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

 Page 1


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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.  
Answer 1:  
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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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.  
Answer 1:  
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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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.  
Answer 2:  
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


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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.  
Answer 1:  
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.  
Answer 2:  
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


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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.  
Answer 1:  
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.  
Answer 2:  
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.  
 
 
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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


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(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.  
Answer 1:  
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.  
  
(www.tiwariacademy.net) 
(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.  
Answer 2:  
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.  
 
 
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(Class – X) 
A Free web support in Education 
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.  
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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.  
Answer 3:  
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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