A man walked 2 mts
towards North, turned
West and walked 5 mts,
then turned North again
and walked 1 mts and
then turned East and
walked 9 kms. How far
away is he from the
starting point ?
Answers
Answered by
0
Answer:
Correct option is
A
5 metre
Let us consider a Cartesian plane thinking positive X-axis as South,
Negative x axis as North, Positive Y-axis as East and Negative as West.
A man walked along negative X-axis 3 m and then parallel to negative Y-axis he walked 2 m.
And then parallel to negative X-axis he walked 1 m and finally walked 5 m parallel to positive Y-axis and he has coordinates of (−4,3).
He makes right angle triangle with base AC=4, side AB=3 and hypotenuse BC.
∴ by Pythagoras theorem,
BC
2
=AB
2
+AC
2
⟹ BC
2
=3
2
+4
2
∴ BC=5.
Hence from origin he is at distance of 5 m.
Thus, option A is the correct.
solution
Explanation:
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Answered by
0
Answer:
5m
Explanation:
please refer the photo for the answer.
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