Given any number b > 2, let l = 2b/(b-2)
then the rectangle with sides of l and b will have an area
= l*b = 2b/(b-2) * b = 2b2/(b-2)
and perimeter = 2*(l+b) = 2*[2b/(b-2) + b]
=2*[2b/(b-2) + b(b-2)/(b-2)] = 2/(b-2)* [2b+b2-2b] = 2b2/(b-2)
So that the area and perimeter have the same numeric value.
Chat with our AI personalities
Yes, it can because a 3 by 6 rectangle has the perimeter of 18 and has the area of 18! :)
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.