ANSWER
[tex]32 \sqrt{3} {ft}^{2} [/tex]
EXPLANATION
The area of a tra-pezoid is half times sum of the bases times the vertical height.
From the diagram ( see attachment) we use the Pythagoras Theorem to obtain,
[tex] {x}^{2} + {(4 \sqrt{3}) }^{2} = {8}^{2} [/tex]
[tex] {x}^{2} + 48 = 64[/tex]
[tex]{x}^{2} = 64 - 48[/tex]
[tex]{x}^{2}=16[/tex]
[tex]x = \sqrt{16} = 4ft[/tex]
This implies that , y=10-4=6ft
The area of tra-pezoid
[tex] = \frac{1}{2} (10 + 6) \times 4 \sqrt{3} [/tex]
[tex] = \frac{1}{2} (16) \times 4 \sqrt{3} [/tex]
[tex] = 8\times 4 \sqrt{3} [/tex]
[tex] = 32 \sqrt{3} {ft}^{2} [/tex]
