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Page 1
91
• Construction of a triangle similar to the given triangle
* To construct a triangle, similar to the given triangle, bearing the
given ratio with the sides of the given triangle.
(i) When vertices are distinct
(ii) When one vertex is common
• Construction of a tangent to a circle.
* To construct a tangent at a point on the circle.
(i) Using centre of the circle.
(ii) Without using the centre of the circle.
* To construct tangents to the given circle from a point outside the circle.
In the previous standard you have learnt the following constructions.
Let us recall those constructions.
• To construct a line parallel to a given line and passing through a
given point outside the line.
• To construct the perpendicular bisector of a given line segment.
• To construct a triangle whose sides are given.
• To divide a given line segment into given number of equal parts
• To divide a line segment in the given ratio.
• To construct an angle congruent to the given angle.
In the ninth standard you have carried out the activity of preparing a map of
surroundings of your school. Before constructing a building we make its plan. The
surroundings of a school and its map, the building and its plan are similar to each
other. We need to draw similar figures in Geography, architecture, machine drawing
etc. A triangle is the simplest closed figure. We shall learn how to construct a triangle
similar to the given triangle.
4 Geometric Constructions
Let’s study.
Let’s recall.
Page 2
91
• Construction of a triangle similar to the given triangle
* To construct a triangle, similar to the given triangle, bearing the
given ratio with the sides of the given triangle.
(i) When vertices are distinct
(ii) When one vertex is common
• Construction of a tangent to a circle.
* To construct a tangent at a point on the circle.
(i) Using centre of the circle.
(ii) Without using the centre of the circle.
* To construct tangents to the given circle from a point outside the circle.
In the previous standard you have learnt the following constructions.
Let us recall those constructions.
• To construct a line parallel to a given line and passing through a
given point outside the line.
• To construct the perpendicular bisector of a given line segment.
• To construct a triangle whose sides are given.
• To divide a given line segment into given number of equal parts
• To divide a line segment in the given ratio.
• To construct an angle congruent to the given angle.
In the ninth standard you have carried out the activity of preparing a map of
surroundings of your school. Before constructing a building we make its plan. The
surroundings of a school and its map, the building and its plan are similar to each
other. We need to draw similar figures in Geography, architecture, machine drawing
etc. A triangle is the simplest closed figure. We shall learn how to construct a triangle
similar to the given triangle.
4 Geometric Constructions
Let’s study.
Let’s recall.
92
Construction of Similar Triangle
To construct a triangle similar to the given triangle, satisfying the condition of
given ratio of corresponding sides.
The corresponding sides of similar triangles are in the same proportion and the
corresponding angles of these triangles are equal. Using this property, a triangle which
is similar to the given triangle can be constructed.
Ex. (1) D ABC ~ D PQR, in D ABC, AB = 5.4 cm, BC = 4.2 cm, AC = 6.0 cm.
AB? PQ = 3?2. Construct D ABC and D PQR.
Construct D ABC of given measure.
D ABC and D PQR are similar.
\ their corresponding sides are proportional.
AB
PQ
=
BC
QR
=
AC
PR
=
3
2
.......... (I)
As the sides AB, BC, AC are known, we can find the lengths of sides
PQ, QR, PR.
Using equation [I]
5.4
PQ
=
4.2
QR
=
6.0
PR
=
3
2
\ PQ = 3.6 cm, QR = 2.8 cm and PR = 4.0 cm
Fig. 4.1
Rough Figure
P Q A B
C
R
Let’s learn.
Page 3
91
• Construction of a triangle similar to the given triangle
* To construct a triangle, similar to the given triangle, bearing the
given ratio with the sides of the given triangle.
(i) When vertices are distinct
(ii) When one vertex is common
• Construction of a tangent to a circle.
* To construct a tangent at a point on the circle.
(i) Using centre of the circle.
(ii) Without using the centre of the circle.
* To construct tangents to the given circle from a point outside the circle.
In the previous standard you have learnt the following constructions.
Let us recall those constructions.
• To construct a line parallel to a given line and passing through a
given point outside the line.
• To construct the perpendicular bisector of a given line segment.
• To construct a triangle whose sides are given.
• To divide a given line segment into given number of equal parts
• To divide a line segment in the given ratio.
• To construct an angle congruent to the given angle.
In the ninth standard you have carried out the activity of preparing a map of
surroundings of your school. Before constructing a building we make its plan. The
surroundings of a school and its map, the building and its plan are similar to each
other. We need to draw similar figures in Geography, architecture, machine drawing
etc. A triangle is the simplest closed figure. We shall learn how to construct a triangle
similar to the given triangle.
4 Geometric Constructions
Let’s study.
Let’s recall.
92
Construction of Similar Triangle
To construct a triangle similar to the given triangle, satisfying the condition of
given ratio of corresponding sides.
The corresponding sides of similar triangles are in the same proportion and the
corresponding angles of these triangles are equal. Using this property, a triangle which
is similar to the given triangle can be constructed.
Ex. (1) D ABC ~ D PQR, in D ABC, AB = 5.4 cm, BC = 4.2 cm, AC = 6.0 cm.
AB? PQ = 3?2. Construct D ABC and D PQR.
Construct D ABC of given measure.
D ABC and D PQR are similar.
\ their corresponding sides are proportional.
AB
PQ
=
BC
QR
=
AC
PR
=
3
2
.......... (I)
As the sides AB, BC, AC are known, we can find the lengths of sides
PQ, QR, PR.
Using equation [I]
5.4
PQ
=
4.2
QR
=
6.0
PR
=
3
2
\ PQ = 3.6 cm, QR = 2.8 cm and PR = 4.0 cm
Fig. 4.1
Rough Figure
P Q A B
C
R
Let’s learn.
93
AB
A¢B
=
BC
BC¢
=
AC
A¢C¢
=
5
3
\ sides of D ABC are longer than corresponding sides of D A¢BC¢ .
\ the length of side BC¢ will be equal to 3 parts out of 5 equal parts of side
BC. So if we construct D ABC, point C¢ will be on the side BC, at a distance
equal to 3 parts from B. Now A¢ is the point of intersection of AB and a line
through C¢, parallel to CA.
For More Information
While drawing the triangle similar to the given triangle, sometimes the lengths
of the sides we obtain by calculation are not easily measureable by a scale. In such
a situation we can use the construction ‘To divide the given segment in the given
number of eqaul parts’.
For example, if length of side AB is
11 6
3
.
cm, then by dividing the line segment
of length 11.6 cm in three equal parts, we can draw segment AB.
Rough Figure
Fig. 4.3
A
A¢
B
C
C¢
Fig. 4.2
A
B
C
4.2
cm
5.4 cm
6.0
cm
P Q
R
2.8
cm
3.6 cm
4.0
cm
If we know the lengths of all sides of D PQR, we can construct D PQR.
In the above example (1) there was no common vertex in the given triangle and
the triangle to be constructed. If there is a common vertex, it is convenient to follow
the method in the following example.
Ex.(2) Construct any D ABC. Construct D A¢BC¢ such that AB : A¢B = 5:3 and
D ABC ~ D A¢BC¢
Analysis : As shown in fig 4.3 , let the points B, A, A¢ and
B, C, C¢ be collinear.
D ABC ~ D A¢BC¢ \ Ð ABC = ÐA¢BC¢
Page 4
91
• Construction of a triangle similar to the given triangle
* To construct a triangle, similar to the given triangle, bearing the
given ratio with the sides of the given triangle.
(i) When vertices are distinct
(ii) When one vertex is common
• Construction of a tangent to a circle.
* To construct a tangent at a point on the circle.
(i) Using centre of the circle.
(ii) Without using the centre of the circle.
* To construct tangents to the given circle from a point outside the circle.
In the previous standard you have learnt the following constructions.
Let us recall those constructions.
• To construct a line parallel to a given line and passing through a
given point outside the line.
• To construct the perpendicular bisector of a given line segment.
• To construct a triangle whose sides are given.
• To divide a given line segment into given number of equal parts
• To divide a line segment in the given ratio.
• To construct an angle congruent to the given angle.
In the ninth standard you have carried out the activity of preparing a map of
surroundings of your school. Before constructing a building we make its plan. The
surroundings of a school and its map, the building and its plan are similar to each
other. We need to draw similar figures in Geography, architecture, machine drawing
etc. A triangle is the simplest closed figure. We shall learn how to construct a triangle
similar to the given triangle.
4 Geometric Constructions
Let’s study.
Let’s recall.
92
Construction of Similar Triangle
To construct a triangle similar to the given triangle, satisfying the condition of
given ratio of corresponding sides.
The corresponding sides of similar triangles are in the same proportion and the
corresponding angles of these triangles are equal. Using this property, a triangle which
is similar to the given triangle can be constructed.
Ex. (1) D ABC ~ D PQR, in D ABC, AB = 5.4 cm, BC = 4.2 cm, AC = 6.0 cm.
AB? PQ = 3?2. Construct D ABC and D PQR.
Construct D ABC of given measure.
D ABC and D PQR are similar.
\ their corresponding sides are proportional.
AB
PQ
=
BC
QR
=
AC
PR
=
3
2
.......... (I)
As the sides AB, BC, AC are known, we can find the lengths of sides
PQ, QR, PR.
Using equation [I]
5.4
PQ
=
4.2
QR
=
6.0
PR
=
3
2
\ PQ = 3.6 cm, QR = 2.8 cm and PR = 4.0 cm
Fig. 4.1
Rough Figure
P Q A B
C
R
Let’s learn.
93
AB
A¢B
=
BC
BC¢
=
AC
A¢C¢
=
5
3
\ sides of D ABC are longer than corresponding sides of D A¢BC¢ .
\ the length of side BC¢ will be equal to 3 parts out of 5 equal parts of side
BC. So if we construct D ABC, point C¢ will be on the side BC, at a distance
equal to 3 parts from B. Now A¢ is the point of intersection of AB and a line
through C¢, parallel to CA.
For More Information
While drawing the triangle similar to the given triangle, sometimes the lengths
of the sides we obtain by calculation are not easily measureable by a scale. In such
a situation we can use the construction ‘To divide the given segment in the given
number of eqaul parts’.
For example, if length of side AB is
11 6
3
.
cm, then by dividing the line segment
of length 11.6 cm in three equal parts, we can draw segment AB.
Rough Figure
Fig. 4.3
A
A¢
B
C
C¢
Fig. 4.2
A
B
C
4.2
cm
5.4 cm
6.0
cm
P Q
R
2.8
cm
3.6 cm
4.0
cm
If we know the lengths of all sides of D PQR, we can construct D PQR.
In the above example (1) there was no common vertex in the given triangle and
the triangle to be constructed. If there is a common vertex, it is convenient to follow
the method in the following example.
Ex.(2) Construct any D ABC. Construct D A¢BC¢ such that AB : A¢B = 5:3 and
D ABC ~ D A¢BC¢
Analysis : As shown in fig 4.3 , let the points B, A, A¢ and
B, C, C¢ be collinear.
D ABC ~ D A¢BC¢ \ Ð ABC = ÐA¢BC¢
94
Note : To divide segment BC, in five
equal parts, it is convenient to draw a ray
from B, on the side of line BC in which
point A does not lie.
Take points T
1
, T
2
, T
3
, T
4
,
T
5
on
the ray such that
BT
1
= T
1
T
2
= T
2
T
3
= T
3
T
4
= T
4
T
5
Join T
5
C and draw lines parallel to T
5
C
through T
1
, T
2
, T
3
, T
4
.
D A¢BC¢ can also be constructed as
shown in the adjoining figure.
What changes do we have to make in
steps of construction in that case ?
Fig. 4.6
A
A¢
B
C
C¢
BA¢
BA
=
BC
BC
¢
=
3
5
i.e,
BA
BA¢
=
BC
BC¢
=
5
3
...... Taking inverse
Steps of construction :
(1) Construct any D ABC.
(2) Divide segment BC in 5 equal parts.
(3) Name the end point of third part of
seg BC as C¢ \ BC¢ =
3
5
BC
(4) Now draw a line parallel to AC
through C¢. Name the point where
the parallel line intersects AB as A¢.
(5) D A¢BC¢ is the required trinangle
similar to D ABC
Fig. 4.5
A
T
1
T
2
T
3
T
4
T
5
B C
Fig. 4.4
A
A¢
C
C¢
B
Let’s think
Page 5
91
• Construction of a triangle similar to the given triangle
* To construct a triangle, similar to the given triangle, bearing the
given ratio with the sides of the given triangle.
(i) When vertices are distinct
(ii) When one vertex is common
• Construction of a tangent to a circle.
* To construct a tangent at a point on the circle.
(i) Using centre of the circle.
(ii) Without using the centre of the circle.
* To construct tangents to the given circle from a point outside the circle.
In the previous standard you have learnt the following constructions.
Let us recall those constructions.
• To construct a line parallel to a given line and passing through a
given point outside the line.
• To construct the perpendicular bisector of a given line segment.
• To construct a triangle whose sides are given.
• To divide a given line segment into given number of equal parts
• To divide a line segment in the given ratio.
• To construct an angle congruent to the given angle.
In the ninth standard you have carried out the activity of preparing a map of
surroundings of your school. Before constructing a building we make its plan. The
surroundings of a school and its map, the building and its plan are similar to each
other. We need to draw similar figures in Geography, architecture, machine drawing
etc. A triangle is the simplest closed figure. We shall learn how to construct a triangle
similar to the given triangle.
4 Geometric Constructions
Let’s study.
Let’s recall.
92
Construction of Similar Triangle
To construct a triangle similar to the given triangle, satisfying the condition of
given ratio of corresponding sides.
The corresponding sides of similar triangles are in the same proportion and the
corresponding angles of these triangles are equal. Using this property, a triangle which
is similar to the given triangle can be constructed.
Ex. (1) D ABC ~ D PQR, in D ABC, AB = 5.4 cm, BC = 4.2 cm, AC = 6.0 cm.
AB? PQ = 3?2. Construct D ABC and D PQR.
Construct D ABC of given measure.
D ABC and D PQR are similar.
\ their corresponding sides are proportional.
AB
PQ
=
BC
QR
=
AC
PR
=
3
2
.......... (I)
As the sides AB, BC, AC are known, we can find the lengths of sides
PQ, QR, PR.
Using equation [I]
5.4
PQ
=
4.2
QR
=
6.0
PR
=
3
2
\ PQ = 3.6 cm, QR = 2.8 cm and PR = 4.0 cm
Fig. 4.1
Rough Figure
P Q A B
C
R
Let’s learn.
93
AB
A¢B
=
BC
BC¢
=
AC
A¢C¢
=
5
3
\ sides of D ABC are longer than corresponding sides of D A¢BC¢ .
\ the length of side BC¢ will be equal to 3 parts out of 5 equal parts of side
BC. So if we construct D ABC, point C¢ will be on the side BC, at a distance
equal to 3 parts from B. Now A¢ is the point of intersection of AB and a line
through C¢, parallel to CA.
For More Information
While drawing the triangle similar to the given triangle, sometimes the lengths
of the sides we obtain by calculation are not easily measureable by a scale. In such
a situation we can use the construction ‘To divide the given segment in the given
number of eqaul parts’.
For example, if length of side AB is
11 6
3
.
cm, then by dividing the line segment
of length 11.6 cm in three equal parts, we can draw segment AB.
Rough Figure
Fig. 4.3
A
A¢
B
C
C¢
Fig. 4.2
A
B
C
4.2
cm
5.4 cm
6.0
cm
P Q
R
2.8
cm
3.6 cm
4.0
cm
If we know the lengths of all sides of D PQR, we can construct D PQR.
In the above example (1) there was no common vertex in the given triangle and
the triangle to be constructed. If there is a common vertex, it is convenient to follow
the method in the following example.
Ex.(2) Construct any D ABC. Construct D A¢BC¢ such that AB : A¢B = 5:3 and
D ABC ~ D A¢BC¢
Analysis : As shown in fig 4.3 , let the points B, A, A¢ and
B, C, C¢ be collinear.
D ABC ~ D A¢BC¢ \ Ð ABC = ÐA¢BC¢
94
Note : To divide segment BC, in five
equal parts, it is convenient to draw a ray
from B, on the side of line BC in which
point A does not lie.
Take points T
1
, T
2
, T
3
, T
4
,
T
5
on
the ray such that
BT
1
= T
1
T
2
= T
2
T
3
= T
3
T
4
= T
4
T
5
Join T
5
C and draw lines parallel to T
5
C
through T
1
, T
2
, T
3
, T
4
.
D A¢BC¢ can also be constructed as
shown in the adjoining figure.
What changes do we have to make in
steps of construction in that case ?
Fig. 4.6
A
A¢
B
C
C¢
BA¢
BA
=
BC
BC
¢
=
3
5
i.e,
BA
BA¢
=
BC
BC¢
=
5
3
...... Taking inverse
Steps of construction :
(1) Construct any D ABC.
(2) Divide segment BC in 5 equal parts.
(3) Name the end point of third part of
seg BC as C¢ \ BC¢ =
3
5
BC
(4) Now draw a line parallel to AC
through C¢. Name the point where
the parallel line intersects AB as A¢.
(5) D A¢BC¢ is the required trinangle
similar to D ABC
Fig. 4.5
A
T
1
T
2
T
3
T
4
T
5
B C
Fig. 4.4
A
A¢
C
C¢
B
Let’s think
95
Ex. (3) Construct D A¢BC¢ similar to D ABC such that AB:A¢B = 5:7
Analysis : Let points B, A, A¢ as well as points B, C, C¢ be collinear.
D ABC ~ D A¢BC¢ and AB : A¢B = 5:7
\ sides of D ABC are smaller than sides of D A¢BC¢
and ÐABC @ ÐA¢BC¢
Let us draw a rough figure with these
considerations. Now
BC
BC'
=
5
7
\ If seg BC is divided into 5 equal parts, then seg BC¢ will be 7 times each
part of seg BC.
\ let us divide side BC of D ABC in 5 equal parts and locate point C¢ at a
distance equal to 7 such parts from B on ray BC. A line through point C¢
parallel to seg AC is drawn it will intersect ray BA at point A¢. According
to the basic proportionality theorem we will get D A¢BC¢ as described.
Steps of construction :
(1) Construct any D ABC.
(2) Divide segment BC into 5 five equal parts. Fix point C¢ on ray BC such that length
of BC¢ is seven times of each equal part of seg BC
(3) Draw a line parallel to side AC, through C¢. Name the point of intersection of the
line and ray BA as A¢.
We get the required D A¢BC¢ similar to D ABC.
Fig. 4.8
A
A¢
B
C
1 2 3 4 5 6
7
C¢
Fig. 4.7
A
A¢
B
C
C¢
Rough Figure
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FAQs on Textbook: Geometric Constructions - Mathematics Class 10 (Maharashtra SSC Board)
| 1. What are geometric constructions and why are they important in mathematics? |  |
Ans. Geometric constructions are methods of creating figures using only a compass and straightedge, without the need for measurements. They are important in mathematics because they help develop logical reasoning, enhance understanding of geometric principles, and illustrate the relationships between different shapes and angles.
| 2. What tools are commonly used for geometric constructions? | |
Ans. The primary tools used for geometric constructions are a compass and a straightedge (ruler without markings). The compass is used to draw arcs and circles, while the straightedge is used to draw straight lines. These tools allow for precise constructions based on fundamental geometric principles.
| 3. How can I construct a perpendicular bisector using a compass and straightedge? | |
Ans. To construct a perpendicular bisector, first draw a line segment. Place the compass point on one endpoint of the segment and draw an arc above and below the line. Without changing the compass width, repeat from the other endpoint to create two intersecting arcs. Draw a line through the intersection points of the arcs. This line is the perpendicular bisector of the segment.
| 4. What is the difference between construction and measurement in geometry? | |
Ans. In geometry, construction refers to creating figures using only a compass and straightedge, focusing on exact relationships and properties. Measurement, on the other hand, involves determining lengths, angles, or areas using numerical values. While both are essential in geometry, construction emphasizes precision through geometric principles without the use of measurements.
| 5. Can geometric constructions be applied in real-life situations? | |
Ans. Yes, geometric constructions have practical applications in various fields such as architecture, engineering, art, and design. They help in creating accurate designs, ensuring structural integrity, and solving problems related to space and form. Understanding geometric constructions allows for better visualization and planning in real-world projects.
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Oct 24, 2025
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