Three-dimensional geometry Questions and Answers

Three-dimensional geometry Questions

1.  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 mid-point of BC

 Calculate:

 (a) QN

 (b) The altitude of P above the base

 (c)The angle between the planes ABQP and ABCD

 (d) (i) Locus P and locus Q meet at X. Mark X

(ii) Construct locus R in which angle BRC is 120^o

(iii) Show that locus inside triangle ABC such that XS  R  

2.

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;

(a) height of the pyramid (b) area of the pentagon  (c) angle between the face VAB and the base of the pyramid  (d) The pyramid is a container filled with orange juice.

Calculate the amount of juice in it.  

(e) find the surface area of the face VCD

3.  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 52.5cm³, find:-

 (i) The height of the pyramid

 (ii) The length of a slanting edge correct to 1decimal place  

 (iii) The angle between AV and CV  

 (iv) The obtuse angle between the edges AB and VD

4.  The figure below is cuboid ABCDEFGH. AB = 12cm, BC=5cm, CF = 6.5cm

 Calculate:

 (a) the length BD  

 (b) the angle AF makes with the base ABCD  

 (c) the angle DHGC makes with the base ABCD

 (d) M is the mid-point of HE. Calculate the length of line MC and the angle line MC makes

  with the base ABCD

5.   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

i) The height OV of the pyramid.

ii) The angle between the plane BMC and base ABCD.

6.  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:  a) the length of the projection of the line VP on the plane PQRS  

   b)the size of the angle between line VP and the plane PQRS  

c) the size of the angle between plane VQR and PQRS

7.  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;

 (a) find the surface area of the monument frustrum

 (b) 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 10m² is painted by ½ litre of paint,

find the cost of painting the monument.  

8.  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

 (a) Determine the height of the pyramid  

 (b) The angle PO makes with base PQRS  

 (c) The angle POS makes with QOR

 (d) The volume of the pyramid

9.  The diagram below shows a right solid pyramid on a square base ABCD of side 12cm and

slanting height of 24cm

 Calculate;

 a) To two decimal place the height (VO) of the pyramid

 b) the volume of the pyramid

 c) the total surface area of the pyramid

10.  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.

  (a) Sketch the general view of the pyramid  

 (b) Calculate:

  (i) The angle between VA and the base  

  (ii) The angle between face VCD and the base

11.  The positions o two towns A and B on earths surface are (60°N, 139°E) and (60°N, 41°W) respectively

 a) Find the difference in longitude between A and B

 b) Given that the radius of the earth is 6370km, calculate the distance between A and B in KM

 c) Another town C is 420km East of town B and on the same latitude A and B find the longitude

of town C

Three dimensional geometry Answers

1.  a)

QN = 12² – 6²

=10.39

b)

QX = ( 108 )² – 6²

=  72

= 8.485

c)

tan  = 8.485

6

  = 54.73⁰

d) tan  = 6

  10

 = 30.96

6 obtuse = 180⁰ – 30.96

10

= 149.04⁰

2.  a) Sin 36⁰ = 5

a

  Where a is the side

  a = 5 = 8.507

  sin 36

h² = 18.2 – 8.507

  = 258.87

H = 16.09 cm

b) ½ ab sin 

  ½ x 8.507² Sin 72 x 5

  = 172.06 cm²

c) Tan 36⁰ = 5

x

  x = 6.882

 Tan  = 16.09

  6.882

   = 66.84²

d) ¹/₃ x 172.06 x 16.09 = 922.8cm³

e) S= 23.2

  23.2 (23.2 – 18.2) (23.2 – 10)

= 87.50cm³

3.  (i) 1 x 4.2 x 7.5h = 52.5

3

h = 52.5 x 3 = 5.0cm

4.2 x 7.5

(ii) AC = 4.2² + 7.5²

= 17.64 + 56.25

= 73.89

= 8.596

AO = 8.596 2 = 4.298

AV = AO² + OV²

= 4.298² + 5²

= 18.47 + 25

= 43.47

= 6.6cm

(iii) Tan = 4.298

5

= 0.8596

 = 40.68^o

AVC = 40.68 x 2

= 81.36

1. Cos = 2.1
   

6.6

= 0.3182

 = 71.45^o Acute angle

obtuse angle = 180^o – 71.45^o

= 108.55^o

4.   (a)

BD² = 122 + 52 = 144 + 25 = 169  

BD = 169 = 13m

(b)  AF² = 13² + 6.52 = 169 + 42.25

=211.25  AF= 211.25 = 14.53cm  B1

tan  = 6.5 = 0.5  M1

  13

   = 26.57^o  A1

 (c) tan
^o= 6.5 = 1.3  M1

5


^o = 52.43  A1

(d) NC² = 2.5² +12² = 150.25

NC = 150,25 = 12.26 B1

 MC² = 6.52 + 150.25

 = 42.75 + 150.25

= 192.5

 MC = 192.5 = 13.87  Ba

tan ^o
6.5 = 0.5302

12.26

^o = 27.93^o  B1

5.  

i) Or = 16² – 5²

= 256 – 25

= 15.198 cm

ii) tan  = 5.066 = 1.2665

4

∴
 51.71⁰

6.  a) Height

AC = AB² + BC²

  = 10² + 10²

 = 200

  = 14.142

OA = ½ AC = 14.14 ² = 7.71

  2

OE = AE² – AO²

= 64 – 59.44 = 4.56

b)i)  Tan = 4.56 = 0.912

5.00

= 65.78°

ii) Tan  = 4.56 = 0.5914

  7.71

 = 30. 6°

c)

< AEC = 30.6 x 2

= 61.2°

7.  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

= √11.3² + 2.75²= 11.6

Slant height of the pyramid cut off

= √ 4.3² + 1.05² = 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 = 113.5

b) ½ litre paint 10m²

4 litres paints 80m²

 113.5m² requires 2 tins

2 x 650 = Kshs.1300/=

8.  (a) PR = 12² + 9² = 144 + 81 = 225 = 15cm

h = 19.52 – 7.52

= 380.25 – 56.25

= 324 = 18

(b) tan = 18 = 2.4

7.5

² tan 2.4 = 67.38^o

(c) tan  = 6 = 1

18 3

  = tan^-1 0.3333

 = 18.43^o

 x OY = 2 x 18.43 = 36.86

(d) Volume = 1 x 12 x 9 x 18   3

  = 648cm³

9.  a)   AC² = 12² + 12² = 288

 AC = 288 = 16.97

VO² = h² = 24² – (16.97)² = 504

2

  h= 504 = 22.45cm

b)  Base area = 12×12 = 144cm

 Volume = ¹/₃ x 144 x 22.45

= 1077.6cm³

c) Slanting surface = 30(30-24) (30-24) (30-12)

  = 139.44cm²

Total curved S.A = 139.44cm² x 4 + 144cm²

= 701.6cm²

10.  (b)

AO = 4.5 x sin 54^o = 3.82cm

 Sin 72^o

= cos^-1
(3.82) = 53.35^o

6.4

(c) Vo = 6.42 – 3.82

  = 5.13

VX = 6.4² – 2.55²

= 5.99cm

= Sin^-1 (^Vo/_Vx) Sin-1(^5.13/_5.99)

= 58.91^o

11.  a) Longitude difference = 139 ° + 41 °

= 180 °

b) Distance along latitude = ^Ø/₃₆₀ x 2  r cos 

= ¹⁸⁰/₃₆₀ x 2 x ²²/₇ x 6370 cos 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

c) Distance = ^{long diff}/₃₆₀ x 2R cos 60°

  420 = ^Ø /₃₆₀ x 2 x ²²/₇ x 6370 cos 60°

   = 420 x 360 x 7

2 x 22 x 6370 cos 60°

 = 7.552°

Longitude of C = 41° – 7.55° = 33.45°N