Draw the triangle formed by the radius, the height, and the slant height 10. Which theorem connects these three lengths?

From your equation involving $r$, $h$, and 10, solve for $h$ in terms of $r$. Remember that height must be positive.

Substitute your expression for $h$ into the cone volume formula so that $V$ depends only on $r$, then match your expression to one of the answer choices.

In Desmos, enter the function for height in terms of radius: h(r) = sqrt(100 - r^2) to represent the Pythagorean relationship $r^2 + h^2 = 100$.

Enter the cone volume in terms of $r$ and $h(r)$: V(r) = (1/3)*pi*r^2*h(r) so Desmos now has the correct volume as a function of $r$.

Pick a convenient radius within $(0,10)$, for example r = 6, and evaluate V(6). Then, for each answer choice, type its formula with r = 6 in Desmos (for example, (pi*6*(10-6)), (1/3)*pi*(100-6^2)^(3/2), etc.) and see which expression gives the same numerical value as V(6).

For any right circular cone with base radius $r$ and height $h$, the volume is

Our goal is to rewrite this so it only uses the variable $r$.

The cone’s cross-section through the center is a right triangle with:

• one leg = radius $r$,
• the other leg = height $h$,
• hypotenuse = slant height $10$.

By the Pythagorean theorem:

From

solve for $h^2$:

Since height is a positive length, take the positive square root:

Substitute $h = \sqrt{100 - r^2}$ into $V = \frac{1}{3}\pi r^2 h$:

So the correct function is $V(r)=\dfrac{1}{3}\pi r^{2}\sqrt{100-r^{2}}$, which matches choice C.