Question

pls ans thiss briefly with numericals

Answer 1

Solution,

Given:

let the point A be the point he stood at first,
let point B be the distance he stood second,
let point C be the distance he stood third,
let point D be the distance he was finally,.

Let "l" be 1 km  unit(rough diagram)

Remember,
           N
            I
W ------l------ E
            I 
           S

He walked 6 km towards south,

. A
l\
l  \
l    \
I      \
I6km\
I. B     \
I           \
I             \
I               \
I                 \
I  12km     \
I  x'             \
I -- -- -- -- -- I  E
I      5km    I
I                  I
I      6km     I
I                  I
I.-- -- -- -- --.I
C   5km   D

To find: distance AE

(AX')²+(EX')² =(AE)²           (pythagoras theorem)

(12)²+(5)²=(AE)²

144+25=(AE)²

169=(AE)²

AE=13km (south-east)

Hope it helps!!