SSC JE Civil Engineering 30-Oct-2020 Shift 2 Paper

Representing the given information on NESW : The shortest distance between home and school is OA.
AB = AC + CB
AB = 30 + 60
AB = 90 m.
OB = 120 m.
Applying Pythagoras theorem in ΔAOB :
${\left(\mathrm{Hypotenuse}\right)}^2$ = ${\left(\mathrm{Base}\right)}^2$ + ${\left(\mathrm{Perpendicular}\right)}^2$
${\left(\mathrm{AO}\right)}^2$ = ${\left(\mathrm{BO}\right)}^2$ + ${\left(\mathrm{AB}\right)}^2$
${\left(\mathrm{AO}\right)}^2$ = 1202 + 902
${\left(\mathrm{AO}\right)}^2$ = 14400 + 8100
${\left(\mathrm{AO}\right)}^2$ = 22500
AO = $\sqrt{22500}$
AO = 150 m.
So, the shortest distance between home and school is 150 m.
Hence, the correct answer is "150 m".