NCERT Solution - Constructions Class 9 Notes | EduRev

Class 9 : NCERT Solution - Constructions Class 9 Notes | EduRev

 Page 1


Cbse-spot.blogspot.com  
1 
  
  
Class IX  Chapter 11 – Constructions   
Maths   
   
  
Exercise 11.1 Question 
1:   
Construct an angle of 90° at the initial point of a given ray and justify the construction.   
Answer:   
The below given steps will be followed to construct an angle of 90°.   
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre,  
which intersects PQ at R.   
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting  
the previously drawn arc at S.   
(iii) Taking S as centre and with the same radius as before, draw an arc intersecting  
the arc at T (see figure).   
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.   
(v) Join PU, which is the required ray making 90° with the given ray PQ.   
Page 2


Cbse-spot.blogspot.com  
1 
  
  
Class IX  Chapter 11 – Constructions   
Maths   
   
  
Exercise 11.1 Question 
1:   
Construct an angle of 90° at the initial point of a given ray and justify the construction.   
Answer:   
The below given steps will be followed to construct an angle of 90°.   
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre,  
which intersects PQ at R.   
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting  
the previously drawn arc at S.   
(iii) Taking S as centre and with the same radius as before, draw an arc intersecting  
the arc at T (see figure).   
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.   
(v) Join PU, which is the required ray making 90° with the given ray PQ.   
Cbse-spot.blogspot.com  
2 
  
   
Justification of Construction:   
We can justify the construction, if we can prove UPQ = 90°.   
For this, join PS and PT.   
   
We have,  SPQ =  TPS = 60°. In (iii) and (iv) steps of this construction, PU was 
drawn as the bisector of  TPS.   
  UPS =  TPS    
Also, UPQ = SPQ + UPS   
= 60° + 30°   
= 90°   
Question 2:   
Construct an angle of 45° at the initial point of a given ray and justify the construction.   
Answer:   
? 
Page 3


Cbse-spot.blogspot.com  
1 
  
  
Class IX  Chapter 11 – Constructions   
Maths   
   
  
Exercise 11.1 Question 
1:   
Construct an angle of 90° at the initial point of a given ray and justify the construction.   
Answer:   
The below given steps will be followed to construct an angle of 90°.   
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre,  
which intersects PQ at R.   
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting  
the previously drawn arc at S.   
(iii) Taking S as centre and with the same radius as before, draw an arc intersecting  
the arc at T (see figure).   
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.   
(v) Join PU, which is the required ray making 90° with the given ray PQ.   
Cbse-spot.blogspot.com  
2 
  
   
Justification of Construction:   
We can justify the construction, if we can prove UPQ = 90°.   
For this, join PS and PT.   
   
We have,  SPQ =  TPS = 60°. In (iii) and (iv) steps of this construction, PU was 
drawn as the bisector of  TPS.   
  UPS =  TPS    
Also, UPQ = SPQ + UPS   
= 60° + 30°   
= 90°   
Question 2:   
Construct an angle of 45° at the initial point of a given ray and justify the construction.   
Answer:   
? 
Cbse-spot.blogspot.com  
3 
  
The below given steps will be followed to construct an angle of 45°.   
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre,  
which intersects PQ at R.   
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting  
the previously drawn arc at S.   
(iii) Taking S as centre and with the same radius as before, draw an arc intersecting  
the arc at T (see figure).   
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.   
(v) Join PU. Let it intersect the arc at point V.   
(vi) From R and V, draw arcs with radius more than RV to intersect each other at W.  
Join PW.   
PW is the required ray making 45° with PQ.   
   
Justification of Construction:   
We can justify the construction, if we can prove WPQ = 45°.   
For this, join PS and PT.   
Page 4


Cbse-spot.blogspot.com  
1 
  
  
Class IX  Chapter 11 – Constructions   
Maths   
   
  
Exercise 11.1 Question 
1:   
Construct an angle of 90° at the initial point of a given ray and justify the construction.   
Answer:   
The below given steps will be followed to construct an angle of 90°.   
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre,  
which intersects PQ at R.   
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting  
the previously drawn arc at S.   
(iii) Taking S as centre and with the same radius as before, draw an arc intersecting  
the arc at T (see figure).   
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.   
(v) Join PU, which is the required ray making 90° with the given ray PQ.   
Cbse-spot.blogspot.com  
2 
  
   
Justification of Construction:   
We can justify the construction, if we can prove UPQ = 90°.   
For this, join PS and PT.   
   
We have,  SPQ =  TPS = 60°. In (iii) and (iv) steps of this construction, PU was 
drawn as the bisector of  TPS.   
  UPS =  TPS    
Also, UPQ = SPQ + UPS   
= 60° + 30°   
= 90°   
Question 2:   
Construct an angle of 45° at the initial point of a given ray and justify the construction.   
Answer:   
? 
Cbse-spot.blogspot.com  
3 
  
The below given steps will be followed to construct an angle of 45°.   
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre,  
which intersects PQ at R.   
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting  
the previously drawn arc at S.   
(iii) Taking S as centre and with the same radius as before, draw an arc intersecting  
the arc at T (see figure).   
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.   
(v) Join PU. Let it intersect the arc at point V.   
(vi) From R and V, draw arcs with radius more than RV to intersect each other at W.  
Join PW.   
PW is the required ray making 45° with PQ.   
   
Justification of Construction:   
We can justify the construction, if we can prove WPQ = 45°.   
For this, join PS and PT.   
Cbse-spot.blogspot.com  
4 
  
   
We have,  SPQ =  TPS = 60°. In (iii) and (iv) steps of this construction, PU was 
drawn as the bisector of  TPS.   
?
 
? 
UPS = 
 
TPS     
Also, UPQ = SPQ + UPS   
= 60° + 30°   
= 90°   
In step (vi) of this construction, PW was constructed as the bisector of  UPQ.   
  WPQ =  UPQ    
Question 3:   
Construct the angles of the following measurements:   
(i) 30° (ii) (iii) 15° Answer:   
 (i)30°   
The below given steps will be followed to construct an angle of 30°.   
Step I: Draw the given ray PQ. Taking P as centre and with some radius, draw an arc 
of a circle which intersects PQ at R.   
Step II: Taking R as centre and with the same radius as before, draw an arc intersecting 
the previously drawn arc at point S.   
? 
? 
Page 5


Cbse-spot.blogspot.com  
1 
  
  
Class IX  Chapter 11 – Constructions   
Maths   
   
  
Exercise 11.1 Question 
1:   
Construct an angle of 90° at the initial point of a given ray and justify the construction.   
Answer:   
The below given steps will be followed to construct an angle of 90°.   
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre,  
which intersects PQ at R.   
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting  
the previously drawn arc at S.   
(iii) Taking S as centre and with the same radius as before, draw an arc intersecting  
the arc at T (see figure).   
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.   
(v) Join PU, which is the required ray making 90° with the given ray PQ.   
Cbse-spot.blogspot.com  
2 
  
   
Justification of Construction:   
We can justify the construction, if we can prove UPQ = 90°.   
For this, join PS and PT.   
   
We have,  SPQ =  TPS = 60°. In (iii) and (iv) steps of this construction, PU was 
drawn as the bisector of  TPS.   
  UPS =  TPS    
Also, UPQ = SPQ + UPS   
= 60° + 30°   
= 90°   
Question 2:   
Construct an angle of 45° at the initial point of a given ray and justify the construction.   
Answer:   
? 
Cbse-spot.blogspot.com  
3 
  
The below given steps will be followed to construct an angle of 45°.   
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre,  
which intersects PQ at R.   
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting  
the previously drawn arc at S.   
(iii) Taking S as centre and with the same radius as before, draw an arc intersecting  
the arc at T (see figure).   
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.   
(v) Join PU. Let it intersect the arc at point V.   
(vi) From R and V, draw arcs with radius more than RV to intersect each other at W.  
Join PW.   
PW is the required ray making 45° with PQ.   
   
Justification of Construction:   
We can justify the construction, if we can prove WPQ = 45°.   
For this, join PS and PT.   
Cbse-spot.blogspot.com  
4 
  
   
We have,  SPQ =  TPS = 60°. In (iii) and (iv) steps of this construction, PU was 
drawn as the bisector of  TPS.   
?
 
? 
UPS = 
 
TPS     
Also, UPQ = SPQ + UPS   
= 60° + 30°   
= 90°   
In step (vi) of this construction, PW was constructed as the bisector of  UPQ.   
  WPQ =  UPQ    
Question 3:   
Construct the angles of the following measurements:   
(i) 30° (ii) (iii) 15° Answer:   
 (i)30°   
The below given steps will be followed to construct an angle of 30°.   
Step I: Draw the given ray PQ. Taking P as centre and with some radius, draw an arc 
of a circle which intersects PQ at R.   
Step II: Taking R as centre and with the same radius as before, draw an arc intersecting 
the previously drawn arc at point S.   
? 
? 
Cbse-spot.blogspot.com  
5 
  
Step III: Taking R and S as centre and with radius more than RS, draw arcs to 
intersect each other at T. Join PT which is the required ray making 30° with the   
 
The below given steps will be followed to construct an angle of .   
(1) Take the given ray PQ. Draw an arc of some radius, taking point P as its centre,  
which intersects PQ at R.   
(2) Taking R as centre and with the same radius as before, draw an arc intersecting  
the previously drawn arc at S.   
(3) Taking S as centre and with the same radius as before, draw an arc intersecting  
the arc at T (see figure).   
(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.   
(5) Join PU. Let it intersect the arc at point V.   
(6) From R and V, draw arcs with radius more than RV to intersect each other at W.  
Join PW.   
(7) Let it intersect the arc at X. Taking X and R as centre and radius more than   
  
given ray PQ.    
    
(   ii   )       
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