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Solution:
There are two right angled triangles:
\[a = \sqrt{4^2-3^2} = \sqrt7\]
By similar triangles:
\[\frac a4 = \frac x3\]
\[\Rightarrow x = \frac {3\sqrt7}4\]
\[\Rightarrow Area = x^2 = \frac {63}{16}\]