Q1. Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC
Steps of construction:
- Draw an angle ABC and a line segment QR.
- With center A and any radius, draw an arc which intersects ∠BACat E and D.
- With Q as a centre and same radius draw an arc which intersects QR at S.
- With S as center and radius equal to DE, draw an arc which intersects the previous arc at T.
- Draw a line segment joining Q and T.
Therefore ∠PQR= ∠BAC
Q2. Draw an obtuse angle. Bisect it. Measure each of the angles so formed.
Steps of construction:
- Draw an angle ∠ABC of 120^{0}.
- With B as a centre and any radius, draw an arc which intersects AB at P and BC at Q.
- With P as center and radius more than half of PQ draw an arc.
- With Q as a center and same radius draw an arc which cuts the previous arc at R.
- Join BR.
Therefore ∠ABR= ∠RBC= 60^{0}
Q3. Using your protractor, draw an angle of 1080. With this given angle as given, draw an angle of 54^{0}.
Steps of construction:
- Draw an angle ABC of 108^{0}.
- With B as the center and any radius draw an arc which intersects AB at P and BC at Q.
- With P as center and radius more than half of PQ draw an arc.
- With Q as the centre and same radius draw an arc which intersects the previous arc at R.
- Join BR.
Therefore ∠RBC= 54^{0}
Q4. Using the protractor, draw a right angle. Bisect it to get an angle of measure 45^{0}.
Steps of construction:
- Draw an angle ABC of 90^{0}.
- With B as the centre and any radius draw an arc which intersects AB at P and BC at Q.
- With P as center and radius more than half of PQ draw an arc.
- With Q as center and same radius draw an arc which intersects the previous arc at R.
- Join RB.
Therefore ∠RBC= 45^{0}
Q5. Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Steps of construction:
- Draw two angles DCA and DCB forming linear pair
- With center C and any radius draw an arc which intersects AC at P and CD at Q and CB at R
- With center P and Q and any radius draw two arcs which intersect each other at S
- Join SC
- With Q and R as center and any radius draw two arcs which intersect each other at T
- Join TC
Therefore ∠SCT= 90^{0}.
Q6. Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line.
Steps of Construction:
- Draw a pair of vertically opposite angle ∠AOC and ∠DOB.
- Keeping O as the center and any radius draw two arcs which intersect OA at P, OC at Q, OB at S and OD at R.
- Keeping P and Q as center and radius more than half of PQ draw two arcs which intersect each other at T.
- Join TO.
- Keeping R and S as center and radius more than half of RS draw two arcs which intersect each other at U.
- Join OU.
Therefore TOU is a straight line
Q7. Using rulers and compasses only, draw a right angle.
Steps of construction:
- Draw a line segment AB.
- Keeping A as the center and any radius draw an arc which intersects AB at C.
- Keeping C as center and the same radius draw an arc which intersects the previous arc at D.
- Keeping D as the center and same radius draw an arc which intersects arc in (2) at E.
- Keeping E and D as center and radius more than half of ED draw arcs which intersect each other at F.
- Join FA.
Therefore ∠FAB= 90^{0}
Q8.Using rulers and compasses only, draw an angle of measure 135^{0}.
Steps of construction:
- Draw a line segment AB and produce BA to C.
- Keeping A as the center and any radius draw an arc which intersects AC at D and AB at E.
- Keeping D and E as center and radius more than half of DE draw arcs which intersect each other at F.
- Join FA which intersects the arc in (2) at G.
- Keeping G and D as center and radius more than half of GD draw arcs which intersect each other at H.
- Join HA.
Therefore ∠HAB= 135^{0}
Q9. Using a protractor, draw an angle of measure 72^{0}. With this angle as given draw angles of measure 360and 540.
Steps of construction:
- Draw an ∠ABC of 72^{0} with the help of a protractor.
- Keeping B as center and any radius draw an arc which intersects AB at D and BC at E.
- Keeping D and E as center and radius more than half of DE draw two arcs which intersect each other at F.
- Join FB which intersects the arc in (2) at G.
- Keeping D and G as center and radius more than half of DG draw two arcs which intersect each other at H
- Join HB
Therefore ∠HBC= 54^{0}
∠FBC= 36^{0}
Q10. Construct the following angles at the initial point of a given ray and justify the construction:
- 45^{0}
- 90^{0}
Answers
- 45^{0}
Steps of construction:
- Draw a line segment AB and produce BA to C.
- Keeping A as the center and any radius draw an arc which intersects AC at D and AB at E.
- Keeping D and E as center and radius more than half of DE draw arcs which intersect each other at F.
- Join FA which intersects the arc in (2) at G.
- Keeping G and E as center and radius more than half of GE draw arcs which intersect each other at H.
- Join HA.
Therefore ∠HAB= 45^{0}
2. 90^{0}
Steps of construction
- Draw a line segment AB.
- Keeping A as the center and any radius draw an arc which intersects AB at C.
- Keeping C as center and the same radius draw an arc which intersects the previous arc at D.
- Keeping D as the center and same radius draw an arc which intersects arc in (2) at E.
- Keeping E and D as center and radius more than half of ED draw arcs which intersect each other at F.
- Join FA.
Therefore ∠FAB= 90^{0}
Q11. Construct the angles of the following measurements:
- 30^{0}
- 75^{0}
- 105^{0}
- 135^{0}
- 15^{0}
ANSWERS:
- 30^{0}
Steps of construction:
- Draw a line segment AB.
- Keeping A as the centre and any radius draw an arc which intersects AB at C.
- Keeping C as center and the same radius draw an arc which intersects the previous arc at D.
- Keeping D and C as center and radius more than half of DC draw arcs which intersect each other at E.
- Join EA.
Therefore ∠EAB= 30^{0}
2. 75^{0}
Steps of construction:
- Draw a line segment AB
- Keeping A as center and any radius draw an arc which intersects AB at C
- Keeping C as center and the same radius draw an arc which intersects the previous arc at D
- Keeping D as center and same radius draw an arc which intersects arc in (2) at E
- Keeping E and D as center and radius more than half of ED, draw arcs intersecting each other at F
- Join FA which intersects arc in (2) at G
- Keeping G and D as center and radius more than half of GD draw arcs intersecting each other at H
- Join HA
Therefore ∠HAB= 75^{0}
3.105^{0}
Steps of construction:
- Draw a line segment AB.
- Keeping A as the center and any radius draw an arc which intersects AB at C.
- Keeping C as center and the same radius draw an arc which intersects the previous arc at D.
- Keeping D as the centre and same radius draw an arc which intersects arc in (2) at E.
- Keeping E and D as center and radius more than half of ED draw arcs which intersect each other at F.
- Join FA which intersects arc in (2) at G.
- Keeping E and G as center and radius more than half of EG draw arcs which intersect each other at H.
- Join HA.
Therefore ∠HAB= 105^{0}
4. 135^{0}
Steps of construction:
- Draw a line segment AB and produce BA to C.
- Keeping A as the center and any radius draw an arc which intersects AC at D and AB at E.
- Keeping D and E as center and radius more than half of DE draw arcs which intersect each other at F.
- Join FA which intersects the arc in (2) at G.
- Keeping G and D as center and radius more than half of GD draw arcs which intersect each other at H
- Join HA.
Therefore ∠HAB= 135^{0}
5. 15^{0}
Steps of construction:
- Draw a line segment AB.
- Keeping A as the centre and any radius draw an arc which intersects AB at C.
- Keeping C as center and the same radius draw an arc which intersects the previous arc at D.
- Keeping D and C as center and radius more than half of DC draw arcs which intersect each other at E.
- Join EA which intersects arc in (2) at F.
- Keeping F and C as center and radius more than half of FC draw arcs which intersect each other at G.
- Join GA.
Therefore ∠GAB= 15^{0}
6.
Steps of construction:
- Draw a line segment AB.
- Keeping A as the center and any radius draw an arc which intersects AB at C.
- Keeping C as center and the same radius draw an arc which intersects the previous arc at D.
- Keeping D as the centre and same radius draw an arc which intersects arc in (2) at E.
- Keeping E and D as center and radius more than half of ED draw arcs which intersect each at F.
- Join FA which intersects arc in (2) at G.
- Keeping G and C as center and radius more than half of GC draw arcs intersecting each other at point H.
- Join HA which intersects the arc in (2) at a point I.
- Keeping I and C as center and radius more than half of IC draw arcs intersecting each other at point J.
- Join JA.
Therefore ∠JAB=