Page 1
216 MATHEMA TICS
11
11.1 Introduction
In Class IX, you have done certain constructions using a straight edge (ruler) and a
compass, e.g., bisecting an angle, drawing the perpendicular bisector of a line segment,
some constructions of triangles etc. and also gave their justifications. In this chapter,
we shall study some more constructions by using the knowledge of the earlier
constructions. You would also be expected to give the mathematical reasoning behind
why such constructions work.
11.2 Division of a Line Segment
Suppose a line segment is given and you have to divide it in a given ratio, say 3 : 2. You
may do it by measuring the length and then marking a point on it that divides it in the
given ratio. But suppose you do not have any way of measuring it precisely, how
would you find the point? We give below two ways for finding such a point.
Construction 11.1 : To divide a line segment in a given ratio.
Given a line segment AB, we want to divide it in the ratio m : n, where both m and
n are positive integers. To help you to understand it, we shall take m = 3 and n = 2.
Steps of Construction :
1. Draw any ray AX, making an acute angle with AB.
2. Locate 5 (= m + n) points A
1
, A
2
, A
3
, A
4
and
A
5
on AX so that AA
1
= A
1
A
2
= A
2
A
3
= A
3
A
4
= A
4
A
5
.
3. Join BA
5
.
4. Through the point A
3
( m = 3), draw a line
parallel to A
5
B (by making an angle equal to
? AA
5
B) at A
3
intersecting AB at the point C
(see Fig. 11.1). Then, AC : CB = 3 : 2.
CONSTRUCTIONS
Fig. 11.1
2022-23
Page 2
216 MATHEMA TICS
11
11.1 Introduction
In Class IX, you have done certain constructions using a straight edge (ruler) and a
compass, e.g., bisecting an angle, drawing the perpendicular bisector of a line segment,
some constructions of triangles etc. and also gave their justifications. In this chapter,
we shall study some more constructions by using the knowledge of the earlier
constructions. You would also be expected to give the mathematical reasoning behind
why such constructions work.
11.2 Division of a Line Segment
Suppose a line segment is given and you have to divide it in a given ratio, say 3 : 2. You
may do it by measuring the length and then marking a point on it that divides it in the
given ratio. But suppose you do not have any way of measuring it precisely, how
would you find the point? We give below two ways for finding such a point.
Construction 11.1 : To divide a line segment in a given ratio.
Given a line segment AB, we want to divide it in the ratio m : n, where both m and
n are positive integers. To help you to understand it, we shall take m = 3 and n = 2.
Steps of Construction :
1. Draw any ray AX, making an acute angle with AB.
2. Locate 5 (= m + n) points A
1
, A
2
, A
3
, A
4
and
A
5
on AX so that AA
1
= A
1
A
2
= A
2
A
3
= A
3
A
4
= A
4
A
5
.
3. Join BA
5
.
4. Through the point A
3
( m = 3), draw a line
parallel to A
5
B (by making an angle equal to
? AA
5
B) at A
3
intersecting AB at the point C
(see Fig. 11.1). Then, AC : CB = 3 : 2.
CONSTRUCTIONS
Fig. 11.1
2022-23
CONSTRUCTIONS 217
Let us see how this method gives us the required division.
Since A
3
C is parallel to A
5
B, therefore,
3
35
AA
AA
=
AC
CB
(By the Basic Proportionality Theorem)
By construction,
3
35
AA 3 AC 3
Therefore,
A A 2 CB 2
=· =
.
This shows that C divides AB in the ratio 3 : 2.
Alternative Method
Steps of Construction :
1. Draw any ray AX making an acute angle with AB.
2. Draw a ray BY parallel to AX by making ? ABY equal to ? BAX.
3. Locate the points A
1
, A
2
, A
3
(m = 3) on AX and B
1
, B
2
(n = 2) on BY such that
AA
1
= A
1
A
2
= A
2
A
3
= BB
1
= B
1
B
2
.
4. Join A
3
B
2
. Let it intersect AB at a point C (see Fig. 11.2).
Then AC : CB = 3 : 2.
Why does this method work? Let us see.
Here ? AA
3
C is similar to ? BB
2
C. (Why ?)
Then
3
2
AA AC
BB BC
= .
Since by construction,
3
2
AA 3
,
BB 2
=
therefore,
AC 3
BC 2
=·
In fact, the methods given above work for dividing the line segment in any ratio.
We now use the idea of the construction above for constructing a triangle similar
to a given triangle whose sides are in a given ratio with the corresponding sides of the
given triangle.
Construction 11.2 : To construct a triangle similar to a given triangle as per
given scale factor.
This construction involves two different situations. In one, the triangle to be
constructed is smaller and in the other it is larger than the given triangle. Here, the
scale factor means the ratio of the sides of the triangle to be constructed with the
corresponding sides of the given triangle (see also Chapter 6). Let us take the following
examples for understanding the constructions involved. The same methods would
apply for the general case also.
Fig. 11.2
2022-23
Page 3
216 MATHEMA TICS
11
11.1 Introduction
In Class IX, you have done certain constructions using a straight edge (ruler) and a
compass, e.g., bisecting an angle, drawing the perpendicular bisector of a line segment,
some constructions of triangles etc. and also gave their justifications. In this chapter,
we shall study some more constructions by using the knowledge of the earlier
constructions. You would also be expected to give the mathematical reasoning behind
why such constructions work.
11.2 Division of a Line Segment
Suppose a line segment is given and you have to divide it in a given ratio, say 3 : 2. You
may do it by measuring the length and then marking a point on it that divides it in the
given ratio. But suppose you do not have any way of measuring it precisely, how
would you find the point? We give below two ways for finding such a point.
Construction 11.1 : To divide a line segment in a given ratio.
Given a line segment AB, we want to divide it in the ratio m : n, where both m and
n are positive integers. To help you to understand it, we shall take m = 3 and n = 2.
Steps of Construction :
1. Draw any ray AX, making an acute angle with AB.
2. Locate 5 (= m + n) points A
1
, A
2
, A
3
, A
4
and
A
5
on AX so that AA
1
= A
1
A
2
= A
2
A
3
= A
3
A
4
= A
4
A
5
.
3. Join BA
5
.
4. Through the point A
3
( m = 3), draw a line
parallel to A
5
B (by making an angle equal to
? AA
5
B) at A
3
intersecting AB at the point C
(see Fig. 11.1). Then, AC : CB = 3 : 2.
CONSTRUCTIONS
Fig. 11.1
2022-23
CONSTRUCTIONS 217
Let us see how this method gives us the required division.
Since A
3
C is parallel to A
5
B, therefore,
3
35
AA
AA
=
AC
CB
(By the Basic Proportionality Theorem)
By construction,
3
35
AA 3 AC 3
Therefore,
A A 2 CB 2
=· =
.
This shows that C divides AB in the ratio 3 : 2.
Alternative Method
Steps of Construction :
1. Draw any ray AX making an acute angle with AB.
2. Draw a ray BY parallel to AX by making ? ABY equal to ? BAX.
3. Locate the points A
1
, A
2
, A
3
(m = 3) on AX and B
1
, B
2
(n = 2) on BY such that
AA
1
= A
1
A
2
= A
2
A
3
= BB
1
= B
1
B
2
.
4. Join A
3
B
2
. Let it intersect AB at a point C (see Fig. 11.2).
Then AC : CB = 3 : 2.
Why does this method work? Let us see.
Here ? AA
3
C is similar to ? BB
2
C. (Why ?)
Then
3
2
AA AC
BB BC
= .
Since by construction,
3
2
AA 3
,
BB 2
=
therefore,
AC 3
BC 2
=·
In fact, the methods given above work for dividing the line segment in any ratio.
We now use the idea of the construction above for constructing a triangle similar
to a given triangle whose sides are in a given ratio with the corresponding sides of the
given triangle.
Construction 11.2 : To construct a triangle similar to a given triangle as per
given scale factor.
This construction involves two different situations. In one, the triangle to be
constructed is smaller and in the other it is larger than the given triangle. Here, the
scale factor means the ratio of the sides of the triangle to be constructed with the
corresponding sides of the given triangle (see also Chapter 6). Let us take the following
examples for understanding the constructions involved. The same methods would
apply for the general case also.
Fig. 11.2
2022-23
218 MATHEMA TICS
Example 1 : Construct a triangle similar to a given triangle ABC with its sides equal
to
3
4
of the corresponding sides of the triangle ABC (i.e., of scale factor
3
4
).
Solution : Given a triangle ABC, we are required to construct another triangle whose
sides are
3
4
of the corresponding sides of the triangle ABC.
Steps of Construction :
1. Draw any ray BX making an acute angle
with BC on the side opposite to the vertex
A.
2. Locate 4 (the greater of 3 and 4 in
3
4
)
points B
1
, B
2
, B
3
and B
4
on BX so that
BB
1
= B
1
B
2
= B
2
B
3
= B
3
B
4
.
3. Join B
4
C and draw a line through B
3
(the
3rd point, 3 being smaller of 3 and 4 in
3
4
) parallel to B
4
C to intersect BC at C'.
4. Draw a line through C ' parallel
to the line CA to intersect BA at A'
(see Fig. 11.3).
Then, ? A'BC' is the required triangle.
Let us now see how this construction gives the required triangle.
By Construction 11.1,
BC 3
CC 1
'
=·
'
Therefore,
BC BC + C C C C 1 4
11
BC BC BC 3 3
'' '
= = + =+ =
'' '
, i.e.,
BC
BC
'
=
3
4
.
Also C'A' is parallel to CA. Therefore, ? A'BC' ~ ? ABC. (Why ?)
So,
A B A CBC3
AB AC BC 4
' '' '
= = =·
Example 2 : Construct a triangle similar to a given triangle ABC with its sides equal
to
5
3
of the corresponding sides of the triangle ABC (i.e., of scale factor
5
3
).
Fig. 11.3
2022-23
Page 4
216 MATHEMA TICS
11
11.1 Introduction
In Class IX, you have done certain constructions using a straight edge (ruler) and a
compass, e.g., bisecting an angle, drawing the perpendicular bisector of a line segment,
some constructions of triangles etc. and also gave their justifications. In this chapter,
we shall study some more constructions by using the knowledge of the earlier
constructions. You would also be expected to give the mathematical reasoning behind
why such constructions work.
11.2 Division of a Line Segment
Suppose a line segment is given and you have to divide it in a given ratio, say 3 : 2. You
may do it by measuring the length and then marking a point on it that divides it in the
given ratio. But suppose you do not have any way of measuring it precisely, how
would you find the point? We give below two ways for finding such a point.
Construction 11.1 : To divide a line segment in a given ratio.
Given a line segment AB, we want to divide it in the ratio m : n, where both m and
n are positive integers. To help you to understand it, we shall take m = 3 and n = 2.
Steps of Construction :
1. Draw any ray AX, making an acute angle with AB.
2. Locate 5 (= m + n) points A
1
, A
2
, A
3
, A
4
and
A
5
on AX so that AA
1
= A
1
A
2
= A
2
A
3
= A
3
A
4
= A
4
A
5
.
3. Join BA
5
.
4. Through the point A
3
( m = 3), draw a line
parallel to A
5
B (by making an angle equal to
? AA
5
B) at A
3
intersecting AB at the point C
(see Fig. 11.1). Then, AC : CB = 3 : 2.
CONSTRUCTIONS
Fig. 11.1
2022-23
CONSTRUCTIONS 217
Let us see how this method gives us the required division.
Since A
3
C is parallel to A
5
B, therefore,
3
35
AA
AA
=
AC
CB
(By the Basic Proportionality Theorem)
By construction,
3
35
AA 3 AC 3
Therefore,
A A 2 CB 2
=· =
.
This shows that C divides AB in the ratio 3 : 2.
Alternative Method
Steps of Construction :
1. Draw any ray AX making an acute angle with AB.
2. Draw a ray BY parallel to AX by making ? ABY equal to ? BAX.
3. Locate the points A
1
, A
2
, A
3
(m = 3) on AX and B
1
, B
2
(n = 2) on BY such that
AA
1
= A
1
A
2
= A
2
A
3
= BB
1
= B
1
B
2
.
4. Join A
3
B
2
. Let it intersect AB at a point C (see Fig. 11.2).
Then AC : CB = 3 : 2.
Why does this method work? Let us see.
Here ? AA
3
C is similar to ? BB
2
C. (Why ?)
Then
3
2
AA AC
BB BC
= .
Since by construction,
3
2
AA 3
,
BB 2
=
therefore,
AC 3
BC 2
=·
In fact, the methods given above work for dividing the line segment in any ratio.
We now use the idea of the construction above for constructing a triangle similar
to a given triangle whose sides are in a given ratio with the corresponding sides of the
given triangle.
Construction 11.2 : To construct a triangle similar to a given triangle as per
given scale factor.
This construction involves two different situations. In one, the triangle to be
constructed is smaller and in the other it is larger than the given triangle. Here, the
scale factor means the ratio of the sides of the triangle to be constructed with the
corresponding sides of the given triangle (see also Chapter 6). Let us take the following
examples for understanding the constructions involved. The same methods would
apply for the general case also.
Fig. 11.2
2022-23
218 MATHEMA TICS
Example 1 : Construct a triangle similar to a given triangle ABC with its sides equal
to
3
4
of the corresponding sides of the triangle ABC (i.e., of scale factor
3
4
).
Solution : Given a triangle ABC, we are required to construct another triangle whose
sides are
3
4
of the corresponding sides of the triangle ABC.
Steps of Construction :
1. Draw any ray BX making an acute angle
with BC on the side opposite to the vertex
A.
2. Locate 4 (the greater of 3 and 4 in
3
4
)
points B
1
, B
2
, B
3
and B
4
on BX so that
BB
1
= B
1
B
2
= B
2
B
3
= B
3
B
4
.
3. Join B
4
C and draw a line through B
3
(the
3rd point, 3 being smaller of 3 and 4 in
3
4
) parallel to B
4
C to intersect BC at C'.
4. Draw a line through C ' parallel
to the line CA to intersect BA at A'
(see Fig. 11.3).
Then, ? A'BC' is the required triangle.
Let us now see how this construction gives the required triangle.
By Construction 11.1,
BC 3
CC 1
'
=·
'
Therefore,
BC BC + C C C C 1 4
11
BC BC BC 3 3
'' '
= = + =+ =
'' '
, i.e.,
BC
BC
'
=
3
4
.
Also C'A' is parallel to CA. Therefore, ? A'BC' ~ ? ABC. (Why ?)
So,
A B A CBC3
AB AC BC 4
' '' '
= = =·
Example 2 : Construct a triangle similar to a given triangle ABC with its sides equal
to
5
3
of the corresponding sides of the triangle ABC (i.e., of scale factor
5
3
).
Fig. 11.3
2022-23
CONSTRUCTIONS 219
Solution : Given a triangle ABC, we are required to construct a triangle whose sides
are
5
3
of the corresponding sides of ? ABC.
Steps of Construction :
1. Draw any ray BX making an acute angle with BC on the side opposite to the
vertex A.
2. Locate 5 points (the greater of 5 and 3 in
5
3
) B
1
, B
2
, B
3
, B
4
and B
5
on BX so that
BB
1
= B
1
B
2
= B
2
B
3
= B
3
B
4
= B
4
B
5
.
3. Join B
3
(the 3rd point, 3 being smaller of 3 and 5 in
5
3
) to C and draw a line through
B
5
parallel to B
3
C, intersecting the extended line segment BC at C'.
4. Draw a line through C ' parallel to CA
intersecting the extended line segment BA at
A' (see Fig. 11.4).
Then A'BC' is the required triangle.
For justification of the construction, note that
? ABC ~ ? A'BC'. (Why ?)
Therefore,
AB AC BC
AB AC BC
= =·
' '' '
But,
3
5
BB BC 3
,
BC BB 5
==
'
So,
BC 5
,
BC 3
'
= and, therefore,
AB AC BC 5
AB AC BC 3
' '' '
= = =·
Remark : In Examples 1 and 2, you could take a ray making an acute angle with AB
or AC and proceed similarly.
EXERCISE 11.1
In each of the following, give the justification of the construction also:
1. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two
parts.
Fig. 11.4
2022-23
Page 5
216 MATHEMA TICS
11
11.1 Introduction
In Class IX, you have done certain constructions using a straight edge (ruler) and a
compass, e.g., bisecting an angle, drawing the perpendicular bisector of a line segment,
some constructions of triangles etc. and also gave their justifications. In this chapter,
we shall study some more constructions by using the knowledge of the earlier
constructions. You would also be expected to give the mathematical reasoning behind
why such constructions work.
11.2 Division of a Line Segment
Suppose a line segment is given and you have to divide it in a given ratio, say 3 : 2. You
may do it by measuring the length and then marking a point on it that divides it in the
given ratio. But suppose you do not have any way of measuring it precisely, how
would you find the point? We give below two ways for finding such a point.
Construction 11.1 : To divide a line segment in a given ratio.
Given a line segment AB, we want to divide it in the ratio m : n, where both m and
n are positive integers. To help you to understand it, we shall take m = 3 and n = 2.
Steps of Construction :
1. Draw any ray AX, making an acute angle with AB.
2. Locate 5 (= m + n) points A
1
, A
2
, A
3
, A
4
and
A
5
on AX so that AA
1
= A
1
A
2
= A
2
A
3
= A
3
A
4
= A
4
A
5
.
3. Join BA
5
.
4. Through the point A
3
( m = 3), draw a line
parallel to A
5
B (by making an angle equal to
? AA
5
B) at A
3
intersecting AB at the point C
(see Fig. 11.1). Then, AC : CB = 3 : 2.
CONSTRUCTIONS
Fig. 11.1
2022-23
CONSTRUCTIONS 217
Let us see how this method gives us the required division.
Since A
3
C is parallel to A
5
B, therefore,
3
35
AA
AA
=
AC
CB
(By the Basic Proportionality Theorem)
By construction,
3
35
AA 3 AC 3
Therefore,
A A 2 CB 2
=· =
.
This shows that C divides AB in the ratio 3 : 2.
Alternative Method
Steps of Construction :
1. Draw any ray AX making an acute angle with AB.
2. Draw a ray BY parallel to AX by making ? ABY equal to ? BAX.
3. Locate the points A
1
, A
2
, A
3
(m = 3) on AX and B
1
, B
2
(n = 2) on BY such that
AA
1
= A
1
A
2
= A
2
A
3
= BB
1
= B
1
B
2
.
4. Join A
3
B
2
. Let it intersect AB at a point C (see Fig. 11.2).
Then AC : CB = 3 : 2.
Why does this method work? Let us see.
Here ? AA
3
C is similar to ? BB
2
C. (Why ?)
Then
3
2
AA AC
BB BC
= .
Since by construction,
3
2
AA 3
,
BB 2
=
therefore,
AC 3
BC 2
=·
In fact, the methods given above work for dividing the line segment in any ratio.
We now use the idea of the construction above for constructing a triangle similar
to a given triangle whose sides are in a given ratio with the corresponding sides of the
given triangle.
Construction 11.2 : To construct a triangle similar to a given triangle as per
given scale factor.
This construction involves two different situations. In one, the triangle to be
constructed is smaller and in the other it is larger than the given triangle. Here, the
scale factor means the ratio of the sides of the triangle to be constructed with the
corresponding sides of the given triangle (see also Chapter 6). Let us take the following
examples for understanding the constructions involved. The same methods would
apply for the general case also.
Fig. 11.2
2022-23
218 MATHEMA TICS
Example 1 : Construct a triangle similar to a given triangle ABC with its sides equal
to
3
4
of the corresponding sides of the triangle ABC (i.e., of scale factor
3
4
).
Solution : Given a triangle ABC, we are required to construct another triangle whose
sides are
3
4
of the corresponding sides of the triangle ABC.
Steps of Construction :
1. Draw any ray BX making an acute angle
with BC on the side opposite to the vertex
A.
2. Locate 4 (the greater of 3 and 4 in
3
4
)
points B
1
, B
2
, B
3
and B
4
on BX so that
BB
1
= B
1
B
2
= B
2
B
3
= B
3
B
4
.
3. Join B
4
C and draw a line through B
3
(the
3rd point, 3 being smaller of 3 and 4 in
3
4
) parallel to B
4
C to intersect BC at C'.
4. Draw a line through C ' parallel
to the line CA to intersect BA at A'
(see Fig. 11.3).
Then, ? A'BC' is the required triangle.
Let us now see how this construction gives the required triangle.
By Construction 11.1,
BC 3
CC 1
'
=·
'
Therefore,
BC BC + C C C C 1 4
11
BC BC BC 3 3
'' '
= = + =+ =
'' '
, i.e.,
BC
BC
'
=
3
4
.
Also C'A' is parallel to CA. Therefore, ? A'BC' ~ ? ABC. (Why ?)
So,
A B A CBC3
AB AC BC 4
' '' '
= = =·
Example 2 : Construct a triangle similar to a given triangle ABC with its sides equal
to
5
3
of the corresponding sides of the triangle ABC (i.e., of scale factor
5
3
).
Fig. 11.3
2022-23
CONSTRUCTIONS 219
Solution : Given a triangle ABC, we are required to construct a triangle whose sides
are
5
3
of the corresponding sides of ? ABC.
Steps of Construction :
1. Draw any ray BX making an acute angle with BC on the side opposite to the
vertex A.
2. Locate 5 points (the greater of 5 and 3 in
5
3
) B
1
, B
2
, B
3
, B
4
and B
5
on BX so that
BB
1
= B
1
B
2
= B
2
B
3
= B
3
B
4
= B
4
B
5
.
3. Join B
3
(the 3rd point, 3 being smaller of 3 and 5 in
5
3
) to C and draw a line through
B
5
parallel to B
3
C, intersecting the extended line segment BC at C'.
4. Draw a line through C ' parallel to CA
intersecting the extended line segment BA at
A' (see Fig. 11.4).
Then A'BC' is the required triangle.
For justification of the construction, note that
? ABC ~ ? A'BC'. (Why ?)
Therefore,
AB AC BC
AB AC BC
= =·
' '' '
But,
3
5
BB BC 3
,
BC BB 5
==
'
So,
BC 5
,
BC 3
'
= and, therefore,
AB AC BC 5
AB AC BC 3
' '' '
= = =·
Remark : In Examples 1 and 2, you could take a ray making an acute angle with AB
or AC and proceed similarly.
EXERCISE 11.1
In each of the following, give the justification of the construction also:
1. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two
parts.
Fig. 11.4
2022-23
220 MATHEMA TICS
2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose
sides are
2
3
of the corresponding sides of the first triangle.
3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose
sides are
7
5
of the corresponding sides of the first triangle.
4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another
triangle whose sides are
1
1
2
times the corresponding sides of the isosceles triangle.
5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ? ABC = 60°. Then construct
a triangle whose sides are
3
4
of the corresponding sides of the triangle ABC.
6. Draw a triangle ABC with side BC = 7 cm, ? B = 45°, ? A = 105°. Then, construct a
triangle whose sides are
4
3
times the corresponding sides of ? ABC.
7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and
3 cm. Then construct another triangle whose sides are
5
3
times the corresponding sides
of the given triangle.
11.3 Construction of T angents to a Circle
You have already studied in the previous chapter that if a point lies inside a circle,
there cannot be a tangent to the circle through this point. However, if a point lies on the
circle, then there is only one tangent to the circle at this point and it is perpendicular to
the radius through this point. Therefore, if you want to draw a tangent at a point of a
circle, simply draw the radius through this point and draw a line perpendicular to this
radius through this point and this will be the required tangent at the point.
You have also seen that if the point lies outside the circle, there will be two
tangents to the circle from this point.
We shall now see how to draw these tangents.
Construction 11.3 : To construct the tangents to a circle from a point outside it.
We are given a circle with centre O and a point P outside it. We have to construct
the two tangents from P to the circle.
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