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/²• This 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•