Question

Construct an isosceles triangle whose base is $8cm$ and altitude $4cm$ and then draw a another triangle whose sides are $1\frac{1}{2}$ times the corresponding sides of the isosceles triangle.

Answer 1

Sol step of construction :
Step I : Draw $BC=8cm$
Step II : Construct $XY$ , the perpendicular bisector of line segment $BC$ meeting $BC$ at $M$.
Step III : Along $MP$ cut -off $MA=4cm$
Step IV : Join $BA$ and $CA$ , then $\Delta ABC$ so obtained is the required $\Delta ABC$
Step V ; Extend $BC$ to $D$ , Such that $BD=12cm(=\frac{3}{2}\times 87cm)$
Step Vi : Draw $DE||CA$ meeting $BA$ produced at $E$ .Then $\Delta EBD$ is the required triangle.
Justification :
Since $DE||CA$
$\therefore \Delta ABC\sim \Delta EBDand\frac{EB}{AB}=\frac{DE}{CA}=\frac{BD}{BC}=\frac{12}{8}=\frac{3}{2}$
Hence we have the new triangle similar to the given triangle whose are $1\frac{1}{2}i.e.\frac{3}{2}$ times the corresponding sides of the isosceles $\Delta ABC$.