Maths 2 Marks Questions

Steps of construction:

1. Draw an angle ABC of 108°.
2. With B as the center and any radius draw an arc which intersects AB at P and BC at Q.
3. With P as center and radius more than half of PQ draw an arc.
4. With Q as the centre and same radius draw an arc which intersects the previous arc at R.
5. Join BR.

Therefore $\angle\text{RBC}=60^\circ$

Steps of construction:

1. Draw an angle ABC of 90°.
2. With B as the centre and any radius draw an arc which intersects AB at P and BC at Q.
3. With P as center and radius more than half of PQ draw an arc.
4. With Q as center and same radius draw an arc which intersects the previous arc at R.
5. Join RB.

Therefore $\angle\text{RBC}=45^\circ$

Steps of construction:

1. Draw a pair of vertically opposite angle $\angle\text{AOC}$ and $\angle\text{DOB}.$
2. Keeping O as the centre and any radius draw two arcs which intersect OA at P, OC at Q, OB at S and OD at R.
3. Keeping P and Q as centre and radius more than half of PQ draw two arcs which intersect each other at T.
4. Join TO.
5. Keeping R and S as centre and radius more than half of RS draw two arcs which intersect each other at U.
6. Join OU.

Therefore TOU is a straight line.

Steps of construction:

1. Draw an angle $\angle\text{ABC}$ of 120°.
2. With B as a centre and any radius, draw an arc which intersects AB at P and BC at Q.
3. With P as center and radius more than half of PQ draw an arc.
4. With Q as a center and same radius draw an arc which cuts the previous arc at R.
5. Join BR.

Therefore $\angle\text{ABR}=\angle\text{RBC} =60^\circ$

1. Draw an angle ABC and a line segment QR.
2. With center A and any radius, draw an arc which intersects $\angle\text{BAC}$ at E and D.
3. With Q as a centre and same radius draw an arc which intersects QR at S.
4. With S as center and radius equal to DE, draw an arc which intersects the previous arc at T.
5. Draw a line segment joining Q and T.

Steps of construction:

1. Draw two angles DCA and DCB forming linear pair.
2. With center C and any radius draw an arc which intersects AC at P and CD at Q and CB at R.
3. With center P and Q and any radius draw two arcs which intersect each other at S.
4. Join SC.
5. With Q and R as center and any radius draw two arcs which intersect each other at T.
6. Join TC.

Therefore $\angle\text{SCT}=90^\circ$

1. Steps of construction:

1. Draw a line segment AB and produce BA to C.
2. Keeping A as the center and any radius draw an arc which intersects AC at D and AB at E
3. Keeping D and E as center and radius more than half of DE draw arcs which intersect each other at F.
4. Join FA which intersects the arc in (2) at G.
5. Keeping G and E as center and radius more than half of GE draw arcs which intersect each other at H.
6. Join HA.

Therefore $\angle\text{HBC}=45^\circ$

2. Steps of construction

1. Draw a line segment AB.
2. Keeping A as the center and any radius draw an arc which intersects AB at C.
3. Keeping C as center and the same radius draw an arc which intersects the previous arc at D.
4. Keeping D as the center and same radius draw an arc which intersects arc in (2) at E.
5. Keeping E and D as center and radius more than half of ED draw arcs which intersect each other at F.
6. Join FA.

Therefore $\angle\text{FAB}=90^\circ$

Steps of construction:

1. Draw a line segment AB.
2. Keeping A as the centre and any radius draw an arc which intersects AB at C.
3. Keeping C as centre and the same radius draw an arc which intersects the previous arc at D.
4. Keeping D as the centre and same radius draw an arc which intersects arc in (2) at E.
5. Keeping E and D as centre and radius more than half of ED draw arcs which intersect each at F.
6. Join FA which intersects arc in (2) at G.
7. Keeping G and C as centre and radius more than half of GC draw arcs intersecting each other at point H.
8. Join HA which intersects the arc in (2) at a point I.
9. Keeping I and C as centre and radius more than half of IC draw arcs intersecting each other at point J.
10. Join JA.

Therefore $\angle\text{JAB}=22\frac{1^\circ}{2}$

Steps of construction:

1. Draw a line segment AB.
2. Keeping A as centre and any radius draw an arc which intersects AB at C.
3. Keeping C as centre and the same radius draw an arc which intersects the previous arc at D.
4. Keeping D as centre and same radius draw an arc which intersects arc in (2) at E.
5. Keeping E and D as centre and radius more than half of ED, draw arcs intersecting each other at F.
6. Join FA which intersects arc in (2) at G.
7. Keeping G and D as centre and radius more than half of GD draw arcs intersecting each other at H.
8. Join HA.

Therefore $\angle\text{HAB}=75^\circ $

Steps of construction:

1. Draw a line segment AB.
2. Keeping A as the centre and any radius draw an arc which intersects AB at C.
3. Keeping C as center and the same radius draw an arc which intersects the previous arc at D.
4. Keeping D and C as center and radius more than half of DC draw arcs which intersect each other at E.
5. Join EA.

Therefore $\angle\text{EAB}=30^\circ $

Steps of construction:

1. Draw a line segment AB.
2. Keeping A as the centre and any radius draw an arc which intersects AB at C.
3. Keeping C as centre and the same radius draw an arc which intersects the previous arc at D
4. Keeping D and C as centre and radius more than half of DC draw arcs which intersect each other at E.
5. Join EA which intersects arc in (2) at F.
6. Keeping F and C as centre and radius more than half of FC draw arcs which intersect each other at G.
7. Join GA.

Therefore $\angle\text{GAB}=15^\circ $

Steps of construction:

1. Draw a line segment AB.
2. Keeping A as the centre and any radius draw an arc which intersects AB at C.
3. Keeping C as centre and the same radius draw an arc which intersects the previous arc at D.
4. Keeping D as the centre and same radius draw an arc which intersects arc in (2) at E.
5. Keeping E and D as centre and radius more than half of ED draw arcs which intersect each other at F.
6. Join FA which intersects arc in (2) at G.
7. Keeping E and G as center and radius more than half of EG draw arcs which intersect each other at H.
8. Join HA.

Therefore $\angle\text{HAB}=105^\circ $