Question

Question 2
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are $\frac{2}{3}$ of the corresponding sides of the first triangle.

Answer 1

Steps of Construction:
Step I: AB = 6 cm is drawn.
Step II: With A as a centre and radius equal to 4 cm, an arc is drawn.
Step III: Again, with B as a centre and radius equal to 5 cm an arc is drawn on the same side of AB intersecting the 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 $A_1,A_2....A_5$
Step VII: $A_{5}B$ is joined. $A_2{B}^{\prime }$ is drawn parallel to $A_{5}B$ and B'C' is drawn parallel to BC.
$\Delta A{B}^{\prime }{C}^{\prime }$ is the required triangle

Justification:
$\angle A\left(Common\right)$
$\angle C=\angle$ C' and $\angle B=\angle$ B' (corresponding angles)
Thus $\Delta AB$'C' ~ $\Delta ABC$ by AAA similarity condition
From the figure,
$\frac{A{B}^{\prime }}{AB}=\frac{AA2}{AA5}=\frac{2}{3}$
$A{B}^{\prime }=\frac{2}{3}AB$
$A{C}^{\prime }=\frac{2}{3}AC$