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Date	May 2018	Marks available	2	Reference code	18M.1.SL.TZ1.T_14
Level	Standard Level	Paper	Paper 1 (with calculator from previous syllabus)	Time zone	Time zone 1
Command term	Calculate	Question number	T_14	Adapted from	N/A

Question

A solid right circular cone has a base radius of 21 cm and a slant height of 35 cm.
A smaller right circular cone has a height of 12 cm and a slant height of 15 cm, and is removed from the top of the larger cone, as shown in the diagram.

Calculate the radius of the base of the cone which has been removed.

[2]

a.

Calculate the curved surface area of the cone which has been removed.

[2]

b.

Calculate the curved surface area of the remaining solid.

[2]

c.

Markscheme

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

$\sqrt {{{15}^2} - {{12}^2}}$ (M1)

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

OR

$\frac{{{\text{radius}}}}{{21}} = \frac{{15}}{{35}}$ (M1)

Note: Award (M1) for a correct equation.

= 9 (cm) (A1) (C2)

[2 marks]

a.

$\pi \times 9 \times 15$ (M1)

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

$= 424\,\,{\text{c}}{{\text{m}}^2}\,\,\,\,\,\left( {135\pi ,\,\,424.115...{\text{c}}{{\text{m}}^2}} \right)$ (A1)(ft) (C2)

Note: Follow through from part (a).

[2 marks]

b.

$\pi \times 21 \times 35 - 424.115...$ (M1)

Note: Award (M1) for their correct substitution into curved surface area of a cone formula and for subtracting their part (b).

$= 1880\,\,{\text{c}}{{\text{m}}^2}\,\,\,\,\,\left( {600\pi ,\,\,1884.95...{\text{c}}{{\text{m}}^2}} \right)$ (A1)(ft) (C2)

Note: Follow through from part (b).

[2 marks]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.1—3d space, volume, angles, distance, midpoints

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Topic 3— Geometry and trigonometry