Question 8

We need to find pairs of whole numbers that multiply to the given area.

Step 1 — Find all possible rectangles for area 24

We want to find two whole numbers that multiply to 24. These numbers will be the length and width of our rectangles. Let us list all the pairs:

• Length = 1 unit, Width = 24 units
• Length = 2 units, Width = 12 units
• Length = 3 units, Width = 8 units
• Length = 4 units, Width = 6 units

Step 2 — Calculate the perimeter for each rectangle

The formula for the perimeter of a rectangle is $2 \times (\text{length} + \text{width})$.

For the 1 unit by 24 units rectangle: $\text{Perimeter} = 2 \times (1 + 24)$ $= 2 \times 25$

$\boxed{50 \text{ units}}$

For the 2 units by 12 units rectangle: $\text{Perimeter} = 2 \times (2 + 12)$ $= 2 \times 14$

$\boxed{28 \text{ units}}$

For the 3 units by 8 units rectangle: $\text{Perimeter} = 2 \times (3 + 8)$ $= 2 \times 11$

$\boxed{22 \text{ units}}$

For the 4 units by 6 units rectangle: $\text{Perimeter} = 2 \times (4 + 6)$ $= 2 \times 10$

$\boxed{20 \text{ units}}$

Step 3 — Identify greatest and least perimeter for area 24

We compare all the perimeters we found: 50, 28, 22, 20.

The greatest perimeter is 50 units. This belongs to the 1 unit by 24 units rectangle.

The least perimeter is 20 units. This belongs to the 4 units by 6 units rectangle.

Step 4 — Find all possible rectangles for area 32 sq cm

Now, let us find two whole numbers that multiply to 32. These will be the length and width for our new rectangles.

• Length = 1 cm, Width = 32 cm
• Length = 2 cm, Width = 16 cm
• Length = 4 cm, Width = 8 cm

Step 5 — Calculate the perimeter for each rectangle with area 32

For the 1 cm by 32 cm rectangle: $\text{Perimeter} = 2 \times (1 + 32)$ $= 2 \times 33$

$\boxed{66 \text{ units}}$

For the 2 cm by 16 cm rectangle: $\text{Perimeter} = 2 \times (2 + 16)$ $= 2 \times 18$

$\boxed{36 \text{ units}}$

For the 4 cm by 8 cm rectangle: $\text{Perimeter} = 2 \times (4 + 8)$ $= 2 \times 12$

$\boxed{24 \text{ units}}$

Step 6 — Identify greatest and least perimeter for area 32

We compare all the perimeters we found: 66, 36, 24.

The greatest perimeter is 66 units. This belongs to the 1 cm by 32 cm rectangle.

The least perimeter is 24 units. This belongs to the 4 cm by 8 cm rectangle.

Step 7 — Predict the shape of rectangles

Let us look at our results for both areas.

For the greatest perimeter, the rectangles were 1 by 24 and 1 by 32. These rectangles are very long and thin. One side is always 1 unit. The other side is the area number itself. So, for any given area, the rectangle with the greatest perimeter will be a 1 unit by (Area) unit rectangle.

For the least perimeter, the rectangles were 4 by 6 and 4 by 8. These rectangles are more "square-like". Their length and width are close to each other. If the area is a perfect square (like 25 or 36), then a square will have the least perimeter. If the area is not a perfect square, the rectangle whose length and width are closest to each other will have the least perimeter.

Answer

a. The rectangle with the greatest perimeter for area 24 square units is the 1 unit by 24 units rectangle. Its perimeter is 50 units. b. The rectangle with the least perimeter for area 24 square units is the 4 units by 6 units rectangle. Its perimeter is 20 units. c. If the area is 32 sq cm: The greatest perimeter is 66 units (for a 1 cm by 32 cm rectangle). The least perimeter is 24 units (for a 4 cm by 8 cm rectangle). We can predict the shape of the rectangle with the greatest perimeter. It will always be a very long and thin rectangle. One side will be 1 unit, and the other side will be the area number. We can predict the shape of the rectangle with the least perimeter. It will always be a rectangle that is almost like a square. Its length and width will be as close to each other as possible.