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i have 9/3sqrt4 but thats not right

\[Area=a \times h/2=a^2 \sqrt3/2=9\sqrt{3}/2\]

so its 9sqrt3/2

Oh, yes, I miss one divider 2. Answer is area is \[9\sqrt3/4\] in^2

\[Area=a \times h/2=a^2\sqrt3/4=9\sqrt{3}/4\]

does anyone know that

I get \[108\sqrt3\]

must check it again

Apothem is 1/3 rd from high, right? Then h=3*6=18 in

\[Area=a \times h /2=|a=2h/\sqrt3|=2h^2/(\sqrt3 \times 2)=324/\sqrt3=108\sqrt3\]