SSC MTS 9 Feb 2026 Shift 3 Paper

Concept:When a chord subtends a 90° angle at the center, the two radii and the chord form a right isosceles triangle. The radius can be found using Pythagoras, and the perpendicular distance from the center to the chord equals half the chord length.Explanation:Let the radius be $r$ and chord $AB = 20$ cm. In $\triangle OAB$, $OA = OB = r$ and $\angle AOB = 90^\circ$.Using Pythagoras: $r^2 + r^2 = 20^2$ ⇒ $2r^2 = 400$ ⇒ $r^2 = 200$ ⇒ $r = \sqrt{200} = 10\sqrt{2}$ cm.The perpendicular from the center $O$ to chord $AB$ is the altitude $d$. In a right isosceles triangle, the altitude to the hypotenuse is half the hypotenuse. So $d = \frac{AB}{2} = \frac{20}{2} = 10$ cm.Alternatively, using the formula $d = \sqrt{r^2 - \left(\frac{AB}{2}\right)^2}$ = $\sqrt{200 - 100} = \sqrt{100} = 10$ cm.Thus the radius is $10\sqrt{2}$ cm and the distance from the center is $10$ cm.Answer:Option D: $10\sqrt{2}$ cm, 10 cm.