Date | May 2009 | Marks available | 2 | Reference code | 09M.1.sl.TZ2.15 |
Level | SL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 15 | Adapted from | N/A |
Question
The right pyramid shown in the diagram has a square base with sides of length 40 cm. The height of the pyramid is also 40 cm.
Find the length of OB.
Find the size of angle OBP.
Markscheme
Note: Unit penalty (UP) applies in this part
\({\rm{PB}} = \frac{1}{2}\sqrt {{{40}^2} + {{40}^2}} = \sqrt {800} = 28.28(28.3)\) (M1)(A1)
Note: Award (M1) for correct substitutions, (A1) for correct answer.
(UP) \({\rm{OB}} = \sqrt {{{40}^2} + {{28.28}^2}} = 49.0{\text{ cm }}\left( {\sqrt {2400} {\text{ cm}}} \right)\) (M1)(A1)(ft) (C4)
Note: Award (M1) for correct substitution, can (ft) from any answer to PB.
[4 marks]
\({\sin ^{ - 1}}\left( {\frac{{40}}{{49}}} \right)\)
OR
\({\cos ^{ - 1}}\left( {\frac{{28.28}}{{49}}} \right)\)
OR
\({\tan ^{ - 1}}\left( {\frac{{40}}{{28.28}}} \right)\) (M1)
= 54.7 (54.8) (A1)(ft) (C2)
Note: Award (M1) for any correct trig. ratio.
In radians = 0.616, award (M1)(A0).
Note: Common error: (a) \(OB = \sqrt {40^2 + 20^2} = 44.7 {\text{ cm}}\). Award (M0)(A0)(M1), (A1)(ft), and (b) angle OBP = 63.4° (63.5°) (M1)(A1)(ft).
[2 marks]
Examiners report
This question was well answered by many candidates although a number lost an accuracy penalty or a unit penalty in this question. A very common error was assuming PB to be 20 cm. The mark-scheme allowed for follow through marks to be awarded in this case. Most candidates could find the angle and very few did not use right angled trigonometry.
This question was well answered by many candidates although a number lost an accuracy penalty or a unit penalty in this question. A very common error was assuming PB to be 20 cm. The mark-scheme allowed for follow through marks to be awarded in this case. Most candidates could find the angle and very few did not use right angled trigonometry.