geometry [complete]

Further Bound Book 2015 - Rahil Chadha

1

Module: Geometry

Chapter 12

Exercise 12A

Properties of parallel lines

y = 180 − 50 = 130° (Supplementary angle)

x = 50° (Corresponding angle with 50)

z = 130° (Corresponding angle with y)

2x − 40 = x + 40

x = 80° (Angles are corresponding thus must be the same)

y = 180 − 40 = 140° (cointerior angle)

x = 40° (supplementary angle to y)

z = 40° (vertically opposite angle to x or alternate angle to 40°)

Terminology has l ittle relevance, you just need to know how to find out the angles using properties of

parallel l ines. This takes practice

Remember a straight l ine needs to add up to 180°

[ 2010]


2

Exercise 12B

Properties of triangles

Types of triangles

Equilateral triangle: where all side lengths of the triangle are equal, AB = BC = CA

Angles of an equilateral triangle are all of magnitude 60°

Isosceles triangle: where two side lengths are the same in a triangle. Angles opposite each of the equal

sides are equal

Right-angled triangle: where a triangle has an angle of 90°

Triangle properties

Sum of interior angles for a triangle will equal 180°: a°+b°+c° = 180°

Exterior angle is equal to sum of two opposite interior angles a°+b° = d°

Bisector of each angle in an equilateral triangle meets opposite side at

right angles and passed through midpoint of that side

Sum of magnitudes of exterior angles of a triangle is equal to 360°,

e° + d° + f° = 360°

Example

a = 75° due to isosceles triangle property

b = 180 − 75 − 75 = 30° because all angles in a triangle add up to 180°

To find angle c we need to figure out what the other angles in the triangle are; 180 − 75 = 105°

c = 180 − 40 − 105 = 35°

Isosceles property,hence y = 40°

z = 180 − 40 − 40 = 100° x = 40° because angle is parallel (see 12A) y + w + x = 180, but we already have x and y

thus, w = 180 − 40 − 40 = 100°

Simply means if you cut a equilateral

triangle in half it will form two right

angled triangles


3

[2011]

[ 2008]

Exercise 12C

Properties of Regular Polygons

A Regular polygon has all sides

of equal length and all angles

are of equal magnitude

Properties of regular polygons:

Polygons with ‘n’ sides can be divided into ‘n’ triangles

Magnitude of each of the interior angles is given by

x° =180n − 360

n

Sum of interior angles is given by formula

S = (180n − 360)°

The 360° rule:

The angles of each triangle must be the same

A full circle equates to 360°, hence if all triangles have the same angle: 360

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠= 𝐴𝑛𝑔𝑙𝑒

An octagon has 8 sides and

thus can be divided into 8

equal triangles

This is an interior angle


4

ABCDE is a regular pentagon

a) Find the value of

i. x ii. y

b) Find the sum of the interior angles of the regular pentagon ABCDE

a) i. 360

5= 72°

ii. 180 − 72 = 108 → 108

2= 54° (Triangle is isoceles)

b) 180(5) – 360 = 540°

The angle sum of a regular polygon is 1260°. How many sides does the polygon have?

1260=180𝑛−360→180𝑛=1620, 𝑛=9 ∴ 9 𝑠𝑖𝑑𝑒𝑠

If the magnitude of each angle of a regular polygon is 135°, how many sides does the polygon have?

(180𝑛−360)

𝑛= 135 → 𝑛 = 8 ∴ 8 𝑠𝑖𝑑𝑒𝑠


5

Exercise 12D

Pythagoras’ Theorem

𝑎2 + 𝑏2 = 𝑐2

Only applies for right-angled triangles

Example. Find the height of the kite

902 + (XY)2 = 1702

(XY)2 = 1702 − 902

(XY)2 = 20800

XY = √20800 = 144.22m

In a circle of centre O, a chord AB is of length 4cm. The radius of the circle is 14cm. Find the distance of the cord

from O.

22 + x 2 = 142

x 2 = 192

x = √192 = 13.86cm

Find the area of a square with a diagonal of length 8√2cm

x 2 + x 2 = (8 √2)2

2x 2 = 128

x 2 = 64 → x = 8cm

Area of Square = 82 = 64cm2

Remember any distance

from the centre to the

outside of the circle is

just the radius length

x

x

8√2


6

[ 2008]

[ 2007]

[ 2007]


7

Exercise 12E

Similar figures

Any square is similar to another

Any circle is similar to another

Rectangles are similar only if their corresponding sides are the same ratio

Triangles are similar if

Corresponding angles in triangles are equal

Corresponding sides are in same ratio

We refer to the scale factor commonly as ‘k’

How to apply similarity:

Always split up your triangles to make finding ratios and lengths easier

Denote unknown sides as ‘x’

Solve using ratios and/or scale factor

Given that AD = 14, ED = 12, BC = 15 and EB =4, find AC, AE and AB

15

12=

y

14→ 17.5 therefore AC = 17.5

x

x+4=

12

15→ 12x + 48 = 15x → 3x = 48 → x = 16, hence AE = 16 and AB = 20

A spotlight is at a height of 0.6m above ground level. A vertical post

1.1m high stands 3m away, and 5m further away there is a vertical

wall. How high up the wall does the shadow reach?

h

0.5=

8

3→ 3h = 4 → h = 1.33m

However we need to still account from the extra 0.6 m therefore 1.93m

Two similar rods 1.3m long have to be hinged together to support a table 1.5m wide. The

rods have been fixed to the floor 0.8m apart. Find the position of the hinge by finding the

value of x

Make sure you set triangles up in correct orientation (angles match up)

x

1.3−x=

1.5

0.8→ 1.5(1.3 − x) = 0.8x → 2.3x = 1.95 → x = 0.848m

14 12

15

x x+4

y

8 3

0.5 h

1.5 0.8

1.3-x x


8

Note: you always have two scale factors, one going UP and one going DOWN

To find the corresponding scale factor, simply inverse:

SF = 3

3−1 =1

3

To avoid confusion, simply use fractions

The horiverti rule

Often in difficult questions, you will be required to form your own similar triangles from the information provided.

Students tend to get the triangles wrong and hence their answers wrong. Since there are a variety of properties that

must be satisfied for two shapes to be ‘similar’ use this rule (see example 2 above)

1. Any right angled triangle cut horizontally will form two similar triangles (hori)

2. Any right angled triangle cut vertically will form two similar triangles (verti)

[ 2009]

[ 2006]

Of course this isn’t restricted to horizontal and vertical

lines. This works as long as the cut is parallel to one of

the sides of the triangle (see question 7 below)


9


10

Exercise 12F

Volume and Surface Areas

Volume of a Cylinder

𝑉 = 𝜋𝑟2ℎ Surface Area of a Cylinder

𝑆 = 2𝜋𝑟2 + 2𝜋𝑟ℎ

Volume of a Cube

𝑉 = 𝑥3

Surface Area of a Cube

𝑆 = 6𝑥2

Volume of a Rectangular Prism

𝑉 = 𝑙𝑤ℎ

Surface Area of a Rectangular Prism

𝑆 = 2(𝑙𝑤 + 𝑙ℎ + 𝑤ℎ)

Volume of a Triangular Prism

𝑉 =1

2𝑏ℎ𝑙

Surface Area of a Triangular Prism

𝑆 = 𝑏ℎ + 𝑏𝑙 + ℎ𝑙 + 𝑙√𝑏2 + ℎ2

Volume of a Pyramid

1

3× 𝑏𝑎𝑠𝑒 𝑎𝑟𝑒𝑎 × 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ℎ𝑒𝑖𝑔ℎ𝑡

1

3𝑥2ℎ

1

3𝜋𝑟2ℎ

Volume & Surface Area of a Sphere

𝑉 =4

3𝜋𝑟3 𝑆 = 4𝜋𝑟2

Reminders

Area of a circle: 𝜋𝑟2

Perimeter of a circle: 2𝜋𝑟

Area of a trapezium: 𝑎+𝑏

2ℎ

Area of a triangle: 1

2𝑏ℎ


11

Composite shapes

Solids made up of multiple shapes

Here make sure you don’t add an extra side area when calculating

total surface area

Example. Find the volume and surface area of each of the solids shown below:

Volume (Rectangular Prism) = 10(base) × 5(height) × 20(width) = 1000cm3

Volume (Cylinder) = π(5)2 × 20 = 1570.79cm3

However this is only half a cylinder, hence → 785.4cm3

Total Volume = 1785.4cm3

Surface Area (Rectangular Prism) = 2(10 × 5) ∗ 2 squares ∗ +2(5 × 20) ∗

2 rectangles ∗ +(10 × 20) ∗ base rectangle ∗ = 500cm2

Surface Area (Cylinder) = 2π(5) 2 + 2π(5)(20) = 785.4cm2

However this is only half a cylinder, hence → 392.7cm2

Total Surface Area = 892.7cm2

Volume (Rectangular Prism) = 4(base) × 4(height) × 10(width) = 160cm3

Volume (Triangular Prism) = (3 × 4 ×1

2) ∗ area of triangle ∗× 10 = 60cm3

Total Volume = 240cm3

Surface Area (Rectangular Prism) = 2(4 × 4) ∗ 2 squares ∗ +2(4 × 10) ∗

2 rectangles ∗ +(4 × 10) ∗ base rectangle ∗ = 152cm2

Surface Area (Triangular Prism) = 6 × 2 ∗ 2 triangles ∗ +10 × 3 ∗ side rectangle ∗

+5 pythagoras theorem used to find this length × 10 ∗ roof rectangle ∗= 92cm2

Total Surface Area = 244cm2

Volume (Square Based Pyramid) =1

3× 2752(base area) ×

175(perpendicular height) = 4411458.33m3

Surface Area (Square Based Pyramid) = 2752 + 4(Triangle Area)

Height of triangle: 137.52 + 1752 = h2 → h = 222.556m

Triangle Area =1

2× 275 × 222.556 = 30601.473m2

Surface Area = 2752 + 4(30601.473) = 198030.9m2

[2008]


12

[ 2008]

[ 2006]

[ 2007]

[ 2007]


13

[ 2008]

[ 2009]

[ 2009]

[ 2010]

[ 2011]


14

[ 2010]


15

Exercise 12G Similarity for Volume & Area

We can derive area and volume of two similar shapes using their length scale factor ‘k’

Scaling areas:

𝑆ℎ𝑎𝑝𝑒 𝐴𝑟𝑒𝑎 = 𝑘2 × 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑠ℎ𝑎𝑝𝑒

Scaling volumes:

𝑆ℎ𝑎𝑝𝑒 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝑘3 × 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑠ℎ𝑎𝑝𝑒

Triangles XBY and ABC are similar. The area of the triangle XBY is 1.8cm2. Determine the area of triangle ABC

Scale factor =7

3= 2.333

Area = (7

3)

2× 1.8 = 9.8cm2

The cones shown are similar. The smaller cone has a diameter of 10cm. The

larger cone has a diameter of 30cm

a) Determine the length scale factor k for scaling up?

b) What is the length scale factor k for scaling down?

c) The height of the larger cone is 45cm. Determine the height of the smaller cone

d) The surface area of the smaller cone is 326.9cm2. Determine the surface area of the

larger cone correct to the nearest cm2

e) The volume of the smaller cone is 392.7cm3. What is the volume of the larger cone

correct to the nearest cm3?

a) Scale factor =30

10= 3

b) Scale factor =10

30=

1

3

c) Height = 45 ×1

3= 15cm

d) Surface Area = 326.9 × 32 = 2942.1cm2 → 2942cm2

e) Volume = 392.7 × 33 = 10602.9cm3 = 10603cm3

An inverted right circular cone of capacity 100m3 is filled up with water to half its

depth. Find the volume of water.

Scale factor =1

2 (Half it′s depth)

(1

2)

3× 100 = 12.5m3

[ 2010]


16

Chapter 13

Exercise 13A

Sine, Cosine and Tangent

Remember SOH, CAH and TOA (Let’s assume x degree)

Also ensure CAS is set on degree, rather than radian

sin(x) =O

A

cos(x) =A

H

tan(x) =O

A or

sin(x)

cos (x)

This is for right-angled triangles only

Right angled triangles are denoted by a square angle

A pendulum swings from the vertical through an angle of 15° of each side

of the vertical. If the pendulum is 90cm long, what is the distance, x cm,

between its highest and lowest points?

cos (15) =A

90→ A = 90cos(15) = 86.93cm

90 − 86.93 = 3.07cm

[ 2011]

O = Opposite (To the angle)

H = Hypotenuse (Longest side)

A = Adjacent (Side not opposite to angle)


17

[ 2006]

[ 2007]

[ 2007]

[ 2006]


18

Exercise 13B

Sine rule

a

sin (A)=

b

sin (B)=

c

sin(C)

When to use sine rule:

Not a right-angled triangle

One side and two angles given or two sides and a non-included angle are given

Labelling convention (stick to it)

Example. Find θ

b

sin(B)=

a

sin(A)→

8

sin(B)=

10

sin (108)

sin(B) =8×sin(108)

10 = 0.760845213 → B = sin−1(0.760845213) = 49.5°

[ 2006]

[2008]

a = 10, b = 8, c = AB

A = 108° B = ___ C = ___

we need to find B


19

[ 2011]


20

Exercise 13C

Cosine Rule

a2 = b2 + c2 − 2bc cos (A)

cos(A) =b2 + c2 − a2

2bc

b2 = a2 + c2 − 2ac cos (B)

cos(B) =a2 + c2 − b2

2ac

c2 = a2 + b2 − 2ab cos (C)

cos(C) =a2 + b2 − c2

2ab

When to use cosine rule:

Not a right-angled triangle

Two sides and an included angle are given or three sides are given

Example. Find AB

c2 = a2 + b2 − 2ab cos (C)

c2 = 42 + 62 − 2(4)(6) cos(20)

c2 = 6.894754202

c = √6.894752402 = 2.6km

[ 2011]

Simple rearrangement of all

formulas

a = 4, b = 6, c = AB

A = ___ B = ___ C = 20°

we need to find c or AB


21

Exercise 13D

Area of a triangle

𝐴 =1

2𝑏 × ℎ

𝐴 =1

2𝑏 × 𝑐 × sin (𝐴)

𝐴 =1

2𝑎 × 𝑐 × sin(𝐵)

𝐴 =1

2𝑎 × 𝑏 × sin (𝐶)

Example. Find Area

A =1

2× n × l × sin (M)

A =1

2× 8.2 × 3.5 × sin (130)

A = 11cm2

When we have three sides we use Heron’s formula:

𝐴 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)

Where 𝑠 =𝑎+𝑏+𝑐

2

S =11+10+2

2= 11.5

A = √11.5(11.5 − 11)(11.5 − 10)(11.5 − 2)

A = 9.05cm2

m = ___, n = 8.2, l = 3.5

M = 130° N = ___, L = ___

We need the area

a = 11, b = 10, c = 2

Area uses the same labelling

convention as sine and cosine

rules. So ‘b’ here does NOT stand

for base as in the first formula


22

In a triangle XYZ, XY = 9 cm and YZ = 13 cm. The area of the triangle is 31cm2. Find two possible values for the

magnitude of XYZ.

A =1

2x × z × sin (Y)

31 =1

2× 13 × 9 × sin (Y)

sin(Y) =31

58.5

Y = sin−1(31

58.5) = 32°

sin(32) = sin(180 − 32) = sin(148) thus 148° is also another possible answer

[ 2006]

[ 2008]

[ 2009]

What you need to know about the unit circle:

sin (x) = sin (180 – x) sin (45) will also equal sin (135)

cos (x) = -cos (180 – x) cos (45) will also equal –cos (135)

this is useful for questions which have more than one answer


23

Chapter 14

Exercise 14A Elevation and depression

Another way to describe where a particular angle l ies in a triangle

A man standing on top of a mountain observes that the angle of depression to the foot of the building is 41°. If

the height of the man above the foot of the building is 500m, find the horizontal distance from the man to the

building

tan(49) =O

500→ O = 500 tan(49) = 575.18m

A tower 110m high stands on top of a hill. From a point A at the foot of the hill the angle of elevation of the

bottom of the tower is 7°, and that of the top is 10°

a) Find the magnitude of the angles TAB, ABT and ATB

b) Use the sine rule to find the length of AB

c) Find CB, the height of the hill

a) ABC = 83° therefore ABT = 97°

ATB = 80°

TAB = 3°

b) 110

sin(3)=

AB

sin(80)→ AB = 2069.87m

c) sin(7) =CB

2069.9→ CB = 252.26m

41°

500m

49° Building Angle of

depression


24

[ 2011]

[ 2009]


25

Bearings

Direction (angle) measured clockwise from north. A full circle is 360°

A ship leaves port A and steams 15km due east. It then turns and steams for 22km due north

a) What is the bearing of the ship from A?

b) What is the bearing of port A from the ship?

a) tan(x) =22

15→ x = tan−1(

22

15) = 55.71°

90 − 55.71 = 034° Bearing

b) 180 + 34.29 = 214° Bearing

The bearing of C from A is 035°

The bearing of B from A is 346°

The distance of C from A is 340km

The distance of B from A is 160km

a) Find the magnitude of angle BAC

b) Use the cosine rule to find the distance of B to C

a) BAC → 35 + 360 − 346 = 49°

b) a2 = 3402 + 1602 − 2(160)(340) cos(49)

a2 = 69820.77765

a = √69820.77765 = 264km ∴ BC = 264km

15km

22km

Port A


26

[ 2007]

[ 2008]

[ 2008]


27

[ 2010]

[ 2006]

Triangulation

Measuring distances to inaccessible points using trigonometry

Two points, A and B, are 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

300m from A and with angles BAC and BCA of 80° and 53°, respectively.

Calculate the distance between A and B

ABC = 180 − 80 − 53 = 47°

c

sin(33)=

300

sin(47)→ Distance between A and B is 227.6m


28

Mixed problems

A yacht starts from point A and sails on a bearing of 035° for 2000m. It then alters its course to one in a

direction with a bearing of 320° and after sailing for 2500m it reaches point B

a) Find the distance AB

b) Find the bearing of B from A

320 − 180 − 35 (use parallel line property) = 105° (AB)2 = 2.52 + 22 − 2(2)(2.5) cos(105) → 12.8381

AB = √12.8381 → 3.583km

3.583

sin(105)=

2.5

sin(A)→ 3.583sin(A) = 2.5 sin(105)

sin(A) =2.5sin(105)

3.583→ A = sin−1(

2.5sin(105)

3.583) → 42.4°

42.4 − 35 = 7.4°

360 − 7.4 = 353°T

A man standing on top of a cliff 50m high is in line with two buoys whose angles of depression are 18° and 20°.

Calculate the distance between the buoys

tan(70) =O

50→ O = 50 tan(70) = 137.3738m

tan(72) =O

50→ O = 50 tan(72) = 153.8841m

Distance = 153.8841 − 137.3738 = 16.5m

2km

2.5km

B

A

35°

320°

Cliff

50m

18° 20°


29

Exercise 14B

Problems in three dimensions

To solve these you need to pick out triangles from a main figure and find the unknown lengths or angles through

trigonometric principles from the previous exercises

VABCD is a right pyramid with a square base. The sides of the base are 8cm in length. The height, VF of pyramid

is 12cm

a) Find EF

b) Find the magnitude of angle VEF

c) Find the length of VE

d) Find the length of a sloping edge

e) Find the magnitude of the angle VAD

f) Find the surface area of the pyramid

a) EF =8

4= 2cm

b)

tan(x) =12

4

x = tan−1(3)

x = 71.57°

c) a2 + b2 = c 2

42 + 122 = (VE)2

(VE)2 = 160 VE = √160 = 12.65cm

d)

42 + 12.652 = (VA)2

(VA)2 = 176.0225

VA = √176.0225 = 13.27cm

e) VAE has same magnitude as VAD

tan(x) =12.65

4= 3.1625

x = tan−1(3.1625) = 72.45°

12

4

x

V

E F

V

E

12.65

A 4

x

f) Surface area = 4 × Area of side triangle + Area of Square

Square = 82 = 64cm2

Triangle =1

2× 8 × 12.65 = 50.6cm2

SA = 64 + 50.6 (4) = 266.4cm2


30


31

Exercise 14C

Contours

Contour maps are a 2D image maps for 3D figures (hills, mountains, etc) used in the geographical field

They can tell us:

The vertical height between two different points

A basic image of what the 3D image would look like

This is known as a cross-sectional diagram

To draw one you must label the vertical axis as vertical

height and draw the image in line with the contours as

shown. This is an artistic component of mathematics

Pay attention to where the black l ine hits the contours

Draw the cross-sectional diagram for the following contour map


32

For this diagram the horizontal distance from A to B is 400m. Find:

a) i. The distance from A to B

ii. The angle of elevation of B from A

The horizontal distance from B to C is 1km. Find

b) i. The distance of B from C

ii. The angle of elevation of B from C

a) B is 150 tall and A is 50m tall. Hence the vertical distance between

them is 100m

(𝐴𝐵)2 = 4002 + 1002 → 170000

𝐴𝐵 = √170000 = 412.31𝑚

Angle of elevation → tan(x) =100

400→ x = 14.03°

B is 150m tall and C is 50m tall. Hence the vertical distance between them is 100m

(𝐶𝐵)2 = 10002 + 1002 → 1010000

𝐶𝐵 = √1010000 = 1005𝑚

Angle of elevation → tan(x) =100

1000→ x = 5.71°

100m

400m A

B

x

100m

1000m

x

B

C


33

[ 2006]

[ 2007]


34

[ 2009]


35

Conversions

Converting between liquid volume and cubic volume:

1ml → 1cm3 , therefore 1L = 1000cm3

Degrees, minutes and seconds

Degrees: °

Minutes: ′ 1

60th of a degree

Seconds:′′1

60th of a minute

Standard conversion Square conversion

Cubic conversion

Liquid volume conversion


36

Extra questions

[ 2007]

[ 2009]

[ 2010]


37

[ 2010]

[ 2010]


38

[ 2011]

[ 2010]

geometry [complete]

Documents