Q6:

In each of the following, give the justification of the construction also: Draw a triangle ABC with side $BC = 7$ cm, $\angle B = 45^{\circ}$, $\angle A = 105^{\circ}$. Then, construct a triangle whose sides are $\frac{4}{3}$ times the corresponding sides of $\triangle ABC$.

Given: A triangle $ABC$ with side $BC = 7$ cm, $\angle B = 45^{\circ}$, and $\angle A = 105^{\circ}$. A scale factor of $\frac{4}{3}$ for the construction of a similar triangle $A'BC'$.

To Find: Construct $\triangle A'BC'$ similar to $\triangle ABC$ such that its sides are $\frac{4}{3}$ of the corresponding sides of $\triangle ABC$, and provide the justification.

Step 1: Determine the third angle of $\triangle ABC$
In $\triangle ABC$, the sum of angles is $180^{\circ}$.
$\angle A + \angle B + \angle C = 180^{\circ}$
$105^{\circ} + 45^{\circ} + \angle C = 180^{\circ}$
$150^{\circ} + \angle C = 180^{\circ}$
$\angle C = 30^{\circ}$

Step 2: Construction Steps
1. Draw a line segment $BC = 7$ cm.
2. At point $B$, construct an angle of $45^{\circ}$ using a compass and ruler.
3. At point $C$, construct an angle of $30^{\circ}$ using a compass and ruler.
4. Let the intersection of these two rays be point $A$. $\triangle ABC$ is now constructed.
5. Draw an acute angle $\angle CBX$ below $BC$.
6. Mark 4 points $B_1, B_2, B_3, B_4$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
7. Join $B_3$ to $C$.
8. Draw a line through $B_4$ parallel to $B_3C$ intersecting the extended line $BC$ at $C'$.
9. Draw a line through $C'$ parallel to $CA$ intersecting the extended line $BA$ at $A'$.
10. $\triangle A'BC'$ is the required triangle.

Step 3: Justification
To prove that $\triangle A'BC' \sim \triangle ABC$ and the ratio of sides is $\frac{4}{3}$:
By construction, $B_4C' \parallel B_3C$.
In $\triangle BB_4C'$, by Basic Proportionality Theorem (Thales Theorem):
$\frac{BC'}{BC} = \frac{BB_4}{BB_3}$
Since $BB_4 = 4$ units and $BB_3 = 3$ units, we have:
$\frac{BC'}{BC} = \frac{4}{3}$
Also, since $A'C' \parallel AC$, $\triangle A'BC' \sim \triangle ABC$ (by AA similarity criterion).
Therefore, the ratios of corresponding sides are equal:
$\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{4}{3}$.

Final Answer: The triangle $\triangle A'BC'$ has been constructed such that its sides are $\frac{4}{3}$ times the sides of $\triangle ABC$, justified by the Basic Proportionality Theorem and AA similarity.