几何体 A 是由一个半径为 $r$ 的半球连接到高度为 $r$ 的圆柱的一端，圆柱的另一端连接一个高度为 $h$ cm 的圆锥而组成的几何体。

1. Find and add the volumes of the hemisphere and cylinder.

Volume of hemisphere:

$\frac{1}{2}\left(\frac{4}{3}\pi r^3\right)=\frac{2}{3}\pi r^3$

Volume of cylinder:

$\pi r^2(r)=\pi r^3$

Total volume:

$\frac{2}{3}\pi r^3+\pi r^3=\frac{5}{3}\pi r^3$

1. Write expressions for the cone volume and the total volume of Solid A.

From part (a), the volume of the hemisphere and cylinder is
$\frac{5}{3}\pi r^3.$
The volume of the cone is
$\frac{1}{3}\pi r^2h.$
Hence the volume of Solid A is
$\frac{5}{3}\pi r^3+\frac{1}{3}\pi r^2h.$

2. Form and solve the equation using the given volume relationship.

Since the volume of the cone is half of the volume of the entire Solid A,
$\frac{1}{3}\pi r^2h=\frac{1}{2}\left(\frac{5}{3}\pi r^3+\frac{1}{3}\pi r^2h\right).$
Multiply by 2:
$\frac{2}{3}\pi r^2h=\frac{5}{3}\pi r^3+\frac{1}{3}\pi r^2h.$
Subtract $\frac{1}{3}\pi r^2h$ from both sides:
$\frac{1}{3}\pi r^2h=\frac{5}{3}\pi r^3.$
Divide by $\frac{1}{3}\pi r^2$:
$h=5r.$

1. Use the given hemisphere volume to find $r^3$.

Volume of hemisphere $= \dfrac{2}{3}\pi r^3$

$\dfrac{2}{3}\pi r^3=54\pi$

$r^3=54\times \dfrac{3}{2}=81$

2. Write the total volume of Solid A using the result $h=5r$ from part (b).

From part (a), volume of hemisphere and cylinder

$=\dfrac{5}{3}\pi r^3$

From part (b), $h=5r$.

Volume of cone

$=\dfrac{1}{3}\pi r^2h$

$=\dfrac{1}{3}\pi r^2(5r)$

$=\dfrac{5}{3}\pi r^3$

Total volume of Solid A

$=\dfrac{5}{3}\pi r^3+\dfrac{5}{3}\pi r^3$

$=\dfrac{10}{3}\pi r^3$

3. Substitute $r^3=81$ to find the volume of Solid A.

$\text{Volume of Solid A}=\dfrac{10}{3}\pi(81)$

$=270\pi$

1. Find the area of the trapezium cross-section in terms of $y$.

Area of trapezium
$= \frac{1}{2}(6+10)y$
$=8y\text{ cm}^2$

2. Write the volume of Solid B in terms of $y$.

Volume of prism
$= \text{cross-sectional area} \times \text{length}$
$=8y \times 20$
$=160y\text{ cm}^3$

3. Equate the volume of Solid B to the volume of Solid A and solve for $y$.

Since Solid A is melted down to form Solid B, their volumes are equal.

From part (c), volume of Solid A $=270\pi\text{ cm}^3$.

$160y=270\pi$
$y=\frac{270\pi}{160}$
$y=\frac{27\pi}{16}$
$y=5.30\text{ cm}\quad \text{(3 s.f.)}$