A right-angled triangle is inscribed in a given circle. What is the area of the given circle (in cm2) ?
I. The base and height of the triangle (in cm) are both the roots of the equation $$x^{2}-23x+120=0$$
II. The sum of the base and height of the triangle is 23 cm.
III. The height of the right-angled triangle is greater than the base of the same.

I : $$x^2 - 23x + 120 = 0$$

=> $$x^2 - 8x - 15x + 120 = 0$$

=> $$x (x - 8) - 15 (x - 8) = 0$$

=> $$(x - 8) (x - 15) = 0$$

=> $$x = 8 , 15$$

Thus, base = 8 cm and height = 15 cm (or vice versa)

=> Hypotenuse of right angled triangle = $$\sqrt{(8)^2 + (15)^2}$$

= $$\sqrt{64 + 225} = \sqrt{289} = 17 cm$$

Since, triangle is inscribed in circle, => Radius of circle = half of hypotenuse

=> $$r = \frac{17}{2} = 8.5$$ cm

$$\therefore$$ Area of circle = $$\pi r^2$$

= $$\frac{22}{7} \times 8.5 \times 8.5 \approx 227 cm^2$$

Thus, I alone is sufficient.

Clearly, we cannot find base and height from statements II or III. Thus, they are insufficient.