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Solution 1:
AC = 10 and AC = 18 so BC = 8
OB = 13 but OB = BC+OC
BC = 8 so OC = 5
By using Pythagoras theorem: h = 12
A = (10)(12)/2 = 60
Solution 2:
AC = CD = AD/2 = 18
\[\color{red} {x = 26\cos\theta}\]
\[\color{red} {h = x\sin\theta = 26\cos\theta\sin\theta}\]
\[\color{blue} {h = AC\tan\theta = 18\tan\theta}\]
\[\color{red} {h} = \color{blue} {h}\]
\[\Rightarrow \color{red} {26\cos\theta\sin\theta} = \color{blue} {18\tan\theta}\]
\[\Rightarrow \color{red} {26\cos\theta\sin\theta} = \color{blue} {18\frac{\sin\theta}{\cos\theta}}\]
\[\Rightarrow \color{red} {13\cos\theta} = \color{blue} {\frac{9}{\cos\theta}}\]
\[\Rightarrow \cos^2\theta =\frac{9}{13}\]
\[\Rightarrow \cos\theta = \frac{3}{\sqrt{13}},⠀\sin\theta = \frac{2}{\sqrt{13}}\]
\[\Rightarrow h = 26\cos\theta\sin\theta = 12\]
Blue Area:
\[Area = \frac{10h}2 = 60\]
