Cones and Spheres

All the formulae for cones and spheres will be given in this assessment. Your task is to select the correct formula, identify the measurements and substitute accurately.

Reading the measurements of a cone

A cone has radius $r$, perpendicular height $h$ and slant height $l$. These are not interchangeable:

• $r$ runs from the centre to the edge of the circular base;
• $h$ runs at right angles from the base to the vertex and is used for volume;
• $l$ runs up the sloping surface and is used for curved surface area.

If the diameter is given, first use $r=\frac{d}{2}$. If $r$ and $h$ are known but $l$ is needed, a cross-section through the centre forms a right-angled triangle, so $l^2=r^2+h^2$.

Volume of a cone

The volume formula is

$V=\frac13\pi r^2h.$

The factor $\frac13$ matters: a cone has one third of the volume of a cylinder with the same circular base and perpendicular height. For a cone with radius $3\text{ cm}$ and perpendicular height $10\text{ cm}$,

$V=\frac13\pi(3)^2(10)=30\pi\text{ cm}^3.$

Use the perpendicular height, not the slant height. Brackets around a substituted radius make the squaring clear.

Surface area of a cone

The curved surface area is

$A_{\text{curved}}=\pi rl.$

A complete closed cone also has a circular base of area $\pi r^2$, so its total surface area is

$A_{\text{total}}=\pi rl+\pi r^2.$

The wording and diagram determine whether the circular base is exposed. For example, if a cone is joined to another solid along its base, that circle becomes internal and is not part of the exterior surface.

Spheres and hemispheres

For a sphere,

$V=\frac43\pi r^3$

and

$A=4\pi r^2.$

The radius is cubed for volume but squared for surface area. A hemisphere has half the volume of a sphere:

$V_{\text{hemisphere}}=\frac23\pi r^3.$

Its curved surface area is half the surface area of a sphere, $2\pi r^2$. If the hemisphere is a separate closed solid, its flat circular face must also be included, giving a total surface area of $3\pi r^2$. If that flat face is joined to another solid, it is internal instead.

Breaking down composite solids

A composite solid is handled by decomposing it into familiar solids. Annotate the diagram and decide whether each part is being added or removed.

For volume:

1. calculate the volume of each non-overlapping part;
2. add the volumes of joined parts;
3. subtract the volume of a hole or cut-out.

For example, a cone of radius $3\text{ cm}$ and height $10\text{ cm}$ joined to a hemisphere of radius $3\text{ cm}$ has volume

$30\pi+\frac12\left(\frac43\pi(3)^3\right)=30\pi+18\pi=48\pi\text{ cm}^3.$

Surface area needs a different check. Add only the surfaces visible from outside. The matching circular faces where the cone and hemisphere meet are internal, so neither is counted. This is why simply adding the two complete surface areas would give too large an answer.

Before calculating, ensure every length uses the same unit. Keep $\pi$ in the working for an exact answer and use the calculator's $\pi$ key only when a decimal answer is required.