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[tex]\cos(28^o)=\cfrac{h}{5}\implies 5\cos(28^o)=h \\\\[-0.35em] ~\dotfill\\\\ \textit{area of a parallelogram}\\\\ A=bh~~ \begin{cases} b=base\\ h=height\\[-0.5em] \hrulefill\\ b=6\\ h=5\cos(28^o) \end{cases}\implies A=6[5\cos(28^o)]\implies A\approx 26.49[/tex]
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