SOLUTION
(17) We want to find the area of the triangle. We need to find the base of this triangle, Let the base be b, Using the Pythagoras theorem (since it is a right-triangle), we have
[tex]\begin{gathered} hyp^2=opp^2+adj^2 \\ (4\sqrt{11})^2=b^2+4^2 \\ 16\times11=b^2+16 \\ 176=b^2+16 \\ b^2=176-16 \\ b=\sqrt{160} \\ b=4\sqrt{10} \end{gathered}[/tex]
We have gotten the base.
The area of the triangle becomes
[tex]\begin{gathered} Area=\frac{1}{2}\times base\times height \\ A=\frac{1}{2}\times4\sqrt{10}\times4 \\ A=2\sqrt{10}\times4 \\ A=8\sqrt{10} \end{gathered}[/tex]
Hence the answer is
[tex]8\sqrt{10}[/tex]
