Haryana Police Constable Model Paper 5

We have to calculate AD with D being the final position and A being the initial position.
We know that AB = 10 m, BC = 6 m and CD = 2 m.
$\mathrm{ED}=\mathrm{BC}=6\mathrm{m}$ .
$\mathrm{BE}=\mathrm{CD}=2\mathrm{m}$ .
$\mathrm{AE}=\mathrm{AB}\mathit{–}\mathrm{BE}=\left(10-2\right)\mathrm{m}=8\mathrm{m}$ .
Using Pythagoras right angled theorem,
${\mathrm{AD}}^2={\mathrm{AE}}^2+{\mathrm{ED}}^2$
${\mathrm{AD}}^2=8^2+6^2=64+36=100\mathrm{m}$ .
$\mathrm{AD}=\sqrt{100}=10\mathrm{m}$ .
Hence, the distance between his final and initial position is $10\mathrm{m}.$