SSC CHSL 20 Dec 2015 Shift 2 Paper

Given : OT is radius of circle = 5 cm and tangent PT = 12 cm
To find : OP = ?
Solution : The radius of a circle intersects the tangent at right angle, ⇒ $\angle \mathit{O}\mathit{T}\mathit{P}=90°$
Thus in ∆OPT,
⇒ ${\left(\mathit{O}\mathit{P}\right)}^2={\left(\mathit{P}\mathit{T}\right)}^2+{\left(\mathit{O}\mathit{T}\right)}^2$
⇒ ${\left(\mathit{O}\mathit{P}\right)}^2={\left(12\right)}^2+{\left(5\right)}^2$
⇒ ${\left(\mathit{O}\mathit{P}\right)}^2=144+25=169$
⇒ $\mathit{O}\mathit{P}=\sqrt{169}=13$cm