Question

A man goes 10m due east and then 24m due north. Find the distance of the man from the starting point.

Mention the property used. Also draw a figure to explain the above.

Answer 1

Its ur ans
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Answer 2

$Starting \:point \: of \: a \: man \red {A}$

$Distance \: travelled \: due \: East (AB) = 10\:m$

$Distance \: travelled \: due \: North (BC)= 24\:m$

$</p><p>[tex] In \: right \: triangle \: ABC ,$

$AC^{2} = AB^{2} + BC^{2}$

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

$\implies AC^{2} = 10^{2} + 24^{2} \\= 100 + 576 \\= 676$

$\implies AC = \sqrt{676} \\= \sqrt{26^{2}} \\= 26 \:m$

Therefore.,

$\red{ Distance \:of \:the \:man\:from \:the } \\\red{ starting \:point } \green {= 26 \:m}$

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