Answer:
[tex]17.3\:\mathrm{m^2}[/tex]
Step-by-step explanation:
Solution 1:
As the angles are marked, the figure forms a 30-60-90 right triangle with the height. Therefore the height will be [tex]\frac{4}{2}\cdot\sqrt{3}=2\sqrt{3}[/tex]. The area of this figure can be given as [tex]b\cdot h[/tex], therefore our answer is [tex]5\cdot 2\sqrt{3}\approx \fbox{$17.3$}[/tex].
Solution 2:
The figure forms two triangles when cut by the shortest diagonal. The area of a triangle can be given by [tex]\frac{1}{2}ab\sin C[/tex]. Since there are 2 of these triangles, the area of the entire figure is [tex]2\cdot\frac{1}{2}\cdot4\cdot 5\cdot \sin 60^{\circ}\approx \fbox{$17.3$}[/tex].
