Similarity and congruency

1

SIMILARITY

Similarity is the quality of being similar. It refers to the closeness of appearance between two or more

objects. It is the relation of sharing properties.

In mathematics figures that are of the same shape but not necessarily of the same size are known as

similar figures. This means one figure could be an enlargement of the other.

Enlarging a figure does not alter the angles. It does change the lengths of the sides but all the lengths

change in the same ratio.

e.g.

e.g.

C

ABC ∼ DEF

A

means “is similar to”

E

D

F

B

10cm 5cm

Similarity and congruency

2

Before you can learn this concept you need to get familiar to some words frequently used in this

chapter. These include:

Corresponding - similar especially in position or purpose; equivalent

Ratio - the relationship between two numbers or quantities (usually expressed as a

quotient)

Proportion - the relation between things (or parts of things) with respect to their

comparative quantity, magnitude, or degree

Congruent - equilateral, equal, exactly the same (size, shape, etc.)

Enlargement - expansion: the act of increasing (something) in size or volume or quantity or

scope

Theorem - an idea accepted as a demonstrable truth

Postulate - Something assumed without proof as being self-evident or generally accepted

Scale factor – reduced ratio of corresponding sides of two similar figures.

SIMILAR POLYGONS

Two polygons with the same shape are called similar polygons. A pair of similar polygons has the

following special properties:

Corresponding angles are equal.

Corresponding sides are in the same proportion.

*Both of the above properties must apply for two polygons to be similar.

Illustration

In figure 1 Quadrilateral ABCD ∼ quadrilateral PQRS.

6cm

2cm

D C

B A

Q

S R

P

FIGURE 1

15cm

6cm

15cm

5cm

5

2cm

5cm

Similarity and congruency

3

This means ∠ A = ∠ P, ∠ B = ∠ Q, ∠ C = ∠ R, ∠ D = ∠ S

ALSO AB = AD = BC = CD = 1

PQ PS QR RS 3

It is possible for a polygon to have one of the above facts true without having the other fact true. The

following two examples show how that is possible.

In Figure 2 , quadrilateral ABCD is not similar to quadrilateral WXYZ.

A 5cm B W 2.5cm X

4cm 4cm 2cm 2 2cm

Z Y

Even though the ratios of corresponding sides are equal( = ), corresponding angles are not

equal (90° 120°, 90° 60°).

In Figure 3 , quadrilateral FGHI is not similar to quadrilateral JKLM.

F 5cm G J 3cm K

3cm 3cm 3cm 3 3cm

Even though corresponding angles are equal, the ratios of each pair of corresponding sides are not equal

( ).

Example 1:

Find the value of x, y and ∠B in the figure below.

Figure 2

60 120

60 120

2.5cm

I H L M

5cm 3cm

Figure 3

5cm D C

Similarity and congruency

4

3 = 5 3 = y angle B = 78 (corresponding angles are same)

7 7 5.5 3 = 35 16.5 = 7y

= 11.67cm y = 2.36cm

Example 2

You have been asked to create a poster to advertise a field trip to see the Liberty Bell. You have a 3.5

inch by 5 inch photo that you want to enlarge. You want the enlargement to be 16 inches wide. How

long will it be?

3.5 = 16 3 = 80 = 22.9’

5

SPECIAL FIGURES

Some figures are always similar. Such figures include: circles, squares, equilateral triangles, regular

pentagons, hexagons, octagons e.t.c. The reason for this is:

Circles are always similar because they the same length of radius throughout the figure and so the

corresponding lengths of two circles will always be proportional.

7cm

3cm

5cm y

5.5cm

78 B

3.5’

5’

16’

Similarity and congruency

5

Squares, equilateral triangles and others are always similar because:

their internal angles are always the same regardless of change in size ( e.g. squares have all

angles of 90 )

the length of their sides are equal(e.g. square has 4 equal sides) and thus corresponding sides of

two such figures are proportional.

Example

T

B C

R

Both triangles are equilateral and so their internal angles are 60 . All 3 sides of the triangle ABC are

5cm while all 3 sides of the triangle RST are 12cm.

Hence ABC ∼ RST

AS all angles are equal (60 ) and all sides have the same proportion ( )

EXERCISE 1.1 (Ans on pg 22)

1. Are the two quadrilaterals similar? If so, state the similarity, and give the scale factor.

5cm 12cm A

S

96 115

88

61

96 115

88

61

12

9

21

18

8

14 6

12

Similarity and congruency

6

2. Find x and y in the figure below.

3. The rectangular patio around a pool is similar to the pool as shown. Calculate the scale factor of

the patio to the pool, and find the ratio of their perimeters.

SIMILAR TRIANGLES

In the case of similar triangles, it isn’t necessary to show that the corresponding angles are equal and

corresponding sides are in the same proportion. Any ONE of the following theorems could apply.

AA Similarity Postulate SAS Similarity Postulate SSS Similarity Postulate

Angle-Angle Similarity - If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.

21ft

24ft 18ft

78 83

Similarity and congruency

7

Side-Side-Side Similarity - If all pairs of corresponding sides of two triangles are proportional, then the triangles are similar.

Side-Angle-Side Similarity - If one angle of a triangle is congruent to one angle of another triangle and the sides that include those angles are proportional, then the two triangles are similar.

Example 1

State whether or not triangles ABC and XYZ are similar

A 8cm C

4.5cm Y

4cm 6cm X 3cm

6cm Z

AB = 4 = 1.333 BC = 6 = 1.333 AND AC = 8 = 1.333

YZ 3 XY 4.5 XZ 6

The three pairs of corresponding sides are in the same ratio. (SSS similarity)

Hence ABC ∼ XYZ

Example 2

Show that ABC ~ PBR. Hence find x and y in the figure below.

12cm

5cm 13cm

30cm

12.5cm

32.5cm

2cm

4.5cm

3cm

3cm

B

Similarity and congruency

8

9cm

Soln; from ABC and PBR we can say;

<C = <R (Alternate Interior Angles), <A=<P (Alternate Interior Angles) and <ABC=<PBR (Vertically Opposite Angles)

Therefore all angles are equal (AAA SIMILARITY)

Hence triangle ABC ~ triangle PBR

For similar triangles:

AB = BC= AC PB BR PR

AC= AB 9 = y y= 18 x 9 y= 10.8 cm PR PB 15 18 15

AC= BC 9 = 4 = 15 x 4 = 6.67 cm PR BR 15 9

Example 3

Sally wishes to find, with a little help from a 6ft pole, the height of an oak tree that stands on level ground. She walks away from the base of the tree, with the Sun directly behind her, until she reaches a point 100ft from its base. At this point she stands the pole on the ground and finds that the top of the shadow of the pole coincides with the top shadow of the tree at appoint a further 10ft away. How tall is the tree?

C

A

P

R

y

B

18cm

15cm

4cm

100

’

10’

6’

Similarity and congruency

9

10 + 100 = 110 length of larger triangle

10 = 6 10 = 660 = 66’ The tree is 66’ tall.

110

THE INTERCEPT THEOREM

“If a line is drawn parallel to one side of a triangle it divides the other two sides in the same ratio.” This

theorem is also known as Side-splitter theorem. Therefore similar triangles can be created by drawing a

segment parallel to one side of a triangle in the triangle.

Example A

If in triangle ABC, PQ is parallel to BC, then

AP = AQ PB QC

So if PB = 6cm, AQ = 3cm and QC = 4cm,

AP = 3 4 AP = 3 6 4AP = 18 AP = 4.5 6 4

THE ANGLE-BISECTOR THEOREM

If a ray bisects an angle of a triangle, then it divides the opposite side into segments that are

proportional to the sides that formed the angle.

Example

Find in the figure below

B 15cm C As BD bisects ABC, the above theorem can be

applied.

12cm 13cm = 12 15 = 156 = 10.4cm 13 15

xcm

z

P

B C

Q

(Intercept theorem)

D

A

Similarity and congruency

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EXERCISE 1.2 (Ans on pg 23)

1. In s ABC and XYZ, ∠A = ∠X and ∠B = ∠Y. AB = 6cm, BC = 5cm and XY = 9cm. Find YZ.

C

Z

xcm

A B

Y X

2. In s ABC and DEF, ∠A = ∠E and ∠B = ∠D. AB = 4cm, DE = 3cm and AC = 6cm. Find EF

when s ABC and EDF are similar.

F C

xcm

A B

D E

3. In the triangle ABC, angle ABC = 90 , and the line through B perpendicular to AC meets AC in D.

Prove that triangles ABC and BDC are equiangular.

4. David is standing next to a very tall building. The building casts a shadow that is 180 feet

long. David is 6 feet tall. His shadow is 10 feet long. How tall is the building?

5. AB is a ladder leaning against a vertical wall BC, with its foot on a level ground such that AC =

2m. XY is a straight pole placed so that it is vertical, with one end X touching the ladder and the

other end Y resting on the horizontal ground. If AY = 0.5m and XY = 2m, find BC and hence find

the length of the ladder.

B

m

A 0.5m Y C

6cm 9cm

5cm

3cm

X

2m

2m

6cm

4cm

4cm

Similarity and congruency

11

6. A straight line parallel to the base BC of the triangle ABC meets AB in X, AC in Y. IF AX: XB = 4: 5, find the ratios (i) AY: YC, (ii) XY: BC.

7. The internal bisector of the angle A of a triangle ABC meets BC at D. IF AB = 7cm, AC = 8cm and BC = 9cm, calculate BD.

8. Find the value of v and w W

9. In the figure below XY is parallel to BC. If AY = 8cm, YC = 6cm and the area of ABC is 98cm2, find the area of AXY.

X Y

10. A closed pair of scissors is 9cm long, the pivot being 5cm from the points. The maximum that

Paul can open the scissors using his thumb and first finger is 6cm. When he does this, what is

the distance between the points?

PERIMETER AND AREA OF SIMILAR FIGURES

When two figures are similar the reduced ratio of the corresponding sides is known as the scale factor

(linear scale factor).

Consider the two similar rectangles below. Let the scale factor be k;

.

.

a

b

ka

kb

1 2

Z X Y

5cm

3cm

4cm

w

v

A

B C

Similarity and congruency

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Perimeter of rectangle 1 = 2a + 2b

Perimeter of rectangle 2 = 2ka + 2kb

perimeter 2 = 2 k (a + b) = k

perimeter 1 2 (a + b)

This shows an important general rule:

“If two similar figures have a scale factor of k, then the ratio of their perimeters is also k.” In

other words if the scale factor of a figure is a: b then the ratio of their perimeters will be a: b .

Area of rectangle 1 = ab

Area of rectangle 2 = ka*kb = k2 ab

Area 2 = k2 ab = k2

Area 1 ab

This shows another important general rule:

“If two similar figures have a scale factor of k, then the ratio of their areas is k2.” In other words

if the scale factor of a figure is a: b then the ratio of their areas will be a2: b2.

Example 1

The above triangles are similar. Use the information given to find they are of the smaller triangle.

50 = 10 100 = 16 50

4 100 = 800

50 = 100 = 8cm

16

Example 2

Two similar triangular buildings have been made on Manchester Avenue. The areas covered by

these two buildings are 45 cm2 and 80 cm2. The sum of their perimeters is 35 cm. The field agent

needs to know the perimeters of each building. It is your job to assist him.

50cm 2

10cm 4cm

2

Similarity and congruency

13

Soln;

Let the scale factor of the 2 triangles be a: b. Area of triangle 1 = a 2 area of triangle 2 b 45 = a 2

80 b

9 = a 2

16 b (square root both sides)

3: 4 = a: b Let 3 = perimeter of 1 and 4 = perimeter of 2

Then, 3 + 4 = 35cm

7 = 35cm

= 5cm

Perimeter of 1 = 3x Perimeter of 2 = 4x

= 3(5) = 4(5)

= 15cm = 20cm

Example 3

The scale factor of a map is 1: 50 000.

a. What is the area scale factor?

b. What distance in m, is represented by 1cm on the map?

c. What area, in m2 is represented by 1cm2 on the map?

d. A field represented on the map by an area of 2.5cm2. Find the area of the field in m2.

Soln;

a) 12: 50 0002

1: 2 500 000 000

b) 1m = 100cm

50,000/100 = 500m

c) 1m2 = 10000cm2

2,500,000,000/ 10,000 = 250,000m2

Similarity and congruency

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d) 2.5 2,500,000,000 = 6,250,000,000cm2

6,250,000,000/ 10,000 = 625,000m2

EXERCISE 1.3 (Ans on pg 23)

1. Find in the following pair of similar figures.

A B P Q

D C S R

2. Find the perimeter of DEF in the figure below.

3. The scale of a map is 1: 10 000. On the map, the area representing a plot of land is 3cm 2. Find the area of the field in m2.

4. My lounge door is 800mm wide and a picture of the door is 20 mm wide. Find the ratio of the

area of the picture of the door to the area of the actual door.

5. The sides of a triangle are 6cm, 7cm and 8cm. The shortest side of a similar triangle is 2cm. Find the lengths of the other two sides, and the ratio of the areas of the two triangles.

6. The triangles ABC and EBD are similar (AC and DE are not parallel). If AB=8cm, BE=4cm and the area of triangle DBE=6cm², find the area of triangle ABC.

7. The perimeters of two similar triangles are in the ratio 3: 4. The sum of their areas is 75 cm2. Find the area of each triangle.

8. The following diagram represents a framework for a factory roof.

A

30cm2 10.8cm2

8’

6’ 10’ 9’

D

E

F

A

C

B

R

Q P

Similarity and congruency

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B C

PQ is parallel to BC and PR is parallel to AC. AQ = 2m, QC = 6m, AP = 3m and PQ = 4m. a) Calculate i) PB ii) BR iii) area APQ area ABC iv) area BPR area ABC

b) If the area of triangle APQ is xm2, express in terms of x

i) area ABC ii) area BPR

SILIMAR 3-D FIGURES

Solid objects/3-D figures can only be similar if one is an accurate enlargement of the other and the corresponding sides are in the same ratio.

Example 1

3cm 12cm

2cm 8cm

5cm 20cm

3 = 2 = 5 = 1 corresponding sides are in the same ratio hence the above cuboids are similar 12 8 20 4

Example 2

The diagrams below represent 3 different sizes of cans used to fill petrol at a petrol pump.

80cm 100cm 140cm

48cm

60cm

84cm

Similarity and congruency

16

a) Write down the ratio of

i) their heights

ii) their base radii.

Hence explain why the shapes are similar.

b) Express the capacity of each can as a multiple of π. Hence find the ratio of their capacities. Is

there a relationship between your answers to (a) and (b)?

Soln;

a) i) 80: 100: 140 ii) 24: 30: 42

4: 5: 7 4: 5: 7

All 3 cans are similar because the ratio of the corresponding sides is the same/constant. (4: 5: 7)

b) Capacity = 2 πrh capacity = 2 πrh capacity = 2 πrh

= 2 π 24 80 = 2 π 30 100 = 2 π 42 140

= 3840 π = 6000 π =11760 π

3840: 6000: 11760

64: 100: 196

Yes, there is a relationship between (a) and (b) which is:

64: 100: 196 = 43: 53: 73

VOLUME AND SURFACE AREA OF 3-D FIGURES

From the example given above we can conclude that “The ratio of the volumes of similar 3-d shapes

is equal to the ratio of the cubes of the corresponding lengths.” If the scale factor of two objects is k

then the ratio of their volumes is k3.

Consider the two figures below

2cm 1cm

1cm 0.5cm

8cm 4cm

Total Surface area of A = 2(2 8) + 2(8 1) + 2(1 2) = 32+16+4 = 52cm

Total Surface area of B = 2(1 4) + 2(4 0.5) + 2(0.5 1) = 8+4+1 = 13cm

A B

Similarity and congruency

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Surface area B = 13 = 1 ratio of corresponding lengths = 1

Surface area A 52 4 2

1 = 1 2

4 2

*So if the scale factor of two objects is k then the ratio of their volumes is k2.

We can also conclude that “The ratio of the surface areas of similar 3-d shapes is equal to the ratio

of the squares of the corresponding lengths.”

Example 1

Two similar spheres made of steel have weights of 64kg and 216kg respectively. If the radius of the

larger sphere is 18cm, find the radius of the smaller sphere.

The ratio of weights will be the same as the ratio of volumes.

Ratio of volumes (k3) = 64 = 8

216 27

k3 = 8

27

k = 3 8 k =2

27 3

Radius of smaller sphere = 2 9 = 6cm 3

Example 2

A student is to make a scale model of a building. The building is 20m high and she decides to make

the model 30cm high.

a) Express, in its simplest form, the ratio of the height of the building to the height of the model.

b) If the total surface area of the model is 6480cm2, find the total surface area of the building.

c) If the volume of the building is 16,000m3, find the total volume of the model.

Soln;

20m = 20 100cm = 2000cm

a) Height of building = 2000 = 200

height of model 30 3

b) Surface area of building = 2002

surface area of model 32

Surface area of building = 40 000

6480

Similarity and congruency

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Surface area of building = 40 000 cm2

9

= 28 800 000cm2 (1m3 = 10 000cm3)

= 2880m2

c) Volume of model = 33

volume of building 2003

volume of model= 27

16 000 8 000 000

volume of model = 16 000 m3

8 000 000

= 0.054 m2

= 54 000cm2

EXERCISE 1.4 (Ans on pg 23)

1. The sides of two cubes are in the ratio 2: 5. What is the ratio of their volumes and surface

areas?

2. Two cylinders have radii 6 cm and 10 cm respectively. If the volume of a smaller cylinder is

400cm³, find the volume of the larger one.

3. Two cones have volumes in the ratio 64: 27. What is the ratio of

(a) their heights (b) their base radii?

4. The cost of baked beans needed to fill a cylindrical can 10cm high is 8p. What is the cost of

the beans needed for a similar can standing 15cm high?

5. The radius of a spherical snowball increases by 60%. Find, correct to the nearest whole

number, the percentage increase in;

a) Its surface area

b) Its volume

6cm

10cm

400cm³

Similarity and congruency

19

6. A cone of height 16cm is cut from a cone of radius 7cm and height 24cm. Find the volume of the

trustrum (truncated cone). (use as )

7. The edge of a wooden cube is 50% longer than the edge of a metal cube. If the volume of the

metal cube is 405cm3, what is the volume of the wooden cube?

CONGRUENCE

Two figures are congruent if one fits exactly on the other. They must be of the same size and the

same shape and hence are sides and angles must be equal. The notation used to show one figure is

congruent to another is ‘ ’.

Two triangles are congruent if they satisfy any one of four conditions:

Side-Angle-Side Congruency Theorem: If two sides and the included angle of one triangle equal the

corresponding parts of the other, then the triangles are congruent. (SAS)

B Q

Angle-Angle-Side Congruency Theorem: If one side and two angles of one triangle equal the

corresponding parts of the other, then the triangles are congruent. (AAS)

A

24cm

C P R

Similarity and congruency

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A

Side-Side-Side Congruency Theorem: If three sides of one triangle equal the corresponding parts of the

other, then the triangles are congruent. (SSS)

Hypotenuse Leg Congruency Theorem: If a right triangle has a leg and hypotenuse equal to the

corresponding parts of the other triangle, then the triangles are congruent. (HL) OR (RHS)

.

Note: The following doesn’t prove that triangles are congruent:

- If three angles in one triangle equal to 3 angles in the second triangle.(SSS)

- two sides and an angle which doesn’t lie btw these sides in one triangle is equal to two sides and a

similarly located angle in the second triangle. (SSA)

Example 1

AC and BD are diameters of a circle with centre O. Prove that triangles AOB and COD are congruent.

In s AOB and COD

AO = CO (both radIi)

BO = DO (both radii)

∠AOB = ∠COD (vertically opposite)

are congruent (SAS)

C B

P

Q R

C

A B Q R

P

C

B

A

D

Similarity and congruency

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Example 2

The diagram below shows the end frame of a garden swing. BD bisects ∠ABC; BE and BF are equal.

Show that triangles BED and BFD are congruent and hence that ED = FD.

As BD bisects ∠ABC, and so ∠BED and ∠BCD are equal

BD is a line shared by both triangles

are congruent as two sides and included angle are equal (SAS)

ED is equal to FD because for two triangles to be congruent both must be of the same size and thus

these two lengths will be equal.

EXERCISE 1.5 (Ans on pg 23)

1. State with reason whether or not the two triangles can be proved to be congruent.

2. Triangle LMN is isosceles with LM=LN; X and Y are points on LM, LN respectively such that LX=LY.

Prove that triangles LMY and LNX are congruent.

3. The diagonals of the quadrilateral XYZW intersect at O. Given that OX = OW and OY = OZ

show that XY = WZ.

A

E

D

B

C

F

60

55 60

65

y

Z

O

X

W

Similarity and congruency

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4. ABCD is part of a regular polygon. Prove that triangles ABC, BCD are congruent, and that AC

= BD.

5. The pattern on a square tile, ABCD, is shown in the diagram. E and F are the midpoints of AB

and BC. By proving that two triangles are congruent, show that DE = DF.

6. Taking the definition of an isosceles triangle as a triangle with two equal sides, prove that

the angles opposite to these sides are equal.

7. ABCDE represents a roof truss. AB = AC and AD = AE. Prove that BD= EC. (Consider triangles

ABD and ACE)

ANSWERS

Exercise 1.1 pg 5

1) Yes;

2) = 28ft; = 83

3)

D

A

C

B

F

E

A

E D B C

A

B C

D

Similarity and congruency

23

Exercise 1.2 pg 10

1) 7.5cm

2) 4.5cm

4) 108ft

5) 8m, 8.25m

6) 4:5 , 4:9

7) 4.2cm

8) v = 5 cm, w = 6 cm

9) 32cm2

10) 7 cm

Exercise 1.3 pg 14

1)

2) 36cm

3) 30 000m2

4)

5) 2 cm, 2 cm, 9:1

6) 24

7) 27cm2, 48cm2

8) (a) (i) 9m (ii) 12m (iii) (iv)

(b) (i) 16 m2 (ii) 9 m2

Exercise 1.4 pg 18

1) 8: 125

2) 1852cm³

3) (a) 4: 3 (b) 4: 3

4) 27p

5) (a) 156% (b) 310%

6) 867cm3

7) 1370cm3 (3 s.f.)

Exercise 1.5 pg 21

1) No