Find the area of the base of the cylinder.

Show that the volume of the metal used in the cylinder is 402 cm³, given correct to three significant figures.

Find the total surface area of the cylinder.

The cylinder was melted and recast into a solid cone, shown in the following diagram. The base radius OB is 6 cm.

Find the height, OC, of the cone.

The cylinder was melted and recast into a solid cone, shown in the following diagram. The base radius OB is 6 cm.

Find the size of angle BCO.

The cylinder was melted and recast into a solid cone, shown in the following diagram. The base radius OB is 6 cm.

Find the slant height, CB.

The cylinder was melted and recast into a solid cone, shown in the following diagram. The base radius OB is 6 cm.

Find the total surface area of the cone.

\( \pi \times 4^2\)     (M1)

= 50.3 (16\(\pi\)) cm²  (50.2654...)     (A1)(G2)

Note: Award (M1) for correct substitution in area formula. The answer is 50.3 cm², the units are required.

[2 marks]

50.265...× 8     (M1)

Note: Award (M1) for correct substitution in the volume formula.

= 402.123...     (A1)
= 402 (cm³)     (AG)

Note: Both the unrounded and the rounded answer must be seen for the (A1) to be awarded. The units are not required

[2 marks]

\(2 \times \pi \times 4 \times 8 + 2 \times \pi \times 4^2\)     (M1)(M1)

Note: Award (M1) for correct substitution in the curved surface area formula, (M1) for adding the area of their two bases.

= 302 cm² (96π cm²) (301.592...)     (A1)(ft)(G2)

Notes: The answer is 302 cm², the units are required. Do not penalise for missing or incorrect units if penalised in part (a). Follow through from their answer to part (a).

[3 marks]

\(\frac{1}{3} \pi \times 6^2 \times \text{OC} = 402\)     (M1)(M1)

Note: Award (M1) for correctly substituted volume formula, (M1) for equating to 402 (402.123…).

\({\text{OC}} = 10.7{\text{ (cm)}}\left( {{\text{10}}\frac{2}{3},{\text{ }}10.6666...} \right)\)     (A1)(G2)

[3 marks]

\(\tan \text{BCO} = \frac{6}{10.66...}\)     (M1)

Note: Award (M1) for use of correct tangent ratio.

\({\text{B}}{\operatorname{\hat C}}{\text{O}} = 29.4^\circ \) (29.3577...)     (A1)(ft)(G2)

Notes: Accept 29.3° (29.2814...) if 10.7 is used. An acceptable alternative method is to calculate CB first and then angle BCO. Allow follow through from parts (d) and (f). Answers range from 29.2° to 29.5°.

[2 marks]

\(\text{CB} = \sqrt{{6^2} + {(10.66...)^2}}\)     (M1)

OR

\(\sin 29.35...^\circ = \frac{6}{\text{CB}}\)     (M1)

OR

\(\cos 29.35...^\circ = \frac{10.66...}{\text{CB}}\)     (M1)

CB = 12.2 (cm) (12.2383...)     (A1)(ft)(G2)

Note: Accept 12.3 (12.2674...) if 10.7 (and/or 29.3) used. Follow through from part (d) or part (e) as appropriate.

[2 marks]

\(\pi \times 6 \times 12.2383... + \pi \times 6^2\)     (M1)(M1)(M1)

Note: Award (M1) for correct substitution in curved surface area formula, (M1) for correct substitution in area of circle formula, (M1) for addition of the two areas.

= 344 cm² (343.785...)     (A1)(ft)(G3)

Note: The answer is 344 cm², the units are required. Do not penalise for missing or incorrect units if already penalised in either part (a) or (c). Accept 345 cm² if 12.3 is used and 343 cm² if 12.2 is used. Follow through from their part (f).

[4 marks]

This question was either very well done – by the majority – or very poorly (but not both). Many incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, since it was that the formulas for cones were not well understood. Further, the idea of “total surface area” was a mystery to many – a slavish reliance of formulas, irrespective of context, led to many errors and a consequent loss of marks.

The invariance of volume for solids and liquids that provided the link in this question was not understood by many, but was felt to be an appropriate subject for an examination.

This question was either very well done – by the majority – or very poorly (but not both). Many incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, since it was that the formulas for cones were not well understood. Further, the idea of “total surface area” was a mystery to many – a slavish reliance of formulas, irrespective of context, led to many errors and a consequent loss of marks.

The invariance of volume for solids and liquids that provided the link in this question was not understood by many, but was felt to be an appropriate subject for an examination.

This question was either very well done – by the majority – or very poorly (but not both). Many incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, since it was that the formulas for cones were not well understood. Further, the idea of “total surface area” was a mystery to many – a slavish reliance of formulas, irrespective of context, led to many errors and a consequent loss of marks.

The invariance of volume for solids and liquids that provided the link in this question was not understood by many, but was felt to be an appropriate subject for an examination.

This question was either very well done – by the majority – or very poorly (but not both). Many incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, since it was that the formulas for cones were not well understood. Further, the idea of “total surface area” was a mystery to many – a slavish reliance of formulas, irrespective of context, led to many errors and a consequent loss of marks.

The invariance of volume for solids and liquids that provided the link in this question was not understood by many, but was felt to be an appropriate subject for an examination.

This question was either very well done – by the majority – or very poorly (but not both). Many incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, since it was that the formulas for cones were not well understood. Further, the idea of “total surface area” was a mystery to many – a slavish reliance of formulas, irrespective of context, led to many errors and a consequent loss of marks.

The invariance of volume for solids and liquids that provided the link in this question was not understood by many, but was felt to be an appropriate subject for an examination.

This question was either very well done – by the majority – or very poorly (but not both). Many incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, since it was that the formulas for cones were not well understood. Further, the idea of “total surface area” was a mystery to many – a slavish reliance of formulas, irrespective of context, led to many errors and a consequent loss of marks.

The invariance of volume for solids and liquids that provided the link in this question was not understood by many, but was felt to be an appropriate subject for an examination.

This question was either very well done – by the majority – or very poorly (but not both). Many incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, since it was that the formulas for cones were not well understood. Further, the idea of “total surface area” was a mystery to many – a slavish reliance of formulas, irrespective of context, led to many errors and a consequent loss of marks.

The invariance of volume for solids and liquids that provided the link in this question was not understood by many, but was felt to be an appropriate subject for an examination.