如图所示，一个直圆锥是从一个实心钢球中切割出来的。 点 O 是半径为 12 cm 的球体中心。

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.