SOLUTION: I have an equilateral triangle with 4(pi) as each of its sides. Using the formula for finding the area of equilateral triangles {[(b^2)(sqrt3)]/4} I got (sqrt3)(pi^2) However, work

Algebra ->  Surface-area -> SOLUTION: I have an equilateral triangle with 4(pi) as each of its sides. Using the formula for finding the area of equilateral triangles {[(b^2)(sqrt3)]/4} I got (sqrt3)(pi^2) However, work      Log On


   

Question 1104761: I have an equilateral triangle with 4(pi) as each of its sides. Using the formula for finding the area of equilateral triangles {[(b^2)(sqrt3)]/4} I got (sqrt3)(pi^2) However, working through the entire problem manually, I got 2(sqrt2)(pi^2). Is one of those solutions right? If not, can you help me find the correct process to get the correct solution? Thank you!
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Make an altitude from a vertex to middle of opposite side. Now the equilateral triangle is two congruent right triangles of the 30-60-90 type. Hypotenuse is 4pi and the short leg is 2pi.


Find the size of the altitude. Call this y.

y%5E2%2B%282pi%29%5E2=%284pi%29%5E2
y%5E2=16pi%5E2-4pi%5E2
y%5E2=12pi%5E2
y=sqrt%2812pi%5E2%29
y=2pi%2Asqrt%283%29 or you could say, y=sqrt%283%29%2A2pi, or 2%2Asqrt%283%29pi=y.

AREA OF THE EQUILATERAL TRIANGLE
%281%2F2%29%2Abase%2Aaltitude
Take any of the 4pi sides as base; you already found altitude.
%281%2F2%29%2A4pi%2A2%2Asqrt%283%29pi
2pi%5E2%2A2%2Asqrt%283%29pi
highlight%284%2Api%5E3%2Asqrt%283%29%29, or into whichever way that commutative property for multiplication allows that you want.

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