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Draw a rectangle with perimeter 30 cm and the least possible area?
If we restrict ourselves to whole numbers, then 1 x 14 will have the least possible area.
What is the least perimeter of a rectangle with an area of 32 ft.?
To find the least perimeter of a rectangle with a fixed area of 32 square feet, we can use the relationship between area and perimeter. For a rectangle, the area ( A = l \times w ) (length times width) and the perimeter ( P = 2(l + w) ). To minimize the perimeter while keeping the area constant, the rectangle should be a square. The side length of a square with an area of 32 ft² is ( \sqrt{32} ), which is approximately 5.66 ft. Thus, the least perimeter is ( 4 \times \sqrt{32} ), which is approximately 22.63 ft.
Is it possible for a rectangle to have a same numerical perimeter and area measure?
Yes.
Is it possible to have a perimeter of 46 and an area of 42 as a rectangle?
No, it is not possible for a rectangle to have a perimeter of 46 and an area of 42 simultaneously. For a rectangle, the perimeter ( P ) is given by ( P = 2(l + w) ), and the area ( A ) is ( A = l \times w ), where ( l ) is the length and ( w ) is the width. Solving these equations shows that the dimensions needed for these values are inconsistent, meaning no such rectangle exists.
Is it possible for a rectangle to have an area of 48 and a perimeter of 28?
You might want to investigate the rectangle that measures [ 6 by 8 ].
What is the least possible perimeter for a rectangle with an area of 169ft2?
52 ft
What is the least possible perimeter for a rectangle with an area of 169 ft?
52 (13•4)
Draw a rectangle with perimeter 30 cm and the least possible area?
If we restrict ourselves to whole numbers, then 1 x 14 will have the least possible area.
In general describe the rectangle that has the least area for a fixed perimeter?
For a fixed perimeter, the area will always be the same, regardless of how you describe the rectangle.
What is the relationship for perimeter and area for rectangle?
There is no relationship between the perimeter and area of a rectangle. Knowing the perimeter, it's not possible to find the area. If you pick a number for the perimeter, there are an infinite number of rectangles with different areas that all have that perimeter. Knowing the area, it's not possible to find the perimeter. If you pick a number for the area, there are an infinite number of rectangles with different perimeters that all have that area.
Draw a rectangle with perimeter 20 cm and the least possible area?
Assuming no fractional dimensions, least possible area would be a rectangle measuring 1cm x 9cm. Area increases to a maximum of 25 sq cm when shape is square, ie 5cm x 5cm.
What is the least perimeter of a rectangle with an area of 32 ft.?
To find the least perimeter of a rectangle with a fixed area of 32 square feet, we can use the relationship between area and perimeter. For a rectangle, the area ( A = l \times w ) (length times width) and the perimeter ( P = 2(l + w) ). To minimize the perimeter while keeping the area constant, the rectangle should be a square. The side length of a square with an area of 32 ft² is ( \sqrt{32} ), which is approximately 5.66 ft. Thus, the least perimeter is ( 4 \times \sqrt{32} ), which is approximately 22.63 ft.
What it the least perimeter of a rectangle with an area of 32 square feet?
47
Is it possible for a rectangle to have a same numerical perimeter and area measure?
Yes.
Is it possible to have a rectangle with the same perimeter and area?
NO, because if you did it would be a square
Is it possible to have a perimeter of 46 and an area of 42 as a rectangle?
No, it is not possible for a rectangle to have a perimeter of 46 and an area of 42 simultaneously. For a rectangle, the perimeter ( P ) is given by ( P = 2(l + w) ), and the area ( A ) is ( A = l \times w ), where ( l ) is the length and ( w ) is the width. Solving these equations shows that the dimensions needed for these values are inconsistent, meaning no such rectangle exists.
What is the larest area possible for any rectangle with the same perimeter?
(p/4)2, where p is the perimeter.
