Question-1 :-  Construct an angle of 900 at the initial point of a given ray and justify the construction.

  Steps of Constructions :   
(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.
          
  Justification of Construction:   
  We can justify the construction, if we can prove ∠UPQ = 90°.   
  For this, join PS and PT.    
  ∠UPS = 1/2 x ∠TPS  = 60 x 1/2 = 30°    
  Also, ∠UPQ = ∠SPQ + ∠UPS   
= 60° + 30°   
= 90°   
    

Question-2 :-  Construct an angle of 450 at the initial point of a given ray and justify the construction.

  Steps of Constructions :
(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/2 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.   
  We have,  ∠SPQ = 1/2 x ∠TPS = 1/2 x 60° = 30°. 
  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 = 1/2 x ∠UPQ = 1/2 x 90 = 45°.
    

Question-3 :-  Construct the angles of the following measurements:
(i) 30°  (ii) 22 by 1/2° (iii) 15°

(i)30°   

  Steps of Constructions :   

(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.   
(II) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point S.   
(III) Taking R and S as centre and with radius more than RS/2, draw arcs to intersect each other at T. 
  Join PT which is the required ray making 30° with the given ray PQ. 
 

(ii) 22 by 1/2° 

  Steps of Constructions :
                 
(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 1/2RV to intersect each other at W.  
  Join PW.   
(VII) Let it intersect the arc at X. Taking X and R as centre and radius more than 1/2RX, draw arcs to intersect each other at Y.    
  Joint PY which is the required ray making 22 by 1/2° with the given ray PQ.    
 

(iii) 15° 

  Steps of Constructions :   

(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.   
(II) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point S.   
(III) Taking R and S as centre and with radius more than RS, draw arcs to intersect each other at T. Join PT.
(IV) Let it intersect the arc at U. Taking U and R as centre and with radius more 1/2 ray making 15° with the given ray PQ.   
 
    

Question-4 :-  Construct the following angles and verify by measuring them by a protractor:
(i) 75°  (ii) 105°  (iii) 135°

(i) 75°

  Steps of constructions :

(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 V. 
  Taking S and V as centre, draw arcs with radius more than 1/2 SV. 
  Let those intersect each other at W. 
  Join PW which is the required ray making 75° with the given ray PQ. 
   
    

(ii) 105°

  Steps of constructions :

(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 V. Taking T and V as centre, draw arcs with   
  radius more than 1/2 TV. Let these arcs intersect each other at W. 
  Join PW which is the required ray making 105° with the given ray PQ. 
 
    

(iii) 135°

  Steps of constructions :

(1) Take the given ray PQ. Extend PQ on the opposite side of Q. Draw a semi-circle of  
  some radius taking point P as its centre, which intersects PQ at R and W.   
(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 V. Taking V and W as centre and with radius   
  more than 1/2 VW, draw arcs to intersect each other at X. 
  Join PX, which is the required ray making 135° with the given line PQ. 

    

Question-5 :-  Construct an equilateral triangle, given its side and justify the construction.

  Let us draw an equilateral triangle of side 5 cm. 
  We know that all sides of an equilateral triangle are equal. 
  Therefore, all sides of the equilateral triangle will be 5 cm. 
  We also know that each angle of an equilateral triangle is 60º.   
  
  Steps of Construction : 
        
(I) Draw a line segment AB of 5 cm length. Draw an arc of some radius, 
  while taking A as its centre. Let it intersect AB at P.   
(II) Taking P as centre, draw an arc to intersect the previous arc at E. Join AE. 
(III) Taking A as centre, draw an arc of 5 cm radius, which intersects extended line segment AE at C. 
  Join AC and BC. ∆ABC is the required equilateral triangle of side 5 cm.

  Justification of Construction:
           
  We can justify the construction by showing ABC as an equilateral triangle i.e., AB = BC = AC = 5 cm and  ∠A =  ∠B =  ∠C = 60°.   
  In ∆ABC, we have AC = AB = 5 cm and  ∠A = 60°.   
  Since AC = AB,   
  ∠B =∠ C (Angles opposite to equal sides of a triangle)   
  In ∆ABC,   
  ∠A + ∠B + ∠C = 180° (Angle sum property of a triangle)   
  60° + ∠C + ∠C = 180°   
  60° + 2 ∠C = 180°   
  2 ∠C = 180° − 60° = 120°   
  ∠C = 60°   
  ∠B = ∠C = 60°   
  We have, ∠A = ∠B = ∠C = 60° ... (1)   
  ∠A = ∠B and ∠A = ∠C   
  BC = AC and BC = AB (Sides opposite to equal angles of a triangle)   
  AB = BC = AC = 5 cm ... (2)   
  From equations (1) and (2), 
  ∆ABC is an equilateral triangle.