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What is similarity?
In geometry, two figures are said to be similar if they have the same shape, though not necessarily the
same size. In other words,one of them is the exact scale model of the other. For example, any two line
segments are similar; any two circles are similar; any two squares are similar. When we talk of two
geometric figures having the same shape,it is always possible to stretch or shrink the first figure in a given
scale, without changing its shape,and to get the second figure. There is one to one correspondence
between the parts of the two figures. The symbol used to denote similarity between two figures is "
"
Eg 1
A
15
cm
12
cm
D
5 cm
B
C
E
6 cm
4 cm
F
2 cm
Triangle ABC and triangle DEF are similar because each side of triangle ABC is three times
as long as each side of triangle DEF.
Eg 2
Real life example: Design engineers and architects most often deal with such figures. a
newly designed structure is first drawn to scale on a paper. The design is much smaller
than the structure itself,but all its parts have the same shape as in the structure.
Blueprints of these drawings are made.The blueprint can be well read by the
contractor.he can determine the true dimensions of any part of the structure represented
in the blueprint on the basis of the scale to which the blueprint is drawn.
S
10cm
R
95°
108°
6cm
U
108°
X
95°
2cm
85°
72°
Z
6cm
T
12 cm
3cm
4 cm
85°
72°
5cm
W
Y
There are some special properties for two figures to be similar. They are as follows:


Their corresponding angles are equal.
The lengths of their corresponding sides are propotional.
For example, consider the following squares.
Thus the squares are similar figures as their corresponding sides are proportional and their
corresponding angles are equal.
Note:


Each side of figure PQRS has been multiplied by 2 to obtain the sides of figure ABCD. The
number 2 is called the scale factor.
Similar figures are equiangular (i.e. the corresponding angles of similar figures are equal).
In similar figures the corresponding angles are equal and the corresponding side are in
the same ratio. Not only are the corresponding angles the same size in similar polygons,
but also the sides are proportional.
Corresponding:
Propotional:
Ratio:
Figures that are always similar are:
Two triangles are similar if:


two pairs of corresponding angles are congruent (therefore the third pair of corresponding angles
are also congruent).
the three pairs of corresponding sides are proportional.
In the case of triangles, if either of the two conditions holds, the other holds automatically. Thus,either of
the two conditions will ensure the similarity of the two triangles.(the ratio of lengths of any two
corresponding sides in similar triangles is always the same irrespective of their actual sides)
Referring to the above figure:
KLM =
EFG,
KML =
EGF and
LKM =
FEG
KL = KM = LM
EF EG FG
Thus, we can say that the triangle KLM and triangle EFG are similar. Symbolically we write the above
relation as
KLM
EFG
Theorems for similar triangles
1. Angle-Angle (AA) Similarity
If two angles in a triangle are congruent to the two corresponding angles in a second triangle, then the two
triangles are similar.
Example 1: Let ABC be a triangle and A'C' a segment parallel to AC. What can you say about triangles ABC
and A'BC'? Explain your answer.
Solution to Example 1:
Since A'C' is parallel to AC, angles BA'C' and BAC are congruent. Also angles BC'A' and BCA are congruent.
Since the two triangles have two corresponding congruent angles, they are similar.
2. Side-Side-Side (SSS) Similarity
If the three sides of a triangle are proportional to the corresponding sides of a second triangle, then the triangles
are similar.
Example 2: Let the vertices of triangles ABC and PQR defined by the coordinates: A(-2,0), B(0,4), C(2,0), P(1,1), Q(0,3), and R(1,1). Show that the two triangles are similar.
Solution to Example 2:
Let us first plot the vertices and draw the triangles.
Since we know the coordinates of the vertices, we can find the length of the sides of the two triangles.
AB = sqrt [ 4 2 + 2 2 ] = 2 sqrt(5)
BC = sqrt [ (-4) 2 + 2 2 ] = 2 sqrt(5)
CA = sqrt [ 4 2 ] = 4
PQ = sqrt [ 2 2 + 1 2 ] = sqrt(5)
QR = sqrt [ (-2) 2 + 1 2 ] = sqrt(5)
RP = sqrt [ 2 2 ] = 2
We now calculate the ratios of the lengths of the corresponding sides.
AB / PQ = 2 , BC / QR = 2 and CA / RP = 2
We can now write.
AB / PQ = BC / QR = CA / RP = 2
The lengths of the corresponding sides are proportional and therefore the two triangles are similar.
3. Side-Angle-Side (SAS) Similarity
If an angle of a triangle is congruent to the corresponding angle of a second triangle, and the lengths
of the two sides including the angle in one triangle are proportional to the lengths of the
corresponding two sides in the second triangle, then the two triangles are similar.
Example 3: Show that triangles ABC and A'BC', in the figure below, are similar.
Solution to Example 3:
Angles ABC and A'BC' are congruent.
Since the lengths of the sides including the congruent angles are given, let us calculate the ratios of
the lengths of the corresponding sides.
BA / BA' = 10 / 4 = 5 / 2
BC / BC' = 5 / 2
The two triangles have two sides whose lengths are proportional and a congruent angle included
between the two sides. The two triangles are similar.
Similar Triangles: Perimeters and Areas
When two triangles are similar, the reduced ratio of any two corresponding sides is called the scale
factor of the similar triangles. In Figure 1 , Δ ABC∼ Δ DEF.
Figure 1 Similar triangles whose scale factor is 2 : 1.
The ratios of corresponding sides are 6/3, 8/4, 10/5. These all reduce to 2/1. It is then said that the
scale factor of these two similar triangles is 2 : 1.
The perimeter of Δ ABC is 24 inches, and the perimeter of Δ DEF is 12 inches. When you compare
the ratios of the perimeters of these similar triangles, you also get 2 : 1. This leads to the following
theorem.
Theorem 60: If two similar triangles have a scale factor of a : b, then the ratio of their perimeters is
a : b.
Example 1: In Figure 2 , Δ ABC∼ Δ DEF. Find the perimeter of Δ DEF
Figure 2 Perimeter of similar triangles.
Figure 3 shows two similar right triangles whose scale factor is 2 : 3. Because GH ⊥ GI and JK ⊥ JL
, they can be considered base and height for each triangle. You can now find the area of each
triangle.
Figure 3 Finding the areas of similar right triangles whose scale factor is 2 : 3.
Now you can compare the ratio of the areas of these similar triangles.
This leads to the following theorem:
Theorem 61: If two similar triangles have a scale factor of a : b, then the ratio of their areas is a2 :
b2.
Example 2: In Figure 4 , Δ PQR∼ Δ STU. Find the area of Δ STU.
Figure 4 Using the scale factor to determine the relationship between the areas of similar triangles.
The scale factor of these similar triangles is 5 : 8.
Example 3: The perimeters of two similar triangles is in the ratio 3 : 4. The sum of their areas is 75
cm2. Find the area of each triangle.
If you call the triangles Δ 1 and Δ2, then
According to Theorem 60, this also means that the scale factor of these two similar triangles is 3 : 4.
Because the sum of the areas is 75 cm 2, you get
Example 4: The areas of two similar triangles are 45 cm 2 and 80 cm2. The sum of their perimeters is
35 cm. Find the perimeter of each triangle.
Call the two triangles Δ 1 and Δ2 and let the scale factor of the two similar triangles be a : b.
a : b is the reduced form of the scale factor. 3 : 4 is then the reduced form of the comparison of the
perimeters.
Reduce the fraction.
Take square roots of both sides.
Congruence
Congruent figures have the same size and the same shape. The corresponding sides and angles of
congruent figures are equal.
Square ABCD is congruent to square EFGH as their corresponding sides and angles are equal.
Note:
Congruent figures are exact duplicates of each other. One could be fitted over the other so that their
corresponding parts coincide.
The concept of congruence applies to figures of any type.
Congruent Triangles
Congruent triangles have the same size and the same shape. The corresponding sides and the
corresponding angles of congruent triangles are equal.
Note:
Principles of Congruent Triangles
The following principles of congruence are used depending on the information given.
1. The side-side-side (SSS) principle
Two triangles are congruent if corresponding sides are equal.
2. The side-angle-side (SAS) principle
Two triangles are congruent if two pairs of corresponding sides and the angle included between the sides
are equal.
3. The angle-side-angle (ASA) principle
Two triangles are congruent if two pairs of corresponding angles and a pair of corresponding sides are
equal.
4. The right angle-hypotenuse-side (RHS) principle
Two right-angled triangles are congruent if the hypotenuses and one pair of corresponding sides are equal.
3D shapes
Similar 3D shapes have the same shape but they can be of different sizes.
Volumes, Surface areas.
When solid objects are similar, one is an accurate enlargement of the other.
If two objects are similar and the ratio of corresponding sides is k, then the ratio of their volumes is k³.
A line has one dimension, and the scale factor is used once.
An area has two dimensions, and the scale factor is used twice.
A volume has three dimensions, and the scale factor is used three times.
eg
2 cm
4 cm
2 cm
4 cm
2 cm
4 cm
For volume the similarity is:
V1/V2 = (K³)1/(K³)2
For surface area the similarity is:
SA1/SA2 = (S²)1/(S²)2
Questions
Problem 1:In the isosceles triangle ABC, BA and BC are congruent. M and N are points on AC such that
MA is congruent to MB and NB is congruent to NC. Show that triangles AMB and CNB are congruent.
Problem 2: ABCD is a parallelogram and BEFC is a square. Show that triangles ABE and DCF are
congruent.
Problem 3: ABCD is a square. C' is a point on BA and B' is a point on AD such that BB' and CC'are
perpendicular. Show that AB'B and BC'C are congruent.
Problem 4: ABC is a triangle and M is the midpoint of AC. I and J are points on BM such that AI and CJ
are perpendicular to BM. Show that triangles AIM and CJM are congruent.
Problem 5
Find the value of x in the following pair of triangles.
Problem 6
Find the value of the height, h m, in the following diagram at which the tennis ball must be hit so that it
will just pass over the net and land 6 metres away from the base of the net.
Problem 7
Adam looks in a mirror and sees the top of a building. His eyes are 1.25 m above ground level, as shown
in the following diagram.
If Adam is 1.5 m from the mirror and 181.5 m from the base of the building, how high is the building?
Problem 8
Find V
60 cm³
5 cm
V
10 cm
SOLUTIONS
Solution to Problem 1:

Since triangle ABC is isosceles and BA and BC are congruent then angles BAM and BCN are
congruent.
Also since MA is congruent to MB, then AMB is an isosceles triangle and angles BAM and ABM are
congruent. NB and NC are also congruent; CNB is an isosceles triangle and angles CBN and BCN are
congruent. In fact all four angles BAM, ABM, CBN and BCN are congruent. Comparing triangles BAM
and CNB, they have corresponding sides AB and BC congruent, corresponding angles BAM and BCN
congruent and corresponding angles ABM and CBN congruent. These two triangles are therefore
congruent. This is the ASA congruent case.
Solution to Problem 2:

In the parallelogram ABCD, BA is congruent to CD. In the square BEFC, EB is congruent to FC.
Since EB is parallel to FC and BA is parallel to CD then angles EBA and FCD are congruent.
Comparing triangles ABE and DCF: angle EBA included between EB and BA in triangle ABE is
congruent to angles FCD included between sides FC and CD. EB is congruent to FC and BA is
congruent to CD. These two triangles are congruent. It is the SAS congruent case.
Solution to Problem 3:

Since ABCD is a square angles CBC' and BAB' are right angles and therefore congruent. Also
side BA is congruent to side BC. BC and AD are parallel and BB' is a transverse, therefore angles
OBC and BB'A are interior alternate angles and are congruent.

Since CC' and BB' are perpendicular, then triangle CBO is rectangle at point O and therefore
size of angle OBC + size of angle BCO = 90 degrees

ABB' is also a right triangle and therefore
size of angle ABB' + size of angle BB'A = 90 degrees

Combine the above equations with the fact that angles OBC and BB'A are congruent, we can
conclude that
size of angle ABB' = size of angle BCC'

Triangles AB'B and BC'C have side BC congruent to side BA; angle BCC' congruent to angle
ABB' and angle BAB' congruent to angle CBC' are congruent. The two triangles are congruent.
This is the ASA congruent case.
Solution to Problem 4:

Since M is the midpoint of AC then AM is congruent to MC. AI and CJ are perpendicular to the
same line BM and are therefore parallel with CA as the transverse. Angles MAI and MCJ are
interior alternate angles and therefore congruent. Angles AMI and JMC are vertical angles and
therefore congruent. Triangles AIM and CJM have one congruent side between two congruent
angles and are therefore congruent. This is the ASA congruent case.
Solution to problem 5
solution to problem 6
Solution:
So, the height at which the ball should be hit is 2.7 m.
Solution to problem 7
So, the height of the building is 150 m.
Solution to problem 8
V=120cm³
Shaleen, you have completely copy-pasted from cliffnotes.. that is not what was expected from you..
You haven’t dealt with many things asked in the question in a proper way like polygons, surface area etc.
Please try to do things on your own to some extent.
GRADE X - ASSESSMENT - SIMILARITY AND CONGRUENCY
Writing
skill
3
creativit
y flow
neatnes
s
1
Drawing
skill
2
pictures
labeling
1
Formattin
g skill
3
page
paragraph
s math
equations
2
Polygon
2
alwayssimilar
polygon
s appln
0
Triangle
2
theorem
s
applns
3D
shapes
3
perimete
r
vol,
SA
1
1
Questions
3
3D
shapes
congruenc
y variety
0
Example
s
2
ex+ans
1
Total
20
7