Question

Kanchan is standing at a point A and his friend,
Kumar, is standing at a point B, which is exactly
in the East direction of Kanchan. Kanchan starts
walking straight in the North direction. At the
same time, Kumar also starts walking straight
at double the speed of Kanchan. Both Kanchan
and Kumar meet each other after Kanchan has
walked 5 km. The distance between A and B is
equal to:​

Answer 1

Answer:

$\sqrt{BC^{2} -AC^{2} } = \sqrt{10^{2} -5^{2} } = \sqrt{75} = 5\sqrt{3}$ units

Step-by-step explanation:

let Kumar & Kanchan have walked to a point C

AC = 5

BC = 10 (double speed i.e. double distance in equal time)

since ABC is a right triangle with BC as hypotenuse:

BC = $\sqrt{BC^{2} -AC^{2} } = \sqrt{10^{2} -5^{2} } = \sqrt{75} = 5\sqrt{3}$