Q. 74.8( 5 Votes )

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Answer :

Steps of Construction:


Step I: BC = 5 cm is drawn.



Step II: Draw perpendicular bisector of BC, Let the bisector intersect BC at O.


Taking point D as center, and 4 cm radius, draw an arc interesting bisector at point A.


Join AB and AC.



Step III: A ray BX is drawn making an acute angle with BC opposite to vertex A.



Step IV: 3 points B1 , B2 and B3 is marked BX.



Step V: B2 is joined with C to form B2C as 2 point is smaller. B3C' is drawn parallel to B2C as 3 point is greater.



Step VI: C'A' is drawn parallel to CA.



Thus, A'BC' is the required triangle formed.


Justification:


Since the scale factor is ,


We need to prove,



By construction,


… (1)


Also, A1C1 is parallel to AC.


So, this will make same angle with BC.


A1C1B = ACB …. (2)


Now,


In ΔA1BC1 and ΔABC


B = B (common)


A1C1B = ACB (from 2)


ΔA1BC1 ΔABC


Since corresponding sides of similar triangles are in same ratio.



From (1)



Hence construction is justified.


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