Q3:

In each of the following, give the justification of the construction also: Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are $\frac{7}{5}$ of the corresponding sides of the first triangle.

Given: A triangle with side lengths $a = 5$ cm, $b = 6$ cm, and $c = 7$ cm. A scale factor of $k = \frac{7}{5}$ for the construction of a similar triangle.

To Find: Construct a triangle similar to the given triangle with sides $\frac{7}{5}$ times the corresponding sides of the first triangle and provide the geometric justification.

Step 1: Construction of the initial triangle $\triangle ABC$

1. Draw a line segment $BC = 5$ cm using a ruler.

2. With $B$ as the center and radius $7$ cm, draw an arc.

3. With $C$ as the center and radius $6$ 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 $\triangle A'BC'$

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

2. Locate $7$ points $B_1, B_2, B_3, B_4, B_5, B_6, B_7$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5 = B_5B_6 = B_6B_7$.

3. Join $B_5$ to $C$.

4. Draw a line through $B_7$ parallel to $B_5C$ intersecting the extended line segment $BC$ at $C'$.

5. Draw a line through $C'$ parallel to $AC$ intersecting the extended line segment $BA$ at $A'$.

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

Step 3: Justification

To justify the construction, we must prove that $\triangle ABC \sim \triangle A'BC'$ and that the ratio of their sides is $\frac{7}{5}$.

By construction, $B_7C' \parallel B_5C$.

In $\triangle BB_7C'$, by the Basic Proportionality Theorem (Thales Theorem):

$\frac{BC'}{BC} = \frac{BB_7}{BB_5} = \frac{7}{5}$ [Since $BB_7 = 7$ units and $BB_5 = 5$ units]

Also, since $A'C' \parallel AC$, $\triangle ABC \sim \triangle A'BC'$ by AA similarity criterion.

Therefore, the ratios of the corresponding sides are equal:

$\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{7}{5}$

This confirms that the sides of $\triangle A'BC'$ are $\frac{7}{5}$ times the corresponding sides of $\triangle ABC$.

Final Answer: The triangle $\triangle A'BC'$ has been constructed such that its sides are $\frac{7}{5}$ of the sides of $\triangle ABC$, justified by the Basic Proportionality Theorem and AA similarity.