Download Example 2

HAPTER 2-GEOMETRY:SIMILARIT
SIMILARITY AND
CONGRUENCE
MATHS PROJECT
MRS. KALYANI
DANISH DHAMANI 10B
26TH MARCH, 10
©DD Productions
1
HAPTER 2-GEOMETRY:SIMILARIT
CHAPTER 2
GEOMETRY: SIMILARITY
CONTENTS:
SIMILAR AND CONGRUENT POLYGONS
SIMILAR AND CONGRUENT TRIANGLES
PERIMETERS AND AREAS OF SIMILAR POLYGONS
SIMILARITY IN 3-D SHAPES
 VOLUME
 SURFACE AREA
©DD Productions
2
HAPTER 2-GEOMETRY:SIMILARIT
SIMILAR and CONGRUENT POLYGONS
Introduction:
Congruent polygons:
In order to be congruent, two figures must be the same size and same shape. Congruent
figures are exact replicas of each other. They have the same shape and the same size.
≅
Sign- “is congruent to”
Similar polygons:
Now consider figures that have the same shape but not the same size. Such figures look 'similar'
but in real are simply proportionate to each other
Similar means to have properties that are alike one another.
*NB. Similar DOES NOT MEAN ‘exactly the same.’
A polygon is a closed figure consisting of sides and angles. Different polygons have different
sides and angles in terms of their numbers and sizes.
Collectively, similar polygons refer to 2 (or more) figures with characteristics that are alike one
another, but not entirely the same. What make them similar are their shapes, but they differ in
sizes.
For two shapes to be similar, they must abide by the following criteria:
Corresponding angles should be equal
The ratios of pairs of corresponding sides should be equal
The symbol used to denote two (or more) similar figures is “∼”, for example, ∆ABC ∼ ∆DEF.
N.B. Students often confuse between similar and congruent polygons. In simpler terms,
congruent polygons are exact same replicas; while similar polygons have proportional sides.
©DD Productions
3
HAPTER 2-GEOMETRY:SIMILARIT
Example 1:
The triangles ABC and DEF are similar; we could also write ABC~DEF.
The following can be said about the two triangles:
1) ∠A=∠D, ∠B=∠E and ∠C=∠F
2)
𝐴𝐵
𝐷𝐸
=
𝐵𝐶
𝐸𝐹
=
𝐶𝐴
8
𝐹𝐷
4
6
10
3
5
= =
As all corresponding angles and the ratio of each pair of corresponding sides is 2:1, we can
conclude that ∆ABC~∆DEF.
Non-similar polygons
Remember, that for polygons to be similar, they must abide by both rules in the criteria. It is
possible to have polygons that abide by only one of them. These polygons cannot be considered
similar polygons.
Example 2: Consider the following quadrilaterals:
Though the ratios of pairs of corresponding sides are the same, but the corresponding angles are
different, so the above quadrilaterals are not similar polygons.
©DD Productions
4
HAPTER 2-GEOMETRY:SIMILARIT
i.e.
𝐴𝐵
𝐸𝐹
=
𝐵𝐶
𝐹𝐻
=
𝐶𝐷
𝐻𝐺
=
𝐷𝐴
𝐺𝐸
but 90°≠ 110°, 90° ≠ 70°
Example 3:
In this case, the corresponding angles are the same, but the ratios of pairs of corresponding
sides are different, so the above quadrilaterals are not similar polygons.
4
2
i.e. 90° = 90° but 3 ≠ 2
How can the idea of SIMILAR POLYGONS be used in real life?
Photo Enlargements/Reductions:

When you have a photograph enlarged/reduced, you make a similar photograph but of
a different size.
As you can see in the above photos, the picture of both is the same, only their sizes differ. They
differ by a scale factor of 3.

This can also be used while photocopying things. You get the similar size (or same) of
the thing you photocopy.
©DD Productions
5
HAPTER 2-GEOMETRY:SIMILARIT
Similar triangles
Generally, to prove that 2 polygons are similar, you must show that all pairs of corresponding
angles are equal and that all ratios of pairs of corresponding sides are equal, but with triangles, this
is not necessary. Two triangles are similar if their corresponding angles are congruent and their
corresponding sides are proportional. It is however not essential to prove all 3 angles of one triangle
congruent to the other, or for that matter all three proportional to the other sides. Out of these if
some particular conditions get satisfied, the rest automatically get satisfied. Those particular
conditions would be sufficient to ensure similarity. These are the two conditions: 1) In 2 triangles if the corresponding angles are congruent, their corresponding sides are equal.
2) If the sides of 2 triangles are proportional then the corresponding angles are congruent.
The following mentioned below, are tests used to show that two triangles are
similar:
1) Postulate 17 (Angle Angle similarity Postulate): If two angles of one triangle are congruent
with the corresponding two angles of another triangle, the two triangles are similar; the sum of all
three angles of a triangle is 180°. Therefore if two angles are congruent the third is automatically
congruent. Therefore the sufficient condition requires only two angles to be congruent.
Example 1:
∠B = ∠E = 80°
∠A + ∠B + ∠C = 180°
∠C = 180° - 80° - 40°
∠C = 60°
But since ∠F = 60°, ∠C =∠F
By Postulate 17 (AA Postulate), the above triangles are similar; the ratio of the pairs of
corresponding sides must be the same.
i.e.
©DD Productions
6
HAPTER 2-GEOMETRY:SIMILARIT
Example 2:
The figure above consists of two similar triangles. How can we prove they are similar? As
mentioned above, for two triangles to be similar, they must have 2 corresponding angles equal to
each other.
∠x = ∠y (Because they are opposite angles)
∠A = ∠C = 50°
Or
∠B = ∠D = 60°(alternate angles)
This proves that by Postulate 17 (AA Postulate), the above figure consists of 2 triangles which are
similar to one another, (∆AOB ∼ ∆COD).
2)Side Angle side Test (postulate 15): If two sides of one triangle are proportionate to the two
corresponding sides of the second triangle and the angles between the two sides of each triangle
are equal the two triangles are similar; if two sides of a triangle are equal, the angles opposite the
sides have equal measures.
i.e.
3)Side Side Side Test: If the three sides of one triangle are proportional to the three
corresponding sides of another triangle, then two triangles are similar.
i.e.
©DD Productions
7
HAPTER 2-GEOMETRY:SIMILARIT
Example 3:
∆ABC and ∆DEF have a one to one correspondence such that
AB =6 DE= 3
AC = 8 DF = 4
BC= 10 EF = 5.
Are the two triangles similar? If so, why?
Solution: AB/DE = AC/DF= BC/EF = 2/1. Yes, the two triangles are similar because their
corresponding sides are proportionate. And the measure of their corresponding lengths varies by a
factor, 2.
Congruent triangles:
Study these two triangles. How are they alike? If we rotated the pink triangle and slid it on top of
the green one, the two would fit on top of each other exactly. All the sides and all the angles are
equal, so these two triangles are congruent.
If two figures are congruent, then their corresponding parts are congruent. Let's look at the
corresponding parts of triangles ABC and DFE.
D
A
B
©DD Productions
C
F
E
8
HAPTER 2-GEOMETRY:SIMILARIT
A triangle has three sides and three angles. If two triangles are congruent, then the sides and
angles that match are called corresponding parts. Here, angle A corresponds to angle D, angle B
corresponds to angle F, and angle C corresponds to angle E. Side AB corresponds to side DF, side BC
corresponds to side FE, and side CA corresponds to side ED.
Congruent figures are named in the order of their corresponding parts. For these triangles, we
say "triangle ABC is congruent to triangle DFE," not "triangle DEF," because vertex A corresponds to
vertex D, vertex B corresponds to vertex F, and vertex C corresponds to vertex E.
∆𝐴𝐵𝐶 ≅ ∆𝐷𝐹𝐸
The lines crossing the sides (hash marks) are another way to keep track of which side
corresponds to another. The side with one hash mark in triangle ABC corresponds to the side with
one hash mark in triangle DFE.
How can we tell if two triangles are congruent? Just by looking at them is not good enough,
because our eyes can play tricks on us. One way is to trace around one of the triangles and place it
over the other. If the angles and sides match exactly, then the two are congruent. But there is
another way to tell. If two triangles have congruent corresponding parts, then the triangles are
congruent triangles.
You can measure the corresponding parts. If they are all congruent, then the triangles are
congruent.
N.B. These principles and rules of congruence hold true for any polygon.
Two triangles are congruent if one of the following conditions applies:
1. Three sides are the same:
The three sides of the first triangle are equal to the three sides of the second triangle (the SSS
rule: Side Side Side).
2. Two sides and one angle are the same:
Two sides of the first triangle are equal to two sides of the second triangle, and the included angle is
equal (the SAS rule: Side Angle Side).
©DD Productions
9
HAPTER 2-GEOMETRY:SIMILARIT
3. Two angles and one side are the same:
Two angles in the first triangle are equal to two angles in the second triangle, and one (similarly
located) side is equal (the AAS rule: Angle Angle Side).
4. Two sides in right-angled triangle are the same:
In a right-angled triangle, the hypotenuse and one other side in the first triangle are equal to the
hypotenuse and corresponding side in the second triangle (the RHS rule: Rightangled, Hypotenuse, Side).
SUMMARY:
DIFFERNCES BETWEEN CONGURANT AND SIMILAR TRIANGLES.
Congruent triangles are equal in all aspects i.e. their angles are equal, their sides are equal
and their areas are equal. Congruent triangles are always similar.
But similar triangles need not be congruent.
To establish the similarity of two triangles, it’s sufficient to satisfy one condition.
1. Their corresponding angles are equal.
2. Their corresponding sides are proportional
If the corresponding angles are equal, the triangles are equiangular.
©DD Productions
10
HAPTER 2-GEOMETRY:SIMILARIT
Truth relating two equiangular triangles.
A famous Greek mathematician Thales gave an important truth relating two equiangular
triangles which is as follows:
The ratio of any two corresponding sides in any two equiangular triangles is always equal.
Double
click!
Height of coloumns by comparing 2 shadows.rm
SPECIAL FIGURES
Some figures are always similar. Such figures include:
Circles: all circles are always 360 ͦ, therefore the length of radii will always be proportional in
two or more circles.
Squares: their internal angles are always the same despite the differing sizes of two different
squares i.e. 90 ͦ.
Equilateral triangles: their internal angles are always the same despite the differing sizes of
two different triangles i.e. 60 ͦ.
Perimeters and areas of similar triangles:
Perimeters:
When 2 polygons are similar, the reduced ratio of any two corresponding sides is the scale factor
of the similar polygons.
Consider the following triangles:
©DD Productions
11
HAPTER 2-GEOMETRY:SIMILARIT
The reduced factor of all sides will be equal (i.e. 8/4 = 6/3 = 4/2). All of these ratios reduce to
2/1. Therefore it is said that the scale factor of these similar polygons is 2: 1.
The perimeters of ∆ABC and ∆DEF are 18cm and 9cm respectively. By comparing the ratios of
the perimeters, you will find that it is also 2: 1. This brings us to our next theorem.
Theorem 60: if two similar polygons have a scale factor of a: b, then the ratio of their perimeters
will also be a: b.
Example 1: consider the following similar triangles (Lengths in cm):
ABC ∼ DEF, find the perimeter of DEF.
By theorem 60;
Thus, the perimeters of two similar triangles are in the ratio of their scale factor.
©DD Productions
12
HAPTER 2-GEOMETRY:SIMILARIT
Areas:
The ratio of the areas of similar polygons is equal to the ratio of the squares of the
corresponding sides’ ratio, i.e. the square of the scale factor.
Consider the following similar rectangles (Lengths in cm) :
The scale factor of these rectangles = 6/4 = 3/2
Area of ABCD = L*W
=2x4
= 8cm2
Area of EFGH = L*W
=6x3
= 18cm2
The ratios of the area = 18/8
= 9/4
=32/22
Therefore, when two similar polygons’ corresponding sides have the ratio of a: b, then the ratio
of their areas will be a2: b2
This is called theorem 61
Example: the following are similar triangles (Lengths in cm):
Area = ½ x base x height
©DD Productions
13
HAPTER 2-GEOMETRY:SIMILARIT
Area ∆ABC = ½ x 15 x 5
= 37.5 cm2
Area ∆MNO= ½ x 9 x 3
= 13.5 cm2
Corresponding sides of the triangles are in a 5:3 proportion.
15 5 5
= =
9
3 3
We found the areas to be 37.5 cm2 and 13.5 cm2. According to rule 4, they should be in the
proportion 52:32 (25:9).
37.5 25 52
=
= 2
13.5
9
3
Circle Areas :
The ratio of two circles’ areas is equal to the ratio of the squares of the radii or diameter.
This can also simply be said as the area of a circle varies directly with square of the radius or
diameter. This means if the radii are multiplied by a positive number, then the area is multiplied by
the square of that number.
Example:
(Lengths in cm)
Area of a circle = pie x radius2
Area of Circle A = pie x 42 = 50.27cm2
Area of Circle B = pie x 32 = 28.27cm2
The ratio of the circles’ radii is 4:3. So the ratio of their areas should be 42:32
50.27
28.27
=
16
9
=
42
32
©DD Productions
14
HAPTER 2-GEOMETRY:SIMILARIT
As mentioned, if the radii are multiplied by a number, then the area is multiplied by the square
of that number. Does this actually happen? Let’s multiply the radii by 3.
Circle A = pie x 122 = 452.39cm2 = 32 (50.27)
Circle B = pie x 92 = 254.47cm2 = 32 (28.27)
Similarity in 3-D Shapes
Everything around us is a 3D object; from our cereal boxes to our pencil cases. When these types
of things are produced in the factories, each product has to be similar with the other one (that is
why two text books of the same subject looks the same, two cereal boxes look the same). The
similarity in 3D shapes has to be taken into account before producing these types of goods. Other
examples could be capsules (tablet), chips packets e.t.c.
We already know that if two shapes are similar their corresponding sides are in the same ratio,
and their corresponding angles are equal, in addition to this, the shape must be the same. E.g. the
two shapes must be only cubes or only cylinders, but not a combination of a cube and a cylinder.
3D shapes can also be similar provided they follow the same rules.
Examples of similar 3D shapes:
Cuboids:
3cm
2cm
6cm
1cm
4cm
2cm
The length of cuboid 1 is twice the length of cuboid 2
The width of cuboid 1 is twice the width of cuboid 2
The height of cuboid 1 twice the height of cuboid 2
Therefore the scale factor is 2:1, all angles are same and so the cuboids are similar.
©DD Productions
15
HAPTER 2-GEOMETRY:SIMILARIT
Cubes :
a
b
a
a
b
b
All cubes are similar without any exception. This is because the length, width and height of a
cube is the same. So the ratio of corresponding sides will also be the same i.e.
Length = width = height
𝑎 𝑎 𝑎
= =
𝑏 𝑏 𝑏
Scale factor = a:b
Volume of similar 3D figures
What is the ratio of their volumes?
Cube1 has a volume of a3
Cube 2 has a volume of b3
The ratio of their volumes is a3:b3
What is the ratio of the corresponding sides (scale factor)?
a: b
This means that for a cube of scale factor a: b the ratio of their volume is a3:b3
©DD Productions
16
HAPTER 2-GEOMETRY:SIMILARIT
This leads to the following conclusion: for any similar 3D shapes with a scale factor of a: b, the
ratio of the volumes will be a3:b3
Example 1:
Look at the shapes below
60cm3
480cm3
5 cm
10 cm
The ratio of corresponding sides is 5:10 which when reduced is 1:2
The ratio of the volume is 60:480 which when reduced you get 1:8
Therefore:
1 = 1 which is the same as ½ = (1/2)3
2 8
This proves our conclusion.
The theorem can also be expressed as follows:
“The ratio of the volumes of similar solids is equal to the ratio of the cubes of the
corresponding sides’ ratio,” i.e.
Example 2: Spheres:
𝑽𝒐𝒍𝒖𝒎𝒆 𝟏
𝑽𝒐𝒍𝒖𝒎𝒆 𝟐
= (𝐚/𝐛)𝟑
Sphere 1
Sphere 2
4
3
Volume = 4/3 x pie x radius3
©DD Productions
17
HAPTER 2-GEOMETRY:SIMILARIT
Volume of sphere 1 = 4/3 x pie x 43 = 268.08cm3
Volume of sphere 2 = 4/3 x pie x 33 = 113.1cm3
According to the theorem of volumes, the ratios should be 43:33 (64:27).
268.08 64 43
=
=
113.1
27 33
What if the radii were multiplied by 2? What would happen?
Sphere 1 = 4/3 x pie x 83 = 2144.66cm3 = 23 (268.08)
Sphere 2 = 4/3 x pie x 63 = 904.78cm3 = 23 (113.1)
To conclude, “if two objects are similar and the ratio of corresponding sides is k, then the ratio of
their volumes is k3.”
Areas of similar 3D figures:
The rule for finding the surface areas of similar 3D figures is the same for finding the area of
similar shapes. I.e. the ratio of the areas of similar 3D shapes is equal to the ratio of the squares of
the corresponding sides’ ratio.
Example 1: Consider the following similar cuboids:
Just like with 2-dimensional figures, the scale factor of any two figures is determined by the ratio
of its corresponding sides.
In the above example, the scale factor is 6/3= 4/2= 2/1
Surface area of figure 1 = 2(LW) + 2(LH) + 2(HW)
= 2(6cm2) + 2(3cm2) +2(2cm2)
= 12cm2 + 6cm2 + 4cm2
= 22cm2
Surface area of figure 2 = 2(LW) + 2(LH) + 2(HW)
= 2(24cm2) + 2(12cm2) +2(8cm2)
= 48cm2 + 24cm2 + 16cm2
= 88cm2
©DD Productions
18
HAPTER 2-GEOMETRY:SIMILARIT
Ratio of surface areas= 88/22
= 4/1
But the scale factor = 2/1
Therefore
4/1=22/12
Just like with 2-dimesional figures, if the scale factor of 2 similar 3-dimensional figures is a: b,
then the ratio of its surface area is a2: b2
Example 2: Similar Cylinders
Area = 2 x pie x radius x height
Area of cylinder 1 = 2 x pie x 6 x 21 = 791.68cm2
Area of cylinder 2 = 2 x pie x 2 x 7 = 87.96cm2
791068 9 32
= = 2
87.96
1 1
Cylinder 1 = 2 x pie x 18 x 63 = 7125.13cm2 = 32 (791.68)
Cylinder 2 = 2 x pie x 6 x 21 = 798.68cm2 = 32 (87.96)
Example 3:
Que. The area of the bases of two similar glasses is in the ratio 4:25. Find the ratio of
their volumes.
©DD Productions
19
HAPTER 2-GEOMETRY:SIMILARIT
Solution:
Your first step is to find the reduced ratio (scale factor)
4:25 is the area ratio that means that it is in the form of (a:b)2
Reduce the ratio to get the scale factor. Reduce it by finding the square root
Your answer will be 2/5
Now use the normal process to find volume ratio
2/5 = (2/5)3
Final answer – 8/125 = 8:125 (ratio of volumes)
Revision Exercise:
1) For each of the following pairs of triangles, state whether they are congruent. If they are,
give a reason for your answer (SSS, SAS, AAS or RHS).
Pair 1
Pair 2
Pair 3
2) Two similar polygons vary in the ratio 3:4. Find:
©DD Productions
20
HAPTER 2-GEOMETRY:SIMILARIT
i) The side of the larger polygon if the corresponding side of the smaller polygon is 9 cm
ii) The perimeter ratio of the two
iii) The area ratio between the two
iv) The new perimeter when the old perimeter is 42cm and the sides are multiplied by 3
v) The new area if the sides of the two polygons are multiplied by 3 and the old area was
108cm2
3) The ratio of the perimeters of two rectangles is in the ratio 5:4. The sum of their areas is
20.5cm2. Find the area of each rectangle.
4) The scale factor of 2 similar triangles is 3:4. The perimeter of triangle 1 is 24 cm. what is the
perimeter of triangle 2?
5) Two similar cylindrical tins have the base radii 6cm and 8 cm respectively. If the area of the
larger tin is 1508cm2, find
i) The height of the larger tin
ii) The area of the smaller tin without using its height.
6) The ratio of the surface areas of 2 cubes is 16:9. The sum of their volumes is 91cm3. Find the
volume of each cube.
7) the ratio of the volumes of two cuboids is 27: 8. The sum of their surface areas is 65cm2. Find
the surface area of each cuboid.
8) Areas of two similar triangles are 144 cm². and 81cm ². If one side of the first triangle is 6 cm
then find the corresponding side of the second triangle.
9) The side of an equilateral triangle ABC is 5 cm. Find the length of the side of another
equilateral ∆PQR whose area is four times area of ∆ABC.
10) The corresponding sides of two similar triangles are 4 cm and 6 cm. Find the ratio of the
areas of the triangles.
11) These two shapes are similar. What is the length of x?
©DD Productions
21
HAPTER 2-GEOMETRY:SIMILARIT
12) Two similar pyramids have volumes of 64cm3 and 343cm3. What is the ratio of their surface
areas?
Answers:
1) Yes. RHS
Yes. SSS
No. The side of length 7cm is not in the same position on both triangles. Therefore, it
is not AAS.
2) i) 12 cm ii) 3:4 iii) 9:16 iv) 126 cm2 v) 972 cm2
3) The areas of the two rectangles are 12.5cm2 and 8cm2 respectively.
4) 28 cm
5) i)30 cm ii) 848.25 cm2
6) Volumes of the cubes are 64cm3 and 27cm3 respectively.
7) The surface areas of the two similar cuboids are 45cm2 and 20cm2 respectively.
8) 4.5 cm
9) 10 cm
10) 0.44
11) x = 2.4cm
12) 16:49
©DD Productions
22
HAPTER 2-GEOMETRY:SIMILARIT
Danish, you have generally done a very good job. You could have done more examples and more
exercised in each case.
Good idea to upload a media file…
GRADE X - ASSESSMENT - SIMILARITY AND CONGRUENCY
Writing
Drawing
skill
skill
3
2
creativity pictures
flow
labeling
neatness
3
2
Formatting
skill
3
page
paragraphs
math
equations
3
3D
Polygon Triangle
shapes
Questions
Examples Total
2
2
3
3
2
20
always- theorems perimeter
3D
ex+ans
similar
applns
vol, SA
shapes
polygons
congruency
appln
variety
2
2
3
3
1
19
You submitted it late. So 2 marks are reduced: your mark is 17!
©DD Productions
23