Q3 of 7 Page 239

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.



More from this chapter

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1

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]

 2

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]

 4

In ΔPQR, ∠PQR = 30°, ∠PRQ = 75°, and QR = 8 cm. Let us draw a rectangle equal in area to that triangle.

 5

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°.