Q2:

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

Given: A triangle with side lengths $a = 4$ cm, $b = 5$ cm, and $c = 6$ cm. We are required to construct a similar triangle whose sides are $\frac{2}{3}$ of the corresponding sides of the original triangle.

To Find: The construction steps and the geometric justification for the similarity of the two triangles.

Step 1: Construction of Triangle ABC

1. Draw a line segment $BC = 6$ cm using a ruler.

2. With $B$ as the center and radius $4$ cm, draw an arc.

3. With $C$ as the center and radius $5$ cm, draw another arc intersecting the previous arc at point $A$.

4. Join $AB$ and $AC$. Thus, $\triangle ABC$ is constructed.

Step 2: Construction of the Similar Triangle

1. Draw a ray $BX$ making an acute angle with $BC$ at $B$.

2. Locate 3 points (since the denominator of $\frac{2}{3}$ is 3) $B_1, B_2, B_3$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3$.

3. Join $B_3$ to $C$.

4. Draw a line through $B_2$ parallel to $B_3C$ intersecting $BC$ at $C'$.

5. Draw a line through $C'$ parallel to $AC$ intersecting $AB$ at $A'$.

6. $\triangle A'BC'$ is the required triangle.

Step 3: Justification

To prove $\triangle A'BC' \sim \triangle ABC$, we observe the following:

1. By construction, $B_2C' \parallel B_3C$. In $\triangle BB_3C$, by the Basic Proportionality Theorem (Thales Theorem):

$\frac{BC'}{C'C} = \frac{BB_2}{B_2B_3} = \frac{2}{1}$

Therefore, $\frac{BC'}{BC} = \frac{BB_2}{BB_3} = \frac{2}{3}$ [Since $BB_2 = 2$ units and $BB_3 = 3$ units].

2. Since $A'C' \parallel AC$, in $\triangle ABC$, $\triangle A'BC' \sim \triangle ABC$ by AA (Angle-Angle) similarity criterion, because:

$\angle B = \angle B$ (Common angle)

$\angle BA'C' = \angle BAC$ (Corresponding angles as $A'C' \parallel AC$)

3. Since the triangles are similar, the ratio of their corresponding sides is equal to the ratio of the segments on the ray $BX$:

$\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{BB_2}{BB_3} = \frac{2}{3}$

Final Answer: The constructed triangle $\triangle A'BC'$ has sides exactly $\frac{2}{3}$ of the corresponding sides of $\triangle ABC$, satisfying the similarity condition.