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.
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.
Consider the triangular prism below.
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
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)
Calculate the length of the side KM shown below given that the triangle is right angled at M.
letting KM = y cm
+ 25 = 169
= 169 - 25
y = 12 cm
There are three main Basic trigonometric rations in a right angled triangle. These are sine, cosine and the tangent.
Calculate the size of the angle θ
in the figure below.
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.
= Opp/ adj = 5/ 7
Tan θ = 0.7143 (look for the angle whole tan = 0.7143
θ = 35.54o
Find the size of the angle θ
In this case the ratio sin is 10 be used since the lines given are opposite and hypotenuse.
= Opp/ Hyp = 3/ 8
Sin θ = 0.375
θ = 22.02
Find the length of the side shown with label y on the figure show below given that the images in right angled at B.
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 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
Solve completely the triangle XYZ in which X = 42o
; x = 8.4 cm and y = 10.2 cm
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
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
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
Solve completely for the following triangle given that P = 6.9cm, Q = 12.3cm and R = 7.6cm
The triangle is drawn and labeled as shown below
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.
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
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.
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.
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
Use the figure below and name
(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
In the figure below ABCDEFGH is a cuboid and K is the mid-point of EH
Calculate the length of lines
By use of Pythagoras theorem
+ ( CH)2
but AC = AB = 8 cm
= 64 + 16
DH = 8.944 cm
(ii) To find AH. We must consider triangle ACH
we first work out the line AC
= 64 + 36 = 100
AC = 10 cm
AH = 10.77 cm
(iii)To find KD we must consider triangle KGD Right angled at G
For Right angled triangle HGK
= 64 + 9
KG = 8.544 cm2
From Right angled triangle KGD
= 73 + 16
KD = 9.433 cm2
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
(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 To calculate angle EAF, SineO (ii)Angle between line AD and DF Therefore , (AD)2 By use of trigonometry (SOHCAHTOA) the angle required is the angle ADF
= 64 + 36 = 100
BD = 10 cm
Use the trigonometric ratio
angle EAF = O
(the angle between lines AF and BD)
We must first calculate the lengths of lines AD and DF
= 100 + 144
AD = 15.62 cm
(DF)2 = 64 + 144
DF = 14.42 cm
To calculate angle EAF,
(ii)Angle between line AD and DF
Therefore , (AD)2
By use of trigonometry (SOHCAHTOA) the angle required is the angle ADF
Angles between a line and a plane
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.
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