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Date November 2010 Marks available 2 Reference code 10N.1.sl.TZ0.12
Level SL only Paper 1 Time zone TZ0
Command term Find Question number 12 Adapted from N/A

Question

The base of a prism is a regular hexagon. The centre of the hexagon is O and the length of OA is 15 cm.

Write down the size of angle AOB.

[1]
a.

Find the area of the triangle AOB.

[3]
b.

The height of the prism is 20 cm.

Find the volume of the prism.

[2]
c.

Markscheme

60°     (A1)     (C1)

[1 mark]

a.

\(\frac{{15 \times \sqrt {{{15}^2} - {{7.5}^2}} }}{2} = 97.4{\text{ c}}{{\text{m}}^2}\)     (97.5 cm2)     (A1)(M1)(A1)


Notes: Award (A1) for correct height, (M1) for substitution in the area formula, (A1) for correct answer.

Accept 97.5 cm2 from taking the height to be 13 cm.


OR

\(\frac{1}{2} \times {15^2} \times \sin 60^\circ  = 97.4{\text{ c}}{{\text{m}}^2}\)     (M1)(A1)(A1)(ft)     (C3)

 

 

Notes: Award (M1) for substituted formula of the area of a triangle, (A1) for correct substitution, (A1)(ft) for answer.

 

Follow through from their answer to part (a).

 

If radians used award at most (M1)(A1)(A0).

 

 

 

[3 marks]

 

 

 

 

b.

97.4 × 120 = 11700 cm3     (M1)(A1)(ft)     (C2)


Notes: Award (M1) for multiplying their part (b) by 120.

 

[2 marks]

c.

Examiners report

This question proved to be difficult for a number of candidates. Most were able find the size of the angle in part a), but many had problems finding the area of the triangle in part b). A significant number of candidates were unable to use the Pythagoras Theorem correctly to find the height of the triangle AOB. Those who used the formula for the area of a triangle \({\text{A}} = \frac{1}{2}{\text{ab}}\sin {\text{C}}\)  were more successful in this part of the question. It was surprising that a great number of candidates were unable to find the volume of the prism – many incorrectly used the formula for calculating volume of a pyramid rather than a hexagonal prism.

a.

This question proved to be difficult for a number of candidates. Most were able find the size of the angle in part a), but many had problems finding the area of the triangle in part b). A significant number of candidates were unable to use the Pythagoras Theorem correctly to find the height of the triangle AOB. Those who used the formula for the area of a triangle \({\text{A}} = \frac{1}{2}{\text{ab}}\sin {\text{C}}\)  were more successful in this part of the question. It was surprising that a great number of candidates were unable to find the volume of the prism – many incorrectly used the formula for calculating volume of a pyramid rather than a hexagonal prism.

b.

This question proved to be difficult for a number of candidates. Most were able find the size of the angle in part a), but many had problems finding the area of the triangle in part b). A significant number of candidates were unable to use the Pythagoras Theorem correctly to find the height of the triangle AOB. Those who used the formula for the area of a triangle \({\text{A}} = \frac{1}{2}{\text{ab}}\sin {\text{C}}\)  were more successful in this part of the question. It was surprising that a great number of candidates were unable to find the volume of the prism – many incorrectly used the formula for calculating volume of a pyramid rather than a hexagonal prism.

c.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.5 » Volume and surface areas of the three-dimensional solids defined in 5.4.
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