| Date | November 2008 | Marks available | 2 | Reference code | 08N.1.sl.TZ0.12 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Calculate | Question number | 12 | Adapted from | N/A |
Question
The diagram shows a pyramid \({\text{VABCD}}\) which has a square base of length \(10{\text{ cm}}\) and edges of length \(13{\text{ cm}}\). \({\text{M}}\) is the midpoint of the side \({\text{BC}}\).
Calculate the length of \({\text{VM}}\).
Calculate the vertical height of the pyramid.
Markscheme
Unit penalty (UP) applies in this question.
\({\text{VM}}^{2} = {13^2} - {5^2}\) (M1)
UP \( = 12{\text{ cm}}\) (A1) (C2)
[2 marks]
Unit penalty (UP) applies in this question.
\({h^2} = {12^2} - {5^2}\) (or equivalent) (M1)
UP \( = 10.9{\text{ cm}}\) (A1)(ft) (C2)
[2 marks]
Examiners report
This question was poorly answered by many of the candidates. Pythagoras was improperly applied and candidates were unable to identify right angled triangles.
This question was poorly answered by many of the candidates. Pythagoras was improperly applied and candidates were unable to identify right angled triangles.
