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 Feb, 2013, 01:03: PM
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 | 28th Feb, 2013, 10:00: PM
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