Four sprinters start running from 4 points which are 4 corners of an imaginary rectangle along straight lines and meet at a point 'O' which falls inside the rectangle as shown in the figure, after the first three sprinters run 55m, 65m and 75m respectively. What is the approximate distance that the fourth sprinter will run to meet the other three sprinters at point O?

From the above figure,

Using Pythagoras theorem

$$In\ \triangle\ AQO\ ,\ 55^2=a^2+c^2$$.....(1)

similarly. 

$$In\ \triangle\ BQO\ ,\ 65^2=a^2+d^2$$.....(2)

$$In\ \triangle\ CSO\ ,\ 75^2=b^2+d^2$$.....(3)

$$In\ \triangle\ DSO\ ,\ m^2=b^2+c^2$$.....(4)

From observation, Eq(1)-Eq(2)  = Eq(4)-Eq(3) = $$c^2-d^2$$

Hence, $$55^2-65^2=m^2-75^2$$

$$3025-4225=m^2-5625$$

$$m^2=4425$$

$$m=\pm\ 66.52$$

As 'm' cannot be negative 

$$m\approx\ \ 66.5$$

Option (A) is correct.