Solution :
we know the area of a rectangle when the diagonal is given
A = l ( [tex] \sqrt{ d^{2} - l^{2} }[/tex] )
the maximum diagonal that can be inscribed on a circle is the Diameter (2r)
so
A = l ( [tex] \sqrt{ 4r^{2} - l^{2} }[/tex] )
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 = [tex] \sqrt{ { r^{2} + r^{2} }[/tex]
= [tex] \sqrt{ 2r^{2} [/tex]
so maximum area = [tex] \sqrt{ 2r^{2} [/tex] X ( [tex] \sqrt{ 4r^{2}-2 r^{2}} [/tex] )
=[tex] \sqrt{ 2r^{2}}^{2} [/tex]
=[tex] 2r^{2} [/tex]
which is a square with side a = [tex] \sqrt{ 2r^{2} [/tex]
hence the maximum area inscribed on a circle is square
