Given the rectangle ABCD, you can identify that it is divided into two Right Triangles:
[tex]\begin{gathered} \Delta ACD \\ \Delta ABD \end{gathered}[/tex]• You can find the length of the rectangle by applying the following Trigonometric Function:
[tex]\cos \alpha=\frac{adjacent}{hypotenuse}[/tex]In this case, you can set up that:
[tex]\begin{gathered} \alpha=30\degree \\ adjacent=CD=AB=l \\ hypotenuse=AD=9 \end{gathered}[/tex]Then, substituting values and solving for "l", you get:
[tex]\begin{gathered} \cos (30\text{\degree})=\frac{l}{9} \\ \\ 9\cdot\cos (30\text{\degree})=l \\ \\ l=\frac{9}{2}\sqrt[]{3}ft \end{gathered}[/tex]• In order to find the width of the rectangle, you can use this Trigonometric Function:
[tex]\sin \alpha=\frac{opposite}{hypotenuse}[/tex]In this case, you can say that:
[tex]\begin{gathered} \alpha=30\degree \\ opposite=AC=BD=w \\ hypotenuse=AD=9 \end{gathered}[/tex]Therefore, substituting values and solving for "w", you get:
[tex]\begin{gathered} \sin (30\degree)=\frac{w}{9} \\ \\ 9\cdot\sin (30\degree)=w \\ \\ w=\frac{9}{2}ft \end{gathered}[/tex]• Now you need to use the following formula for calculating the area of a rectangle:
[tex]A=lw[/tex]Where "l" is the length and "w" is the width.
Substituting the length and the width of the given rectangle into the formula and evaluating, you get:
[tex]A=(\frac{9}{2}\sqrt[]{3}ft)(\frac{9}{2}ft)[/tex][tex]A\approx35.1ft^2[/tex]Hence, the answer is: Option C.
