#### Answer

A=36$\pi$$\approx$113

#### Work Step by Step

The diameter of the inscribed circle is equal to the height of the isosceles trapezoid.
h$^2$=13$^2$-($\frac{1}{2}$(18-8))$^2$
h$^2$=169-25=144
h=12
r=.5h
r=6
A=$\pi$r$^2$
A=36$\pi$$\approx$113

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A=36$\pi$$\approx$113

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