General Mathematics Paper 2,Nov/Dec. 2010
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 Main
Weakness/Remedies
Strengths
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Question 9

OABCD is a right pyramid with a rectangular base ABCD. Its vertical height is OG. If IABI = 6em, I BC/ = 8cm and each slant edge is 13cm,
(a) calculate, correct to one decimal placer the
(i)vertical height IOGI;
(ii)angle between a slant edge and the base ABCD;
(iii)angle between the triangle OAB and the base ABCD;
(b) find the volume of the pyramid.

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Observation

According to the report, this question was quite unpopular among the candidates. Very few candidates attempted it. However, it was observed that candidates performed better in part (a)(i) and (b) of this question than in part (a)(ii) and (iii). They could not calculate angles between edges and faces in three dimensional problems correctly. Generally, candidates' performance was reportedly very poor.
From the diagram, IAGI = ½/ACI = ½(√/ABP+/BCP ) = ½(√J62+ 82) = 5cm. Similarly,
IOGl² + IAGl² = IAO/²This implies that IOG/²= /AO/²- /AGl² = 132² - 52² = 144.
IOG/ = 144 = 12cm. Angle between a slant edge and the base ABCD is OAG which can
be obtained as follows: /AG/ = cos(OAG). i.e. 5/13 = cos(OAG). Therefore, OAG =
/AO/
cos-¹( 15) = 67.4°. If X is the mid point of AB, then the angle between faces OAB and ABCD
13
can be obtained as follows: Let the angle be a. Then tan a = /OG/= 12.Simplifying this gave
/GX/     4
the value of a as 71.6°. The volume of the pyramid = 1/3(base area x height) = 1/3 x 6 x 8 x 12

= 192cm3•