Yes, it can because a 3 by 6 rectangle has the perimeter of 18 and has the area of 18! :)
Chat with our AI personalities
No. Here are four rectangles with the same perimeter:1 by 6 . . . . . perimeter = 14, area = 62 by 5 . . . . . perimeter = 14, area = 103 by 4 . . . . . perimeter = 14, area = 1231/2 by 31/2 . . perimeter = 14, area = 121/4With all the same perimeter . . . -- The nearer it is to being square, the more area it has.-- The longer and skinnier it is, the less area it has. If somebody gives you some wire fence and tells you to put it uparound the most possible area, your first choice is to put it up ina circle, and your second choice is to put it up in a square. Rectanglesare out, if you can avoid them.
This browser is hopeless for drawing but consider the following two rectangles: a*b and (a+1)*(b-1). Their perimeter will be 2a+2b but unless a = b-1, their area will be different.
Yes. Say there are two rectangles, both with perimeter of 20. One of the rectangles is a 2 by 8 rectangle. The area of this rectangle is 2 x 8 which is 16. The other rectangle is a 4 by 6 rectangle. It has an area of 4 x 6 which is 24.
Infinite amounts.
* It is unclear if the question is asking about two rectangles, each with a perimeter of 16, or two rectangles whose perimeters sum to 16. This answer assumes the former.Other than the 4x4 square, which coincidentally has both a perimeter and area of 16, some examples would be:1 x 7 rectangle : perimeter 16 in. , area 7 sq. in2 x 6 rectangle : perimeter 16 in., area 12 sq. in3 x 5 rectangle: perimeter 16 in., area 15 sq. inYou can calculate that for a given perimeter, the largest area is found in the square with a side measurement of P/4, i.e. the length and the width are the same.