Question

a man walks 10 kilometres due north then he turns right and walked 12 kilometre again he turns right and walked 5km how far is he from starting point​

Answer 1

$Distance \:walked \:due \:north (OA) = 10\:km$

$Distance \: walked \:due \: East (AB) = 12\:km$

$Distance \: walked \:due \: South (BC) = 5\:km$

$Draw \: Perpendicular \:CD \:to \: AO$

$In \: \triangle ODC , \angle {ODC} = 90\degree$

/* By Phythagorean theorem */

$OC^{2} = OD^{2} + DC^{2}$

$= 5^{2} + 12^{2}\\= 25 + 144 \\= 169$

$\implies OC = \sqrt{169} \\= \sqrt{13^{2}} \\= 13 \:km$

$\red{ Distance \: from \: initial \: point }\\\red{ ending\: point } \\= OC \\\green { = 13 \: km }$

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