Question

Draw a triangle for 4cm, 5cm , 6cm and construct a trianglr similiar whose sides are 3/2 of the corresponding sides of the first triangle​

Answer 1

Step-by-step explanation:

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