A conical tank with a radius of 4 ft and a height of 12 ft is filled with liquid. How much liquid has poured from the tip of the cone if the water level is 9 ft from the tip?

Volume of a cone:
#1/3pir^2h#

We will use #12# ft as the height, as the water is only filled up to the height of #9ft#

#1/3pi(4ft)^2(12ft)= 64 pift^3 water#

This is the amount of water in the cone initially

Now when the water goes down to 9ft of height, the volume of water in the cone is:

#r/h=4/12=x/9#
#x=3 ft#

#1/3pi(3ft)^2(9ft)= 27pift^3#

This is the volume of water in the cone after water has been poured out.

So the amount of water poured out:
#64pift^3-27pift^3= 37pift^3approx 116.24 ft^3 water#

The original cone has height 12 feet and radius 4 feet. If 9 feet of height is lost by being poured out we need to find the volume of the smaller cone and subtract it from the larger one.
Using similar shapes to find the radius of the smaller cone.

#(radius)/(height)# will be the same for both cones

#4/12=1/3# so #=> r/9=1/3# , the smaller radius is 3 feet

Volume of a cone is #1/3pir^2h#

Volume of liquid remaining is

#(1/3xxpixx4^2xx12)-(1/3xxpixx3^2xx9)#

#=64pi - 27pi#

#=37pi#

#~~116.23892818#

#~~116.24 feet^3#

The large cone has height 12 feet and radius 4 feet, the smaller cone has height 9 feet and radius 3 feet ( by using similar shapes)

#[radius]/(height)# #=># #4/12=r/9 => r=3#

Volume of a frustum is #1/3pi(R^2+Rr+r^2)h#

Where #R# is the radius of the base, #r# is the radius of the top and #h# is the height of the frustum

#=> 1/3xxpixx(4^2+4xx3+3^2)xx3#

#=>pi/3(16+12+9)xx3#

#=>37pi#

#~~116.23892818#

#~~116.24 feet^3#