Question

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

Answer 1

Steps of construction:

1. Draw a triangle $\Delta ABC$ in which $AB=4cm,BC=6cm$ and $AC=4cm$.

2. Draw a ray $AX$ making an acute angle with $B$ on the side $AB$ opposite to the vertex $C$.

3. Mark three points $A_1,A_2$ and $A_3$ on the ray $AX$ so that, $A{A}_1=A_1{A}_2=A_2{A}_3$

4. Join $B{A}_3$

5. Since, required triangle is $\frac{2}{3}$,draw a line through $A_2$ and parallel to $B{A}_3$, which meets $AB$ at $B^{\prime }$.

6. Draw a line through $B^{\prime }$ and parallel to $BC$.

i.e., $BC\parallel B^{\prime }{C}^{\prime }$

Hence, $A{B}^{\prime }{C}^{\prime }$ is required triangle similar to triangle $ABC$.

Justification :

In $△AB{C}^{\prime }$ and $ABC$

$\Rightarrow \angle A^{\prime }=\angle A$ ( Common )

$\Rightarrow \angle B^{\prime }=\angle B$ ( Corresponding angles )

$\therefore △A{B}^{\prime }{C}^{\prime }\sim △ABC$ ( by A.A similarities )

$\therefore \frac{A{B}^{\prime }}{AB}=\frac{B^{\prime }{C}^{\prime }}{BC}=\frac{A{C}^{\prime }}{AC}=\frac{2}{3}$

Hence, solved.