SSC CGL 16 Aug 2015 Shift 2 Paper

AB = 10 cm.
∴ AF = FB = 5 cm.
CD = 24 cm.
∴ CE = DE = 12 cm.
Let OE = x cm
∴ OF = (17 – x) cm
From D ODE,
OD = $\sqrt{\mathit{O}{\mathit{E}}^2+\mathit{D}{\mathit{E}}^2}$
= $\sqrt{{\mathit{x}}^2+{12}^2}$ ... (i)
From Δ OAF
OA = $\sqrt{\mathit{O}{\mathit{F}}^2+\mathit{A}{\mathit{F}}^2}$
= $\sqrt{{\left(17-\mathit{x}\right)}^2+5^2}$ ... (ii)
Since OA = OD
∴ $\sqrt{{\mathit{x}}^2+{12}^2}$ = $\sqrt{{\left(17-\mathit{x}\right)}^2+5^2}$
⇒ ${\mathit{x}}^2$ + 144 = 289 - 34x + ${\mathit{x}}^2$ + 25
⇒ 34x = 289 + 25 - 144 = 170
⇒ x = $\frac{170}{34}$ = 5
∴ From equation (i),
OD = $\sqrt{{\mathit{x}}^2+{12}^2}$ = $\sqrt{5^2+144}$
= $\sqrt{169}$
= 13 cm