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Question

Draw a parallelogram ABCD in which BC=5cm,AB=3cmand angle ABC=60°, divide it into triangles BCD and ABD by the diagonal BD. Construct the triangle BD'C' similar to triangle BDC with scale factor 43. Draw the line segment D'A' parallel to DA where A' lies on extended side BA. IsA'BC'D' a parallelogram?


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Solution

Step 1. Draw a line AB=3cm and then draw a ray BY making an acute ABY=60°.

Step 2. Considering B as center and radius 5cm, draw an arc cutting the point C on BYand then draw a ray AZ making an acute ZAX'=60°BY||AZ,YBX'=ZAX'=60°

Step 3. With centre A and radius 5cm, draw an arc cutting the point D on AZ and then Join CD

Thus parallelogram ABCD is constructed.

Step 8. Join BD, the diagonal of parallelogram ABCD.

Step 9. Draw a ray BX downwards making an acute CBX.

Step 10. Locate four points B1,B2,B3,B4 on BX, such that BB1=B1B2=B2B3=B3B4

Step 11. Join B4C' and from B3C draw a line B4C'||B3C intersecting the extended line segment BC at C'.

Step 12. Draw C'D'||CD intersecting the extended line segment BD at D'. Then, D'BC'is the required triangle whose sides are 43 of the corresponding sides of BDC.

Step 13. Now draw a line segment D'A'||DA, where A' lies on the extended side BA.

Step 14. Finally, we observe thatA'BC'D' a parallelogram in which in which BC=5cm,AB=3cmand angle ABC=60°, divide it into triangles BCD and ABD by the diagonal BD.


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