What is the area of an equilateral triangle whose vertices lie on a circle with radius 2?

1 Answer
Rui D.
Jan 22, 2016

3*sqrt(3)~=5.1963⋅√3≅5.196

Explanation:

Refer to figure below

I created this figure using MS Excel

The figure represents an equilateral triangle inscribed in a circle, where ss stands for the triangle's sides, hh stands for the height of the triangle, and RR stands for the circle's radius.

We can see that triangles ABE, ACE and BCE are congruents, that's why we can say that angle E hat C D=(A hat C D)/2=60^@/2=30^@EˆCD=AˆCD2=60∘2=30∘.

We can see in triangle_(CDE)△CDE that
cos 30^@=(s/2)/Rcos30∘=s2R => s=2*R*cos 30^@=cancel(2)*R*sqrt(3)/cancel(2) => s=sqrt(3)*R

In triangle_(ACD) we cant see that
tan 60^@=h/(s/2) => h=s*tan 60^@/2 => h=sqrt(3)/2*s=sqrt(3)/2*sqrt(3)*R => h=(3R)/2

From the formula of the area of the triangle:
S_triangle=(base*height)/2
We get
S_triangle=(s*h)/2=(sqrt(3)R*(3R)/2)/2=(3*sqrt(3)*R^2)/4=(3*sqrt(3)*cancel(2^2))/cancel(4)=3*sqrt(3)

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