A square is inscribed in a quarter circle in such a way that two of its adjacent vertices on the radius are equidistant from the centre and other two vertices lie on the circumference. If the side of square is √(5/2) cm, then what is the radius (in cm) of the circle?

OB is the radius of circle = $$r$$ cm

Side of square = $$x=\sqrt{\frac{5}{2}}$$ cm, => Diagonal AB = $$\sqrt2x$$ cm

=> $$AB = \sqrt2\times\sqrt{\frac{5}{2}}=\sqrt5$$ cm ------------(i)

Also, OC = OA = $$a$$ 

In, $$\triangle$$ OAC,

=> $$a^2+a^2=x^2$$

=> $$2a^2=(\sqrt{\frac{5}{2}})^2$$

=> $$a^2=\frac{5}{4}$$ -----------(ii)

In $$\triangle$$ OAB,

=> $$(OB)^2=(OA)^2+(AB)^2$$

=> $$r^2=a^2+(\sqrt5)^2$$

=> $$r^2=\frac{5}{4}+5$$

=> $$r^2=\frac{5+20}{4}=\frac{25}{4}$$

=> $$r=\sqrt{\frac{25}{4}}=\frac{5}{2}$$

=> $$r=2.5$$ cm

=> Ans - (B)