The diagram below shows a gift souvenir formed by joining the plane face of a solid hemisphere of radius 3 cm and a solid frustum, which is obtained by cutting a right circular cone of radius 12 cm into two portions by a plane parallel to its base. The height of the frustum is given by $x$ cm.

1. Find the height of the small cone cut off using similar triangles

Let the height of the small cone cut off be $h$ cm.

Since the cut is parallel to the base, the small cone is similar to the original cone.

$\frac{h}{24}=\frac{3}{12}$

$h=24\times \frac{3}{12}=6$

2. Subtract the small cone height from the original cone height to find the frustum height

The height of the frustum is

$x=24-6=18$

1. Find the volume of the frustum

For a frustum with height $h=18$, base radius $R=12$ and top radius $r=3$,
$V=\frac{1}{3}\pi h(R^2+Rr+r^2).$
$V_{\text{frustum}}=\frac{1}{3}\pi(18)(12^2+12\times3+3^2)$
$=6\pi(144+36+9)$
$=1134\pi\text{ cm}^3.$

2. Find the volume of the solid hemisphere

Volume of a hemisphere:
$V=\frac{2}{3}\pi r^3.$
With $r=3$,
$V_{\text{hemisphere}}=\frac{2}{3}\pi(3^3)$
$=18\pi\text{ cm}^3.$

3. Add the volumes to get the volume of the souvenir

$V_{\text{souvenir}}=1134\pi+18\pi$
$=1152\pi\text{ cm}^3$
$=3619.\ldots\text{ cm}^3.$

1. Find the slant height of the frustum

The frustum has height $18\text{ cm}$, lower radius $12\text{ cm}$ and upper radius $3\text{ cm}$.

Difference in radii $=12-3=9\text{ cm}$.

Using Pythagoras' theorem,
$s=\sqrt{18^2+9^2}=\sqrt{405}=9\sqrt5\text{ cm}.$

2. Find the curved surface area of the frustum

Curved surface area of a frustum:
$\pi(R+r)s$
where $R=12$, $r=3$ and $s=9\sqrt5$.

$\text{CSA of frustum}=\pi(12+3)(9\sqrt5)=135\sqrt5\pi\text{ cm}^2.$

3. Find the exposed surface area of the hemisphere

Only the curved surface of the hemisphere is exposed.

Curved surface area of hemisphere:
$2\pi r^2=2\pi(3)^2=18\pi\text{ cm}^2.$

4. Find the exposed circular base area of the frustum

The lower circular base of the frustum is exposed.

$\text{Base area}=\pi(12)^2=144\pi\text{ cm}^2.$

5. Add all exposed surface areas and round the answer

$\text{Total surface area}=135\sqrt5\pi+18\pi+144\pi$
$=(135\sqrt5+162)\pi$
$=1457.4\ldots\text{ cm}^2$

Correct to 3 significant figures,
$\text{Total surface area}=1460\text{ cm}^2.$