We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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.

michaelcai Oct 24, 2017

#1**0 **

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 Oct 24, 2017

#3**+2 **

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%

CPhill Oct 24, 2017