectictar's answer is correct if the tank is a cylinder....however....a cone is a little different because of the variable cross-sections
We can use similar triangles here......call the radius of the tank r and the height h
So.....if the water is filled to 3/4 of the inverted cone's height.....we have
r / h = x / [3/4] h where x = the radius of the cone formed at the 3/4 water level
So.....x = 3/4 r
So.....the volume of the water when the tank is full is pi * r^2 * h / 3
And the volume of water at the 3/4 level is [pi * [ (3/4) r ] ^2 * [ (3/4)] h ] / 3 =
[ 27/64] pi* r^2 * h / 3
So....the percent of the cone filled with water is 27/64 ≈ .4219 ≈ 42.2%