﻿ Three Dimensional Geometry | Mathematics Form 4

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Butterfly Chameleon #### Three Dimensional Geometry - Mathematics Form 4

(i) Common solids and Nets of Solids All solids are three dimensional

This is because they have length and width, this makes them to occupy space.

Examples

This can be best demonstrated by the following nets and their respective solids.

Geometric properties of common solids

Solids have length, area and volume since measurements on them are said to be three dimensional. A vertex on a solid is a point where three or more edges meet. An edge is a line where faces meet. A plane / face is a set of points in a flat surface and extends indefinitely in all directions. In many solids some face, line or line and faces are parallel while others are not.

Example

Consider the triangular prism below.
a) Name the line(s) which are parallel i. AB ii. BC iii. BF b) Name a face parallel to face ABF c) Name the line where face ABCD and ABF intersect. d) Name the point of intersection of i. Line DA and face ABF ii. Line DC and face BCEF.

Solution

a) . Line(s) parallel to (i). AB is DC (ii). BC are AD and FE (iii). BF is CE b) Face parallel to ABF is DCE c) Line of intersection of ABCD and ABF is AB. d) Point of intersection of (i) Line DA and face ABF is A (ii) Line DC and face BCEF is C

1.Pythagorus theorem

This theorem states that in any right-angled triangle the sum of the squares on the two shorter sides equals to the square of the hypothenuse; i.e. AB2 + BC2 = AC2 ( Height 2 + Base 2 = Hypothenuse2)

Example:
Calculate the length of the side KM shown below given that the triangle is right angled at M.

Solution:

(KM)2
+ (ML)2
= (KL)2

letting KM = y cm
y
2
+ 52
= 132
y
2
+ 25 = 169
y
2
= 169 - 25
y
2
= 144
y = 12 cm

TRIGONOMETRIC RATIOS
There are three main Basic trigonometric rations in a right angled triangle. These are sine, cosine and the tangent.
Example 1
Calculate the size of the angle θ
in the figure below.

Solution
The side 5 is on the opposite side of the angle θ
and side 7 is the adjacent side: the ratio tan θ
will be used.
Tan
θ
= Opp/ adj = 5/ 7

Tan θ = 0.7143 (look for the angle whole tan = 0.7143
θ = 35.54o

Example 2
Find the size of the angle θ
shown below.

Solution
In this case the ratio sin is 10 be used since the lines given are opposite and hypotenuse.
Sine
θ
= Opp/ Hyp = 3/ 8

Sin θ = 0.375
θ = 22.02

Example 3
Find the length of the side shown with label y on the figure show below given that the images in right angled at B.
Solution
The relationship between the side 12cm and the angle 300 given is that so the side is the adjacent side and y is the hypotenuse of the of the triage therefore we use the ratio cosine
i.e. Cos θ = Adj/Hyp ; Cos 30
o
= 12/y

Y = 12/Cos 30o ; y = 12/0.866

Y = 13.85 cm

SINE AND COSINE RULE

The sine and cosine rules are used to solve for the sides and angles of triangle which are not right angles.
The sine rule: consider the non-right angled triangle ABC below.
The sine Rule will given by
a/Sin A = b/ = c/sin C = 2R where R is the Radius of the circumcircle of the triangle.

The sine Rule is used to
(i) Determine the Radius of the circumcircle
(ii) Solve for a triangle completely given two
a) Angles and a side
b) Sides and the angle opposite on of the sides

Example
Solve completely the triangle XYZ in which X = 42o
; x = 8.4 cm and y = 10.2 cm

Solution :
The triangle is drawn as shown below

Using sine rule
x /Sin x = y/Sin y = z/sin z = 2R

But we do not have R therefore

8.4/Sin 42 = 10.2/Sin y = 8.4 Sin y =10.2 sin 42

Sin Y = (10.2 x sin 42)/8.4

Sin y = 0.8125 - y = 54.34
Therefore :
Angle XYZ = Angle Y = 54.34o
Angle XZY = 180 - (42 + 54.34) = 180 - 96.34
Angle XZY = Angle z = 83.66o
Using sine rule Again implies that Z/Sin Z = x / Sin x
Z = 8.4/Sin 83.66 = 8.4/sin 42 which implies that

Z = ( 8.4 sin 83.66)/ Sin 42

Z = 12.48 cm

NB: The triangle has been solved completely since all its angles and sides are now known

The Cosine rule
By use of the non-right angled triangle shown below

The Cosine rule
a2 = b2 + c2 - 2bc cos A

Or

b2 = a2 + c2 - 2ac Cos B
c2 + a2 + b2 - 2ab Cos B

All these alternative forms of cosine Rule are used to find the :
(i) The third side of a triangle given two sides and an included angle
(ii) Three angle given three sides
Example
Solve completely for the following triangle given that P = 6.9cm, Q = 12.3cm and R = 7.6cm

Solution
The triangle is drawn and labeled as shown below
By use of the cosine rule

p2 = q2 + r2 - 2qr cos P

6.92 = (12.3)2 + (7.6)2 - 2 (12.3) (7.6) Cos P

47.61 = 151.29 + 57.76 -186.96 cos P

47.61 = 209.05 - 186.96 Cos P

-161.44/-186.96 = -186.96 Cos P

Cos P = -161.44/-186.96

Cos P = 0.8635

P = 30.29o

By use of cosine Rule still

R2 = p2 + q2 - 2(p)(q) Cos R
(7.6)2 =6.92 + 12.32 - 2(6.9) 912.30 Cos R
57.76 = 47.61 + 151.29 - 169.74 Cos R
57.76 = 198.90 - 169.74 Cos R
141.14 = -169.74 Cos R
Cos R = -141.14/ -169.74 which implies that Cos R = 0.8315
R = 33.75o

From the triangle PQR then therefore Angle Q can be obtained by
Q = 180 - (30.29 + 33.75)
Q =180 - 64.04
Q = 115.96o

The triangle has been solved completely since all its angles and sides are now known.

OBJECTIVE

By the end of the lesson you should be able to:

1. State the geometry properties of common solids.

2. Identify the projection of a line on to a plane.

3. Identify skew lines.

4. Calculate the length between two points in three dimensional geometry.

5. Identify and calculate the angle between

i. Two lines

ii. A line and a plane

iii. Two planes.

THREE DIMENSIONAL GEOMETRY

In our everyday life we are in touch with 3 dimensional geometry. Every object one sees and touches has 3 dimension: Length, width and height. Look at any room and items in it. These can be described in 3 dimension.
3 dimensional geometry is important in many careers such as architecture.

Identifying the projections of a line onto a plane
Projections
If the plane WXYZ is lit from an external source as shown the shadow of line AB forms on the plane as AC and is known as the projection of the line AB onto the plane, WXYZ.

NOTE:

Point B is projected on C.
The angle between the line AB and AC is taken to be the angle between AB and plane WXYZ.
In general, the angle between a line and a plane is the angle between the line and its projection on the plane.

Example
The figure given is a right pyramid with square base PQRS, O is the centre of the base.
(a) Name the projection of the following on the plane PQRS
(i) VP (ii) VS (iii) VO
(b) Name projection of VQ onto plane VRP.
(c) Name projection of VP on plane VQS.
(d) Name the angle between VS and the base PQRS

Solution
(a) (i) PO (ii) SO (iii) point O
(b) VO
(c) PO
(d) Angle VSO

Skew Lines
The lines HG and AD do not intersect and are not parallel. These lines do not lie on the same plane such lines are skew Lines.

The angle between the skew lines is found by translating one of the lines to the plane containing the other.

To get the angle between HG and AD

translate line HG to plane ABCD such that it lies at line AB on the plane, then the angle between the two is angle DAB = 90o
.

Example
Use the figure below and name
(i) Lines which are skew to GB
(ii) Angles between EG and DC
(iii) Angles between BD and AH

Solution

(i) Skew lines to GB are AD, DC, HE, EF (blink lines GB, AD, DC, HE, EF)
(ii) Angle FEG or CDB
(iii) Angle EDB and angle GBD

Calculating Lengths And Angles In A Solid

In three dimensional geometry unknown lengths and angles can be determined by solving tringles (i.e. determining the unknown sides and angles of triangles as illustrated in the background information)

Calculating the length of a line

Example
In the figure below ABCDEFGH is a cuboid and K is the mid-point of EH
Calculate the length of lines
(i) DH
(ii) AH
(iii) KD

Solution
By use of Pythagoras theorem

(i) (DH)2
= (DC)2
+ ( CH)2

but AC = AB = 8 cm
Therefore,
(DH)
2
= 82
+ 42

(DH)
2
= 64 + 16
(DH)
2
= 80

DH = 8.944 cm

(ii) To find AH. We must consider triangle ACH
Therefore,
we first work out the line AC

(AC)2
= (AB)2
+ (BC)2

(AC)
2
= 82
+ 62

(AC)
2
= 64 + 36 = 100
AC = 10 cm

Therefore, (AH)
2
= (AC)2
+ (CH)2

(AH)
2
= 102
+ 42
= 116
AH = 10.77 cm

(iii)To find KD we must consider triangle KGD Right angled at G
For Right angled triangle HGK

(KG)2
= (HG)2
+ (AK)2

(KG)
2
= 82
+ 32

(KG)
2
= 64 + 9
(KG)
2
= 73
KG = 8.544 cm
2

From Right angled triangle KGD

(KD)2
= (KG)2
+ (GD)2

(KD)
2
= 8.5442
+ 42

(KD)
2
= 73 + 16
(KD)
2
= 89
KD = 9.433 cm
2

Calculating The Length And Angles Between
(a) Two Lines
The figure below shows a wedge and its given measurements.

Calculate the angles between lines
(i) AF and BD
(ii) AD and DF
(iii) FD and ED

Solution
(i) Lines AF and BD are skewed lines. Therefore, we must translate (move) line BD to where AE is, to form angle EAF
or move line AF to where BC is, to form angle DBC.

Therefore , moving line BD
(BD)2
= (DC)2
+ (BC)2

(BD)
2
= 82
+ 62

(BD)
2

= 64 + 36 = 100
BD = 10 cm

To calculate angle EAF,

Use the trigonometric ratio

SineO
= 0.8
angle EAF =
O
= 53.13o
(the angle between lines AF and BD)

(ii)Angle between line AD and DF

We must first calculate the lengths of lines AD and DF

Therefore , (AD)2
= (DB)2
+ (AB)2

(AD)
2
= 102
+ 122

(AD)
2
= 100 + 144
(AD)
2
= 244
AD = 15.62 cm
(DF)
2
= (DF)2
+ (CF)2

(DF)
2
= 82
+ 122

(DF)2 = 64 + 144
(DF)
2
= 208
DF = 14.42 cm

By use of trigonometry (SOHCAHTOA) the angle required is the angle ADF
Therefore,

Tan O = 0.4161;
O = 22.59o (this is the angle between line AD and DF)

(iii)Angle between line FD and ED

The angle between line FD and ED is angle FDE.

By use of trigonometry

Cos O = 0.8322
O= 33.68o (this is the angle between line FD and ED)

Angles between a line and a plane
Example
The figure along side is a square based pyramid. The triangular faces of the pyramid are isosceles triangles of slanting edges 10 cm each.
(i) Find the angle between VA and plane ABCD
(ii) If m is the mid-point of line BC find the angle between VM and the plane ABCD.

Solution
(i) The angle between line VA and plane ABCD is the angle VAD

(AC)2
= (AB)2
+ (BC)2

(AC)
2
= 82
+ 82

(AC)
2
= 64 + 64
(AC)
2
= 128
AC = 11.32 cm
AD =1/2 AC
Therefore, AD = 5.66 cm

By use of trigonometry

CosO
= 0.566;
O
= 55.53O
(this is the angle between lines VA and plane ABCD)

(ii) Since M is the mid-point of BC then MC = BM = 4 cm.
(VM)2 = (VC)2 + (CM)2
(VM)2 = 102 - 42
(VM)2 = 100 - 16
(VM)2 = 84
VM = 9.165 cm
Angle between VM and the plane ABCD is angle VMO(blink angle between line VM and plane ABCD)

By use of trigonometry angle VMO

CosO
= 0.4364;
O
= 64.12O

NOTE:

An angle between a line and a plane is the size of the angle between the line and its projection on that given plane.

Angle between a plane and another plane
Example

The figure below is a cuboid 8.5 cm long, 6 cm wide and 3.5 cm high. Calculate the angle between:
(i) Plane KLQP and MNPQ
(ii) KLQP and KLMN
(iii) KMQS and MNPQ

Solution
(i) The angle between planes KLQP and MNPQ is given by (formed on line PQ)

(ii) The angle between planes KLQP and KLMN is angle formed on line KL (intersection of the plane)

(iii) Angle between planes KMQS and MNPQ is the angle formed on the line QM (intersection of the planes)

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