Question

Construct a triangle with sides $5cm,6cm$ and $7cm$ and then another triangle whose sides are $\frac{7}{5}$ of the corresponding sides of the first triangle.

Answer 1

Steps of Construction:
Step I:

Construct a A triangle ABC with sides $5cm,6cm$ and $7cm$ is drawn.

Step II:

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

$7$ points, $B_1,B_2,B_3,B_4,B_5,B_6$ and $B_7$ are marked on $BX$.
Step IV:

Point $B_5$ is joined with $C$ to draw $B_{5}C$.

Step V:

$B_7{C}^{\prime }$ is drawn parallel to $B_{5}C$ and $C^{\prime }{A}^{\prime }$ is parallel to $CA$.
Thus $A^{\prime }B{C}^{\prime }$ is the required triangle.

Justification
$\Delta A{B}^{\prime }{C}^{\prime }\Delta ABC$ by $AAA$ similarity condition
$\therefore \frac{AB}{A^{\prime }B}=\frac{AC}{A^{\prime }{C}^{\prime }}=\frac{BC}{B{C}^{\prime }}$

and $\frac{BC}{B{C}^{\prime }}=\frac{BB5}{BB7}=\frac{5}{7}$

$\therefore \frac{A^{\prime }B}{AB}=\frac{A^{\prime }{C}^{\prime }}{AC}=\frac{B{C}^{\prime }}{BC}=\frac{7}{5}$