24
5
Yes.Yes.Yes.Yes.
9X2
A 3 x 8 rectangle
It is area, not perimeter!
Area = 22*9 = 198 square feet Perimeter = 22+22+9+9 = 62 feet
Multiply for the area, add for the perimeter.
a - 28 p - 22
Yes. For instance, the rectangle measuring 1 by 10 has a perimeter of 22 and an area of 10, whereas the rectangle measuring 4 by 4 has a perimeter of 16 and an area of 16.
The area of a rectangle is not sufficient to determine its shape and therefore its perimeter. For example, each of the following rectangles has an area of 22 square units. But the perimeter, P, varies considerably. sqrt(22)*sqrt(22) : P = 4*sqrt(22) 2*11 : P = 26 1*22 : P = 46 0.1*220 : P = 440.2 0.01*2200 : P = 4400.02 0.001*22000 : P = 44000.002 As you may begin to see, there is no limit to the perimeter.
When all of the linear dimensions are doubled . . .-- the perimeter is also doubled-- the area is multiplied by 22 = 4.
To find the dimensions of a rectangle with an area of 24 and a perimeter of 22, we can use the formulas for area (A = length × width) and perimeter (P = 2(length + width)). Let the length be ( l ) and the width be ( w ). From the area, we have ( lw = 24 ), and from the perimeter, ( 2(l + w) = 22 ), simplifying to ( l + w = 11 ). Solving these equations simultaneously, we find the dimensions are ( l = 8 ) and ( w = 3 ) (or vice versa).
To find the area of a shape given its perimeter of 22, we need to know the specific shape. For example, if we consider a square, the perimeter is 4 times the side length, so each side would be 5. The area would then be (5^2 = 25) square units. For a rectangle with a perimeter of 22, various combinations of length and width could yield different areas, so more information is needed to provide a specific area.
5
6x5
Yes.Yes.Yes.Yes.
9X2