Question

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.

Answer 1

i have 9/3sqrt4 but thats not right

Answer 2

P=9in
a=9/3=3in
High of triangle (all corners are 60 degrees):\[h=a \times \sin60^o=a \sqrt3 /2\]
\[Area=a \times h/2=a^2\sqrt3/2=9\sqrt{3}/2}\]

Answer 3

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

Answer 4

so its 9sqrt3/2

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

does anyone know that

Answer 10

I get \[108\sqrt3\]

Answer 11

must check it again

Answer 12

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

Answer 13

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