A school canteen sells ice cream in cones. Each cone has radius 3 cm and height 11 cm. A scoop of ice cream is a sphere of radius 3 cm, and it is placed on ...

Three standard formulae are needed here: the volume of a cone \(V = \frac{1}{3}\pi r^2 h\), the curved surface area of a cone \(A = \pi r l\) where \(l\) is the slant height, and the volume of a sphere \(V = \frac{4}{3}\pi r^3\). The final part compares two of them.

(a) Volume of the cone. [2]
Radius 3 cm, vertical height 11 cm:

\[V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3} \times \pi \times 3^2 \times 11 = \tfrac{1}{3} \times \pi \times 9 \times 11 = 33\pi\] \[V = 103.67\ldots = 104\ \text{cm}^3\ \text{(3 s.f.)}\]

(b) Slant height and curved surface area. [3]
The slant height is the distance from the rim to the point, and it is the hypotenuse of a right-angled triangle whose legs are the radius and the vertical height:

\[l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 11^2} = \sqrt{9 + 121} = \sqrt{130} = 11.4017\ldots = 11.4\ \text{cm}\]

Then

\[A = \pi r l = \pi \times 3 \times 11.4017\ldots = 107.47\ldots = 107\ \text{cm}^2\ \text{(3 s.f.)}\]

The vertical height 11 cm and the slant height 11.4 cm are close but not the same, and the curved surface formula needs the slant one. Marks: one for the Pythagoras step, one for the slant height, one for the surface area.

(c) Volume of the scoop. [2]
A sphere of radius 3 cm:

\[V = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3} \times \pi \times 3^3 = \tfrac{4}{3} \times \pi \times 27 = 36\pi\] \[V = 113.09\ldots = 113\ \text{cm}^3\ \text{(3 s.f.)}\]

Note \(r^3\), not \(r^2\): the radius is cubed for a sphere.

(d) Will the cone overflow? [2]
Compare the melted ice cream with the space inside the cone:

\[\text{scoop } 113.09\ldots\ \text{cm}^3 \ >\ \text{cone } 103.67\ldots\ \text{cm}^3\]

The scoop has the larger volume, so the cone does overflow. The excess is

\[113.09\ldots - 103.67\ldots \approx 9.4\ \text{cm}^3\]

which is roughly 9% more than the cone can hold.

The reason is worth noticing. The cone and the sphere have the same radius, but the sphere of radius 3 cm has volume \(36\pi\) while the cone needs a height of \(h\) with \(\frac{1}{3}\pi \times 9 \times h = 36\pi\), that is \(h = 12\) cm, to match it. The cone here is only 11 cm tall, so it falls just short.

Examination point: use the unrounded volumes for the comparison in part (d). Rounded to 3 significant figures the two values are 113 cm³ and 104 cm³, which still gives the right conclusion, but in a closer question rounding first can reverse the decision.