How many rectangles are possible for 28 square units?

Answer 1

Oh, dude, let me break it down for you. So, for 28 square units, you could have 6 rectangles: a 1x28 rectangle, a 2x14 rectangle, a 4x7 rectangle, a 7x4 rectangle, a 14x2 rectangle, and a 28x1 rectangle. Six rectangles, man. That's it. Like, not a million, just six.

Answer 2

Infinitely many.

Select any number between 0 and sqrt(28) and call it W.

Then let L = 28/W.

A rectangle with length L units and width W units will have an area of LxW = (28/W)*W = 28 square units. And, since W could be any one of infinitely many numbers between 0 and sqrt(28), there are infinitely many possible answers.

Infinitely many.

Select any number between 0 and sqrt(28) and call it W.

Then let L = 28/W.

A rectangle with length L units and width W units will have an area of LxW = (28/W)*W = 28 square units. And, since W could be any one of infinitely many numbers between 0 and sqrt(28), there are infinitely many possible answers.

Infinitely many.

Select any number between 0 and sqrt(28) and call it W.

Then let L = 28/W.

A rectangle with length L units and width W units will have an area of LxW = (28/W)*W = 28 square units. And, since W could be any one of infinitely many numbers between 0 and sqrt(28), there are infinitely many possible answers.

Infinitely many.

Select any number between 0 and sqrt(28) and call it W.

Then let L = 28/W.

A rectangle with length L units and width W units will have an area of LxW = (28/W)*W = 28 square units. And, since W could be any one of infinitely many numbers between 0 and sqrt(28), there are infinitely many possible answers.

Answer 3

Infinitely many.

Select any number between 0 and sqrt(28) and call it W.

Then let L = 28/W.

A rectangle with length L units and width W units will have an area of LxW = (28/W)*W = 28 square units. And, since W could be any one of infinitely many numbers between 0 and sqrt(28), there are infinitely many possible answers.