a person goes 12m due East and then 5m due north.what is the distance from his starting point
Answers
Answered by
7
It is not a lengthy questions first of all ,
Now to solve it first draw the diagram ,we will see that it will form a triangle so
let start point a ,second point b and the last point c .
so
AB^2 + BC^2 = CA^2
12^2+5^2 = CA^2
144+25 = CA^2
169 = CA^2
13 = CA Answer
Now to solve it first draw the diagram ,we will see that it will form a triangle so
let start point a ,second point b and the last point c .
so
AB^2 + BC^2 = CA^2
12^2+5^2 = CA^2
144+25 = CA^2
169 = CA^2
13 = CA Answer
Answered by
17
Answer:
ABC is right angled triangle ,
By Pythagoras property of triangle
AC² = BC² + AB²
AC² = 12² + 5²
AC²= 144 + 25
AC² = 169
AC²= 13²
AC = 13 cm
Shortest distance from starting point = 13cm
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