Q7:

In each of the following, give the justification of the construction also: Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are $\frac{5}{3}$ times the corresponding sides of the given triangle.

Given: A right-angled triangle $ABC$ with sides $AB = 3\text{ cm}$ and $BC = 4\text{ cm}$, where $\angle B = 90^\circ$.

To Find: Construct a triangle $A'BC'$ similar to $\triangle ABC$ such that its sides are $\frac{5}{3}$ times the corresponding sides of $\triangle ABC$.

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

1. Draw a line segment $BC = 4\text{ cm}$.

2. At point $B$, construct an angle of $90^\circ$ using a compass and ruler.

3. Cut an arc of $3\text{ cm}$ on the perpendicular ray from $B$ to mark point $A$.

4. Join $AC$. Thus, $\triangle ABC$ is formed.

Step 2: Construction of the similar triangle

1. Draw an acute angle $\angle CBX$ below the line $BC$.

2. Mark 5 points (since the numerator of $\frac{5}{3}$ is 5) $B_1, B_2, B_3, B_4, B_5$ on ray $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5$.

3. Join $B_3$ to $C$ (since the denominator is 3).

4. Draw a line through $B_5$ parallel to $B_3C$ 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

By construction, $B_5C' \parallel B_3C$.

In $\triangle BB_5C'$, by Basic Proportionality Theorem (Thales Theorem):

$\frac{BC'}{BC} = \frac{BB_5}{BB_3} = \frac{5}{3}$ [Since $BB_5 = 5$ units and $BB_3 = 3$ units].

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

Therefore, the ratio of corresponding sides is:

$\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{5}{3}$.

This confirms that the sides of $\triangle A'BC'$ are $\frac{5}{3}$ times the corresponding sides of $\triangle ABC$.

Final Answer: The triangle $A'BC'$ has been constructed such that its sides are $\frac{5}{3}$ times the sides of $\triangle ABC$, with $A'B = 5\text{ cm}$ and $BC' = 6.67\text{ cm}$ approximately.