## Task 3 - V00903

From India, after the mathematician Bhaskara (1114 AD):

A lotus flower protrudes 4 feet above the surface of a pond. Driven by the wind, it disappears 16 feet from its original position under the water.

How deep was the pond?
Let $x$ be the length of the lotus flower. Then $x^{2}=(x-4)^{2}+16^{2}$ with $x=34$. Since the lotus flower protrudes $4 \mathrm{~m}$ above the pond, the pond is $30 \mathrm{~m}$ deep.