Q5:

In each of the following, give the justification of the construction also: Draw a triangle ABC with side $BC = 6$ cm, $AB = 5$ cm and $\angle ABC = 60^{\circ}$. Then construct a triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the triangle ABC.

Given: A triangle $ABC$ with side $BC = 6$ cm, $AB = 5$ cm, and $\angle ABC = 60^{\circ}$.

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

Visual Representation of the Construction:

Steps of Construction:

Step 1: Draw a line segment $BC = 6$ cm.

Step 2: At point $B$, construct an angle of $60^{\circ}$ using a compass and ruler. Draw a ray $BY$ such that $\angle CBY = 60^{\circ}$.

Step 3: From ray $BY$, cut off a segment $BA = 5$ cm. Join $AC$. Thus, $\triangle ABC$ is constructed.

Step 4: Draw an acute angle $\angle CBX$ below the line segment $BC$.

Step 5: Locate 4 points $B_1, B_2, B_3, B_4$ on ray $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.

Step 6: Join $B_4$ to $C$.

Step 7: Through $B_3$, draw a line parallel to $B_4C$ intersecting $BC$ at $C'$.

Step 8: Through $C'$, draw a line parallel to $CA$ intersecting $AB$ at $A'$.

Step 9: $\triangle A'BC'$ is the required triangle.

Justification:

To justify the construction, we must prove that $\triangle A'BC' \sim \triangle ABC$ and the ratio of their sides is $\frac{3}{4}$.

1. By construction, $B_3C' \parallel B_4C$.

2. In $\triangle BB_4C$, since $B_3C' \parallel B_4C$, by the Basic Proportionality Theorem (Thales Theorem):

$\frac{BC'}{BC} = \frac{BB_3}{BB_4}$ [Since $B_3$ is the 3rd point and $B_4$ is the 4th point on the ray]

$\frac{BC'}{BC} = \frac{3}{4}$ --- (Equation 1)

3. Since $A'C' \parallel AC$, in $\triangle ABC$, $\triangle A'BC' \sim \triangle ABC$ by AA similarity criterion.

4. Therefore, the ratio of corresponding sides is equal:

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

5. Substituting the value from Equation 1:

$\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{3}{4}$

Final Answer: The constructed triangle $A'BC'$ has sides exactly $\frac{3}{4}$ of the sides of $\triangle ABC$, satisfying the condition of similarity.