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Draw a triangle ABC with side BC=3.5 cm,angle B=60 degrees,altitude AD=2.5 cm . Then,draw another triangle AB'C' similar to triangle ABC such that each side of AB'C' is 2 times that of the corresponding side of triangle ABC.Please write the steps of construction.

Asked by MANISHA MOHANTY 27th February 2013, 1:03 PM
Answered by Expert

Steps of construction:

1.  Draw a line segment BC = 3.5 cm.

2. With B and C as centre, draw a perpendicular bisector (PDQ) of BC.

3. With D as centre and radius 2.5 cm, draw an arc to cut DP at A.

4. Join AB and AC. Thus, ABC is the required triangle.

5. Draw a ray AX making an acute angle with AB.

6. Mark points A1, A2 on AX such that AA1 = A1A2.

7. Join A1B.

8. Draw a line parallel to A1B and passing from A2. Let it meet the extended line AB at B'.

9. Draw a line parallel to BC and passing through B'. Let it meet the extended line AC at C'.

10. AB'C' is the required triangle similar to ABC.

Answered by Expert 28th February 2013, 10:00 PM
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