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Solution :
we know the area of a rectangle when the diagonal is given

A = l (  \sqrt{ d^{2} - l^{2} } )
the maximum diagonal that can be inscribed on a circle is the Diameter (2r)
A = l (  \sqrt{ 4r^{2} - l^{2} } )

now l can be computed using the circles 1 quad ( draw a circle  with diameters at right angle , join the diameters ends with the adjacent diameter ends . Now take any one of the four quads in the drawn circle .)
where it is a right angled triangle with sides l and height and base r

by applying pythagroas  theorem

l =  \sqrt{ { r^{2} + r^{2} }
  =  \sqrt{ 2r^{2}

so maximum area =  \sqrt{ 2r^{2} X (  \sqrt{ 4r^{2}-2 r^{2}}   )
                           = \sqrt{ 2r^{2}}^{2}
                           = 2r^{2}

which is a square with side  a =  \sqrt{ 2r^{2}

hence the maximum area inscribed on a circle is square

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