Maths 4 Marks Questions

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 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 D as center and radius more than half of GD draw arcs which intersect each other at H.
6. Join HA.

Therefore $\angle\text{HAB}=135^\circ$

1. Construct a line segment BC of 7cm.
2. At the point B, draw $\angle\text{XBC}=60^\circ$
3. Keeping B as centre and radius 12cm draw an arc which intersects XB at D.
4. Join DC.
5. Draw the perpendicular bisector of DC which intersects DB at A.
6. Join AC.

Steps of construction:

1. Draw a line segment XY of 12cm.
2. Draw $\angle\text{DXY}=45^\circ$ and $\angle\text{EYX}=60^\circ.$
3. Draw the angle bisectors of $\angle\text{DXY}$ and $\angle\text{EYX}$ which intersect each other at A.
4. Draw the perpendicular bisector of AX and AY which intersect XY at B and C respectively.
5. Join AB and AC.

Hence $\triangle\text{ABC}$ is the required triangle.

Steps of construction:

1. Draw an $\angle\text{ABC}$ of 720 with the help of a protractor.
2. Keeping B as center and any radius draw an arc which intersects AB at D and BC at E.
3. Keeping D and E as center and radius more than half of DE draw two arcs which intersect each other at F.
4. Join FB which intersects the arc in (2) at G.
5. Keeping D and G as center and radius more than half of DG draw two arcs which intersect each other at H.
6. Join HB.

Therefore $\angle\text{HBC}=54^\circ,\angle\text{FBC}=36^\circ$

Steps of construction:

1. Draw a line segment AB.
2. With A as centre and radius more than half of AB draw arcs, one on each side of AB.
3. With B as the centre and same radius draw arcs cutting the previous arcs at points P and Q respectively.
4. Join P and Q which intersects AB at C.
5. With A as centre and radius more than half of AC draw arcs, one on each side of AC.
6. With C as the centre and same radius draw arcs cutting the previous arcs at R and S respectively.
7. Join points R and S which intersects AC at D.

Therefore $\text{AD}=\big(\frac{1}{2}\big)\text{(AB)} $

Steps of construction:

1. Draw a line segment AB.
2. With A as centre and radius more than half of AB draw arcs, one on each side of AB.
3. With B as the centre and same radius draw arcs cutting the previous arcs at points P and Q respectively.
4. Join P and Q which intersects AB at C.
5. With A as centre and radius more than half of AC draw arcs, one on each side of AC.
6. With C as the centre and same radius draw arcs cutting the previous arcs at R and S respectively.
7. Join points R and S which intersects AC at D.

Therefore $\text{AD}=\frac{3}{4}\text{(AB)}$

Steps of construction:

1. Draw a line segment AB and produce BA to C.
2. Keeping A as the centre and any radius draw an arc which intersects AC at D and AB at E.
3. Keeping D and E as centre 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 D as centre and radius more than half of GD draw arcs which intersect each other at H
6. Join HA.

Therefore $\angle\text{HAB}=135^\circ $

Steps of construction:

1. Draw a line segment AB of 11cm.
2. Draw $\angle\text{DAB}=30^\circ$ and $\angle\text{FBA}=90^\circ.$
3. Draw the angle bisectors of $\angle\text{DAB}$ and $\angle\text{EBA}$ which intersect each other at A.

4. Draw the perpendicular bisector of XA and XB which intersect AB at Y and Z respectively

5. Join XY and XZ.

Hence $\triangle\text{XYZ}$ is the required triangle.

Steps of construction:

1. Draw a line segment XY of 6.4cm.
2. Draw $\angle\text{DXY}=60^\circ$ and $\angle\text{EYX}=45^\circ.$
3. Draw the angle bisectors of $\angle\text{DXY}$ and $\angle\text{EYX}$ which intersect each other at A.
4. Draw the perpendicular bisector of AX and AY which intersect XY at B and C respectively.
5. Join AB and AC.

Hence $\triangle\text{ABC}$ is the required triangle.

Steps of construction:

1. Draw a line segment BC of 6cm.
2. Take mid-point O of side BC.
3. With center B and D and radii 6cm and 4cm, draw two arcs which intersect each other at A.
4. Join AB, AD and AC.

Hence $\triangle\text{ABC}$ is the required triangle.

Steps of Construction:

1. Draw a line segment XY of 10cm.
2. Draw $\angle\text{DXY}=90^\circ$ and $\angle\text{EYX}=60^\circ.$
3. Draw the angle bisectors of $\angle\text{DXY}$ and $\angle\text{EYX}$ which intersect each other at A.
4. Draw the perpendicular bisector of AX and AY which intersect XY at B and C respectively.
5. Join AB and AC.

Hence $\triangle\text{ABC}$ is the required triangle.

Steps of construction:

1. Construct a line segment BC of 6cm.
2. At the point B, draw $\angle\text{XBC}=90^\circ.$
3. Keeping B as center and radius 10cm draw an arc which intersects XB at D.
4. Join DC.
5. Draw the perpendicular bisector of DC which intersects DB at A.
6. Join AC.

Hence $\triangle\text{ABC}$ is the required triangle.

Steps of Construction:

1. Construct a line segment BC of 3.4cm.
2. At the point B, draw $\angle\text{XBC}=45^\circ$
3. Keeping B as centre and radius 1.5cm draw an arc which intersects XB at D.
4. Join DC.
5. Draw the perpendicular bisector of DC which intersects DB at A.
6. Join AC.

Hence $\triangle\text{ABC}$ is the required triangle.

Steps of Construction:

1. Construct a line segment BC of 3.6cm.
2. At the point B, draw $\angle\text{XBC}=60^\circ.$
3. Keeping B as center and radius 4.8cm draw an arc which intersects XB at D.
4. Join DC.
5. Draw the perpendicular bisector of DC which intersects DB at A.
6. Join AC.

Hence $\triangle\text{ABC}$ is the required triangle.

Steps of Construction:

1. Construct a line segment BC of 4.5cm.
2. At the point B, draw $\angle\text{XBC}=45^\circ.$
3. Keeping B as centre and radius 5.6cm draw an arc which intersects XB at D.
4. Join DC.
5. Draw the perpendicular bisector of DC which intersects DB at A.
6. Join AC.

Hence $\triangle\text{ABC}$ is the required triangle.