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I really need help with this-example-1
User Cuizizhe
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check the picture below.

the radius, so-called because if we circumscribe the triangle in a circle, that'd be the radius of the circumcircle, is 3 inches, as you see there in the picture in red.

now, bear in mind that the triangle is an equilateral one, and therefore, 180/3 = 60, meaning all its interior angles are congruently 60°.

if we run the radius of the triangle, like there, it will cut one of those interior angles in half, namely the radius is an angle bisector, and if we run a perpendicular line to the bottom side, like the dashed one there, we end up with a 30-60-90 triangle, to which we can apply the 30-60-90 rule, as you see it there.

since we know what half of a side of the triangle is, as you see it there in blue, then we can use that and plug it in at


\bf \textit{area of an equilateral triangle}\\\\ A=\cfrac{s^2√(3)}{4}~~ \begin{cases} s=length~of\\ \qquad a~side\\ -------\\ s=\stackrel{(3√(3))/(2)+(3√(3))/(2)}{3√(3)} \end{cases}\implies A=\cfrac{(3√(3))^2√(3)}{4} \\\\\\ A=\cfrac{(3^2√(3^2))√(3)}{4}\implies A=\cfrac{(9\cdot 3)√(3)}{4}\implies A=\cfrac{27√(3)}{4}
I really need help with this-example-1
User Dvtoever
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