Q4:

In each of the following, give the justification of the construction also: Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are $1\frac{1}{2}$ times the corresponding sides of the isosceles triangle.

Given: An isosceles triangle with base $BC = 8$ cm and altitude $AD = 4$ cm. A scale factor of $1\frac{1}{2} = \frac{3}{2}$ for the construction of a similar triangle.

To Find/Construct: An isosceles triangle $ABC$ and a similar triangle $A'BC'$ such that the sides of $A'BC'$ are $\frac{3}{2}$ times the sides of $\triangle ABC$.

Step 1: Construction of the Isosceles Triangle $ABC$

1. Draw a line segment $BC = 8$ cm.

2. Construct the perpendicular bisector of $BC$ to find the midpoint $D$.

3. From $D$, mark a point $A$ on the perpendicular bisector such that $AD = 4$ cm.

4. Join $AB$ and $AC$. Thus, $\triangle ABC$ is the required isosceles triangle.

Step 2: Construction of the Similar Triangle $A'BC'$

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

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

3. Join $B_2$ to $C$ (since the denominator is $2$).

4. Draw a line through $B_3$ parallel to $B_2C$ 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 the ratio of their sides is $\frac{3}{2}$.

By construction, $AC \parallel A'C'$.

In $\triangle ABC$ and $\triangle A'BC'$:

$\angle ABC = \angle A'BC'$ (Common angle)

$\angle BCA = \angle BC'A'$ (Corresponding angles since $AC \parallel A'C'$)

Therefore, by $AA$ similarity criterion, $\triangle ABC \sim \triangle A'BC'$.

Since the triangles are similar, the ratio of their corresponding sides is equal:

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

From the construction of parallel lines using the Basic Proportionality Theorem (or Thales' Theorem) on $\triangle BB_3C'$:

$\frac{BC'}{BC} = \frac{BB_3}{BB_2} = \frac{3}{2}$

Since $\frac{BC'}{BC} = \frac{3}{2}$, it follows that $\frac{A'B}{AB} = \frac{A'C'}{AC} = \frac{3}{2}$.

This confirms that the sides of $\triangle A'BC'$ are $1\frac{1}{2}$ times the corresponding sides of $\triangle ABC$.

Final Answer: The constructed triangle $A'BC'$ has sides exactly $1.5$ times the sides of the isosceles triangle $ABC$ with base $8$ cm and altitude $4$ cm.