Let the dimensions be x and y:
2x+2y = 66 => 2y = 66-2x => y = 33-x
xy = 216 => x(33-x) = 216 => 33x-x2-216 = 0 => x2-33x+216 = 0
Solving the above quadratic equation gives x a value of 9 or 24
Therefore the dimensions are 9 feet by 24 feet
Check: (2*9)+(2*24) = 66 feet and 9*24 = 216 square feet
There is no limit to the size of the perimeter.
Length = 9 Width = 9 Your rectangle is a square.
The dimensions of the rectangle are 5 cm by 4 cm
The dimensions work out as 7 units and 15 units
the area of a rectangleis 100 square inches. The perimeter of the rectangle is 40 inches. A second rectangle has the same area but a different perimeter. Is the secind rectangle a square? Explain why or why not.
what are the dimensions of the rectangle with this perimeter and an area of 8000 square meters
What are the dimensions of a rectangle that has a perimeter of 56 units and an area of 96 square units
There is no limit to the size of the perimeter.
Length = 9 Width = 9 Your rectangle is a square.
The dimensions of the rectangle are 5 cm by 4 cm
Type your answer here... give the dimensions of the rectangle with an are of 100 square units and whole number side lengths that has the largest perimeter and the smallest perimeter
The dimensions work out as 7 units and 15 units
The greatest area that a rectangle can have is, in fact, attained when it is a square. A square with perimeter of 16 cm must have sides of 4 cm and so an area of 4*4 = 16 cm2.
4 x 24
the area of a rectangleis 100 square inches. The perimeter of the rectangle is 40 inches. A second rectangle has the same area but a different perimeter. Is the secind rectangle a square? Explain why or why not.
To find the minimum perimeter of a rectangle with an area of 84 square units, we need to consider the dimensions of the rectangle. The perimeter ( P ) is given by ( P = 2(l + w) ), where ( l ) is the length and ( w ) is the width. The area constraint ( lw = 84 ) can be optimized by setting ( l ) and ( w ) equal, which occurs at a square. The dimensions for a rectangle with area 84 that minimizes the perimeter are ( l = 12 ) and ( w = 7 ), resulting in a minimum perimeter of ( P = 2(12 + 7) = 38 ) units.
Yes, the perimeter of a rectangle can be larger than its area. For example, consider a rectangle with dimensions 1 unit by 1 unit, which has a perimeter of 4 units and an area of 1 square unit. As the rectangle's dimensions change, especially when one dimension is much larger than the other, the perimeter can exceed the area even more significantly.