To draw a perpendicular to \(\bar{PQ}\) using ruler and set-square, we use the following steps:
Step I Draw a line segment \(\bar{PQ}\) and take a point R, outside of \(\bar{PQ}\).
                    
Step II Place a set-square on \(\bar{PQ}\) such that one arm of its right angle aligns along \(\bar{PQ}\).
                     
Step III Place a ruler along the edge opposite to the right angle of the set-square.
                      
Step IV Hold the ruler fixed. Slide the set-square along the ruler till the point R touches the other arm of the set square.
                      
Step V Join RS along the edge through R meeting \(\bar{PQ}\) at S.
                       
Hence, \(\bar{RS}\bot\bar{PQ}\)

To draw \(\bar{XY}\) perpendicular on 1and then a perpendicular to \(\bar{XY}\) at Y, we use the following steps:
Step I Firstly, draw a line 1 and take point X on it.
Step II With X as centre and a convenient radius, draw an arc intersecting the line 1 at two points A and B.
Step III With A and B as centres and radius greater than \(\bar{XA}\), draw two arcs, which cut each other at C.
Step IV Join XC and produce it to Y. Then, \(\bar{XY}\) is perpendicular to l.
StepV With Y as centre and a convenient radius, draw an arc intersecting \(\bar{XY}\) at two points E and D.
StepVI With E and D as centres and a radius greater than \(\bar{YD}\), draw two arcs, which cut each other at F.
StepVII Join \(\bar{YF}\), then \(\bar{YF}\) is perpendicular to \(\bar{XY}\) at Y.
i.e. YF \(\bot\) XY and YX \(\bot\) l.
                               

To find out the axis of symmetry of \(\bar{AB}\) of length 7.3 cm, we use the following steps:
Step I Draw a line segment \(\bar{AB}\) of length 7.3 cm.
                   
Step II With A as centre, using compasses, draw a circle, whose radius is more than half the length of \(\bar{AB}\).
                     
Step III With the same radius and with B as centre, draw another circle using compasses. It cut the previous circle at C and D.
                    
Step IV Join \(\bar{CD}\) . It cuts \(\bar{AB}\) at O.
Then, \(\bar{CD}\) is the perpendicular bisector of AB.
Also, it is the axis of symmetry.

On measuring, \(\angle\)BOD = \(\angle\)AOD = 30^o.

On measuring, \(\angle\)BOA = 90^o.

On measuring, \(\angle\)COA = 120^o.

To construct an angle of measure 135°, we use the following steps of construction:
Step I Draw any line PQ and take a point O on it.
                
Step II Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line at A and D.
                  
Step III Without disturbing the radius of the compasses, draw an arc with A as centre, which cuts the first arc at B.
                  
Step IV Again, with the same radius draw an arc with B as centre which cuts the first arc (as drawn in Step II) at C. Join OC by dotted line. Then, we get \(\angle\) COQ = l20°.
                  
Step V Now, with C and D as centre and radius more than half of length CD, draw arcs which cut each other at E.
Join OE by dotted line, we get \(\angle\)EOQ = 150°, Also, OE cuts the arc CD at I.
                   
Step VI With C and I as centre and radius more than half of length CI, draw arcs which cur each other at F.
Join OF. Then, we get \(\angle\)FOQ = 135°.
                       

Here, we use the following steps of construction:
Step I Draw \(\bar{AB}\) of any length. Place the centre of the protractor at A and the zero edge along \(\bar{AB}\).
                  
Step II Start with zero near B, mark point C at 45°.
                    
Step III Join AC. Then, \(\angle\)BAC is an angle of measure 45°.
                   
Step IV Now, with A as centre and using compasses, draw an arc that cuts both rays of \(\angle\)A at P and Q.
                
Step V With P as centre, draw (in the interior of \(\angle\)A) an arc whose radius is more than half the length of PQ.
                 
StepVI With the same radius and with Q as centre, draw another arc in the interior of \(\angle\)A. Let the two arcs intersect at D. Join \(\bar{AD}\) . Then, \(\bar{AD}\) is the required bisector of \(\angle\)A, i.e. angle bisector of angle 45°.
                  

Step 1. Draw a line segment \(\bar { AB } \) of length 12.8 cm.
Step 2. With A as centre, using compass, draw two arcs on either side of AB. The radius of this arc should be more than half of the length of \(\bar { AB } \).
Step 3. With the same radius and with B as centre, draw another arc using compass. Let it cut the previous arc at C and D.
Step 4. Join \(\bar { CD } \). It cuts \(\bar { AB } \) at E. Then \(\bar { CD } \) is the perpendicular bisector of the line segment \(\bar { AB } \).
Step 5. With A as centre, using compass, draw a circle. The radius of this circle should be more than half of the length of AE.
Step 6. With the same radius and with E as centre, draw another circle using compass. Let it cut the previous circle at F and G.
Step 7. Join \(\bar {FG} \) . It cuts \(\bar {AE} \) at H. Then \(\bar {FG} \) is the perpendicular bisector of the line segment \(\bar {AE} \).
Step 8. With E as centre, using compass, draw a circle. The radius of this circle should be more than half of the length of EB.
Step 9. With the same radius and with B as centre, draw another circle using compass. Let it cut the previous circle at I and J.
Step 10. Join \(\bar {IJ} \) It cuts \(\bar {EB} \) at K. Then \(\bar {IJ} \) is the perpendicular bisector of the line segment \(\bar {EB} \).
Now, the points H, E and K divide AB into four equal parts, i.e.,
\(\bar { AH } =\bar { HE } =\bar { EK } =\bar { KB } \)
By measurement,
\(\bar { AH } =\bar { HE } =\bar { EK } =\bar { KB } \)=3.2 cm.

Take the following numbers:
9
9 x 2 = 18
9 x 3 = 27
9 x 4 = 36
9 x 5 = 45
9 x 6 = 54
9 x 7 = 63
9 x 8 = 72
9 x 9 = 81
9 x 10 = 90
9 x 11 = 99
9 x 12 = 108
9 x 13 = 117
9 x 14 = 126
9 x 15 = 135
Divide these numbers by 16. The remainders we get are as follows: 9, 2,11, 4, 13, 6, 15, 8, 1, ......
We se that
9 + 2 = 11
9 + 4 = 13
9 + 6 = 15
9 + 8 = 17 = 1 (17 -16 = 1)
9 + 10 = 19 = 3 (19 - 16 = 3)
9 + 12 = 21 = 5 (21 - 16 = 5)
9 + 14 = 23 = 7 (23 - 16 = 7)
Also, the remainders at the odd places are as follows:
9, 11, 13, 15, 15, 17 = 1,
19 = 3, 21 = 5, 23 = 7
The remainders at the even places are as follows:
2, 4, 6, 8, 10, 12, 14
Now, take the following numbers:
12
12 x 2 = 24
12 x 3 = 36
12 x 4 = 48
12 x 5 = 60
12 x 6 = 72
12 x 7 = 84
12 x 8 = 96
12 x 9 = 108
12 x 10 = 120
12 x 11 = 132
12 x 12 = 144
12 x 13 = 156
Divide these numbers by 14. The remainders we get are as follows:
12, 10, 8, 6, 4, 2, 0, 12, 10, 8, 6, 4
We see that the remainders go on reducing by 2. On reaching at 0, we start to get them again in the same sequence.