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Properties of parallel lines
The angles 1-3, 2-4, 6-8 and 5-7 are vertically opposite
angles.
Vertically opposite angles are equal.
The angles 2-6, 1-5, 7-3 and 8-4 are corresponding
angles.
Corresponding angles are equal.
The angles 3-4 and 4-6 are alternate angles.
Alternate angles are equal.
The angles 3-6 and 4-5 are cointerior angles.
Cointerior angles are supplementary (180∘).
Example:Find the values of the pronumerals.
Properties of triangles
1. a ° , b° and c° are the magnitudes of the interior
angles of the triangle ABC.
d° is the magnitude of an exterior angle at C. A
triangle is said to be a right-angled triangle if it has
one angle of magnitude 90°. 2. The sum of the
magnitudes of the interior angles of a triangle is
equal to
a ◦ + b◦ + c◦ = 180° . 3. The magnitude of an
exterior angle is equal to the sum of the magnitudes
of the two opposite interior angles. b° + a ° = d °.
4. A triangle is said to be equilateral if all its sides are of
the same length: AB = BC = CA.
5. The angles of an equilateral triangle are all of
magnitude 60◦ .
6.The bisector of each of the angles of an equilateral triangle
meets the opposite side at right angles and passes through the
midpoint of that side.
7. A triangle is said to be isosceles if it has two sides of
equal length. If a triangle is isosceles, the angles
opposite each of the equal sides are equal.
8. The sum of the magnitudes of the exterior
angles of a triangle is equal to 360°: e° +d°+f°=360°
Example 1: Find the values of the
pronumerals.
Solution: 20° +22°+x°=180° (sum
angles△=180°)
∴42° +x°=180° or x =138°
138°+y°=180° (sum angles=180°) ∴y
=42°
Example: Find the values of the pronumerals.
Solution: 100°+2x°=180° (sum
angles△=180°)
∴2x° =80° or x =40°
Properties of regular polygons
A regular polygon has all sides of equal length and all angles
of equal magnitude. A polygon with n sides can be divided into
n triangles.
The angle sum of the interior angles of an n-sided polygon is given by the
formula: S = [180(n − 2)]◦ = (180n − 360)◦
The magnitude of each of the interior angles of an n-sided polygon is given
by: x = (180n − 360)◦
n
interior angle
exterior angle
The sum of the exterior angles of a regular polygon is 360◦ .
Example1: The diagram opposite shows a regular octagon.
a Show that x = 45.
b Find the size of angle y.
Solution
a x = 360°÷ 8 = 45°
b BOC is isosceles
∧
∧
then OBC and OCB are equal
x° + 2y°= 180°
45° + 2y° = 180°
2y°= 180° - 45°
2y°= 135°
y° = 67.5°
Example:Find the sum of the interior angles of an 8-sided polygon(octagon).
Solution
Use the formula x°= (180n-360)°
x°= 180×8-360 = 1080 °
Pythagoras theorem
Example 1: Find the value, correct to two decimal places, of the unknown
length for the triangle below.
Solution: x2 = 5.32 + 6.12
√x2 = √( 5.32 + 6.12)
x = 8.08cm
Example 2: Find the value, correct to two decimal places, of the unknown length for the
triangle below.
Solution: 8.62 = y2 + 5.62
y2 = 8.62 - 5.62
y = √( 8.62 - 5.62)
y = 6.53cm
Similar figures
Shapes are similar when they have the same shape but not the same size.
Similar Triangles
Two triangles are similar if one of the following conditions holds:
1.The corresponding angles in the triangles are equal
A = A′, B = B′, C = C′
2. The ratio of the corresponding sides is equal
A′B′ = B′C′ = A′C′ =k
AB
BC
AC
k is the scale factor.
If AB = 2, BC = 3, AC =4 and A′B′ = 6, B′C′ = 9, A′C′= 12
then 6 = 9 = 12 = 3= k
2 3 4
3.Two pairs of corresponding sides have the same ratio and the included angles are equal
Example 1: Find the value of length of side AC in
△ABC,correct to two decimal places.
Solution: Prove that the triangles are similar.
B= B′ and 5 = 4
3 =4
6.25 5
3.75 5
k=4
5
(Two pairs of corresponding sides of equal ratio and the
included angles equal).
Use k to find x
x
=5
3.013
6.25
x = 3.013 x 5 = 2.4103 ≈2.41cm
6.25
Example 2:
Find the value of length of side AB in △ABC.
Solution: Prove that the triangles ACB and
AYX are similar.
A is a common angle
∧ ∧
C=Y
( corresponding angles)
∧ ∧
B=X
( corresponding angles)
Use the ratio to find x
x = 3
x + 6 3+ 2.5
x = 3
x+6 5.5
5.5x = 3(x + 6)
5.5x = 3x + 18
2.5x = 18
x = 18
2.5
x = 7.2cm
cross multiply
Volumes and surface areas
A prism is a solid which has a constant cross-section. Examples are cubes,
cylinders, rectangular prisms and triangular prisms
Cross section is the side of the prism that does not change.
volume of prism =area of cross-section × height (or length)
V=A×h
Example 1:Find the volume of the cylinder which has radius 3 cm and height 4cm,
correct to two decimal places.
Solution: The cross section is the circular part of the
cylinder
cross section = πr2
Volume of cylinder = πr2h =
π(3)24 = 113. 10 cm3
The formulas for determining the volumes of some ‘standard’ prisms are given here.
Volume of a pyramid
Pyramids are shapes where the outer surfaces are triangular
and converge at a point.The base of the pyramid can be
square,triangular...
Volume of pyramid = 1 × base area × perpendicular height
3
For the square pyramid shown: V = 1 x2 h
3
Example: Find the volume of this hexagonal pyramid with a base area of 40cm2
and a height of 20 cm.Give the answer correct to one decimal place.
Solution
V = 1 ×A×h = 1 ×40×20 =266.7cm3
3
3
Example: Find the volume of this square pyramid with a square base
with each edge 10 cm and a height of 27 cm.
Solution
V = 1 ×A×h = 1 ×10×10 x 27 = 900 cm3
3
3
Volume of cone
Cones are shapes where the outer surfaces are rounded and converge at a point.
The base of the cone is circular.
The formula for finding the volume of a cone can be stated as:
Volume of cone = 1 × base area × height
3
Volume of cone = 1 × π r2 × height
3
Volume of a sphere
The formula for the volume of a sphere is:
V = 4 π r3
where r is the radius of the sphere.
3
Example: Find the volume of the sphere with radius 4cm.
Solution
Volume of sphere = 4 π r3 = 4 × π × 43 = 268.08cm3
3
3
Composite shapes
Using the shapes above, new shapes can be made.The volumes of these can be found
by summing the volumes of the component solids.
Example: A hemisphere is placed on top of a cylinder to form a capsule. The radius of
both the hemisphere and the cylinder is 5 mm. The height of the cylinder is also 5 mm.
What is the volume of the composite solid in cubic millimeters, cor rect to two decimal
places?
Solution:
Volume of the composite = volume of cylinder+volume of hemisphere
= π r 2 h + 1 (4 π r3)
2 3
= π 52 × 5 + 1 (4 π 53)
2 3
= 654.50 mm3
Surface area of three-dimensional shapes
The surface area of a solid can be found by calculating and totalling the area of each of its
surfaces. The net of the cylinder in the diagram demonstrates how this can be done.
Here are some more formulas for the surface areas of some solids.
Example:Find the surface of the right square pyramid shown if the square base has each
edge 10 cm in length and the isosceles triangles each have height 15 cm.
Solution:
TSA = 4 × 1 10 × 15 + 10 × 10 = 300 + 100 =
400cm2
2
Lengths, Areas,Volumes and Similarity
Definition 1: When two shapes are similar the ratio of their sides is k.
where k = scale factor
Ratio of lengths = length 1 =3 = k
length 2 4
Definition 2: When two shapes are similar the ratio of their areas is k2.
Ratio of areas = area 1 = π× 32=(3)2 = 9 =k2
area 2 π× 42 (4)2 16
Definition 3: When two shapes are similar the ratio of their volumes is k3.
Ratio of volumes = of volume 1= 4/3π× 33 = (3)3 = 27 =k3
of volume 2 4/3π× 43 (4)3 64
Example 1:
The two triangles shown are similar. The base of the smaller triangle has a length
of 10 cm. Its area is 40 cm2.
The base of the larger triangle has a length of 25 cm. Determine its area.
Solution:
Ratio of lengths = length small = 10 = 2 = k
length big
25 5
k2 = 4
1
25
Ratio of areas = k2 = 40
2
x
1=2 and 4 = 40
25 x
x = 25 ×40 = 250 cm2
4
Example 2: The two cuboids shown are similar solids. The height of the larger cuboid is 6
cm. Its volume is 120 cm3. The height of the smaller cuboid is 1.5 cm. Determine its
volume.
Solution:
Ratio of lengths = length small = 1.5 = 1 = k
length big
6
4
k3 = 1
1
64
Ratio of volume= k3 = x
2
120
1=2 and 1 = x
64 120
x = 120 = 1.875 cm3
64
Trigonometric Ratios
Example 1: Find the value of x correct to two decimal places.
Solution : sin 29.6°= x
80
x = 80sin 29.6°
x = 39.52 cm
Example 2: Find the length of the hypotenuse correct to two decimal places.
Solution : cos 15°= 10
AB
AB = 10
cos 15°
x = 10.35 cm
Example 3: Find the magnitude of ∠ABC.
tan x° = 11
3
x° = tan-1 11
3
x°= 74.74°
When the triangles are not right-angled then the pythagoras theorem and the trigonometric
ratios are not appropriate for finding unknown lenghts and angles. The appropriate rules to use
are the sine rule and the cosine rule.
The sine rule
We use the sine rule when we are given:
1. Two sides and the non- included angle
2. One side and two angles
Considering that the lower- case letters represent the sides of the triangle and the
upper- case represent angles, then the following formula states the sine rule.
a = b = c
sinA sinB sinC
The cosine rule
We use the cosine rule when is given:
1. Two sides and the included angle of the
triangle.
2. All sides.
Example 1: Find the length of AB.
Solution: Two angles and a side -sine rule
b
= c
sin B
sinC
10 = c
sin 70° sin 31°
c = 10 sin 31° = 5.48cm
sin 70°
Example 2: For triangle ABC, find the length of AB in centimetres,
correct to two decimal places.
Solution: Two sides and the included angle - cosine rule
AB2 = 102 + 52- 2×5×10 cos67°
AB = √(100 + 25 - 100 cos67°)
AB = 9.27cm
Example 3: Find the magnitude of angle B.
Solution: Three sides- cosine rule
cosB = 62 + 122 - 152
2×6×12
B = cos-1(62 + 122 - 152)
2×6×12
B =cos-1(-45/144) = 108.21°
Area of the triangle
It is known that the area of a triangle is given by the
formula
Area = 1 × base length ×height
2
A = 1 bh
2
When the triangle is not a right angled and the 2 sides and the included angle is given
then the following formula can be used
A = 1 × side a × side b × sin (angle in between a and b)
2
Example 1: Find the area of triangle ABC. Give your answer correct to two decimal places.
Solution:
The triangle is not right angled and we are
given 2 sides and the included angle.
We can use the formula A =1/2 ac sinB
A= 1/2 × 6.5 × 7.2 × sin 140° = 15.04°
Heron’s Formula
When the 3 sides of the triangle are given, then Heron’s Formula can be used to
determine the area of the triangle:
Example 1: Find the area of the triangle with sides 6 cm, 4 cm and 4 cm. Give your answer
correct to two decimal places.
Solution:
s = (6+4+4) = 7
2
A = √7(7 - 6)(7 - 4)(7 - 4) = √ 7 × 1 × 3 × 3 = √63 = 7.94cm2
Angles of elevation and depression
The angle of elevation is the angle between the horizontal and a direction above the
horizontal.
The angle of depression is the angle between the horizontal and a direction below the
horizontal.
Example 1: The pilot of a helicopter flying at 400 m observes a small boat at an angle of
depression of 1.2◦ . Draw a diagram and calculate the horizontal distance of the boat to the
helicopter, correct to the nearest 10 metres.
Solution: The horizontal distance of the boat to the
helicopter is AB. The angle of depression H = B (
alternate).
tan(1.2◦) = 400
AB
AB = 400
tan(1.2◦)
AB = 19 095.80 m
Example 2: The light on a cliff-top lighthouse, known to be 75 m above sea level, is
observed from a boat at an angle of elevation of 7.1◦ . Draw a diagram and calculate the
distance of the boat from the lighthouse, to the nearest metre.
Solution: tan (7.1◦) = 75
AB
AB = 75
tan (7.1◦)
AB = 602 m
Example 3: From a point A, a man observes that the angle of elevation of the summit of a
hill is 10◦ . He then walks towards the hill for 500 m along flat ground. The summit of the hill
is now at an angle of elevation of 14◦ . Draw a diagram and find the height of the hill above
the level of A, to the nearest metre.
Solution: We need to find HC
To find HC we need to find HB
3 angles and a side are given- sine rule
HB
= 500
sin(10◦) sin(4◦)
HB = 500 sin(10◦) = 1244.67 m
sin(4◦)
sin(14◦) = HC
1244.67
HC = 1244.67 sin(14◦) = 301.11m
BEARINGS
Bearings are used to indicate direction.
The three-figure bearing the True Bearing is the direction measured clockwise from north and
starts from 0° to 360°.
For the Compass Bearing we seperate the plane in 4 sections of 90°.
Example:
A True Bearing is 30°.
Compass Bearing = N 30°E
B True Bearing = 150°
Compass Bearing = E 30°B
C True Bearing = 210°
Compass Bearing = S 30°W
D True Bearing = 330°
Compass Bearing = W60°N
Example: The road from town A runs due west for 14 km to town B. A television mast is
located due south of B at a distance of 23 km. Draw a diagram and calculate the distance of
the mast from the centre of town A, to the nearest kilometre. Find the bearing of the mast
from the centre of the town.
Solution: AT2 = 142 + 232
AT = √( 142 + 232 ) = 26.93 km
For Bearing find θ
tan θ = 23
14
θ = tan-1(23/14) = 58.67°
Bearing = 270°- 58.67°= 211.33°
The bearing of the mast from A is 211.33°T
Example: A yacht starts from a point A and sails on a bearing of 038◦ for 3000m. It then
alters its course to a bearing of 318◦, and after sailing for 3300 m it reaches a point B.
a Find the distance AB,correct to the nearest metre.
b Find the bearing of B from A, correct to the nearest degree.
Solution: a 2 sides and the included angle are givencosine rule
AB = √(33002 +30002 - 2 × 3300 × 3000 cos 100°)
AB = 4830 m
b To find the bearing of B from A we need to find angle A
We may use sine and cosine rule
3300 = 4830
sinA° sin100°
A°= sin-1 (3300sin100° ) = 42.28°
4830
The bearing of B from A = 360◦ −(42.29◦- 38◦ ) = 355.71◦
The bearing of B from A is 356◦T , to the nearest degree.
Example : Two points, A and B, are on opposite sides of a lake so that the distance between
them cannot be measured directly. A third point, C, is chosen at a distance of 100 m from A
and with angles BAC and BCA of 65◦ and 55◦ , respectively. Calculate the distance between A
and B, correct to two decimal places.
Solution: 2 angles and a sidesine rule
AB = 100
sin55° sin60°
AB = 100sin55°
sin60°
AB = 94.59m
Problems in three dimensions
Problems in three dimensions are solved by picking out triangles from a main figure and
finding lengths and angles through these triangles.
Example 1:ABCDEFGH is a cuboid. Find:
a distance DB
b distance HB
c the magnitude of angle HBD
d distance HA
e the magnitude of angle HBA
Solution:
a.By selecting the appropriate triangle from the
diagram and by using the Pythagoras
Theorem:
DB2 = 82 + 102
DB = √(82 + 102) = √164 = 12.81 cm
b. Using the information from a
HB2 = 72 + √1642
HB = √(49 + 164)
HB = √213 = 14.59 cm
c. tanθ = 7
√164
θ = tan-1 7
√164
θ = 28.66◦
d. Use the triangle HAD
HA2 = 82 + 72
HA = √(82 + 72 ) = √113 cm
e. 3 sides-cosine rule
cosB = 102 +√2132 -√1132
2×10×√213
cosB =0.68518
B =46.75◦
Example 2: The diagram shows a pyramid with a square base.The base has sides 6 cm
long and the edges VA, VB,VC,VD are each10 cm long.
a Find the length of DB.
b Find the length of BE.
c Find the length of VE.
d Find the magnitude of angle VBE.
Give all answers correct to two decimal places.
Solution:
a DB2 = 62 + 62
DB = √( 62 + 62 )
DB = 8.49cm
b BE = 8.49/2 = 4.24 cm
c VE = √( 102 - 4.242) = 9.06 cm
d sin (VBE) = 9.06
10
VBE = sin-1(9.06/10) = 64.90°
Example 3: A communications mast is erected at the corner A of a rectangular courtyard
ABCD whose sides measure 60 m and 45 m. If the angle of elevation of the top of the mast
from C is 12◦ , find:
a the height of the mast
b the angle of elevation of the top of the mast from B (where AB = 45 m)
Give answers cor rect to two decimal places.
Solution :
a We need to find AC first.
AC = √(452 + 602) = 75m
Then we can use AC to find the height of the mast.
tan(12°) = HA
75
HA = 75tan (12°) = 15.94m
b tanθ = 15.94
45
θ = tan-1( 15.94/45) = 19.51◦
Contour maps
The diagram above is called contour map.
The lines are called contour lines and they represent different hights above sea
level.
The map gives the horizontal scaled distance between the lines and not the actual
distance.
The real/cross-sectional representation of the contour map above is as follows:
To find the distance between B and C, first determine from the diagram the
horizontal distance B′C′. Suppose this distance is 80 m.Then triangle BCH
in the first diagram can be used to find the distance between B and C and
the average slope between B and C.
A cross-sectional profile can be drawn from a contour map for a given cross-section AB.
This is illustrated below. The horizontal distance that is represented on the crosssectional profile is the real distance.The distance AB on the contour map is 6cm
then the distance on the cross-sectional profile will be 600cm which is 6m.
scale 1:100
CONVERSION OF UNITS
When we are converting from smaller units to larger we divide
When we are converting from larger to smaller units we
multiply
Example 1: Convert 10cm to m
Answer: There are 100 cm in a m
10cm /100 = 0.1m
Conversion of units of length
a 1 cm (mm) = 1 × 10 = 10 mm
b 1 m (cm) = 1 × 100 = 100 cm
c 1 km (m) = 1 × 1000 = 1000km
d 1 mm(cm) = 1/10 = 0.1 cm
e 1 cm ( m) = 1/100 = 0.01 m
f 1m (km) = 1/1000 = 0.001 km
Conversion of units of area
1 cm2 (mm2) = 1 × 102 = 100 mm2
1 m2 (cm2) = 1 × 1002 = 10000 cm2 = 104 cm2
1 km2 (m2) = 1 × 10002 = 1000000km2 = 106 km2
1 mm2(cm2) = 1/102 = 0.01 cm2 = 10-2 cm2
1 cm2 (m2) = 1/1002 = 0.0001 m2 = 10-4 m2
1m2 (km2) = 1/10002 = 0.000001 km2= 10-6
Example:
a Convert 156000 m2 to km2
Answer: 156000 m2 / 10002 = 156000/1000000 = 0.156 km2
b Convert 20000mm2 to cm2
Answer: 20000mm2 /102 = 20000/100 = 200 cm2
Conversion of units of volume
1000mm3 (cm3) = 1000/103 = 1000/1000 = 1 cm3
1000000cm3(m3) = 1000000/1003 = 1000000/1000000 = 1 m3
1 litre = 1000cm3
1000 litres = 1m3