Question

Construct a $△ABC$ in which $AB=4cm,BC=5cm$ and $AC=6cm.$ Now, construct a triangle similar to $△ABC$ such that each of its sides is two-third of the corresponding sides of $△ABC.$ Also, prove your assertion.

Answer 1

Steps:

1. draw base $AB=4cm$

2. with A as center, with a scale of 5cm as radius draw an arc and with B as center, with a scale of 6cm as radius draw an arc.

3. let C be the point of intersection of the above 2 lines.

4. draw a ray AX making an acute angle with line AB on the opposite side if the vertex C.

5. mark 3 points $A1,A2,A3$ such tha $A1A2=A2A3=A3A1$

6. join $A3B$ and draw a parallel line $B^{\prime }A2$

7. draw a line parallel to $BC$ to intersect $AC$ at $C^{\prime }$

Thus $A{B}^{\prime }{C}^{\prime }$ is a required triangle.

Justification;

By construction,

$\frac{A{B}^{\prime }}{AB}=\frac{AA2}{AA3}=\frac{2}{3}$

also $B^{\prime }{C}^{\prime }\parallel BC$

$\therefore \angle A{B}^{\prime }{C}^{\prime }=\angle ABC$

In $△A{B}^{\prime }{C}^{\prime },△ABC$

$\angle A=\angle A$

$\angle A{B}^{\prime }{C}^{\prime }=\angle ABC$

$△A{B}^{\prime }{C}^{\prime }\sim △ABC$

$\Rightarrow \frac{A{B}^{\prime }}{AB}=\frac{A{C}^{\prime }}{AC}=\frac{B^{\prime }{C}^{\prime }}{BC}$

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