Question

A man walks 9km east and take right walk 4 km and again he take right 6 km ..How far he from starting point?

Answer 1

this is your answer friend

Answer 2

$Distance \: walked \: due \:East (AB) = 6\:km$

$Distance \: walked \: due \: North (BC)= 4\:km$

$CD = 6 \:km$

$Join \: A\:to \: D\: and \: draw \: perpendicular \\DE \:to \: AB .$

$In \: \triangle AED , \: \angle E = 90\degree$

$AD^{2} = DE^{2} + AE^{2}$

$\blue { ( By \:Phythagorean\: theorem )}$

$\implies AD^{2} = 4^{2} + 3^{2} \\= 16 + 9 \\= 25$

$\implies AD = \sqrt{25} \\= 5 \: km$

Therefore.,

$\red { Distance \:from \: starting \: point \:to }\\\red{ \:ending\: point } \green { = 5 \: km}$

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