17M.1.SL.TZ1.T_9

pestleMathematicsAASLPaper 117M· sl-3-1-3d-space-volume-angles-distance-midpointssource ↗

A type of candy is packaged in a right circular cone that has volume ${\text{100 c}}{{\text{m}}^{\text{3}}}$ and vertical height 8 cm.

M17/5/MATSD/SP1/ENG/TZ1/09

Find the radius, $r$ , of the circular base of the cone.

[2]

Find the slant height, $l$ , of the cone.

[2]

Find the curved surface area of the cone.

[2]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$100 = \frac{1}{3}\pi {r^2}(8)$ (M1)

Note: Award (M1) for correct substitution into volume of cone formula.

$r = 3.45{\text{ (cm) }}\left( {3.45494 \ldots {\text{ (cm)}}} \right)$ (A1) (C2)

[2 marks]

${l^2} = {8^2} + {(3.45494 \ldots )^2}$ (M1)

Note: Award (M1) for correct substitution into Pythagoras’ theorem.

$l = 8.71{\text{ (cm) }}\left( {8.71416 \ldots {\text{ (cm)}}} \right)$ (A1)(ft) (C2)

Note: Follow through from part (a).

[2 marks]

$\pi \times 3.45494 \ldots \times 8.71416 \ldots$ (M1)

Note: Award (M1) for their correct substitutions into curved surface area of a cone formula.

$= 94.6{\text{ c}}{{\text{m}}^2}{\text{ }}(94.5836 \ldots {\text{ c}}{{\text{m}}^2})$ (A1)(ft) (C2)

Note: Follow through from parts (a) and (b). Accept $94.4{\text{ c}}{{\text{m}}^2}$ from use of 3 sf values.

[2 marks]

Examiners’ report

[N/A]