the perimeter of a square and a rectangle are equal which has greater area
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Assume a square of side s and rectangle with one side as a and b.
Since the perimeters are equal
4s = 2(a+b)
or b = 2s-a.
The sides of rectangle can be rewritten as a and 2s-a.
Areas are
s² and (2s-a)a
If they are equal
s² = 2sa - a²
s² -2sa + a² = 0
(s-a)² = 0
s=a
so only when s = a, both perimeter and area are equal. This is the trivial case where square is also taken as rrctanɡle also.
In all other cases, no area of rectanɡle amd square are not equal.
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