Fatema Shariff, 10 B

SIMILARITY AND CONGRUENCY

Similarity What is Similarity? Similar Polygons Similar Triangles Area of Similar Triangles Perimeter of Similar Polygons

Congruency

Similar 3-D Shapes

1

x

TS

V

R

U

y + 2

18 4

5

3

B C

E

A

D

Fatema Shariff, 10 B

SimilarityWhat is similarity?Similarity is a relation between two shapes.

The word ‘similar’ means ‘having characteristics in common’.

Similar Polygons Let us take a real life example: you take a photograph of an ancient monument and you want to enlarge the photograph from the original one. When you enlarge the photograph, it is ensured that the shape of the photograph remains the same from the original only the size differs – the size changes but the shape remains the same.

For tw o figures to be similar, they have to: be of the same type, e.g. both rectangles All the corresponding sides must have the same ratio. This is called the scale factor of

the shapes The corresponding angles of the shapes must be congruent.

When we say ‘congruent’ we mean equal, equilateral or exactly the same (shape or size).‘Corresponding’ means any pair of angles/sides in similar location.

The symbol used to note that two shapes are similar is ‘~’.

Example 1Polygon RSTUV is similar to polygon ABCDE.

a) Find the scale factor of polygon RSTUV to polygon ABCDEb) Find the values of x and y

Answer:

Scale Factor:

184

=92

2

1 cm

2 cm AA 2 cm

3 cm

B

Fatema Shariff, 10 B

Value of x:

x3=18

4

4 x4

=544

x=13.5

Value of y:

184

= y+25

90=4 y+8

4 y4

=824

y=20.5

Example 2

The above rectangles are not similar even though their angles are congruent because their corresponding sides are not in the same ratio.

22

≠ 13

Similar TrianglesTo prove the similarity of two triangles, the following conditions must be fulfilled:

Corresponding angles should be equal Corresponding sides should be in the same ratio

3

B

A

C

A’

B’ C’

Fatema Shariff, 10 B

Example 1

Referring to the above figure:

KLM = EFG, KML = EGF and LKM = FEG

KLEF

= KMEG

= LMFG

Thus, we can say that the triangle KLM and triangle EFG are similar. Symbolically we write the above relation as KLM EFG because

It is important to note that, as in the case of congruency, similar triangles also should be written in the correct correspondence of their vertices i.e. it would be incorrect to write triangle KLM is similar to triangle GFE.

Example 2

4

RI

S

N

G

B

T

LA

Fatema Shariff, 10 B

A ' B 'AB

= B' C 'BC

= A ' C 'AC

AA Similarity (Angle-Angle)This theorem states that if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.By knowing this we also know that equilateral triangles are always similar as in an equilateral triangle, all angles are equal which satisfies the Angle – Angle theorem.

ExampleGiven: ΔING ∼ ΔRNSGS = 6 inGN = 3 inRI = (x + 5) inIN = (x + 1) inFind the value of x.

Answer:Since ΔING ∼ ΔRNS, corresponding sides have a proportional relationship.

We can write the following proportion: NGNS

= ¿NR

By substitution we have the following: 39= ( x+1 )

(2 x+6 )

By simplifying 39 we get the following: 13

=(x+1)(2 x+1)

Ifab= c

d , thenad=bc. Therefore, 3 ( x+1 )=1 (2 x+6 )

Then we get, 3 x−2 x=6−3

Therefore x=3

SSS Similarity (Side-Side-Side)This theorem states that if the measures of the corresponding sides of two triangles are proportional, then the triangles are similar.

ExampleGiven: Right triangle BLT with right angle ∠BTLAltitude TAAL = 6 cmTL = 10 cmFind: BL

5

50°

50°

2.5 m4.5 m

Fatema Shariff, 10 B

Answer:Right triangle BLT is similar to right triangle TLA as follows:A ∠BTL ≅ ∠TAL (congruent right angles)A ∠TLB ≅ ∠TLB (reflexive property)The measures of corresponding sides of similar triangles are in proportion.∆ BLT ≅ ∆ TLA

So TLAL

=BLTL

i.e. BL=1006

=16.67

SAS Similarity (Side-Angle-Side)This theorem states that if the measures of two sides of a triangle are proportional to the measures of two corresponding sides of another triangle and the included angles are congruent, then the triangles are similar.

ExampleJustify how the below figure is similar by SAS Similarity Theorem.

Answer:The triangles are similar by the SAS similarity theorem. The measures oftwo sides of one triangle are in proportion with the measures of thecorresponding sides of the other triangle as shown below.

52.5

= 94.5

=21

The included angle between the sides of one triangle is congruent to theincluded angle between the corresponding sides of the other triangle.

6

9 m5 m

E

D

B

A

C

6

6

2

f

e

3

ZYYY

X

W

3

45

v

w

A B6

D CE

F x

1

1

Fatema Shariff, 10 B

Similarity of triangles is more special than similarity in other polygons and triangles are the only shapes with 3 angles. When there are two triangles, we know that when 2 angles are equal then the triangles are equal but for other polygons if we consider only 2 corresponding angles then it is not necessary for the polygons to be similar as a polygon can have many angles and sides.

Exercise 1 (Similar polygons and triangles)Find the sides marked with letters:

1)

2)

3) From rectangle ABCD a square is cut off to leave the rectangle BCEF. Rectangle BCEF is similar to ABCD. Find x and hence state the ratio of the sides of rectangle ABCD. ABCD is called the Golden Rectangle and is an important shape in architecture.

4) A tree of height 4 m casts a shadow of length 6.5 m. Find the height of a house casting a shadow 26 m long.

7

A’

C’B’

c’ b’

a’

A

B Ca

b

Fatema Shariff, 10 B

Area of similar triangles

Areaof ABC=12

ab sin c

Areaof A' B' C'=12

a ' b ' sin c '

Area of ABCArea of A ' B ' C '

=

12

ab sin c

12

a ' b ' sin c '

Area of ABCArea of A ' B ' C '

= aba ' b'

Area of ABCArea of A ' B ' C '

= aa '

× bb '

Areaof ABCArea of A ' B' C '=k ×k

Area of ABCArea of A ' B ' C '

=k2

Area of ABCArea of A ' B ' C '

=( side 1side 2 )

2

Here you will notice that, as mentioned earlier, the area of the similar figures is equal to the square of their corresponding sides.

8

8 cm

16 cm

A

3 cm

6 cm18 cm²

58 cm²5 cm

20 cm10 cm

3 cmA

Fatema Shariff, 10 B

Perimeter of similar polygonsSuppose that a triangle with sides  a ,  b , and  c  has been scaled by  s to get a similar triangle with corresponding sides  A ,  B , and  C .Thus, A=sa and B=sb andC=sc.

Let's investigate the relationship between the perimeters of these two triangles:

perimeter of original triangle= a+b+c

Perimeter of scaled triangleA+B+C

sa+sb+scs (a+b+c )

s( perimeter of originaltriangle)

Thus, the perimeter ends up being scaled by the same factor that scales the sides.A similar calculation shows that this result is indeed true for polygons in general: The ratio of the

perimeters of two similar polygons is equal to the ratio of the corresponding sides.

Exercise 2 (Areas and perimeter of similar polygons and triangles)Find the unknown area (question 1 & 2):

1)

2)

9

3 cm

4 cm²9 cm²

z

y

36 cm²

6 cm

4 cm²

G

H

IF J

E

16

6

8

X

Fatema Shariff, 10 B

Find the unknown length marked for the following pairs of similar shapes (question 3 & 4):3)

4)

5) The isosceles triangles below are similar.

a) What is the similarity of triangle EFG to triangle HIJ? Simplify your ratio.b) What is the value of x? Use mathematics to explain how you determined your answer.

Use words, symbols, or both in your explanation. c) What is the perimeter of each triangle?

d) What is perimeter of ∆ EFGperimeter of ∆ HIJ ? Simplify your ratio.

CongruencyFigures that are congruent are equal in both size and shape.

Congruent shapes can be similar but similar shapes cannot be congruent.

In our daily life, we come across many objects with same shape and size. Objects, which have the same shape and size, are called congruent objects. The relation of two objects being congruent is called congruence. The term congruent means that two or more things are the exact same size and shape.

10

Fatema Shariff, 10 B

In the above figure, the quadrilaterals ABCD and EFGH are exactly equal with their corresponding angles and sides, hence they are congruent.

The corresponding parts of congruent triangles are equal. It can be shown that the converse is also true i.e. if there is a correspondence between the vertices of the two triangles so that the corresponding parts of two triangles are equal, then the two triangles are congruent.

ExampleFind out if the two triangles XYZ and KLM, below, are congruent.

Answer :

If triangle KLM covers the triangle XYZ exactly, we say that the two triangles are congruent, otherwise not. The matching parts of the triangle are also called the corresponding parts of the two triangles.

Since triangle XYZ is congruent to triangle KLM then their corresponding parts are congruent. If two figures are congruent, then their corresponding parts are congruent too. It is important to note that triangle XYZ is congruent to triangle KLM and we can say that triangle KLM is congruent to XYZ since vertex X corresponds to vertex K, vertex Y corresponds to vertex L, and vertex Z corresponds to vertex M.

XYZ KLM

11

D

E

B

CA

FG

Rr

Fatema Shariff, 10 B

Exercise 3 (Congruency)1) Identify pairs of congruent shapes below:

Similar 3-D ShapesSimilar solids have the exactly same shape but not necessarily the same size. To determine whether two solids are similar or not, you can compare the measurements of the corresponding linear measurements.In two similar polyhedral, the corresponding faces are similar and all the corresponding edges are proportional.We can state whether the two solids are similar by getting the ratio of 1:1 for the ratio of the corresponding measurements.For two solids to be congruent:

The corresponding angles have to be congruent The corresponding edges have to be congruent The areas of the corresponding faces have to be equal The volumes have to be equal

If two solids are similar with a scale factor ofa :b, then the surface areas will have a ratio of a2:b2 and the volumes will have a ratio ofa3:b3.

rR

=k= side 1side 2

12

V = ?6V = 504

Fatema Shariff, 10 B

volume1volume2

=

43

π r3

43

π R3

volume1volume2

= r3

R3 =k3

volume1volume2

=( side 1side 2 )

3

Example 1

v 1v 2

=( side1side 2 )

3

50v

=( 46 )

3

64 v64

=1080064

v=168.75 cm2

Relationship between the ‘surface area – the sides’ and the ‘volume – the sides’

volume 1volume 2

=( side1

side 2 )3

∧Area1

Area2=( side1

side2 )2

13

3.1

V88

6.2

270y7 10

V

Radius: 12 cm

4.5

Radius: 1.2 cm

Fatema Shariff, 10 B

side 1side 2

=( volume1

volume 2 )23∧side 1

side 2=√ area1

area2

(( Area1Area2 )

12)

2

=(( volume1volume2 )

13)

2

Area 1Area 2

=( volume 1volume 2 )

23

Exercise 4 (Volumes of 3-D Shapes)1) Find the unknown volume V.

2) Two similar cylindrical tins have base radii of 6 cm and 8 cm respectively. If the capacity of the larger tin is 252 cm³, find the capacity of the small tin.

‘3) Find the height y of the cylinder.

4)

14

Fatema Shariff, 10 B

Answers Exercise 1 (page 7)

1) e = 9 cm, f = 4.5 cm

2) v = 513 cm, w = 6

23 cm

3) 0.618; 1.678:14) 16 m

Exercise 2 (page 9)1) 128 cm²2) 14.5 cm²3) 4.5 cm4) 18 cm

5) a) 168

∨2

b) x = 3c)∆ EFG=36∧∆ HIJ=18

d)3618 or 2

Exercise 3 (page 11)1) A and G, B and E

Exercise 4 (page 14)1) V = 11 cm³2) V = 106.3 cm³3) y = 21 cm4) v = 4500 cm³

GRADE X - ASSESSMENT - SIMILARITY AND CONGRUENCY

15

Fatema Shariff, 10 B

Writing skill

Drawing skill

Formatting skill Polygon Triangle

3D shapes Questions

Examples Total

3 2 3 2 2 3 3 2 20creativity flow

neatness

pictures labeling

page paragraph

s math equations

always-similar

polygons appln

theorems

applns

perimeter vol,

SA

3D shapes

congruency variety

ex+ans

3 2 3 1 2 3 2 2 18

16