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Find the area of an equilateral triangle (regular 3-gon) with the given measurement. 9-inch perimeter A = sq. in Help me my answer needs to be in square root.

Mathematics
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i have 9/3sqrt4 but thats not right
P=9in a=9/3=3in High of triangle (all corners are 60 degrees):\[h=a \times \sin60^o=a \sqrt3 /2\] \[Area=a \times h/2=a^2\sqrt3/2=9\sqrt{3}/2}\]
\[Area=a \times h/2=a^2 \sqrt3/2=9\sqrt{3}/2\]

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Other answers:

so its 9sqrt3/2
in case of an equilateral triangle, perimeter = 3 * side so side = perimeter/3 so each side = 9 in/3 = 3 in root(3) Area of an equilateral triangle = --------- * s² where s = length of side 4 root(3) 9 root(3) so area = --------- * (3cm)² = ---------- cm² 4 4
Oh, yes, I miss one divider 2. Answer is area is \[9\sqrt3/4\] in^2
\[Area=a \times h/2=a^2\sqrt3/4=9\sqrt{3}/4\]
Find the area of an equilateral triangle (regular 3-gon) with the given measurement. 6-inch apothem A = sq. in. for this one i havwe 144sqrt2
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
\[Area=a \times h /2=|a=2h/\sqrt3|=2h^2/(\sqrt3 \times 2)=324/\sqrt3=108\sqrt3\]

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