Question

if the area of a rectangle inscribed in a circle is maximum, then show that the rectangle becomes a square   

Answer 1

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)
so
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