A man jogs 24m due east and then 10m due north everyday.How far is he away from his initial point
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hey there ,,,
if you make the diagram of his walking path you gotta a triangle from these points . let us see how ;-
let he started from a point A and go towards the point B which is in east about 24m and stopped and then move toward north about 10m and stopped at point C .
now , join the point A,B,C you will get a right triangle .
now find the third side which is AC
so , by Pythagoras theorem
ac² = ab² + bc²
ac² = 24² + 10²
ac² = 676m²
ac = √676m² = 26 m
hence he is 26m away from its initial point .
hope it helps !!!
be brainly ♥
if you make the diagram of his walking path you gotta a triangle from these points . let us see how ;-
let he started from a point A and go towards the point B which is in east about 24m and stopped and then move toward north about 10m and stopped at point C .
now , join the point A,B,C you will get a right triangle .
now find the third side which is AC
so , by Pythagoras theorem
ac² = ab² + bc²
ac² = 24² + 10²
ac² = 676m²
ac = √676m² = 26 m
hence he is 26m away from its initial point .
hope it helps !!!
be brainly ♥
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