2022-01-18

Two semicircles having parallel diameters

This question was uploaded on 18/01/22 on social media accounts.

Solution:

Here, r = x/2, R = (2a+b)/2

In red triangle:
\[\color{red} {h^2=R^2-r^2}\]

In blue triangle:
\[\color{blue} {h^2=r^2-(R-a)^2}\]

Comparing both equations:
\[\color{fuchsia} {\color{red}{R^2-r^2}=\color{blue}{r^2-(R-a)^2}}\]
\[\Rightarrow\color{fuchsia} {2r^2=R^2+(R-a)^2}\]
\[\Rightarrow\color{fuchsia} {2\left(\frac x2\right)^2=\left(\frac{2a+b}{2}\right)^2+\left(\frac{2a+b}{2}-a\right)^2}\]
\[\Rightarrow\color{fuchsia} {x^2=2a^2+b^2+2ab}\]
\[\Rightarrow\color{fuchsia} {x=\sqrt{2a^2+b^2+2ab}}\]
or
\[\Rightarrow\color{fuchsia} {x=\sqrt{(a+b)^2-a^2}}\]
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