The diagram shows a right circular cone cut from a solid steel sphere. Point O is the centre of the sphere with radius 12 cm.

1. Find the radius of the base of the cone from its circumference.

Let the radius of the base of the cone be $r$ cm.

$2\pi r = 50$

$r = \frac{50}{2\pi} = \frac{25}{\pi}$

2. Use Pythagoras' theorem to find the distance from the sphere centre to the base plane.

In the right triangle formed by the centre $O$, a point on the rim of the base, and the axis of the cone, the hypotenuse is the radius of the sphere, $12$ cm.

Let the distance from $O$ to the base plane be $x$ cm.

$x^2 + r^2 = 12^2$

$x^2 + \left(\frac{25}{\pi}\right)^2 = 144$

$x = \sqrt{144 - \left(\frac{25}{\pi}\right)^2}$

$x = 8.982\ldots$

3. Find the height of the cone and round to 3 significant figures.

The apex is $12$ cm above $O$, and the base plane is $x$ cm below $O$.

$\text{height} = 12 + x$

$= 12 + 8.982\ldots$

$= 20.982\ldots$

$= 21.0\text{ cm} \quad \text{to 3 significant figures}$

1. Find the radius of the base of the cone from its circumference.

Given circumference $=50\text{ cm}$,
$2\pi r=50$
$r=\frac{50}{2\pi}=\frac{25}{\pi}$

2. Find the curved surface area of the cone.

From part (a), the height of the cone is $21.0\text{ cm}$.

Slant height:
$l=\sqrt{21.0^2+\left(\frac{25}{\pi}\right)^2}$

Curved surface area of cone:
$\pi rl=\pi\left(\frac{25}{\pi}\right)\sqrt{21.0^2+\left(\frac{25}{\pi}\right)^2}$
$=528.220\ldots$
$\approx 528\text{ cm}^2$

1. Find the volume of steel remaining after the original cone is cut from the sphere.

From earlier parts, the cone cut from the sphere has base circumference $50\text{ cm}$ and height $21.0\text{ cm}$.

Radius of its base:
$r=\frac{50}{2\pi}=\frac{25}{\pi}$

Volume of sphere:
$\frac{4}{3}\pi(12)^3=2304\pi$

Volume of cone cut out:
$\frac{1}{3}\pi\left(\frac{25}{\pi}\right)^2(21.0)=\frac{4375}{\pi}$

Remaining steel volume:
$2304\pi-\frac{4375}{\pi}=5845.6\ldots\text{ cm}^3$

2. Use similarity to find the volume ratio of the top, centre and bottom layers of the new cone.

The cone has three equal heights: $h$, $h$, $h$.

Since similar cones have volumes in the ratio of the cubes of their corresponding lengths,
$V_{\text{top cone}}:V_{\text{cone up to second layer}}:V_{\text{whole cone}}=1^3:2^3:3^3=1:8:27$

So the three layer volumes are:
$\text{top}:\text{centre}:\text{bottom}=1:(8-1):(27-8)=1:7:19$

The top and bottom layers are made from steel, so
$\text{steel}:\text{acrylic}=(1+19):7=20:7$

3. Calculate the volume of acrylic used for the centre layer.

Since the remaining steel forms the top and bottom layers, the acrylic volume is
$\frac{7}{20}\times 5845.6\ldots=2045.9\ldots$

Therefore, the volume of acrylic used is
$2050\text{ cm}^3$
to 3 significant figures.