What is the area of a rectangle with a perimeter of 16cm?

Answer 1

Any area you want greater than 0 cm2 and less than or equal to 16 cm2.

perimeter_of_rectangle = 2 x (width + length)

16 cm = 2 x (width + length)

⇒ width + length = 8 cm

⇒ length = 8 cm - width

Neither dimension can be negative, so they are both greater than 0 cm and less than 8 cm.

Area_of_rectangle = width x length

⇒ area = width x (8 cm - width)

= 8 x width - width2 cm2

This has a minimum value when width = 0 cm or 8cm of 0 cm2, so the area of the perimeter must be greater than 0 cm2 (as the width can be neither 0 cm nor 8 cm).

This has a maximum value when width = 4 cm:

max_area = 8 x 4 - 42 cm2 = 16 cm2.

when the width is between 0 cm and 4 cm (and so the length is between 8 cm and 4 cm) the area is between 0 cm2 and 16 cm2

When the width = 4 cm, the length is also 4 cm, that is the rectangle is a square.

Answer 2

Well, well, well, aren't we getting fancy with our geometry questions? If the perimeter of a rectangle is 16cm, that means the sum of all its sides is 16cm. Since a rectangle has two pairs of equal sides, we can divide 16cm by 4 to get each side length, which is 4cm. To find the area, you simply multiply the length and width, so the area of this rectangle would be 16 square centimeters.

Answer 3

To find the area of a rectangle with a given perimeter, we need to use the formula for the perimeter of a rectangle, which is 2(length + width) = 16cm. Since the perimeter is given as 16cm, we have 2(length + width) = 16. If we divide by 2, we get length + width = 8. Without knowing the specific dimensions of the rectangle, we cannot determine the exact area.

Answer 4

Oh, isn't that just delightful? To find the area of a rectangle, we need to know the length and width. With a perimeter of 16cm, we could have different lengths and widths that add up to 8cm each. So, the area could vary depending on the specific dimensions of the rectangle. Just remember, there are endless possibilities to create something beautiful with math!