Q. 74.8( 5 Votes )
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.
Since the scale factor is ,
We need to prove,
Also, A1C1 is parallel to AC.
So, this will make same angle with BC.
∴ ∠A1C1B = ∠ACB …. (2)
In ΔA1BC1 and ΔABC
∠ B = ∠ B (common)
∠A1C1B = ∠ACB (from 2)
Since corresponding sides of similar triangles are in same ratio.
Hence construction is justified.
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