2021-12-16

A semicircle and a quarter circle in a square

This question was uploaded on 16/12/21 on social media accounts.

Solution:

By Pythagoras theorem:
\[AB = Y\frac{\sqrt5}{2}\]
By similar triangle ABC and ACD:
\[\frac{AB}{BC} = \frac{AC}{CD}\]
\[\Rightarrow \frac{Y\sqrt5/2}{Y/2} = \frac{Y}{X/2}\]
\[\Rightarrow \frac{X}{Y} = \frac{2}{\sqrt5} = \frac{2}{5}\sqrt5\]
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