Answer:
[tex]32\sqrt3[/tex]
Step-by-step explanation:
We need first to find the measure of the red horizontal line, call it x. The triangle i filled is a right triangle, so we can apply pythagoras theorem,
[tex]8^2 = x^2+(4\sqrt3)^2 \rightarrow 64=x^2+(16\times3) \\x^2=64-48 \rightarrow x^2=16 \rightarrow x=4[/tex]
(we are talking lengths of segment so we take only the positive root!)
At this point we can find the length of the smaller base of the trapezoid (which is [tex]10-4=6[/tex]) and apply the formula, or split the figure in a rectangle of sides 6 and [tex]4\sqrt3[/tex] and a triangle of sides [tex]4\sqrt3[/tex] and 4.
With the trapezoid formula
[tex]A=\frac12(B+b)h = \frac12(10+6)4\sqrt3 = 32\sqrt3[/tex]
With the sum of figures:
[tex]A= bh+\frac12 xh=6\times4\sqrt3 +\frac12\times4\times4\sqrt3=24\sqrt3+8\sqrt3=32\sqrt3[/tex]
