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Mathematics 76 Online
OpenStudy (anonymous):

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.

OpenStudy (amistre64):

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OpenStudy (amistre64):

area = base*height /2

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

we could just as easily go with sides of x

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

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