Question: curved surface area of a cone = $\pi rl$, where $r$ is the radius and $l$ is the slant height.
surface area of a sphere = $4\pi r^2$, where $r$ is the radius.
A hemisphere is joined to a cone to form the shape below.
The surface area of the shape is $33\pi$ cm$^2$
Calculate the perpendicular height of the cone.
Graph/shape: top-right/center area shows a green solid made of a hemisphere joined to an inverted cone; the hemisphere radius is labeled $3$ cm with a horizontal double-headed arrow across the top circular join.
1. **State the problem:** We have a shape formed by joining a hemisphere and a cone. The total surface area of this shape is $33\pi$ cm$^2$. The radius of the hemisphere (and cone base) is $r=3$ cm. We need to find the perpendicular height $h$ of the cone.
2. **Recall formulas:**
- Curved surface area of cone = $\pi r l$, where $l$ is the slant height.
- Surface area of a sphere = $4\pi r^2$.
- Surface area of hemisphere = half the surface area of sphere = $2\pi r^2$.
3. **Surface area of the shape:**
The shape's surface area includes:
- Curved surface area of cone: $\pi r l$
- Curved surface area of hemisphere: $2\pi r^2$
Note: The base of the cone and the flat circular face of the hemisphere are joined, so that area is not exposed.
Given total surface area:
$$\pi r l + 2\pi r^2 = 33\pi$$
4. **Substitute $r=3$:**
$$\pi \times 3 \times l + 2\pi \times 3^2 = 33\pi$$
$$3\pi l + 18\pi = 33\pi$$
5. **Divide both sides by $\pi$:**
$$3l + 18 = 33$$
6. **Solve for $l$:**
$$3l = 33 - 18$$
$$3l = 15$$
$$l = \frac{15}{3} = 5$$
7. **Find perpendicular height $h$ of the cone:**
Using Pythagoras theorem in the cone:
$$l^2 = r^2 + h^2$$
$$5^2 = 3^2 + h^2$$
$$25 = 9 + h^2$$
$$h^2 = 25 - 9 = 16$$
$$h = \sqrt{16} = 4$$
**Final answer:** The perpendicular height of the cone is $4$ cm.