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There are infinitely many such rectangles.

Consider any positive number, B which is less than 6 gronks. Let L be 12-B gronnks. Then L > 6 and so no ordered pairs (L, B) will be equal to a (B, L).


Any rectangle with length L gronks and breadth B gronks will have a perimeter of 2*(L+B) = 2*12 = 24 gronks.


Since the choice of B was arbitrary there are infinitely many choices for B and since each value of B gives a unique rectangle, there are infinitely many rectangles.



As many as you want.

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9y ago
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9y ago

There are infinitely many such rectangles.

Consider any positive number, B which is less than 6 gronks. Let L be 12-B gronnks. Then L > 6 and so no ordered pairs (L, B) will be equal to a (B, L).


Any rectangle with length L gronks and breadth B gronks will have a perimeter of 2*(L+B) = 2*12 = 24 gronks.


Since the choice of B was arbitrary there are infinitely many choices for B and since each value of B gives a unique rectangle, there are infinitely many rectangles.

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9y ago

As many as you want.

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Q: How many different rectangles have a perimeter of 24 gronks?
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