This question was uploaded on 22/11/21 on social media accounts.
Solution 1:
Here radius is 7 due to 60 degrees angle and diameter is 14.
Also, (b = 2a) and (a.b = 2a2)
Due to semicircle, we can get two right angled triangles.
\[x = 7\sqrt3\]
\[y = \sqrt{14^2-a^2}\]
Now by using Ptolemy theorem:
\[(a)(7)+(2a)(14) = (7\sqrt3)(\sqrt{14^2-a^2})\]
\[\Rightarrow a = \sqrt{21}\]
\[\Rightarrow a.b = 2a^2 = 42\]
Solution 2:
\[(a+a)^2 + (a\sqrt3)^2 = (7\sqrt3)^2\]
\[\Rightarrow a = \sqrt{21}\]
\[\Rightarrow a.b = 2a^2 = 42\]
