2021-12-27

Equilateral triangle inscribed in a quarter circle

This question was uploaded on 27/12/21 on social media accounts.

Solution 1:

Value of x:
\[x+x\sqrt3 = r = 4\]
\[\Rightarrow x = \frac4{\sqrt3+1} = 2(\sqrt3-1)\]

Value of s:
\[\Rightarrow s = 2x = 4(\sqrt3-1)\]

Blue Area:
\[A = \frac{\sqrt3}4s^2 = \frac{\sqrt3}4\cdot4^2(\sqrt3-1)^2\]
\[\Rightarrow A = (16\sqrt3-24)\]


Solution 2:

Value of a:
\[a = 4\sin(45) = 2\sqrt2\]

Value of s:
\[\sin(75) = \frac as\]
\[\Rightarrow s = \frac a{\sin(75)}\]
\[\Rightarrow s = (2\sqrt2)\left(\frac {2\sqrt2}{\sqrt3+1}\right)\]
\[\Rightarrow s = 4(\sqrt3-1)\]

Blue Area:
\[A = \frac{\sqrt3}4s^2 = \frac{\sqrt3}4\cdot4^2(\sqrt3-1)^2\]
\[\Rightarrow A = (16\sqrt3-24)\]
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