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## Introduction

### Extension : History of Geometry

Work in pairs or groups and investigate the history of the development of geometry in the last 1500 years. Describe the various stages of development and how different cultures used geometry to improve their lives.

The works of the following people or cultures should be investigated:

1. Islamic geometry (c. 700 - 1500)
1. Thabit ibn Qurra
2. Omar Khayyam
3. Sharafeddin Tusi
2. Geometry in the 17th - 20th centuries (c. 700 - 1500)

## Right Pyramids, Right Cones and Spheres

A pyramid is a geometric solid that has a polygon base and the base is joined to a point, called the apex. Two examples of pyramids are shown in the left-most and centre figures in Figure 1. The right-most figure has an apex which is joined to a circular base and this type of geometric solid is called a cone. Cones are similar to pyramids except that their bases are circles instead of polygons.

Surface Area of a Pyramid

Figure 2
Khan academy video on solid geometry volumes

The surface area of a pyramid is calculated by adding the area of each face together.

### Exercise 1: Surface Area

If a cone has a height of hh and a base of radius rr, show that the surface area is πr2+πrr2+h2πr2+πrr2+h2.

#### Solution

1. Step 1. Draw a picture :

2. Step 2. Identify the faces that make up the cone :

The cone has two faces: the base and the walls. The base is a circle of radius rr and the walls can be opened out to a sector of a circle.

This curved surface can be cut into many thin triangles with height close to aa (aa is called the slant height). The area of these triangles will add up to 12×12×base××height(of a small triangle) which is 12×2πr×a=πra12×2πr×a=πra

3. Step 3. Calculate aa :

aa can be calculated by using the Theorem of Pythagoras. Therefore:

a = r 2 + h 2 a = r 2 + h 2
(1)
4. Step 4. Calculate the area of the circular base :
A b = π r 2 A b = π r 2
(2)
5. Step 5. Calculate the area of the curved walls :
A w = π r a = π r r 2 + h 2 A w = π r a = π r r 2 + h 2
(3)
6. Step 6. Calculate surface area A :
A = A b + A w = π r 2 + π r r 2 + h 2 A = A b + A w = π r 2 + π r r 2 + h 2
(4)

Volume of a Pyramid: The volume of a pyramid is found by:

V = 1 3 A · h V = 1 3 A · h
(5)

where AA is the area of the base and hh is the height.

A cone is like a pyramid, so the volume of a cone is given by:

V = 1 3 π r 2 h . V = 1 3 π r 2 h .
(6)

A square pyramid has volume

V = 1 3 a 2 h V = 1 3 a 2 h
(7)

where aa is the side length of the square base.

### Exercise 2: Volume of a Pyramid

What is the volume of a square pyramid, 3cm high with a side length of 2cm?

#### Solution

1. Step 1. Determine the correct formula :

The volume of a pyramid is

V = 1 3 A · h V = 1 3 A · h
(8)

where AA is the area of the base and hh is the height of the pyramid. For a square base this means

V = 1 3 a · a · h V = 1 3 a · a · h
(9)

where aa is the length of the side of the square base.

2. Step 2. Substitute the given values :
= 1 3 · 2 · 2 · 3 = 1 3 · 12 = 4 cm 3 = 1 3 · 2 · 2 · 3 = 1 3 · 12 = 4 cm 3
(10)

We accept the following formulae for volume and surface area of a sphere (ball).

Surface area = 4 π r 2 Volume = 4 3 π r 3 Surface area = 4 π r 2 Volume = 4 3 π r 3
(11)

### Surface Area and Volume

1. Calculate the volumes and surface areas of the following solids: *Hint for (e): find the perpendicular height using Pythagoras. Click here for the solution
2. Water covers approximately 71% of the Earth's surface. Taking the radius of the Earth to be 6378 km, what is the total area of land (area not covered by water)?
3. A triangular pyramid is placed on top of a triangular prism. The prism has an equilateral triangle of side length 20 cm as a base, and has a height of 42 cm. The pyramid has a height of 12 cm.
1. Find the total volume of the object.
2. Find the area of each face of the pyramid.
3. Find the total surface area of the object.

## Similarity of Polygons

In order for two polygons to be similar the following must be true:

1. All corresponding angles must be congruent.
2. All corresponding sides must be in the same proportion to each other. Refer to the picture below: this means that the ratio of side AEAE on the large polygon to the side PTPT on the small polygon must be the same as the ratio of side ABAB to side PQPQ, BC/QRBC/QR etc. for all the sides.

If
1. A^=P^A^=P^; B^=Q^B^=Q^; C^=R^C^=R^; D^=S^D^=S^; E^=T^E^=T^ and
2. ABPQ=BCQR=CDRS=DEST=EATPABPQ=BCQR=CDRS=DEST=EATP
then the polygons ABCDE and PQRST are similar.

### Exercise 3: Similarity of Polygons

Polygons PQTU and PRSU are similar. Find the value of xx.

#### Solution

1. Step 1. Identify corresponding sides :

Since the polygons are similar,

P Q P R = T U S U x x + ( 3 - x ) = 3 4 x 3 = 3 4 x = 9 4 P Q P R = T U S U x x + ( 3 - x ) = 3 4 x 3 = 3 4 x = 9 4
(12)

## Triangle Geometry

### Proportion

Two line segments are divided in the same proportion if the ratios between their parts are equal.

A B B C = x y = k x k y = D E E F A B B C = x y = k x k y = D E E F
(13)
the line segments are in the same proportion the line segments are in the same proportion
(14)

If the line segments are proportional, the following also hold

1. C B A C = F E D F C B A C = F E D F
2. A C · F E = C B · D F A C · F E = C B · D F
3. ABBC=DEFEABBC=DEFE and BCAB=FEDEBCAB=FEDE
4. ABAC=DEDFABAC=DEDF and ACAB=DFDEACAB=DFDE

#### Proportionality of triangles

Triangles with equal heights have areas which are in the same proportion to each other as the bases of the triangles.

h 1 = h 2 area A B C area D E F = 1 2 B C × h 1 1 2 E F × h 2 = B C E F h 1 = h 2 area A B C area D E F = 1 2 B C × h 1 1 2 E F × h 2 = B C E F
(15)

• A special case of this happens when the bases of the triangles are equal: Triangles with equal bases between the same parallel lines have the same area.
(16)
• Triangles on the same side of the same base, with equal areas, lie between parallel lines.
IfareaABC=areaBDC,IfareaABC=areaBDC,
(17)
(18)

Theorem 1 Proportion Theorem: A line drawn parallel to one side of a triangle divides the other two sides proportionally.

Given:ABC with line DE BC

R.T.P.:

A D D B = A E E C A D D B = A E E C
(19)

Proof: Draw h1h1 from E perpendicular to AD, and h2h2 from D perpendicular to AE.

Draw BE and CD.

area ADE area BDE = 1 2 A D · h 1 1 2 D B · h 1 = A D D B area ADE area CED = 1 2 A E · h 2 1 2 E C · h 2 = A E E C but area BDE = area CED (equal base and height) area ADE area BDE = area ADE area CED A D D B = A E E C DE divides AB and AC proportionally. area ADE area BDE = 1 2 A D · h 1 1 2 D B · h 1 = A D D B area ADE area CED = 1 2 A E · h 2 1 2 E C · h 2 = A E E C but area BDE = area CED (equal base and height) area ADE area BDE = area ADE area CED A D D B = A E E C DE divides AB and AC proportionally.
(20)

Similarly,

A D A B = A E A C A B B D = A C C E A D A B = A E A C A B B D = A C C E
(21)

Following from Theorem "Proportion", we can prove the midpoint theorem.

Theorem 2 Midpoint Theorem: A line joining the midpoints of two sides of a triangle is parallel to the third side and equal to half the length of the third side.

Proof: This is a special case of the Proportionality Theorem (Theorem "Proportion"). If AB = BD and AC = AE, and AD = AB + BD = 2AB AE = AC + CB = 2AC then DE BC and BC = 2DE.

Theorem 3 Similarity Theorem 1: Equiangular triangles have their sides in proportion and are therefore similar.

Given:ABC and DEF with A^=D^A^=D^; B^=E^B^=E^; C^=F^C^=F^

R.T.P.:

A B D E = A C D F A B D E = A C D F
(22)

Construct: G on AB, so that AG = DE, H on AC, so that AH = DF

Proof: In 's AGH and DEF

AG = DE (const.) AH = D ( const. ) A ^ = D ^ ( given ) AGH DEF ( SAS ) A G ^ H = E ^ = B ^ G H BC ( corres. 's equal ) AG AB = A H A C ( proportion theorem ) DE AB = D F A C ( AG = DE ; AH = DF ) ABC | | | DEF AG = DE (const.) AH = D ( const. ) A ^ = D ^ ( given ) AGH DEF ( SAS ) A G ^ H = E ^ = B ^ G H BC ( corres. 's equal ) AG AB = A H A C ( proportion theorem ) DE AB = D F A C ( AG = DE ; AH = DF ) ABC | | | DEF
(23)
##### Tip:
|||||| means “is similar to"

Theorem 4 Similarity Theorem 2: Triangles with sides in proportion are equiangular and therefore similar.

Given:ABC with line DE such that

A D D B = A E E C A D D B = A E E C
(24)

Proof: Draw h1h1 from E perpendicular to AD, and h2h2 from D perpendicular to AE.

Draw BE and CD.

area ADE area BDE = 1 2 A D · h 1 1 2 D B · h 1 = A D D B area ADE area CED = 1 2 A E · h 2 1 2 E C · h 2 = A E E C but A D D B = A E E C (given) area ADE area BDE = area ADE area CED area BDE = area CED D E B C (same side of equal base DE, same area) A D ^ E = A B ^ C (corres 's) and A E ^ D = A C ^ B area ADE area BDE = 1 2 A D · h 1 1 2 D B · h 1 = A D D B area ADE area CED = 1 2 A E · h 2 1 2 E C · h 2 = A E E C but A D D B = A E E C (given) area ADE area BDE = area ADE area CED area BDE = area CED D E B C (same side of equal base DE, same area) A D ^ E = A B ^ C (corres 's) and A E ^ D = A C ^ B
(25)
(26)
A D E | | | A B C (AAA) A D E | | | A B C (AAA)
(27)

Theorem 5 Pythagoras' Theorem: The square on the hypotenuse of a right angled triangle is equal to the sum of the squares on the other two sides.

Given: ABC with A^=90A^=90

Required to prove:BC2=AB2+AC2BC2=AB2+AC2

Proof:

Let C ^ = x D A ^ C = 90 - x ( 's of a ) D A ^ B = x A B ^ D = 90 - x ( 's of a ) B D ^ A = C D ^ A = A ^ = 90 Let C ^ = x D A ^ C = 90 - x ( 's of a ) D A ^ B = x A B ^ D = 90 - x ( 's of a ) B D ^ A = C D ^ A = A ^ = 90
(28)
ABD | | | CBAand CAD | | | CBA ( AAA ) ABD | | | CBAand CAD | | | CBA ( AAA )
(29)
A B C B = B D B A = A D C A and C A C B = C D C A = A D B A A B C B = B D B A = A D C A and C A C B = C D C A = A D B A
(30)
A B 2 = C B × B D and A C 2 = C B × C D A B 2 = C B × B D and A C 2 = C B × C D
(31)
A B 2 + A C 2 = C B ( B D + C D ) = C B ( C B ) = C B 2 i . e . B C 2 = A B 2 + A C 2 A B 2 + A C 2 = C B ( B D + C D ) = C B ( C B ) = C B 2 i . e . B C 2 = A B 2 + A C 2
(32)
##### Exercise 4: Triangle Geometry 1

In GHI, GH LJ; GJ LK and JKKIJKKI = 5353. Determine HJKIHJKI.

###### Solution
1. Step 1. Identify similar triangles :
L I ^ J = G I ^ H J L ^ I = H G ^ I ( Corres . s ) L I J | | | G I H ( Equiangular s ) L I ^ J = G I ^ H J L ^ I = H G ^ I ( Corres . s ) L I J | | | G I H ( Equiangular s )
(33)
L I ^ K = G I ^ J K L ^ I = J G ^ I ( Corres . s ) L I K | | | G I J ( Equiangular s ) L I ^ K = G I ^ J K L ^ I = J G ^ I ( Corres . s ) L I K | | | G I J ( Equiangular s )
(34)
2. Step 2. Use proportional sides :
H J J I = G L L I ( L I J | | | G I H ) and G L L I = J K K I ( L I K | | | G I J ) = 5 3 H J J I = 5 3 H J J I = G L L I ( L I J | | | G I H ) and G L L I = J K K I ( L I K | | | G I J ) = 5 3 H J J I = 5 3
(35)
3. Step 3. Rearrange to find the required ratio :
H J K I = H J J I × J I K I H J K I = H J J I × J I K I
(36)

We need to calculate JIKIJIKI: We were given JKKI=53JKKI=53 So rearranging, we have JK=53KIJK=53KI And:

J I = J K + K I = 5 3 K I + K I = 8 3 K I J I K I = 8 3 J I = J K + K I = 5 3 K I + K I = 8 3 K I J I K I = 8 3
(37)

Using this relation:

= 5 3 × 8 3 = 40 9 = 5 3 × 8 3 = 40 9
(38)
##### Exercise 5: Triangle Geometry 2

PQRS is a trapezium, with PQ RS. Prove that PT ·· TR = ST ·· TQ.

###### Solution
1. Step 1. Identify similar triangles :
P 1 ^ = S 1 ^ ( Alt . s ) Q 1 ^ = R 1 ^ ( Alt . s ) P T Q | | | S T R ( Equiangular s ) P 1 ^ = S 1 ^ ( Alt . s ) Q 1 ^ = R 1 ^ ( Alt . s ) P T Q | | | S T R ( Equiangular s )
(39)
2. Step 2. Use proportional sides :
P T T Q = S T T R ( P T Q | | | S T R ) P T · T R = S T · T Q P T T Q = S T T R ( P T Q | | | S T R ) P T · T R = S T · T Q
(40)
##### Triangle Geometry
1. Calculate SV
2. CBYB=32CBYB=32. Find DSSBDSSB.
3. Given the following figure with the following lengths, find AE, EC and BE. BC = 15 cm, AB = 4 cm, CD = 18 cm, and ED = 9 cm.
4. Using the following figure and lengths, find IJ and KJ. HI = 26 m, KL = 13 m, JL = 9 m and HJ = 32 m.
5. Find FH in the following figure.
6. BF = 25 m, AB = 13 m, AD = 9 m, DF = 18m. Calculate the lengths of BC, CF, CD, CE and EF, and find the ratio DEACDEAC.
7. If LM JK, calculate yy.

## Co-ordinate Geometry

### Equation of a Line between Two Points

Figure 28
Khan academy video on point-slope and standard form

There are many different methods of specifying the requirements for determining the equation of a straight line. One option is to find the equation of a straight line, when two points are given.

Assume that the two points are (x1;y1)(x1;y1) and (x2;y2)(x2;y2), and we know that the general form of the equation for a straight line is:

y = m x + c y = m x + c
(41)

So, to determine the equation of the line passing through our two points, we need to determine values for mm (the gradient of the line) and cc (the yy-intercept of the line). The resulting equation is

y - y 1 = m ( x - x 1 ) y - y 1 = m ( x - x 1 )
(42)

where (x1;y1)(x1;y1) are the co-ordinates of either given point.

#### Finding the second equation for a straight line

This is an example of a set of simultaneous equations, because we can write:

y 1 = m x 1 + c y 2 = m x 2 + c y 1 = m x 1 + c y 2 = m x 2 + c
(43)

We now have two equations, with two unknowns, mm and cc.

y 2 - y 1 = m x 2 - m x 1 m = y 2 - y 1 x 2 - x 1 y 1 = m x 1 + c c = y 1 - m x 1 y 2 - y 1 = m x 2 - m x 1 m = y 2 - y 1 x 2 - x 1 y 1 = m x 1 + c c = y 1 - m x 1
(44)

Now, to make things a bit easier to remember, substitute (Reference) into Equation 41:

y = m x + c = m x + ( y 1 - m x 1 ) y - y 1 = m ( x - x 1 ) y = m x + c = m x + ( y 1 - m x 1 ) y - y 1 = m ( x - x 1 )
(45)

#### Tip:

If you are asked to calculate the equation of a line passing through two points, use:

m = y 2 - y 1 x 2 - x 1 m = y 2 - y 1 x 2 - x 1
(46)

to calculate mm and then use:

y - y 1 = m ( x - x 1 ) y - y 1 = m ( x - x 1 )
(47)

to determine the equation.

For example, the equation of the straight line passing through (-1;1)(-1;1) and (2;2)(2;2) is given by first calculating mm

m = y 2 - y 1 x 2 - x 1 = 2 - 1 2 - ( - 1 ) = 1 3 m = y 2 - y 1 x 2 - x 1 = 2 - 1 2 - ( - 1 ) = 1 3
(48)

and then substituting this value into

y - y 1 = m ( x - x 1 ) y - y 1 = m ( x - x 1 )
(49)

to obtain

y - y 1 = 1 3 ( x - x 1 ) . y - y 1 = 1 3 ( x - x 1 ) .
(50)

Then substitute (-1;1)(-1;1) to obtain

y - ( 1 ) = 1 3 ( x - ( - 1 ) ) y - 1 = 1 3 x + 1 3 y = 1 3 x + 1 3 + 1 y = 1 3 x + 4 3 y - ( 1 ) = 1 3 ( x - ( - 1 ) ) y - 1 = 1 3 x + 1 3 y = 1 3 x + 1 3 + 1 y = 1 3 x + 4 3
(51)

So, y=13x+43y=13x+43 passes through (-1;1)(-1;1) and (2;2)(2;2).

#### Exercise 6: Equation of Straight Line

Find the equation of the straight line passing through (-3;2)(-3;2) and (5;8)(5;8).

##### Solution
1. Step 1. Label the points :
( x 1 ; y 1 ) = ( - 3 ; 2 ) ( x 2 ; y 2 ) = ( 5 ; 8 ) ( x 1 ; y 1 ) = ( - 3 ; 2 ) ( x 2 ; y 2 ) = ( 5 ; 8 )
(52)
2. Step 2. Calculate the gradient :
m = y 2 - y 1 x 2 - x 1 = 8 - 2 5 - ( - 3 ) = 6 5 + 3 = 6 8 = 3 4 m = y 2 - y 1 x 2 - x 1 = 8 - 2 5 - ( - 3 ) = 6 5 + 3 = 6 8 = 3 4
(53)
3. Step 3. Determine the equation of the line :
y - y 1 = m ( x - x 1 ) y - ( 2 ) = 3 4 ( x - ( - 3 ) ) y = 3 4 ( x + 3 ) + 2 = 3 4 x + 3 4 · 3 + 2 = 3 4 x + 9 4 + 8 4 = 3 4 x + 17 4 y - y 1 = m ( x - x 1 ) y - ( 2 ) = 3 4 ( x - ( - 3 ) ) y = 3 4 ( x + 3 ) + 2 = 3 4 x + 3 4 · 3 + 2 = 3 4 x + 9 4 + 8 4 = 3 4 x + 17 4
(54)
4. Step 4. Write the final answer :

The equation of the straight line that passes through (-3;2)(-3;2) and (5;8)(5;8) is y=34x+174y=34x+174.

### Equation of a Line through One Point and Parallel or Perpendicular to Another Line

Another method of determining the equation of a straight-line is to be given one point, (x1;y1)(x1;y1), and to be told that the line is parallel or perpendicular to another line. If the equation of the unknown line is y=mx+cy=mx+c and the equation of the second line is y=m0x+c0y=m0x+c0, then we know the following:

If the lines are parallel, then m = m 0 If the lines are perpendicular, then m × m 0 = - 1 If the lines are parallel, then m = m 0 If the lines are perpendicular, then m × m 0 = - 1
(55)

Once we have determined a value for mm, we can then use the given point together with:

y - y 1 = m ( x - x 1 ) y - y 1 = m ( x - x 1 )
(56)

to determine the equation of the line.

For example, find the equation of the line that is parallel to y=2x-1y=2x-1 and that passes through (-1;1)(-1;1).

First we determine mm, the slope of the line we are trying to find. Since the line we are looking for is parallel to y=2x-1y=2x-1,

m = 2 m = 2
(57)

The equation is found by substituting mm and (-1;1)(-1;1) into:

y - y 1 = m ( x - x 1 ) y - 1 = 2 ( x - ( - 1 ) y - 1 = 2 ( x + 1 ) y - 1 = 2 x + 2 y = 2 x + 2 + 1 y = 2 x + 3 y - y 1 = m ( x - x 1 ) y - 1 = 2 ( x - ( - 1 ) y - 1 = 2 ( x + 1 ) y - 1 = 2 x + 2 y = 2 x + 2 + 1 y = 2 x + 3
(58)

### Inclination of a Line

In Figure 31(a), we see that the line makes an angle θθ with the xx-axis. This angle is known as the inclination of the line and it is sometimes interesting to know what the value of θθ is.

Firstly, we note that if the gradient changes, then the value of θθ changes (Figure 31(b)), so we suspect that the inclination of a line is related to the gradient. We know that the gradient is a ratio of a change in the yy-direction to a change in the xx-direction.

m = Δ y Δ x m = Δ y Δ x
(59)

But, in Figure 31(a) we see that

tan θ = Δ y Δ x m = tan θ tan θ = Δ y Δ x m = tan θ
(60)

For example, to find the inclination of the line y=xy=x, we know m=1m=1

tan θ = 1 θ = 45 tan θ = 1 θ = 45
(61)

#### Co-ordinate Geometry

1. Find the equations of the following lines
1. through points (-1;3)(-1;3) and (1;4)(1;4)
2. through points (7;-3)(7;-3) and (0;4)(0;4)
3. parallel to y=12x+3y=12x+3 passing through (-1;3)(-1;3)
4. perpendicular to y=-12x+3y=-12x+3 passing through (-1;2)(-1;2)
5. perpendicular to 2y+x=62y+x=6 passing through the origin
2. Find the inclination of the following lines
1. y=2x-3y=2x-3
2. y=13x-7y=13x-7
3. 4y=3x+84y=3x+8
4. y=-23x+3y=-23x+3 (Hint: if mm is negative θθ must be in the second quadrant)
5. 3y+x-3=03y+x-3=0
3. Show that the line y=ky=k for any constant kk is parallel to the x-axis. (Hint: Show that the inclination of this line is 00.)
4. Show that the line x=kx=k for any constant kk is parallel to the y-axis. (Hint: Show that the inclination of this line is 9090.)

## Transformations

### Rotation of a Point

When something is moved around a fixed point, we say that it is rotated about the point. What happens to the coordinates of a point that is rotated by 9090 or 180180 around the origin?

#### Investigation : Rotation of a Point by 90∘90∘

Complete the table, by filling in the coordinates of the points shown in the figure.

 Point xx-coordinate yy-coordinate A B C D E F G H

What do you notice about the xx-coordinates? What do you notice about the yy-coordinates? What would happen to the coordinates of point A, if it was rotated to the position of point C? What about point B rotated to the position of D?

#### Investigation : Rotation of a Point by 180∘180∘

Complete the table, by filling in the coordinates of the points shown in the figure.

 Point xx-coordinate yy-coordinate A B C D E F G H

What do you notice about the xx-coordinates? What do you notice about the yy-coordinates? What would happen to the coordinates of point A, if it was rotated to the position of point E? What about point F rotated to the position of B?

From these activities you should have come to the following conclusions:

• 90 clockwise rotation: The image of a point P(x;y)(x;y) rotated clockwise through 90 around the origin is P'(y;-x)(y;-x). We write the rotation as (x;y)(y;-x)(x;y)(y;-x).
• 90 anticlockwise rotation: The image of a point P(x;y)(x;y) rotated anticlockwise through 90 around the origin is P'(-y;x)(-y;x). We write the rotation as (x;y)(-y;x)(x;y)(-y;x).
• 180 rotation: The image of a point P(x;y)(x;y) rotated through 180 around the origin is P'(-x;-y)(-x;-y). We write the rotation as (x;y)(-x;-y)(x;y)(-x;-y).

#### Rotation

1. For each of the following rotations about the origin: (i) Write down the rule. (ii) Draw a diagram showing the direction of rotation.
1. OA is rotated to OA'' with A(4;2) and A''(-2;4)
2. OB is rotated to OB'' with B(-2;5) and B''(5;2)
3. OC is rotated to OC'' with C(-1;-4) and C''(1;4)
2. Copy ΔΔXYZ onto squared paper. The co-ordinates are given on the picture.
1. Rotate ΔΔXYZ anti-clockwise through an angle of 90 about the origin to give ΔΔX''Y''Z''. Give the co-ordinates of X'', Y'' and Z''.
2. Rotate ΔΔXYZ through 180 about the origin to give ΔΔX''''Y''''Z''''. Give the co-ordinates of X'''', Y'''' and Z''''.

### Enlargement of a Polygon 1

When something is made larger, we say that it is enlarged. What happens to the coordinates of a polygon that is enlarged by a factor kk?

#### Investigation : Enlargement of a Polygon

Complete the table, by filling in the coordinates of the points shown in the figure. Assume each small square on the plot is 1 unit.

 Point xx-coordinate yy-coordinate A B C D E F G H

What do you notice about the xx-coordinates? What do you notice about the yy-coordinates? What would happen to the coordinates of point A, if the square ABCD was enlarged by a factor 2?

#### Investigation : Enlargement of a Polygon 2

In the figure quadrilateral HIJK has been enlarged by a factor of 2 through the origin to become H'I'J'K'. Complete the following table using the information in the figure.

 Co-ordinate Co-ordinate' Length Length' H = (;) H' = (;) OH = OH' = I = (;) I' = (;) OI = OI' = J = (;) J' = (;) OJ = OJ' = K = (;) K' + (;) OK = OK' =

What conclusions can you draw about

1. the co-ordinates
2. the lengths when we enlarge by a factor of 2?

We conclude as follows:

Let the vertices of a triangle have co-ordinates S(x1;y1)(x1;y1), T(x2;y2)(x2;y2), U(x3;y3)(x3;y3). S'T'U' is an enlargement through the origin of STU by a factor of cc (c>0c>0).

• STU is a reduction of S'T'U' by a factor of cc.
• S'T'U' can alternatively be seen as an reduction through the origin of STU by a factor of 1c1c. (Note that a reduction by 1c1c is the same as an enlargement by cc).
• The vertices of S'T'U' are S'(cx1;cy1)(cx1;cy1), T'(cx2,cy2)(cx2,cy2), U'(cx3,cy3)(cx3,cy3).
• The distances from the origin are OS' = ccOS, OT' = ccOT and OU' = ccOU.

#### Transformations

1. Copy polygon STUV onto squared paper and then answer the following questions.
1. What are the co-ordinates of polygon STUV?
2. Enlarge the polygon through the origin by a constant factor of c=2c=2. Draw this on the same grid. Label it S'T'U'V'.
3. What are the co-ordinates of the vertices of S'T'U'V'?
2. ABC is an enlargement of A'B'C' by a constant factor of kk through the origin.
1. What are the co-ordinates of the vertices of ABC and A'B'C'?
2. Giving reasons, calculate the value of kk.
3. If the area of ABC is mm times the area of A'B'C', what is mm?
1. What are the co-ordinates of the vertices of polygon MNPQ?
2. Enlarge the polygon through the origin by using a constant factor of c=3c=3, obtaining polygon M'N'P'Q'. Draw this on the same set of axes.
3. What are the co-ordinates of the new vertices?
4. Now draw M”N”P”Q” which is an anticlockwise rotation of MNPQ by 90 around the origin.
5. Find the inclination of OM”.

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