We draw a line segment PQ of length 5 cm. We take an external point A of line segment. Let us draw parallel line through the point A to line segment PQ. [Let us draw three alternative process]
Method 1:
Steps of construction:
1. We draw a line segment PQ of length 5cm.
2. We take an external point A of line segment.
3. Draw an arc from A that intersects PQ at two different points B and C.
4. Draw perpendicular bisector of BC. It passes through A.
5. Draw perpendicular of this line through A.
6. This line is parallel to PQ.
Method 2:
Steps of Construction:
1. We draw a line segment PQ of length 5cm.
2. We take an external point A of line segment.
3. Draw a bigger arc from A cutting PQ at a point.
4. Mark it as B.
5. With the same radius and B as center, draw another arc on PQ as C.
6.From C, with the same radius draw another arc cutting the first arc at D.
7. Join D to A and extend the line.
8. This line is parallel to PQ.
Method 3:
Steps of Construction:
1. We draw a line segment PQ of length 5cm.
2. We take an external point A of line segment.
3. We draw a transverse line passing through A and cutting PQ at B.
4. We draw the angle at A equal to angle at B.
5. The corresponding drawn line is parallel to PQ.
We draw a triangle with length of sides 5 cm, 8 cm and 11 cm and draw a parallelogram equal in area to that triangle and having an angle 60°. [Let us write instruction process and proof]
Steps of Construction:
1. We draw a triangle ABC of given dimensions.
2. We draw a perpendicular bisector of side AB which intersects AB at D.
3. Join C to D.
4. Draw a line parallel to AB through C by completing the parallelogram ACQB.
5. From D, construct ∠GDB=60°.
6. The line cuts parallel line through C at E.
7. From E, cut an arc of length DB on the parallel line.
8. Mark the point as F.
9. Join F to B.
10. DBFE is the required parallelogram.
We draw a triangles which AB = 6 cm, BC = 9 cm, ∠ABC = 55°, let us draw a parallelogram equal in area to that triangle having an angle 60° and length of one side is of AC.
Steps of Construction:
1. We draw a triangle ABC of given dimensions.
2. We draw a perpendicular bisector of side AB which intersects AC at D.
3. Join B to D.
4. Draw a line parallel to AC through B by completing the parallelogram AQBC.
5. From D, construct ∠GDA=60°.
6. The line cuts parallel line through B at E.
7. From E, cut an arc of length DA on the parallel line.
8. Mark the point as F.
9. Join F to A.
10. ADEF is the required parallelogram.
In ΔPQR, ∠PQR = 30°, ∠PRQ = 75°, and QR = 8 cm. Let us draw a rectangle equal in area to that triangle.
Steps of Construction:
1. We draw a triangle PQR of given dimensions.
2. We draw a perpendicular bisector of side QR which intersects QR at D.
3. Join P to D.
4. Draw a line parallel to AC through B by completing the parallelogram AQBC.
5. As we have to draw a rectangle. The perpendicular bisector through D can be used.
6. The perpendicular bisector cuts parallel line through P at E.
7. From E, cut an arc of length DR on the parallel line.
8. Mark the point as F.
9. Join F to R.
10. EFRD is the required rectangle.
Draw an equilateral triangle with length of side 6.5 cm and let us draw a parallelogram equal in area to that triangle and having an angle 45°.
Steps of Construction:
1. We draw a triangle ABC of given dimensions.
2. We draw a perpendicular bisector of side AB which intersects AB at D.
3. Join C to D.
4. Draw a line parallel to AB through C by completing the parallelogram ACQB.
5. From D, construct ∠GDB=45°.
6. The line cuts parallel line through C at E.
7. From E, cut an arc of length DB on the parallel line.
8. Mark the point as F.
9. Join F to B.
10. DBFE is the required parallelogram.
Length of each equal sides of an isosceles triangle in 8 cm and length of base is 5 cm. Let us draw a parallelogram equal in area to that triangle and having one angle of parallelogram is equal to one of equal angle of isosceles triangle and one side is of equal side.
Steps of Construction:
1. We draw a triangle ABC of given dimensions.
2. We draw a perpendicular bisector of side BC which intersects BC at D.
3. Join A to D.
4. Draw a line parallel to BC through A by completing the parallelogram ABCQ.
5. From D, construct ∠EDC=∠ABC.
6. ABDE is the required parallelogram.
Let us draw an isosceles triangle whose equal sides are of length 8 cm. and angle between them is 30°. Let us draw a rectangle whose area is equal to the above isosceles triangle.
Steps of Construction:
1. We draw a triangle ABC of given dimensions.
2. We draw a perpendicular bisector of side AB which intersects AB at D.
3. Join C to D.
4. Draw a line parallel to AB through C by completing the parallelogram ACQB.
5. As we have to draw a rectangle of same area. We can use the perpendicular bisector drawn through D.
6. The perpendicular bisector cuts parallel line through C at E.
7. From E, cut an arc of length DB on the parallel line.
8. Mark the point as F.
9. Join F to B.
10. DBFE is the required rectangle.