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Similarity and congruency
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.
ABC ∼
e.g.
DEF
means “is similar to”
B
E
C
10cm
5cm
F
D
A
e.g.
1
Similarity and congruency
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.
15cm
P
A
5cm
5
6cm
6cm
2cm
2cm
D
B
Q
5cm
C
S
15cm
FIGURE 1
2
R
Similarity and congruency
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
120 ͦ
4cm
4cm 2cm
60 ͦ
Z
D
5cm
X
60 ͦ
2
2cm
120 ͦ
Y
2.5cm
C
Figure 2
Even though the ratios of corresponding sides are equal(5/2.5 = 4/2), corresponding angles are not
equal (90° ≠ 120°, 90° ≠ 60°).

In Figure 3 , quadrilateral FGHI is not similar to quadrilateral JKLM.
F
5cm
3cm
I
G
J
3cm
3cm 3cm
3
H
5cm
M
3cm
K
3cm
L
Figure 3
Even though corresponding angles are equal, the ratios of each pair of corresponding sides are not equal
(3/3 ≠ 5/3).
Example 1:
Find the value of x, y and ∠B in the figure below.
3
Similarity and congruency
𝑥
78 ͦ
y
B
5cm
5.5cm
3cm
7cm
angle B = 78 ͦ (corresponding angles are same)
3=5
3=y
7 𝑥
3𝑥 = 35
7 5.5
16.5 = 7y
𝑥 = 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?
5’
3.5’
16’
3.5 = 16
5
𝑥
3𝑥 = 80
𝑥 = 22.9’
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.
4
Similarity and congruency
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
A
12cm
5cm
S
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 (5/12)
EXERCISE 1.1 (Ans on pg 22)
1. Are the two quadrilaterals similar? If so, state the similarity, and give the scale factor.
12
96 ͦ
115 ͦ
8
9
18
88 ͦ
6
115 ͦ
96 ͦ
14
88 ͦ
21
61 ͦ
12
5
61 ͦ
Similarity and congruency
2. Find x and y in the figure below.
21ft
𝑥
18ft
78 ͦ
24ft
𝑦
83 ͦ
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.
6
Similarity and congruency
Side-Side-Side Similarity - If all pairs of corresponding sides of two triangles are proportional, then the
triangles are similar.
30cm
13cm
5cm
12.5cm
32.5cm
12cm
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.
4.5cm
3cm
2cm
3cm
Example 1
State whether or not triangles ABC and XYZ are similar
A
8cm
C
4.5cm
4cm
6cm
X
3cm
6cm
B
AB = 4 = 1.333
BC = 6 = 1.333 AND AC = 8 = 1.333
YZ
XY 4.5
3
Y
Z
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.
7
Similarity and congruency
P
C
4cm
9cm
18cm
B
y
15cm
A
𝑥
R
Soln; from ∆𝑠 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
PR PB
9=y
15 18
y= 18 x 9
15
AC= BC
PR BR
9=4
15 𝑥
𝑥 = 15 x 4
9
y= 10.8 cm
𝑥 = 6.67 cm
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?
𝑥
6’
100
’
10’
8
Similarity and congruency
10 + 100 = 110
10 = 6
110 𝑥
length of larger triangle
𝑥 = 66’
10𝑥 = 660
∴ The tree is 66’ tall.
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
z
P
Q
AP = AQ
PB QC
B
(Intercept theorem)
C
So if PB = 6cm, AQ = 3cm and QC = 4cm,
AP = 3
6 4
4×AP = 3×6
4AP = 18
AP = 4.5
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 𝑥 in the figure below
B
15cm
C
As BD bisects
ABC, the above theorem can be
applied.
12cm
13cm
xcm
𝑥 = 12
13 15
D
A
9
15𝑥 = 156
𝑥 = 10.4cm
Similarity and congruency
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
5cm
xcm
A
B
6cm
Y
9cm
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
6cm
xcm
A
D
4cm
B
E
3cm
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
𝑥m
X
2m
A 0.5m Y
2m
C
10
Similarity and congruency
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
5cm
w
4cm
X
3cm
Y
Z
v
9. In the figure below XY is parallel to BC. If AY = 8cm, YC = 6cm and the area of
find the area of AXY.
A
X
ABC is 98cm 2,
Y
B
C
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
.
.
1
2
ka
b
kb
11
Similarity and congruency
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.”
if the scale factor of a figure is a: b then the ratio of their areas will be a2: b2.
In other words
Example 1
10cm
2
4cm
50cm
The above triangles are similar. Use the information given to find they are of the smaller triangle.
2
50 = 10
𝑥
4
50 = 100
𝑥 16
100 × 𝑥 = 16×50
100 𝑥 = 800
𝑥 = 8cm
Example 2
Two similar triangular buildings have been made on Manchester Avenue. The areas covered by
these two buildings are 45 cm 2 and 80 cm 2. 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.
12
Similarity and congruency
Soln;
Let the scale factor of the 2 triangles be a: b.
Area of triangle 1 =
area of triangle 2
45 =
80
2
2
a
b
9= a
16 b
a
b
2
(square root both sides)
3: 4 = a: b
Let 3𝑥 = perimeter of ∆1
and 4𝑥 = perimeter of ∆ 2
Then, 3𝑥 + 4𝑥 = 35cm
7𝑥 = 35cm
𝑥 = 5cm
Perimeter of ∆1 = 3x
= 3(5)
= 15cm
Perimeter of ∆2 = 4x
= 4(5)
= 20cm
Example 3
The scale factor of a map is 1: 50 000.
a.
b.
c.
d.
What is the area scale factor?
What distance in m, is represented by 1cm on the map?
What area, in m2 is represented by 1cm2 on the map?
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
13
Similarity and congruency
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)
𝐴𝐵
1. Find
𝑃𝑄
in the following pair of similar figures.
A
B
P
Q
10.8cm
30cm2
D
C
2
S
R
2. Find the perimeter of ∆ DEF in the figure below.
D
A
6’
10’
9’
B
E
8’
C
F
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 m 2.
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 t riangles.
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.
P
A
Q
R
14
Similarity and congruency
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
3=2=5=1
12 8 20 4
20cm
corresponding sides are in the same ratio hence the above cuboids are similar
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
15
Similarity and congruency
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
4: 5: 7
ii) 24: 30: 42
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
= 2×π×24×80
= 2×π×30×100
= 3840 π
= 6000 π
capacity = 2 πrh
= 2×π×42×140
=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
A
B
1cm
8cm
0.5cm
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
16
Similarity and congruency
Surface area B = 13 = 1
Surface area A 52 4
1= 1
4 2
ratio of corresponding lengths = 1
2
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
27
k =2
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
17
Similarity and congruency
Surface area of building = 40 000 × 6480 cm2
9
= 28 800 000cm2
= 2880m2
3
c) Volume of model = 3
volume of building 2003
(1m3 = 10 000cm3)
volume of model= 27
16 000
8 000 000
volume of model = 16 000 × 27 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.
10cm
6cm
400cm³
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
18
Similarity and congruency
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 𝜋 as
22
7
)
24cm
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
A
Q
C
P
R
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)
19
Similarity and congruency
A
P
B
Q
C
R
Side-Side-Side Congruency Theorem: If three sides of one triangle equal the corresponding parts of the
other, then the triangles are congruent. (SSS)
C
P
Q
R
B
A
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.
A
In ∆s AOB and COD
D
B
AO = CO (both radIi)
BO = DO (both radii)
∠AOB = ∠COD (vertically opposite)
∴ ∆𝑠 𝐴𝑂𝐵 𝑎𝑛𝑑 𝐶𝑂𝐷 are congruent (SAS)
C
20
Similarity and congruency
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.
B
E
F
D
C
A
As BD bisects ∠ABC, and so ∠BED and ∠BCD are equal
BD is a line shared by both triangles
∴ ∆𝑠 𝐵𝐸𝐷 𝑎𝑛𝑑 𝐵𝐹𝐷 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.
65 ͦ
60 ͦ
60 ͦ
55 ͦ
ͦ
2. Triangle Type
LMN isequation
isosceleshere.
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.
y
X
O
W
Z
21
Similarity and congruency
4. ABCD is part of a regular polygon. Prove that triangles ABC, BCD are congruent, and that AC
= BD.
B
C
D
A
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.
A
E
B
F
D
C
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)
A
B
D
E
C
ANSWERS
Exercise 1.1 pg 5
1) Yes; 3⁄2
2) 𝑥 = 28ft; 𝑦 = 83 ͦ
3) 3⁄2 ; 3⁄2
22
Similarity and congruency
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
1
2
8) v = 5 3 cm, w = 63 cm
9) 32cm 2
1
10) 72 cm
Exercise 1.3 pg 14
1)
3
5
2) 36cm
3) 30 000m 2
4)
1
1600
1
2
5) 23 cm, 23 cm, 9:1
6) 24
7) 27cm2, 48cm2
8) (a) (i) 9m
(ii) 12m (iii)
1
16
(iv)
9
16
(b) (i) 16𝑥m2 (ii) 9𝑥m2
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
1) No
pg 21
23
Similarity and congruency
GRADE X - ASSESSMENT - SIMILARITY AND CONGRUENCY
Writing
skill
Drawing
skill
Formattin
g skill
Polygon
Triangle
3D
shapes
3
creativit
y flow
neatness
2
pictures
labeling
3
page
paragraph
s math
equations
2
alwayssimilar
polygon
s appln
2
theorem
s
applns
3
perimete
r
vol,
SA
3
2
3
2
2
3
Questions
3
3D
shapes
congruenc
y variety
3
Example
s
2
ex+ans
2
Total
20
20
Excellent work done! Keep it up! Your examples and exercise questions are very good.
24