A square is inscribed inside the ellipse`x^2/a^2+y^2/b^2=1`, the length of the side of the square is

A square is inscribed inside the ellipse (x^(2))/( a^(2)) +(y^(2))/( b^(2)) =1 then the length of the side of the square is

If a square is inscribed in a circle of diameter 4 cm, what will be the length of a side of the square?

If a square is inscribed in the circle x^(2) + y^(2) + 2gx +2fy + c= 0 then the length of each side of the square is

a square is inscribed in the circle x^2 + y^2- 10x - 6y + 30 = 0. One side of the square is parallel to y = x + 3, then one vertex of the square is :

a square is inscribed in the circle x^2 + y^2- 10x - 6y + 30 = 0. One side of the square is parallel to y = x + 3, then one vertex of the square is :

a square is inscribed in the circle x^2 + y^2- 10x - 6y + 30 = 0. One side of the square is parallel to y = x + 3, then one vertex of the square is :

a square is inscribed in the circle x^(2)+y^(2)-10x-6y+30=0. One side of the square is parallel to y=x+3, then one vertex of the square is:

The area of the parallelogram inscribed in the ellipse x^(2)/a^(2)+y^(2)/b^(2)=1 , whose diaonals are the conjugate diameters of the ellipse is given by