SSC CGL 9 Aug 2015 Shift 1 Paper

OA = OB = OC = OD = radius(r)
∠OAB = 90°; AB = b, CD = a
From ΔOAB,
$\mathit{O}{\mathit{A}}^2+\mathit{O}{\mathit{B}}^2$ = $\mathit{A}{\mathit{B}}^2$
⇒ ${\mathit{r}}^2+{\mathit{r}}^2$ = ${\mathit{b}}^2$ ⇒ $2{\mathit{r}}^2$ = $\mathit{b}{6}^2$
⇒ ${\mathit{r}}^2$ = $\frac{{\mathit{b}}^2}{2}$
⇒ r = $\frac{\mathit{b}}{\sqrt{2}}$ ... (i)
In ΔOCD
∠COD = 60° ;
∴ ∠OCD = ∠ODC = 60°
∴ From equations (i) and (ii),
$\frac{\mathit{b}}{\sqrt{2}}$ = a ⇒ b = $\sqrt{2}$ a