Questions
 The figure below represents a plan of a roof with a rectangular base ABCD. AB = 20cm and BC=12cm. the ridge PQ = 8cm is centrally placed. The faces ADP and BCQ are equilateral triangles. N is the midpoint of BC
Calculate: QN
 The altitude of P above the base
 The angle between the planes ABQP and ABCD
 Locus P and locus Q meet at X. Mark X
 Construct locus R in which angle BRC is 120o
 Show that locus inside triangle ABC such that XS ≥R

The diagram alongside shows a right pyramid whose base is a regular pentagon of side 10cm.VA=B=VC=VD=VE=18.2cm and O is the centre of the pyramid. Calculate; height of the pyramid
 area of the pentagon
 angle between the face VAB and the base of the pyramid
 The pyramid is a container filled with orange juice. Calculate the amount of juice in it.
 find the surface area of the face VCD
 The diagram below shows a right pyramid on a rectangular base ABCD measuring 7.5cm by 4.2cm.
If the volume of the pyramid is 52.5cm^{3}, find: The height of the pyramid
 The length of a slanting edge correct to 1decimal place
 The angle between AV and CV
 The obtuse angle between the edges AB and VD
 The figure below is cuboid ABCDEFGH. AB = 12cm, BC=5cm, CF = 6.5cm.
Calculate: the length BD
 the angle AF makes with the base ABCD
 the angle DHGC makes with the base ABCD
 M is the midpoint of HE. Calculate the length of line MC and the angle line MC makes with the base ABCD
 The figure below is a right pyramid with a rectangular base ABCD and vertex V.
O is the centre of the base and M is a point on OV such that OM = ^{1}/_{3 }OV, AB = 8 cm, BC = 6 cm
and VA = VB=VD = VC = 15 cm. Find ; The height OV of the pyramid.
 The angle between the plane BMC and base ABCD.
 The figure below represents a right pyramid with vertex V and a rectangular base PQRS, VP=VQ=VR=VS=18cm, PQ=16cm and QR=12cm. M and O are the midpoints of QR and PR
respectively.
Find the length of the projection of the line VP on the plane PQRS
 the size of the angle between line VP and the plane PQRS
 the size of the angle between plane VQR and PQRS
 Mayoni Municipal Council wishes to construct a monument on the grounds. The monument is designed to be in the shape of a frustrum of a right pyramid. The base of the frustrum is a square of side 5.5meters while the top of the frustrum is a square of side 2.1cm
If the perpendicular distance between faces ABCD and EFGH is 7cm; find the surface area of the monument frustrum
 The monument is to be painted on all surface excluding the base. Paint is sold in 4 litre tins each costing Kshs.640. It is estimated that an area 10m^{2 }is painted by ½ litre of paint, find the cost of painting the monument.
 The figure below is a pyramid of a rectangular base PQRS of length 12cm and width 9cm. The slanting edge has a length of 19.5cm
 Determine the height of the pyramid
 The angle PO makes with base PQRS
 The angle POS makes with QOR
 The volume of the pyramid
 The diagram below shows a right solid pyramid on a square base ABCD of side 12cm and slanting height of 24cm
Calculate; To two decimal place the height (VO) of the pyramid
 the volume of the pyramid
 the total surface area of the pyramid
 The base of a pyramid consists of a regular pentagon ABCDE, 4.5cm a side. The vertex of the pyramid is V and VA = VB = VC = VD = VE = 6.4cm.
 Sketch the general view of the pyramid
 Calculate:
 The angle between VA and the base
 The angle between face VCD and the base
 The positions o two towns A and B on earths surface are (60°N, 139°E) and (60°N, 41°W) respectively
 Find the difference in longitude between A and B
 Given that the radius of the earth is 6370km, calculate the distance between A and B in Km
 Another town C is 420 km East of town B and on the same latitude A and B find the longitude of town C
Answers


 Sin 36^{o} = 5
a
Where a is the side
a = 5 = 8.507
sin 36
h^{2 }= 18.2 – 8.507
= 258.87
H = 16.09 cm  ½ ab sin θ
½ x 8.5072 Sin 72 x 5
= 172.06 cm^{2}  ^{}Tan 36^{o }= 5
x
x = 6.882
Tan θ= 16.09
6.882
θ = 66.842  ^{1}/_{3 }x 172.06 x 16.09 = 922.8cm^{3}
 ^{}S= 23.2
√[23.2 (23.2 – 18.2) (23.2  10)]
= 87.50cm^{3}
 Sin 36^{o} = 5



 Or = 16^{2 } 5^{2}
= 256 – 25
= 15.198 cm  tan θ= 5.066 = 1.2665
4
∴ θ=51.71^{o}
 Or = 16^{2 } 5^{2}

 Let length of cut off pyramid be meters
Then 7 + h = 5.5
H 2.1
14.7 + 2.1h = 5.5
3.4h = 14.7
h = 4.3
Slant height of big pyramid
= √[11.3^{2 }+ 2.75^{2}]= 11.6
Slant height of the pyramid cut off
= √[4.3^{2 }+ 1.05^{2}]= 4.4m
Area of EFCD = ½ x11.6 x 5.5 – ½ x 4.4 x 2.1
= 27.28 m
Total surface area = 4 x 27.28 + 2.1 x 2.1 = 113.  ½ litre paint 10m^{2}
4 litres paints 80m^{2}
∴ 113.5m^{2}requires 2 tins
2 x 650 = Kshs.1300/=
 Let length of cut off pyramid be meters

 AC^{2 }= 12^{2 }+ 12^{2 }= 288
∴ AC = √288 = 16.97
VO2 = h2 = 242 – (16.97)^{2 }= 504
2
h= √504 = 22.45cm  Base area = 12x12 = 144cm
∴ Volume =^{1}/_{3 }x 144 x 22.45
= 1077.6cm^{3}  Slanting surface = √30(3024) (3024) (3012)
= 139.44cm^{2}
Total curved S.A = 139.44cm^{2 }x 4 + 144cm^{2}
= 701.6cm^{2}
 AC^{2 }= 12^{2 }+ 12^{2 }= 288

 Longitude difference = 139° + 41°
= 180 °  Distance along latitude = ^{Ø}/_{360 }x 2πrcos θ
=^{180}/_{360 }x 2 x ^{22}/_{7 }x 6370cos 60°
= 22 x 910 x 0.5
= 10,010 Km
Or via north pole (great circle)
Latitude difference = 60°
Distance = ^{60}/_{360 }x 2 x ^{22}/_{7 }x 6370
= 6673.33 Km  Distance = long diff/360 x 2πR cos 60°
420 = ^{Ø}/_{360 }x 2 x ^{22}/_{7 }x 6370 cos 60°
Ø = 420 x 360 x 7
2 x 22 x 6370 cos 60°
= 7.552°
Longitude of C = 41°  7.55° = 33.45°N
 Longitude difference = 139° + 41°
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