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# NCERT Solutions, Chapter 11: Constructions, Class 10 (Mathematics) Class 10 Notes | EduRev

## Class 10 : NCERT Solutions, Chapter 11: Constructions, Class 10 (Mathematics) Class 10 Notes | EduRev

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Exercise 11.1
Question 1:
Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure
the two parts. Give the justification of the construction.
A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.
Step 1 Draw line segment AB of 7.6 cm and draw a ray AX making an acute
angle with line segment AB.
Step 2 Locate 13 (= 5 + 8) points, A
1
, A
2
, A
3
, A
4
…….. A
13
, on AX such
that AA
1
= A
1
A
2
= A
2
A
3
and so on.
Step 3 Join BA
13
.
Step 4 Through the point A
5
, draw a line parallel to BA
13
(by making an
angle equal to ?AA
13
B) at A
5
intersecting AB at point C.
C is the point dividing line segment AB of 7.6 cm in the required ratio of
5:8.
The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7
cm respectively.

Page 2

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1

Exercise 11.1
Question 1:
Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure
the two parts. Give the justification of the construction.
A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.
Step 1 Draw line segment AB of 7.6 cm and draw a ray AX making an acute
angle with line segment AB.
Step 2 Locate 13 (= 5 + 8) points, A
1
, A
2
, A
3
, A
4
…….. A
13
, on AX such
that AA
1
= A
1
A
2
= A
2
A
3
and so on.
Step 3 Join BA
13
.
Step 4 Through the point A
5
, draw a line parallel to BA
13
(by making an
angle equal to ?AA
13
B) at A
5
intersecting AB at point C.
C is the point dividing line segment AB of 7.6 cm in the required ratio of
5:8.
The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7
cm respectively.

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2
Justification
The construction can be justified by proving that
By construction, we have A
5
C || A
13
B. By applying Basic proportionality
theorem for the triangle AA
13
B, we obtain
… (1)
From the figure, it can be observed that AA
5
and A
5
A
13
contain 5 and 8 equal
divisions of line segments respectively.
… (2)
On comparing equations (1) and (2), we obtain
This justifies the construction.
Question 2:
Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar
to it whose sides are 2/3of the corresponding sides of the first triangle.
Give the justification of the construction.
Step 1
Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5
cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius.
These arcs will intersect each other at point C. Now, AC = 5 cm and BC =
6 cm and ?ABC is the required triangle.
Page 3

(Class – X)

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1

Exercise 11.1
Question 1:
Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure
the two parts. Give the justification of the construction.
A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.
Step 1 Draw line segment AB of 7.6 cm and draw a ray AX making an acute
angle with line segment AB.
Step 2 Locate 13 (= 5 + 8) points, A
1
, A
2
, A
3
, A
4
…….. A
13
, on AX such
that AA
1
= A
1
A
2
= A
2
A
3
and so on.
Step 3 Join BA
13
.
Step 4 Through the point A
5
, draw a line parallel to BA
13
(by making an
angle equal to ?AA
13
B) at A
5
intersecting AB at point C.
C is the point dividing line segment AB of 7.6 cm in the required ratio of
5:8.
The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7
cm respectively.

(Class – X)
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2
Justification
The construction can be justified by proving that
By construction, we have A
5
C || A
13
B. By applying Basic proportionality
theorem for the triangle AA
13
B, we obtain
… (1)
From the figure, it can be observed that AA
5
and A
5
A
13
contain 5 and 8 equal
divisions of line segments respectively.
… (2)
On comparing equations (1) and (2), we obtain
This justifies the construction.
Question 2:
Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar
to it whose sides are 2/3of the corresponding sides of the first triangle.
Give the justification of the construction.
Step 1
Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5
cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius.
These arcs will intersect each other at point C. Now, AC = 5 cm and BC =
6 cm and ?ABC is the required triangle.
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Step 2
Draw a ray AX making an acute angle with line AB on the opposite side of
vertex C.
Step 3
Locate 3 points A
1
, A
2
, A
3
(as 3 is greater between 2 and 3) on line AX such
that AA
1
= A
1
A
2
= A
2
A
3
.
Step 4
Join BA
3
and draw a line through A
2
parallel to BA
3
to intersect AB at point
B'.
Step 5
Draw a line through B' parallel to the line BC to intersect AC at C'.
?AB'C' is the required triangle.

Page 4

(Class – X)

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1

Exercise 11.1
Question 1:
Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure
the two parts. Give the justification of the construction.
A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.
Step 1 Draw line segment AB of 7.6 cm and draw a ray AX making an acute
angle with line segment AB.
Step 2 Locate 13 (= 5 + 8) points, A
1
, A
2
, A
3
, A
4
…….. A
13
, on AX such
that AA
1
= A
1
A
2
= A
2
A
3
and so on.
Step 3 Join BA
13
.
Step 4 Through the point A
5
, draw a line parallel to BA
13
(by making an
angle equal to ?AA
13
B) at A
5
intersecting AB at point C.
C is the point dividing line segment AB of 7.6 cm in the required ratio of
5:8.
The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7
cm respectively.

(Class – X)
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2
Justification
The construction can be justified by proving that
By construction, we have A
5
C || A
13
B. By applying Basic proportionality
theorem for the triangle AA
13
B, we obtain
… (1)
From the figure, it can be observed that AA
5
and A
5
A
13
contain 5 and 8 equal
divisions of line segments respectively.
… (2)
On comparing equations (1) and (2), we obtain
This justifies the construction.
Question 2:
Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar
to it whose sides are 2/3of the corresponding sides of the first triangle.
Give the justification of the construction.
Step 1
Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5
cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius.
These arcs will intersect each other at point C. Now, AC = 5 cm and BC =
6 cm and ?ABC is the required triangle.
(Class – X)

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3

Step 2
Draw a ray AX making an acute angle with line AB on the opposite side of
vertex C.
Step 3
Locate 3 points A
1
, A
2
, A
3
(as 3 is greater between 2 and 3) on line AX such
that AA
1
= A
1
A
2
= A
2
A
3
.
Step 4
Join BA
3
and draw a line through A
2
parallel to BA
3
to intersect AB at point
B'.
Step 5
Draw a line through B' parallel to the line BC to intersect AC at C'.
?AB'C' is the required triangle.

(Class – X)
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4
Justification
The construction can be justified by proving that
By construction, we have B’C’ || BC
? ?A = ?ABC (Corresponding angles)
In ?AB'C' and ?ABC,
? = ?ABC (Proved above)
? = ?BAC (Common)
? ? ~ ?ABC (AA similarity criterion)
… (1)
In ?AA
2
B' and ?AA
3
B,
?A
2
AB' = ?A
3
AB (Common)
?AA
2
B' = ?AA
3
B (Corresponding angles)
? ?AA
2
B' ~ ?AA
3
B (AA similarity criterion)
From equations (1) and (2), we obtain
This justifies the construction.
Page 5

(Class – X)

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1

Exercise 11.1
Question 1:
Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure
the two parts. Give the justification of the construction.
A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.
Step 1 Draw line segment AB of 7.6 cm and draw a ray AX making an acute
angle with line segment AB.
Step 2 Locate 13 (= 5 + 8) points, A
1
, A
2
, A
3
, A
4
…….. A
13
, on AX such
that AA
1
= A
1
A
2
= A
2
A
3
and so on.
Step 3 Join BA
13
.
Step 4 Through the point A
5
, draw a line parallel to BA
13
(by making an
angle equal to ?AA
13
B) at A
5
intersecting AB at point C.
C is the point dividing line segment AB of 7.6 cm in the required ratio of
5:8.
The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7
cm respectively.

(Class – X)
A Free web support in Education
2
Justification
The construction can be justified by proving that
By construction, we have A
5
C || A
13
B. By applying Basic proportionality
theorem for the triangle AA
13
B, we obtain
… (1)
From the figure, it can be observed that AA
5
and A
5
A
13
contain 5 and 8 equal
divisions of line segments respectively.
… (2)
On comparing equations (1) and (2), we obtain
This justifies the construction.
Question 2:
Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar
to it whose sides are 2/3of the corresponding sides of the first triangle.
Give the justification of the construction.
Step 1
Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5
cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius.
These arcs will intersect each other at point C. Now, AC = 5 cm and BC =
6 cm and ?ABC is the required triangle.
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3

Step 2
Draw a ray AX making an acute angle with line AB on the opposite side of
vertex C.
Step 3
Locate 3 points A
1
, A
2
, A
3
(as 3 is greater between 2 and 3) on line AX such
that AA
1
= A
1
A
2
= A
2
A
3
.
Step 4
Join BA
3
and draw a line through A
2
parallel to BA
3
to intersect AB at point
B'.
Step 5
Draw a line through B' parallel to the line BC to intersect AC at C'.
?AB'C' is the required triangle.

(Class – X)
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4
Justification
The construction can be justified by proving that
By construction, we have B’C’ || BC
? ?A = ?ABC (Corresponding angles)
In ?AB'C' and ?ABC,
? = ?ABC (Proved above)
? = ?BAC (Common)
? ? ~ ?ABC (AA similarity criterion)
… (1)
In ?AA
2
B' and ?AA
3
B,
?A
2
AB' = ?A
3
AB (Common)
?AA
2
B' = ?AA
3
B (Corresponding angles)
? ?AA
2
B' ~ ?AA
3
B (AA similarity criterion)
From equations (1) and (2), we obtain
This justifies the construction.
(Class – X)

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5

Question 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.
Give the justification of the construction.
Step 1
Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6
cm and 5 cm radius respectively. Let these arcs intersect each other at point
C. ?ABC is the required triangle having length of sides as 5 cm, 6 cm, and
7 cm respectively.
Step 2
Draw a ray AX making acute angle with line AB on the opposite side of
vertex C.
Step 3
Locate 7 points, A
1
, A
2
, A
3
, A
4
A
5
, A
6
, A
7
(as 7 is greater between 5and 7),
on line AX such that AA
1
= A
1
A
2
= A
2
A
3
= A
3
A
4
= A
4
A
5
= A
5
A
6
= A
6
A
7
.
Step 4
Join BA
5
and draw a line through A
7
parallel to BA
5
to intersect extended
line segment AB at point B'.
Step 5
Draw a line through B' parallel to BC intersecting the extended line segment
AC at C'. ?AB'C' is the required triangle.

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