(i)

Given: Rectangle $ABCD$ has sides $a$, $b$; rectangle $PQRS$ has sides $2a$, $2b$.

Formula:

$$\text{Area of rectangle} = \text{length} \times \text{width}$$

Substitution:

$$\text{Area of } ABCD = a \times b = ab$$ $$\text{Area of } PQRS = 2a \times 2b = 4ab$$

Calculation:

$$\frac{\text{Area of } PQRS}{\text{Area of } ABCD} = \frac{4ab}{ab} = 4$$

Answer: The area of $PQRS$ is 4 times the area of $ABCD$.

Yes, 4 copies of rectangle $ABCD$ will fit into rectangle $PQRS$. We can arrange them in a $2 \times 2$ grid: two copies along the side of length $2a$ and two copies along the side of length $2b$.

∴ 4 copies of rectangle $ABCD$ fit into rectangle $PQRS$ in a $2 \times 2$ grid.

(ii)

Given: $\Delta ABC$ has sides $a$, $b$, $c$; $\Delta PQR$ has sides $2a$, $2b$, $2c$.

Formula: Using Heron's formula,

$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}, \quad s = \frac{a+b+c}{2}$$

Substitution:

Semi-perimeter of $\Delta ABC$:

$$s = \frac{a+b+c}{2}$$ $$\text{Area}(\Delta ABC) = \sqrt{s(s-a)(s-b)(s-c)}$$

For $\Delta PQR$, semi-perimeter:

$$S = \frac{2a+2b+2c}{2} = 2s$$

Calculation:

$$\text{Area}(\Delta PQR) = \sqrt{2s(2s-2a)(2s-2b)(2s-2c)}$$ $$= \sqrt{16\,s(s-a)(s-b)(s-c)}$$ $$= 4 \times \text{Area}(\Delta ABC)$$

Answer: $\Delta PQR$ has 4 times the area of $\Delta ABC$.

Yes, 4 copies of $\Delta ABC$ will fit into $\Delta PQR$. By joining the midpoints of the sides of $\Delta PQR$, we divide it into 4 congruent triangles, each with sides $a$, $b$, $c$.

∴ 4 copies of $\Delta ABC$ fit into $\Delta PQR$ by joining the midpoints of $\Delta PQR$.

(iii)

Given: $\Delta ABC$ has sides $a$, $b$, $c$; $\Delta PQR$ has sides $3a$, $3b$, $3c$.

Formula: Using Heron's formula,

$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}, \quad s = \frac{a+b+c}{2}$$

Substitution:

For $\Delta PQR$ with sides $3a$, $3b$, $3c$, the semi-perimeter is:

$$S = \frac{3a+3b+3c}{2} = 3s$$

Calculation:

$$\text{Area}(\Delta PQR) = \sqrt{3s(3s-3a)(3s-3b)(3s-3c)}$$ $$= \sqrt{81\,s(s-a)(s-b)(s-c)}$$ $$= 9 \times \text{Area}(\Delta ABC)$$

Answer: $\Delta PQR$ has 9 times the area of $\Delta ABC$.

Yes, 9 copies of $\Delta ABC$ will fit into $\Delta PQR$. By dividing each side of $\Delta PQR$ into 3 equal parts and drawing lines parallel to the sides through these points, we get 9 congruent triangles, each with sides $a$, $b$, $c$.

∴ 9 copies of $\Delta ABC$ fit into $\Delta PQR$ by dividing each side into 3 equal parts.