2017/18 AHSMC Part 1 Problem 3

Let $D$ be the point on the road closest to the pole. Then the distance between the pole and $D$ is $12$ m. At first, Mark is $$\sqrt{37^2 - 12^2} = \sqrt{1369 - 144} = \sqrt{1225} = 35$$ m from $D$. When he is $20$ m from the pole, Mark is $$\sqrt{20^2 - 12^2} = \sqrt{400 - 144} = \sqrt{256} = 16$$ m from $D$. Hence, Mark walked $35 - 16 = 19$ m. The answer is (B). (Notice that Mark could pass point $D$ when he is $20$ metres from the pole. In this case Mark has to walk $35 + 16 = 51$ m.)