SSC CHSL 20 Dec 2015 Shift 2 Paper

Given : OB is the radius of circle = 5 cm and OC = 4 cm
To find : AB = ?
Solution : The line from the centre of the circle to the chord bisects it at right angle.
⇒ AC = BC = $\frac{1}{2}$AB
In ∆OBC,
⇒ ${\left(\mathit{B}\mathit{C}\right)}^2={\left(\mathit{O}\mathit{B}\right)}^2-{\left(\mathit{O}\mathit{C}\right)}^2$
⇒ ${\left(\mathit{B}\mathit{C}\right)}^2={\left(5\right)}^2-{\left(4\right)}^2$
⇒ ${\left(\mathit{B}\mathit{C}\right)}^2=25-16=9$
⇒ $\mathit{B}\mathit{C}=\sqrt{9}=3$cm
∴ AB = 2 × BC
= 2 × 3 = 6 cm