SSC CGL 8 Aug 2017 Shift 1 Paper

According to the question, ∠DCE = 45° and CD = $10\sqrt{2}$ cm
∴ CB $=\frac{10\sqrt{2}}{2}=5\sqrt{2}\mathrm{cm}$ [∵ CB = DB]
∵ ∠OCE = 90°
∴ ∠OCD = 90°-45° =45°
In ΔOCB,
$\cos 45°=\frac{\mathrm{BC}}{\mathrm{OC}}=\frac{5\sqrt{2}}{\mathrm{OC}}$
⇒ $\frac{1}{\sqrt{2}}=\frac{5\sqrt{2}}{\mathrm{OC}}$
∴ OC = 10cm
and $\tan 45°=\frac{\mathrm{OB}}{\mathrm{CB}}=\frac{\mathrm{OB}}{5\sqrt{2}}$
⇒ $1=\frac{\mathrm{OB}}{5\sqrt{2}}$
∴ $\mathrm{OB}=5\sqrt{2}\mathrm{cm}$
Again, in ΔACB,
${\left(\mathit{A}\mathit{C}\right)}^2={\left(\mathit{A}\mathit{B}\right)}^2+{\left(\mathit{B}\mathit{C}\right)}^2$
$={\left(10+5\sqrt{2}\right)}^2+{\left(5\sqrt{2}\right)}^2$
$=100+50+100\sqrt{2}+50$
$=200+100\times 1.41$
$=200+141=341$
∴ $\mathit{A}\mathit{C}=\sqrt{341}=18.5\mathrm{cm}$