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Mohamed Bhanji
10B
Similarity:
Figures which have the same shape are called similar figures. Two shapes are Similar if the
only difference is size (and possibly the need to turn or flip one around).
Example:
One shape can become another using Resizing (also called dilation, contraction, compression,
enlargement or even expansion), then the shapes are Similar:
These Shapes are Similar!
There may be Turns, Flips or Slides, Too!
Sometimes it can be hard to see if two shapes are Similar, because you may need to turn, flip or
slide one shape as well as resizing it.
Mohamed Bhanji
10B
Rotation
Turn!
Reflection
Flip!
Translation
Slide!
Examples
These shapes are all Similar:
Resized
Resized and Reflected
Resized and Rotated
When two shapes are similar, then there are some special properties:
1. Their corresponding sides are in the same ratio.
2. Their corresponding angles are equal
This can make life a lot easier when solving geometry puzzles, as in this example:
Mohamed Bhanji
10B
Example: What is the missing length here?
Notice that the red triangle has the
same angles as the main triangle ...
... they both have one right angle,
and a shared angle in the left corner
In fact you could flip over the red triangle, rotate it a little, then resize it and it would fit
exactly on top of the main triangle. So they are similar triangles.
So the line lengths will be in proportion, and we can calculate:
? = 80 × (130/127) = 81.9
The polygons below are similar.
We can verify that if one polygon is similar to another polygon and this second polygon is
similar to the first polygon, Then the first polygon is similar to the second one.
In the two quadrilaterals (a square and a rectangle) shown below, the corresponding angles
are equal, but their corresponding sides are not in the same ratio.
Similarly, we may note that in that in the two quadrilaterals (a square and a rhombus) shown
below, corresponding sides are in the same ratio, but their corresponding angles are not
equal. So again, the two polygons are not similar.
A
B
D
A
D
C
C
B
Mohamed Bhanji
10B
SIMILARITY OF TRIANGLES.
Two triangles are said to be similar, if their corresponding angles are equal and
corresponding sides are proportional.
A= D, B= E and C=F
If ∆s ABC and DEF are similar, then
And AB = BC = AC
DE EF DF
A
B
D
C
E
F
THREE SIMILARITY POSTUALATERS FOR TRAINGLES.
1) SAS-postulate: if two triangles have a pair of corresponding angles equal and the
sides including them proportional, then the triangles are similar.
if in
s ABC and DEF
A
D
A=D
And AB = BC
DE EF
THEN
B
C
E
F
ABC ∆DEF
AAA-postulate: (Angle-Angle-Angle) This is when, two triangles have the pairs of
corresponding angles equal, the triangles are similar.
A= D AND B= E
Then ∆ABC~ ∆DEF
Mohamed Bhanji
10B
Proof: take two triangles ABC and DEF such that A = D, B = E and C = F
Cut DP=AB and DQ= AC and join PQ.
D
A
P
B
C
E
Q
F
IN ∆’S ABC and DPQ,
AB=DP
AC=DQ
BAC =PDQ
∆ABC  ∆DPQ
│ SAS congruence criterion.
This gives
B = P
B= E
P = E
But these form a pair of corresponding angles.
PQ║EF
DP = DQ
PE QF
By basic proportionality Theorem
PE = QF TAKING RECIPROCALS
DP DQ
PE +1 = QF +1
DP
DQ
PE + DP = QF + DQ
DP
DQ
DE= DF
DP DQ
DE= DF
AB AC
AB = AC
DE DF
SIMILARLY,
AB = BC and so AB = BC = AC
DE EF
DE EF DF
Mohamed Bhanji
10B
THEREFORE ∆ABC ~ ∆DEF
SSS-postulate: if two triangles have their three pairs of corresponding sides proportional,
the triangles are similar.
D
A
IF IN ∆S ABC and DEF
AB = BC = AC
DE
EF
DF
THEN ∆ABC~∆ DEF
B
C
E
F
AREAS OF SIMLAR TRIANGLES
THEOREM: the ratio of the areas of two similar triangles is equal to the square of the ratio
of their corresponding sides.
Given: two triangles ABC and PQR such that ∆ABC ~ ∆PQR
To prove.
ar(ABC) = AB
ar(PQR)
PQ
2
= BC
QR
ar ( ABC)=1 BC x AM
2
And
ar(PQR)=1 QR x PN
2
So,
ar (ABC) = 1 x BC x AM
2
ar (PQR) = 1 x QR x PN
2
= BC x AM
QR x PN
2
= CA
RP
2
Mohamed Bhanji
10B
Now, in ∆ABM and ∆PQN,
B = Q
(As ∆ABC - ∆ PQR)
M = N
(Each is of 90º)
So, ∆ABM - ∆PQN
│ AA similarity criterion
Therefore, AM = AB
PN PQ
….(2)
│Corresponding sides of two similar triangles are proportional
Also, ∆ABC - ∆PQR
(Given)
So, AB = BC = CA
PQ QR RP
…(3)
│. . Corresponding sides of two similar triangles are proportional
Therefore, ar(ABC) = AB x AM
ar (PQR) = PQ PN
[From (1) and (3)]
= AB x AB
PQ PQ
= AB
PQ
[From (2)]
2
Now using (3), we get
ar (ABC) = AB
ar (PQR)
PQ
2
x BC
QR
2
x CA
RP
2
EXAMPLES
2
Example 1.
Solution: ∆ABC ~ ∆PQR
 ar (∆ABC) =
Ar (∆PQR)
BC
QR
and 121 cm2.
│ Given
2
│The ration of the areas of two similar triangles is equal to the square of the ratio of their
Corresponding sides.
Mohamed Bhanji
10B
64 = BC
121 15.4
2
8
11
2
2
=
BC
15.4
Taking square
root on both sides
BC = 8 x 15.4
11
BC = 11.2 cm.
Example 2. D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ABC.
Find the ration of the areas ∆DEF and ∆ABC.
Solution : D, E and F are respectively the mid-points of side AB, BC and CA of ∆ABC.
=
1
2 BC
B
and
AB = BC = CA
DE EF
FD
. . The corresponding sides of two similar triangles
Are proportional.
Form first two of (2),
AB = BC
DE EF
. . AM and DN are the medians.
= 2BM
2EN
= BN
EN
Also,ABM =  DEF
….(3)
….(4)
Example: Prove that the area of an equilateral triangle described on one side of a square is
equal to half the area of the equilateral triangle described on one of its diagonals.
Solution ; Give ABCD is a square whose one diagonal is AC. ∆APC and ∆BQC are two
equilateral triangles described on the diagonal AC and side BC of the square ABCD.
Mohamed Bhanji
10B
To prove. ar (∆BQC) = 1 ar (∆APC)
2
P
A
B
Q
D
C
Proof : . . ∆APC and ∆BQC are both equilateral triangles
 ∆APC ~ ∆BQC
│ AAA similarity criterion
 ar(∆APC) = AC
ar(∆BQC) BC
2
The ratio of the areas of two similar triangles is
equal to the square of the ratio of their corresponding
sides.
= AC2
BC2
= √2 BC
BC
2
│  Diagonal = √2 side
=2
ar(∆BQC) =1 ar(∆APC).
2
Example 3: A line PQ is drawn parallel to the base BC of ∆ABC which meets sides AB and
AC at points P and Q respectively. If
AP = 1 PB; find the value of :
3
Area of ∆ABC
(1)
Area of ∆APQ
(2)
Area of ∆ABC
Area of trapezium PBCQ
Mohamed Bhanji
10B
Solution :
AP=1 PB
3
AP:PB= 1:3
AP+PB = AB
A
AP= 1 AB
1+3
=
1 AB
4
P
B
Q
Q
C
In ∆s APQ and ABC, we have
APQ= corresponding ABC
AQP = corresponding ACB
By AA-similarity
∆APQ ~ ∆ ABC
i.
the ratio of the corresponding of similar
triangles
Area of ∆ABC = AB2
Area of ∆APQ
AP2
to the ratio of their corresponding sides.
 Area of ∆ABC = AB2 = 16
Area of ∆APB 1AB2
1
16
Area of ∆APQ = AP2
Area of ∆ABC AB2
1AB2
16 = 1
AB2 16
Area of ∆APQ
= 1
Area of ∆ABC-Area of ∆APQ
Area of ∆APQ
= 1
Area of trapezium PBCQ
16-1
Mohamed Bhanji
10B
DIFFERNCES BETWEEN CONGURANT AND SIMILAR TRIANGLES.
Congruent triangles are equal in all respects 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
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.
It is believed that he had used the Basic proportionality Theorem (now known as Thales
Theorem) for the same.
BASIC PROPOTIONALITY THEOREM.
If a line is drawn parallel to one side of the triangle to intersect the other two sides is
distinpoints, the other two sides are divided in the same ratio.
Given: A triangle ABC in which a line parallel to side BC intersects the other two sides AB
and AC at D and E respectively.
To prove: AD = AE
DB
EC
CONSTRUCTION: Join BE and CD and then draw DM  AC and EN  AB
Proof: Area of ∆ADE= 1 Base x height
Mohamed Bhanji
10B
2
=1AD X EN
2
ar(ADE)= 1 DB x EN
2
ar( ADE)=1EC x DM
2
Therefore, ar(ADE)
ar(BDE)
1AD x EN
2
1AD x EN
2
= AD
DB
CRITERIA FOR SIMILARITY OF TRIANGLES
1. SYMBOL OF SIMILARITY
A
D
B
C
E
F
Here we can see that A corresponds to E and C corresponds to F. Symbolically, we write
similarity of the two triangles as ‘∆ABC~ ∆DEF’ and read it ‘as triangle ABC is similar to
Triangle DEF’. So the symbol ~ stands for ‘is similar to’.
Minimum essential requirements of the two triangles.
Now, we will examine that for checking the similarity of two triangles, say ABC and DEF,
whether we should always look for all the equality relations for the corresponding angles.
A= D, B =E, C=F
AB = BC = CA
DE
EF
FE
We may recall that, we have some criteria for congruency of two triangles
involving only two pairs of corresponding parts or elements of the two triangles. Here also,
we shall try to arrive certain criteria for the similarity of two triangles involving relationship
between less numbers of pairs of corresponding part of the two triangles, instead of all six
Mohamed Bhanji
10B
pairs of corresponding parts.
EXAMPLE: draw the line segments BC and EF of two different lengths, 3cm and 5 cm
respectively. Then, at the points B and C respectively, construct angles PBC and QCB of
some must measure, say, 60º and 40°. Also at the points E and F, construct angles REF and
SFE of 60º and 40º respectively.
R
P
S
Q
60º 40º
º
B
D
40º
º
60º
C
E
F
Let rays BP and CQ intersect each other at A and rays ER and FS intersect each other
at D. In the two triangles ABC and DEF, we see that B= E, C=F AND A =D
That is corresponding angles of these two triangles are equal. Regarding CA are equal to 0.6.
FD
BA= 3 = 0.6
EF 5
Thus AB = BC = CA
DE EF DF
EXAMPLES
Example 1. State which pairs of triangles in figure are similar. Write down the similarity
criterion used by you for answering the questions and also write pairs of similar triangles in
symbolic form.
i.
iv.
60º
70
60º
80 40
80
6
ii.
3
5
2
2.5
4
iii.
2.5
70 5
6
70
40
70
Mohamed Bhanji
10B
Solution. (i) In ∆ABC and PQR
A = P
∆ABC ~ ∆PQR
B = Q
AAA similarity criterion.
C =R
ii. In ∆ABC and ∆QPR
AB = BC = CA
QR RP PQ
iii. in ∆MNL and ∆QPR
ML = MN
QR QP
NML =PQR
And
∆MNL ~ ∆PQR
SAS similarity criterion
iv.in ∆DEF and ∆PQR
D = P (=70º)
E = Q (=80º)
F = R (=30º)
Areas of similar shapes:
The following rectangles are similar and their ratio of corresponding sides is y. WXYZ is of
length b and width a.
B
Y
X
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑊𝑋𝑌𝑍 = 𝑎 × 𝑏
a
ya
𝑎𝑟𝑒𝑎 𝑜𝑓 𝐴𝐵𝐶𝐷 = 𝑦𝑎 × 𝑦𝑏 = 𝑦 2 𝑎𝑏
𝑎𝑟𝑒𝑎 𝐴𝐵𝐶𝐷 𝑦 2 𝑎𝑏
∴
=
= 𝑦2
𝑎𝑟𝑒𝑎 𝑊𝑋𝑌𝑍
𝑎𝑏
W
b
Z
A
Yb
“If two figures are similar and their sides are in the ratio y then their areas will be in the
ratio y2”.
Example:
∆𝐴𝐶𝐸~∆𝐵𝐶𝐷
C
D
Mohamed Bhanji
10B
If the area of ∆ 𝐵𝐶𝐷 = 6 cm2, find the
area of ∆ 𝐴𝐶𝐸.
C
2 cm
3 cm
B
D
Solution:
𝐵𝐶
2
A
Ratio of corresponding sides = 𝐴𝐶 = 3
2
4
Ratio of areas = (3)2 = 9 =
4
9
𝑎𝑟𝑒𝑎 ∆ 𝐵𝐶𝐷
E
Area ∆ 𝐴𝐶𝐸 = 𝑥
𝑎𝑟𝑒𝑎 ∆ 𝐴𝐶𝐸
6
=𝑥
𝑥=
6×9
4
1
= 13 2
Perimeters of similar shapes:
The following rectangles are similar and their ratio of corresponding sides is 2.
Perimeter of ABCD = (2 + 3) × 2 = 10 𝑐𝑚
Perimeter of WXYZ = (4 + 6) × 2 = 20 𝑐𝑚
∴
Perimeter WXYZ 20
=
=2
Perimeter ABCD 10
C
B
2 cm
A
4 cm
W
3 cm
Y
X
6 cm
Z
D
“If the ratio of corresponding sides of two similar figures is y, then the ratio of their
perimeters is also y.”
More Examples:
1. Ally washes a napkin in hot water and it shrinks by 10%. What is the percentage reduction
in its area?
Mohamed Bhanji
10B
Solution:
Ratio of sides =
100−10
100
9
90
9
= 100 = 10
81
Ratio of areas = (10)2 = 100
∴ The reduction in area = 100 − 81 = 19%
2. On heating, a metallic strip expands by 30%. What is the percentage of increase in the
surface area?
Solution:
130
13
Ratio of corresponding sides = 100 = 10
13
169
Ratio of areas = (10)2 = 100
Percentage increase in area =169 − 100 = 69%
3D OBJECTS
3D OBJECTS ARE THOSE OBCTS THAT HAS 3 DIFFERENT DIMENSIONS. Almost
everything around us is 3d objects.
ARE 3D SHAPES EVER GEOMETRICALLY SIMILAR?
This is one question that mathematicians have been working on, and finally they reached a
point where it was proved that Yes, 3D shapes are similar.
SIMILARITIES OF 3D OBJECTS
Two figures having the same shape not necessarily the same size are similar figures. Similar
figures have the SAME shape but not the same size as spheres of different radii and cubes
and cuboids of different sides.
We say these types of object 3D objects because they have sides which are equal to each
other.
For example: a cube and a cuboids has got 6 sides but its length, width and height are
repetitions of their opposite sides.
Same applies to a sphere because it has the repetitions of its radii.
Mohamed Bhanji
10B
Similar shapes can also be congruent. We know that all congruent figures are similar but not
all similar figures are congruent.
e.g
SURFACE AREAS OF SIMILAR 3D OBJECTS
THEOREM: the ratio of the surface areas of two similar 3D objects is equal to the square of
the ratio of their corresponding sides.
4cm
A
8cm
3cm
8cm
B
6cm
16cm
Lets first find out the surface areas first and then we will compare them and find out whether
they are similar or not.
Surface area of a cuboid = S = 2lb + 2lh + 2hb
we got this formula
of surface area
Surface area of cuboid A = 2 × (3×4) + 2 × (4×8) + 2 × (3×8)
because the opposite
S = 2×12 + 2×32 + 2×24
rectangles making
S = 24 + 64 + 48
the cuboid are equal.
S = 136cm2
So we multiply the
by 2
Surface area of cuboid A = 2 × (6×8) + 2 × (8×16) + 2 × (6×16)
S = 2×48 + 2×128 + 2×96
S = 96 + 256 + 192
S=544cm2
Mohamed Bhanji
10B
Now we will find the scale factor of one of the
Scale factor =new measurement
old measurement
= 136 =1
544 4
=1:4
Therefore the scale factor (k) = 1:4
Let check whether the cuboids above are similar.
Surface area of cuboid A = k2
Surface area of cuboid B
136cm2 = 1
544cm2 4
1= 1 2
4
4
2
A cube is a very simple form of cuboid which has edges that are all the same length. All the
faces are therefore identical squares. Dice are good examples of cubes.
If l=3
S = 6l2
S= 6(6)
S= 36cm2
If l=6
S = 6l2
S=6(36)
S=216 cm2
Scale factor =new measurement
old measurement
= 3 =1
6 2
=1:2
Therefore the scale factor (k) = 1:2
Mohamed Bhanji
10B
Surface area of cube 1= k2
Surface area of cube 2
36 cm2 = 1 2
216 cm2 2
1= 1 2
4
2
Volumes of similar 3-D shapes:
A and B are two similar 3-D shapes. Their ratio of corresponding sides is 𝑘.
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐴 = 𝑎𝑏𝑐
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐵 = 𝑘𝑎 × 𝑘𝑏 × 𝑘𝑐 = 𝑘 3 𝑎𝑏𝑐
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐵
=
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐴
𝑘 3 𝑎𝑏𝑐
𝑎𝑏𝑐
= 𝑘3
kb
b
kc
c
ka
a
“If the ratio of the corresponding sides of two 3-D objects is 𝑘, then the ratio of their
volumes is 𝑘 3 .”
Example:
There are two similar boxes. The height of the small box is 3 cm and the height of the big
one is 7cm. If the volume of the smaller box is 120 cm3, find the volume of the bigger box?
Ratio of sides =
3
7
3
27
Ratio of volumes = (7)3 = 343
27
120
=
343
𝑥
𝑥=
343×120
27
4
= 1524 9 𝑐𝑚3
volume of bigger box = 𝑥
Mohamed Bhanji
10B
Congruency:
Figures which have the same size and shape are called Congruent.
Two shapes are congruent if you can Turn, Flip and/or Slide one so it fits exactly on the other.
In this example the shapes are congruent (you only need to flip one over and move it a little)
Congruent or Similar?
The two shapes need to be the same size to be congruent. (If you need to resize one shape to
make it the same as the other, the shapes are called Similar)
If you ...
Then the shapes are ...
... only Rotate, Reflect and/or
Translate
Congruent
... need to Resize
Similar
Congruent? Why such a funny word that basically means "equal"? Probably
because they would only be "equal" if laid on top of each other. Anyway it
comes from Latin congruere, "to agree". So the shapes "agree"
Congruency in triangles:
For two triangles to be congruent there are certain conditions.
1. Side, Side, Side (SSS)
If three sides of a triangle are equal with three sides of another triangle, then the two triangles are
congruent.
Y
L
∆ 𝐾𝐿𝑀 ≡ ∆ 𝑋𝑌𝑍 (𝑆𝑆𝑆)
K
M
X
Z
Mohamed Bhanji
10B
2. Side, Angle, Side (SAS)
If two sides from one triangle are equal to two sides from another triangle and the angle between
these two pairs of sides are equal, then the two triangles are congruent.
A
B
U
S
∆ 𝐵𝐴𝑆 ≡ ∆ 𝑄𝑈𝑍 (𝑆𝐴𝑆)
Q
Z
3. Angle, Side, Angle (ASA)
If two angles from a triangle are equal to two corresponding angles from another triangle, and
the side between the pairs of angles is equal in both triangles, then they are congruent.
B
A
Q
C
∆ 𝐴𝐵𝐶 ≡ ∆ 𝑃𝑄𝑅 (𝐴𝑆𝐴)
R
P
4. Right Angle, Hypotenuse, Side (RHS)
In right-angled triangles, if the longest sides are equal and another side from the first triangle is
equal to a side of the second triangle, then the triangles are congruent.
L
A
∆ 𝐶𝐴𝑋 ≡ ∆ 𝐾𝐿𝑀 (𝑅𝐻𝑆)
C
X
K
M
5. Angle, Angle, Side (AAS)
If two angles and the non-included side of a triangle are equal to the corresponding parts of
another triangle, then the two triangles are congruent.
Mohamed Bhanji
10B
∆ 𝐴𝐵𝐶~∆ 𝑅𝑆𝑇 (𝐴𝐴𝑆)
S
B
A
C
T
R
SUMMARY
Similarity of Triangles : Three Postulates
SAS – Postulate: If two triangles have a pair of corresponding angles equal and
the sides including them proportional, than the triangles are similar.
AA – Postulate: If two triangles have two pairs of corresponding angles equal,
the triangles are similar.
SSS-Postulate: If two triangles have their three pairs of corresponding sides
proportional, the triangles are similar.
Relation between the Areas of Two Similar Triangles:
Basic Proportionality Theorem and Conversely: A line drawn parallel to any side of a
triangle, divides the other two sides proportionally. (Basic Proportionality Theorem)
In the give figure, DE ║ BC
A
 AD = AE  DE ║ BC
BD CE
D
B
E
C
Conversely, If a line divides two sides of a triangle proportionally, the line is parallel to
the third side.
Similarity as a size Transformation: In a size transformation, a given figure is enlarged
(or reduced) by a scale factor K, such that the resulting figure is similar to the given
figure.
If k > 1, then the transformation is the given figure.
Mohamed Bhanji
10B
If k < 1, then the transformation is reduction and
If k = 1, then the transformation is an identity transformation.
For a scale factor k, the
Measure of the length of the resulting plane figure
= k times the corresponding length of the given plane figure
Area of the resulting plane figure
= k2 x (area of the given plane figure)
In case of solids:
= k3 x (volume of the given solid)