Question

Construct a triangle of side $4cm,5cmand6cm$ and then a triangle similar to it whose side are $\frac{2}{3}$ of the corresponding sides of the first triangle.

Answer 1

Step I : Draw a line segment $BC=6cm$
Step II : Draw an are with $B$ as center and radius equal to $5cm$
Step III : Draw an are , with $C$ as center and radius equal to $4cm$ intersecting the previous drawn are at $A.$
Step IV : Join $AB$ and $AC$ , then $△ABC$ is the required triangle.
Step V : Below $BC$ , make an acute angle $CBX$.
Step VI : Along $BX$ , mark off three points at equal distance : $B_1,B_2,B_3$ such that $B{B}_1=B_1{B}_2=B_2{B}_3$.
Step VII : Join $B_{3}C$
Step VIII : From $B_2$ draw $B_{2}D||B_{3}C$ meeting $BC$ at $D$.
Step IX : From $D$ , Draw $ED||AC$, meeting $BAE$. Then we have $\Delta EDB$ which is three required triangle.
Justification :
Since $DE||CA$
$\Delta ABC\sim \Delta EBDand\frac{EB}{AB}=\frac{BD}{BC}=\frac{DW}{CA}=\frac{2}{3}$
Hence , we have the new $\Delta EBD$ similar to the given $\Delta ABC$, whose sides are equal to $\frac{2}{3}rd$ of the corresponding sides of $\Delta ABC$.