can anyone explain, in detail, the answer to this question? write the area "A" of an equilateral trinagle as a function of is side length x.

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area = base*height /2

height = 2x sin(60) = sqrt(3)/2 base = 2x

area of the triangle = 1/2 * base * height = 1/2 * x * height

we could just as easily go with sides of x

x*x*sqrt(3) ---------- = x^2 sqrt(3)/4 2*2

you can calculate the height using trigonometry: tan 60 = height / (x/2)

so height = sqrt3 * 2 / x now plug this into the formula for the area

somethings amiss there. there shouldnt be an "x" in the denominator that I can tell

tan(60) = sqrt(3) sqrt(3) = 2h/x x sqrt(3) ------- = h ; is better 2

if you keep x/2 in the denom; then you dont need to reciprocate it; you simply multiply by x/2

\[\sqrt{3}=\frac{h}{x/2}\] \[\sqrt{3}(x/2)=\frac{h(x/2)}{x/2}\] \[\sqrt{3}(x/2)=h\] \[\frac{x\sqrt{3}}{2}=h\]

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