Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are (2/3) of the corresponding sides of it.



Steps of Construction:


Step I: AB = 6 cm is drawn.


Step II: With A as a centre and radius equal to 4cm, an arc is draw.


Step III: Again, with B as a centre and radius equal to 5 cm an arc is drawn on same side of AB intersecting previous arc at C.


Step IV: AC and BC are joined to form ΔABC.


Step V: A ray AX is drawn making an acute angle with AB below it.


Step VI: 5 equal points (sum of the ratio = 2 + 3 =5) is marked on AX as A1 A2....A5


Step VII: A5B is joined. A2B' is drawn parallel to A5B and B'C' is drawn parallel to BC.


ΔAB'C' is the required triangle


Justification:


A(Common)


C = C' and B = B' (corresponding angles)


Thus ΔAB'C' ~ ΔABC by AAA similarity condition


From the figure,


AB'/AB = AA2/AA5 = 2/3


AB' =2/3 AB


AC' = 2/3 AC


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