<PRE><B>Question 1133099</B>
&lt;br&gt;
Let the radius and height of the cone be r and x.  The volume of the cone is then&lt;br&gt;
{{{(1/3)(pi)(r^2)(x)}}}&lt;br&gt;
When the cone is resting on its base, the water is in the shape of a frustum of a cone, with the surface of the water 8cm from the vertex.&lt;br&gt;
View the volume of water as the volume of the whole cone, minus the volume of the cone that is NOT filled with water.&lt;br&gt;
Those two cones are similar; their heights are in the ratio 8:x; so their volumes are in the ratio (8:x)^3 = 512:x^3.  Then the fraction of the volume of the whole cone that is filled with water is&lt;br&gt;
{{{1-512/x^3}}}&lt;br&gt;
When the cone is resting on its vertex, the water is in the shape of a cone that is similar to the whole cone, with the surface of the water 2cm from the base.&lt;br&gt;
The ratio of the heights of those two cones is (x-2):x; so their volumes are in the ratio (x-2)^3:x^3.  The fraction of the volume of the whole cone that is filled with water now is&lt;br&gt;
{{{(x-2)^3/x^3}}}&lt;br&gt;
We have two expressions that are both the fraction of the whole cone that is filled with water; so those expressions must be equal.&lt;br&gt;
{{{1-512/x^3 = (x-2)^3/x^3}}}
{{{x^3-512 = (x-2)^3}}}
{{{x^3-512 = x^3-6x^2+12x-8}}}
{{{6x^2-12x-504 = 0}}}
{{{x^2-2x-84 = 0}}}&lt;br&gt;
The quadratic does not factor; the quadratic formula gives x = 1+sqrt(85) = 10.220 to 3 decimal places.&lt;br&gt;
ANSWER: The height of the cone, to 3 decimal places, is 10.220cm.</PRE>