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Textbook: Similarity
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Page 1
1
• Ratio of areas of two triangles • Basic proportionality theorem
• Converse of basic proportionality theorem • Tests of similarity of triangles
• Property of an angle bisector of a triangle • Property of areas of similar triangles
• The ratio of the intercepts made on the transversals by three parallel lines
Ex. In D ABC, AD is the height and BC
is the base.
In D PQR, PS is the height and
QR is the base
A( D ABC)
A( D PQR)
=
1
2
´ BC ´ AD
1
2
´ QR ´ PS
A
P
B Q D
S
C R
Fig. 1.1 Fig. 1.2
Let’s recall.
We have studied Ratio and Proportion. The statement, ‘the numbers a and b are in
the ratio
m
n
’ is also written as, ‘the numbers a and b are in proportion m:n.’
For this concept we consider positive real numbers. We know that the lengths of
line segments and area of any figure are positive real numbers.
We know the formula of area of a triangle.
Area of a triangle =
1
2
Base ´ Height
Let’s learn.
Ratio of areas of two triangles
Let’s find the ratio of areas of any two triangles.
1
Similarity
Let’s study.
Page 2
1
• Ratio of areas of two triangles • Basic proportionality theorem
• Converse of basic proportionality theorem • Tests of similarity of triangles
• Property of an angle bisector of a triangle • Property of areas of similar triangles
• The ratio of the intercepts made on the transversals by three parallel lines
Ex. In D ABC, AD is the height and BC
is the base.
In D PQR, PS is the height and
QR is the base
A( D ABC)
A( D PQR)
=
1
2
´ BC ´ AD
1
2
´ QR ´ PS
A
P
B Q D
S
C R
Fig. 1.1 Fig. 1.2
Let’s recall.
We have studied Ratio and Proportion. The statement, ‘the numbers a and b are in
the ratio
m
n
’ is also written as, ‘the numbers a and b are in proportion m:n.’
For this concept we consider positive real numbers. We know that the lengths of
line segments and area of any figure are positive real numbers.
We know the formula of area of a triangle.
Area of a triangle =
1
2
Base ´ Height
Let’s learn.
Ratio of areas of two triangles
Let’s find the ratio of areas of any two triangles.
1
Similarity
Let’s study.
2
\
A(D ABC)
A(D PQR)
=
BC ´ AD
QR ´ PS
Hence the ratio of the areas of two triangles is equal to the ratio of the
products of their bases and corrosponding heights.
Base of a triangle is b
1
and height is h
1
. Base of another triangle is b
2
and
height is h
2
. Then the ratio of their areas =
b
1
´ h
1
b
2
´
h
2
Suppose some conditions are imposed on these two triangles,
Condition 1 ? If the heights of both triangles are equal then-
A(D ABC)
A(D PQR)
=
BC ´ h
QR ´ h
=
BC
QR
\
A(D ABC)
A(D PQR)
=
b
1
b
2
Property ? The ratio of the areas of two triangles with equal heights is equal
to the ratio of their corresponding bases.
Condition 2 ? If the bases of both triangles are equal then -
Fig. 1.4 Fig. 1.3
A P
B
Q
D
S
C R
h h
Fig. 1.5
C
A D
P
Q B
h
2
h
1
A(D ABC)
A(D APB)
=
AB ´ h
1
AB ´ h
2
A(D ABC)
A(D APB)
=
h
1
h
2
Property ? The ratio of the areas of two triangles with equal bases is equal to the ratio
of their corresponding heights.
Page 3
1
• Ratio of areas of two triangles • Basic proportionality theorem
• Converse of basic proportionality theorem • Tests of similarity of triangles
• Property of an angle bisector of a triangle • Property of areas of similar triangles
• The ratio of the intercepts made on the transversals by three parallel lines
Ex. In D ABC, AD is the height and BC
is the base.
In D PQR, PS is the height and
QR is the base
A( D ABC)
A( D PQR)
=
1
2
´ BC ´ AD
1
2
´ QR ´ PS
A
P
B Q D
S
C R
Fig. 1.1 Fig. 1.2
Let’s recall.
We have studied Ratio and Proportion. The statement, ‘the numbers a and b are in
the ratio
m
n
’ is also written as, ‘the numbers a and b are in proportion m:n.’
For this concept we consider positive real numbers. We know that the lengths of
line segments and area of any figure are positive real numbers.
We know the formula of area of a triangle.
Area of a triangle =
1
2
Base ´ Height
Let’s learn.
Ratio of areas of two triangles
Let’s find the ratio of areas of any two triangles.
1
Similarity
Let’s study.
2
\
A(D ABC)
A(D PQR)
=
BC ´ AD
QR ´ PS
Hence the ratio of the areas of two triangles is equal to the ratio of the
products of their bases and corrosponding heights.
Base of a triangle is b
1
and height is h
1
. Base of another triangle is b
2
and
height is h
2
. Then the ratio of their areas =
b
1
´ h
1
b
2
´
h
2
Suppose some conditions are imposed on these two triangles,
Condition 1 ? If the heights of both triangles are equal then-
A(D ABC)
A(D PQR)
=
BC ´ h
QR ´ h
=
BC
QR
\
A(D ABC)
A(D PQR)
=
b
1
b
2
Property ? The ratio of the areas of two triangles with equal heights is equal
to the ratio of their corresponding bases.
Condition 2 ? If the bases of both triangles are equal then -
Fig. 1.4 Fig. 1.3
A P
B
Q
D
S
C R
h h
Fig. 1.5
C
A D
P
Q B
h
2
h
1
A(D ABC)
A(D APB)
=
AB ´ h
1
AB ´ h
2
A(D ABC)
A(D APB)
=
h
1
h
2
Property ? The ratio of the areas of two triangles with equal bases is equal to the ratio
of their corresponding heights.
3
Activity :
Fill in the blanks properly.
(i) (ii)
?????????????? Solved Examples ?????????????
Ex. (1)
A( D ABC)
A( D APQ)
=
´
´
=
A(D LMN)
A(D DMN)
=
´
´
=
Fig. 1.6
A
B
P Q R
C
Fig.1.7
(iii) M is the midpoint of
seg AB and seg CM is a median
of D ABC
\
A(D AMC)
A(D BMC)
=
= =
State the reason.
Fig. 1.8
C
A B M
In adjoining figure
AE ^ seg BC, seg DF ^ line BC,
AE = 4, DF = 6 , then find
A( D ABC)
A( D DBC)
.
Fig.1.9
A
D
B E F C
Solution ?
A( D ABC)
A( D DBC)
=
AE
DF
.......... bases are equal, hence areas proportional to
heights.
=
4
6
=
2
3
D
P
L
M N
Q
Page 4
1
• Ratio of areas of two triangles • Basic proportionality theorem
• Converse of basic proportionality theorem • Tests of similarity of triangles
• Property of an angle bisector of a triangle • Property of areas of similar triangles
• The ratio of the intercepts made on the transversals by three parallel lines
Ex. In D ABC, AD is the height and BC
is the base.
In D PQR, PS is the height and
QR is the base
A( D ABC)
A( D PQR)
=
1
2
´ BC ´ AD
1
2
´ QR ´ PS
A
P
B Q D
S
C R
Fig. 1.1 Fig. 1.2
Let’s recall.
We have studied Ratio and Proportion. The statement, ‘the numbers a and b are in
the ratio
m
n
’ is also written as, ‘the numbers a and b are in proportion m:n.’
For this concept we consider positive real numbers. We know that the lengths of
line segments and area of any figure are positive real numbers.
We know the formula of area of a triangle.
Area of a triangle =
1
2
Base ´ Height
Let’s learn.
Ratio of areas of two triangles
Let’s find the ratio of areas of any two triangles.
1
Similarity
Let’s study.
2
\
A(D ABC)
A(D PQR)
=
BC ´ AD
QR ´ PS
Hence the ratio of the areas of two triangles is equal to the ratio of the
products of their bases and corrosponding heights.
Base of a triangle is b
1
and height is h
1
. Base of another triangle is b
2
and
height is h
2
. Then the ratio of their areas =
b
1
´ h
1
b
2
´
h
2
Suppose some conditions are imposed on these two triangles,
Condition 1 ? If the heights of both triangles are equal then-
A(D ABC)
A(D PQR)
=
BC ´ h
QR ´ h
=
BC
QR
\
A(D ABC)
A(D PQR)
=
b
1
b
2
Property ? The ratio of the areas of two triangles with equal heights is equal
to the ratio of their corresponding bases.
Condition 2 ? If the bases of both triangles are equal then -
Fig. 1.4 Fig. 1.3
A P
B
Q
D
S
C R
h h
Fig. 1.5
C
A D
P
Q B
h
2
h
1
A(D ABC)
A(D APB)
=
AB ´ h
1
AB ´ h
2
A(D ABC)
A(D APB)
=
h
1
h
2
Property ? The ratio of the areas of two triangles with equal bases is equal to the ratio
of their corresponding heights.
3
Activity :
Fill in the blanks properly.
(i) (ii)
?????????????? Solved Examples ?????????????
Ex. (1)
A( D ABC)
A( D APQ)
=
´
´
=
A(D LMN)
A(D DMN)
=
´
´
=
Fig. 1.6
A
B
P Q R
C
Fig.1.7
(iii) M is the midpoint of
seg AB and seg CM is a median
of D ABC
\
A(D AMC)
A(D BMC)
=
= =
State the reason.
Fig. 1.8
C
A B M
In adjoining figure
AE ^ seg BC, seg DF ^ line BC,
AE = 4, DF = 6 , then find
A( D ABC)
A( D DBC)
.
Fig.1.9
A
D
B E F C
Solution ?
A( D ABC)
A( D DBC)
=
AE
DF
.......... bases are equal, hence areas proportional to
heights.
=
4
6
=
2
3
D
P
L
M N
Q
4
Solution ? Point A is common vertex of
D ABD, D ADC and D ABC
and their bases are collinear.
Hence, heights of these three
triangles are equal
BC = 15, DC = 6 \ BD = BC - DC = 15 - 6 = 9
A( D ABD)
A( D ABC)
=
BD
BC
.......... heights equal, hence areas proportional to
bases.
=
9
15
=
3
5
A( D ABD)
A( D ADC)
=
BD
DC
.......... heights equal, hence areas proportional to
bases.
=
9
6
=
3
2
Ex. (3)
Fig. 1.10
Fig. 1.11
D
C
B
P
A
Ex. (2) In D ABC point D on side BC is such that DC = 6, BC = 15.
Find A(D ABD) : A(D ABC) and A(D ABD) : A(D ADC) .
A D
C B P
c ABCD is a parallelogram. P is any
point on side BC. Find two pairs of
triangles with equal areas.
Solution ? c ABCD is a parallelogram.
\ AD || BC and AB || DC
Consider D ABC and D BDC.
Both the triangles are drawn in two parallel lines. Hence the distance
between the two parallel lines is the height of both triangles.
In D ABC and D BDC, common base is BC and heights are equal.
Hence, A(D ABC) = A(D BDC)
In D ABC and D ABD, AB is common base and and heights are equal.
\ A(D ABC) = A(D ABD)
Page 5
1
• Ratio of areas of two triangles • Basic proportionality theorem
• Converse of basic proportionality theorem • Tests of similarity of triangles
• Property of an angle bisector of a triangle • Property of areas of similar triangles
• The ratio of the intercepts made on the transversals by three parallel lines
Ex. In D ABC, AD is the height and BC
is the base.
In D PQR, PS is the height and
QR is the base
A( D ABC)
A( D PQR)
=
1
2
´ BC ´ AD
1
2
´ QR ´ PS
A
P
B Q D
S
C R
Fig. 1.1 Fig. 1.2
Let’s recall.
We have studied Ratio and Proportion. The statement, ‘the numbers a and b are in
the ratio
m
n
’ is also written as, ‘the numbers a and b are in proportion m:n.’
For this concept we consider positive real numbers. We know that the lengths of
line segments and area of any figure are positive real numbers.
We know the formula of area of a triangle.
Area of a triangle =
1
2
Base ´ Height
Let’s learn.
Ratio of areas of two triangles
Let’s find the ratio of areas of any two triangles.
1
Similarity
Let’s study.
2
\
A(D ABC)
A(D PQR)
=
BC ´ AD
QR ´ PS
Hence the ratio of the areas of two triangles is equal to the ratio of the
products of their bases and corrosponding heights.
Base of a triangle is b
1
and height is h
1
. Base of another triangle is b
2
and
height is h
2
. Then the ratio of their areas =
b
1
´ h
1
b
2
´
h
2
Suppose some conditions are imposed on these two triangles,
Condition 1 ? If the heights of both triangles are equal then-
A(D ABC)
A(D PQR)
=
BC ´ h
QR ´ h
=
BC
QR
\
A(D ABC)
A(D PQR)
=
b
1
b
2
Property ? The ratio of the areas of two triangles with equal heights is equal
to the ratio of their corresponding bases.
Condition 2 ? If the bases of both triangles are equal then -
Fig. 1.4 Fig. 1.3
A P
B
Q
D
S
C R
h h
Fig. 1.5
C
A D
P
Q B
h
2
h
1
A(D ABC)
A(D APB)
=
AB ´ h
1
AB ´ h
2
A(D ABC)
A(D APB)
=
h
1
h
2
Property ? The ratio of the areas of two triangles with equal bases is equal to the ratio
of their corresponding heights.
3
Activity :
Fill in the blanks properly.
(i) (ii)
?????????????? Solved Examples ?????????????
Ex. (1)
A( D ABC)
A( D APQ)
=
´
´
=
A(D LMN)
A(D DMN)
=
´
´
=
Fig. 1.6
A
B
P Q R
C
Fig.1.7
(iii) M is the midpoint of
seg AB and seg CM is a median
of D ABC
\
A(D AMC)
A(D BMC)
=
= =
State the reason.
Fig. 1.8
C
A B M
In adjoining figure
AE ^ seg BC, seg DF ^ line BC,
AE = 4, DF = 6 , then find
A( D ABC)
A( D DBC)
.
Fig.1.9
A
D
B E F C
Solution ?
A( D ABC)
A( D DBC)
=
AE
DF
.......... bases are equal, hence areas proportional to
heights.
=
4
6
=
2
3
D
P
L
M N
Q
4
Solution ? Point A is common vertex of
D ABD, D ADC and D ABC
and their bases are collinear.
Hence, heights of these three
triangles are equal
BC = 15, DC = 6 \ BD = BC - DC = 15 - 6 = 9
A( D ABD)
A( D ABC)
=
BD
BC
.......... heights equal, hence areas proportional to
bases.
=
9
15
=
3
5
A( D ABD)
A( D ADC)
=
BD
DC
.......... heights equal, hence areas proportional to
bases.
=
9
6
=
3
2
Ex. (3)
Fig. 1.10
Fig. 1.11
D
C
B
P
A
Ex. (2) In D ABC point D on side BC is such that DC = 6, BC = 15.
Find A(D ABD) : A(D ABC) and A(D ABD) : A(D ADC) .
A D
C B P
c ABCD is a parallelogram. P is any
point on side BC. Find two pairs of
triangles with equal areas.
Solution ? c ABCD is a parallelogram.
\ AD || BC and AB || DC
Consider D ABC and D BDC.
Both the triangles are drawn in two parallel lines. Hence the distance
between the two parallel lines is the height of both triangles.
In D ABC and D BDC, common base is BC and heights are equal.
Hence, A(D ABC) = A(D BDC)
In D ABC and D ABD, AB is common base and and heights are equal.
\ A(D ABC) = A(D ABD)
5
Ex.(4)
In adjoining figure in D ABC, point D is
on side AC. If AC = 16, DC = 9 and
BP ^ AC, then find the following ratios.
(i)
A( D ABD)
A( D ABC)
(ii)
A( D BDC)
A( D ABC)
(iii)
A( D ABD)
A( D BDC)
B
P
D
A
C
Fig. 1.12
Solution ? In D ABC point P and D are on side AC, hence B is common vertex of
D ABD, D BDC, D ABC and D APB and their sides AD, DC, AC and AP are
collinear. Heights of all the triangles are equal. Hence, areas of these triangles
are proportinal to their bases. AC = 16, DC = 9
\ AD = 16 - 9 = 7
\
A( D ABD)
A( D ABC)
=
AD
AC
=
7
16
. . . . . . . . triangles having equal heights
A( D BDC)
A( D ABC)
=
DC
AC
=
9
16
. . . . . . . . triangles having equal heights
A( D ABD)
A( D BDC)
=
AD
DC
=
7
9
. . . . . . . . triangles having equal heights
Remember this !
• Ratio of areas of two triangles is equal to the ratio of the products of their
bases and corresponding heights.
• Areas of triangles with equal heights are proportional to their
corresponding bases.
• Areas of triangles with equal bases are proportional to their
corresponding heights.
Practice set 1.1
1. Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height
is 6. Find the ratio of areas of these triangles.
Read MoreFAQs on Textbook: Similarity
| 1. What is the definition of similarity in geometry? |  |
Ans. Similarity in geometry refers to the relationship between two shapes that have the same shape but may differ in size. If two figures are similar, their corresponding angles are equal, and the lengths of their corresponding sides are proportional.
| 2. How can we determine if two triangles are similar? | |
Ans. Two triangles can be determined to be similar by using one of the following criteria: Angle-Angle (AA) criterion, where if two angles of one triangle are equal to two angles of another triangle, the triangles are similar; Side-Angle-Side (SAS) criterion, where if one angle of a triangle is equal to one angle of another triangle and the sides including those angles are in proportion, the triangles are similar; or Side-Side-Side (SSS) criterion, where if the corresponding sides of two triangles are in proportion, the triangles are similar.
| 3. Can similarity be applied to figures other than triangles? | |
Ans. Yes, similarity can be applied to various geometric figures beyond triangles, including quadrilaterals, circles, and polygons. For instance, two rectangles can be similar if their corresponding sides are in the same ratio, and two circles are always similar since they have the same shape regardless of their size.
| 4. What role does the concept of scale factor play in similarity? | |
Ans. The scale factor is a crucial concept in similarity, representing the ratio of the lengths of corresponding sides of two similar figures. It helps in resizing the figure while maintaining its shape. For example, if the scale factor is 2, every side of the original figure will be doubled in length in the similar figure.
| 5. How is similarity used in real-life applications? | |
Ans. Similarity has various real-life applications, such as in architecture, where scaled drawings of buildings maintain the same proportions as the actual structures. It is also used in map-making, where a smaller representation of a geographical area maintains the same shape and proportions as the actual land, making it easier to navigate and understand spatial relationships.
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