Quandaries and Queries

secondary (12th grade)


I am having difficulty figuring out this problem.  Thank you in advance, any help is much appreciated.

Water is poured into a tank in the shape of an inverted right circular cone.  The height of the tank is 8 m and its radius at the top is 4 m.

a. Draw and label a picture to represent this situation.  (I know how to do this)

b. Identify all variable quantities. (h = 8m, r = 4m)

c. Find an equation that relates the variable quantities, and reduce the number of variable quantities to two.

I was thinking about the equation V = 1/3 pi r2 h, which is the Volume of a cone, but I am stumped as to how I am supposed to "reduce the number of variable quantities to two." Can you point me in the right direction?

Thank you,


Hi Adrienne,

Here is my diagram.

The important variables are r and h, the height and radius of the water in the cone. Then, as you said, the volume of the water in the cone is V = 1/3 pi r2 h cubic meters.

To reduce the number of variables look back at the diagram. I see a triangle, which I have labeled ABC, with |AB| = 8 meters and |CA| = 4 meters.

Inside is the triangle DBE with |DB| = h meters and |ED| = r meters.

The triangles ABC and DBE are similar so

 |AB|/|CA||DB|/|ED| That is  8/4h/r so h = 2 r.

If you substitute this into the equation for the volume of the water in the cone you will have an expression with the only variables being V and r.


Go to Math Central