No — these are given on the exam formula sheet. The skill being tested is selecting and applying the right one, not memorising it:

A cone's radius ($r$), perpendicular height ($h$), and slant height ($l$) form a right-angled triangle, so Pythagoras applies directly: $r^2+h^2=l^2$. Surface-area questions very often give you $h$ but need $l$ — spotting this hidden Pythagoras step is the key skill.

Example: $r=9\text{ cm}$, $h=12\text{ cm}$. $l=\sqrt{9^2+12^2}=\sqrt{225}=15\text{ cm}$. Total surface area $=\pi(9)(15)+\pi(9^2)=135\pi+81\pi=216\pi\text{ cm}^2$.

Treat the formula as an equation to solve, not just a one-way calculation.

Example: A sphere has volume $288\pi\text{ m}^3$. Find $r$.

1. $288\pi=\tfrac43\pi r^3$
2. Divide by $\pi$: $288=\tfrac43r^3$
3. Multiply by 3, divide by 4: $r^3=216$
4. Cube root: $r=6\text{ m}$.

Calculate each solid's volume (or surface area) separately, then combine — but for surface area, exclude any internal flat face where the two solids join.

Example: Cone ($r=5,h=12$) joined to a hemisphere of the same radius.

• $V_{cone}=\tfrac13\pi(5^2)(12)=100\pi$; $V_{hemi}=\tfrac23\pi(5^3)=\tfrac{250}{3}\pi$; total volume $=\tfrac{550}{3}\pi\text{ cm}^3$.
• Slant height: $l=\sqrt{5^2+12^2}=13$. Curved areas only: $\pi(5)(13)+2\pi(5^2)=65\pi+50\pi=115\pi\text{ cm}^2$ (the shared circular base is internal and excluded).

Worked Example

A solid metal sphere of radius $3.5\text{ cm}$ is dropped into a rectangular tank ($12\text{ cm}\times11\text{ cm}$ base) of water and fully submerges. Find the rise in water level, to 3 s.f.

1. Volume of sphere: $V=\tfrac43\pi(3.5)^3\approx179.59\text{ cm}^3$.
2. This volume raises the water as a thin cuboid of the same base: $12\times11\times x = 179.59$.
3. $132x=179.59 \Rightarrow x\approx1.36\text{ cm}$.

"Calculate" or "work out" want a numerical answer; "leave your answer in terms of $\pi$" means keep $\pi$ symbolic rather than evaluating it; "show that" means the result is already given, and the marks are for a clear, logically ordered derivation reaching it. Spotting these words before you start tells you exactly what form your final line of working needs to take, and prevents losing marks for giving a decimal when an exact multiple of $\pi$ was required (or vice versa).

Radius or diameter given — have you halved the diameter if needed? Perpendicular height or slant height — does the formula you're using need the other one via Pythagoras? Have you avoided premature rounding? And do your final units match what's asked (cm vs cm² vs cm³)? These four checks catch the large majority of errors made on this topic.