Question 2

The area of the path is the space between the outer rectangle and the inner rectangle. We find it by subtracting the inner area from the outer area.

Step 1 — Finding the area of the path with given dimensions

We need to know the length and width of both the outer rectangle ABCD and the inner rectangle EFGH.

Let us assign some values to these measurements. Let the length of the outer rectangle ABCD be $L = 10 \text{ units}$. Let the width of the outer rectangle ABCD be $W = 8 \text{ units}$. Let the length of the inner rectangle EFGH be $l = 6 \text{ units}$. Let the width of the inner rectangle EFGH be $w = 4 \text{ units}$.

First, we calculate the area of the outer rectangle ABCD. $Area_{\text{ABCD}} = L \times W$ $Area_{\text{ABCD}} = 10 \times 8$ $Area_{\text{ABCD}} = 80 \text{ square units}$

Next, we calculate the area of the inner rectangular park EFGH. $Area_{\text{EFGH}} = l \times w$ $Area_{\text{EFGH}} = 6 \times 4$ $Area_{\text{EFGH}} = 24 \text{ square units}$

Now, we find the area of the path by subtracting the inner area from the outer area. $Area_{\text{path}} = Area_{\text{ABCD}} - Area_{\text{EFGH}}$ $Area_{\text{path}} = 80 - 24$

$\boxed{56 \text{ square units}}$

The formula for the area of the path is: $Area_{\text{path}} = (L \times W) - (l \times w)$

Step 2 — Finding the area of the path with path width given

If the width of the path along each side is given, we can find its area. We need the length and width of the inner rectangular park EFGH, and the uniform width of the path.

Let us assign values to these measurements. Let the length of the inner rectangle EFGH be $l = 6 \text{ units}$. Let the width of the inner rectangle EFGH be $w = 4 \text{ units}$. Let the width of the path be $x = 2 \text{ units}$.

We can find the dimensions of the outer rectangle ABCD using these values. The length of the outer rectangle $L$ will be the inner length plus twice the path width. $L = l + 2x$ $L = 6 + 2(2)$ $L = 6 + 4$ $L = 10 \text{ units}$

The width of the outer rectangle $W$ will be the inner width plus twice the path width. $W = w + 2x$ $W = 4 + 2(2)$ $W = 4 + 4$ $W = 8 \text{ units}$

Now, we can find the area of the path by subtracting the inner area from the outer area, as in Step 1. $Area_{\text{path}} = (L \times W) - (l \times w)$ $Area_{\text{path}} = (10 \times 8) - (6 \times 4)$ $Area_{\text{path}} = 80 - 24$

$\boxed{56 \text{ square units}}$

Alternatively, we can break the path into four separate rectangular regions. Imagine two long strips along the length of the inner park, and two shorter strips along its width. Two rectangles with dimensions $l \times x$ (top and bottom of the inner park). Two rectangles with dimensions $(w + 2x) \times x$ (left and right sides, including the corners). $Area_{\text{path}} = 2(l \times x) + 2((w + 2x) \times x)$ $Area_{\text{path}} = 2lx + 2wx + 4x^2$ Let us use our assigned values: $l = 6$, $w = 4$, $x = 2$. $Area_{\text{path}} = 2(6)(2) + 2(4)(2) + 4(2)^2$ $Area_{\text{path}} = 24 + 16 + 4(4)$ $Area_{\text{path}} = 24 + 16 + 16$

$\boxed{56 \text{ square units}}$

The formula for the area of the path using these measurements is: $Area_{\text{path}} = 2lx + 2wx + 4x^2$ This formula can also be written as: $Area_{\text{path}} = (l + 2x)(w + 2x) - lw$

Step 3 — Effect of moving the outer rectangle

The area of the path is the difference between the area of the outer rectangle ABCD and the area of the inner rectangular park EFGH. $Area_{\text{path}} = Area_{\text{ABCD}} - Area_{\text{EFGH}}$ When the outer rectangle is moved, its dimensions (length and width) do not change. So, its area ($Area_{\text{ABCD}}$) remains the same. Similarly, the inner rectangular park EFGH also keeps its original dimensions. So, its area ($Area_{\text{EFGH}}$) remains the same. Since both $Area_{\text{ABCD}}$ and $Area_{\text{EFGH}}$ do not change, their difference, which is the area of the path, will also not change. The relative position of the inner rectangle within the outer rectangle does not affect the total area of the path.

Answer

(i) We need the length and width of the outer rectangle ABCD, and the length and width of the inner rectangle EFGH. Possible values: Outer length $L = 10 \text{ units}$, Outer width $W = 8 \text{ units}$, Inner length $l = 6 \text{ units}$, Inner width $w = 4 \text{ units}$. The area of the path is $\mathbf{56 \text{ square units}}$. The formula for the area is $Area_{\text{path}} = (L \times W) - (l \times w)$. (ii) Yes, we can find its area. We need the length ($l$) and width ($w$) of the inner rectangular park EFGH, and the width of the path ($x$). Possible values: Inner length $l = 6 \text{ units}$, Inner width $w = 4 \text{ units}$, Path width $x = 2 \text{ units}$. The area of the path is $\mathbf{56 \text{ square units}}$. The formula for the area is $Area_{\text{path}} = 2lx + 2wx + 4x^2$ or $Area_{\text{path}} = (l + 2x)(w + 2x) - lw$. (iii) No, the area of the path does not change. The dimensions (and thus areas) of the outer rectangle ABCD and the inner rectangular park EFGH remain constant, so their difference (the path area) also remains constant.