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. The length of side $A^{\prime }{C}^{\prime }$ (in $cm.$) is:

Answer 1

Steps of construction:

1. Construct $△ABC$, such that $AB=4cm,BC=5cm$ and $CA=6cm$.

2. Draw any ray $BX$ making an acute angle with $BC$ (below the side $BC$).

3. Mark three points $B_1,B_2,B_3$ on $BX$ such that $B{B}_1=B_1{B}_2=B_2{B}_3$.

4. Join $B_{3}C$

5. Through $B_2$ draw $B_2{C}^{\prime }\parallel B_{3}C$ and let it intersect $BC$ at $C^{\prime }$.

6. Through $C^{\prime }$ draw $C^{\prime }{A}^{\prime }\parallel CA$ and let it intersect $BA$ at $A^{\prime }$.

$△A^{\prime }B{C}^{\prime }$ is the required triangle similar to the given $△ABC$ whose side are $\frac{2}{3}$ of the

corresponding sides of the $△ABC$.

$A^{\prime }{C}^{\prime }=$ $\frac{2\times 6}{3}=4$