Question

a man goes 10m due south and then 24m due east. find the shortest distance from starting point

Answer 1

ACC

ORDING TO QUESTION,
AB=10m and BC =24m
Now, shortest distance between them is AC

According to Pythagoras Theourum
${h}^{2} = {p}^{2} + {b}^{2}$
${h}^{2} = {10}^{2} + {24}^{2}$
${h}^{2} = 100 + 576$
${h}^{2} = 676$
$h = \sqrt{676}$
$h = 26$
therefore shortest distance is 26 m

Answer 2

as AB = 10m
BC = 24m
so the shortest distance according to Pythagoras theorum would be
as A^2 + B^2 = C^2
here A = 10 m
B= 24m
C= shortest distance covered (displacement)
so, A^2 =10*10 =100
B^2 =24*24 = 576
so 100+576 = 676= C^2
so C= 26