An equilateral triangle is inscribed in a circle of radius 3. Determine the area of the equilateral triangle?

Answer 1

You can subdivide the triangle into 6 equal triangles. First draw lines from the center to each vertex. This length is 3. Now draw lines from the center to the midpoint of each side to make 6 right triangles. They would each be 30-60-90 triangles, with a hypotenuse of 3 (the radius of the circle).

Remember the ratio of the sides of 30-60-90 triangle is 1 : 2 : sqrt(3). In your case your triangle has sides 1.5 : 3 : 1.5 sqrt(3). The legs are therefore 1.5 and 1.5*sqrt(3). (You can also use the Pythagorean theorem to derive this result.)

The area of a triangle is (base * height) / 2, so the area of one triangle would be 1.5 x 1.5 x sqrt(3) / 2. But you have 6 of them.

So the total area would be:
1.5 x 1.5 x sqrt(3) x 3

That works out to 6.75 x sqrt(3)

That's approximately 11.6913 sq. units.

Let's round it off and say 11.69 sq. units.

Answer 2

The center of the circle will also be the point where the medians of the triangle intersect, which is the centroid. The line from the centroid to one corner is a radius of the circle, so it has a length of 3. That line is two thirds of the total length of the median, which is an altitude. So the altitude of the triangle is 3 / (2/3) = 4.5. This creates two right triangles that are 30-60-90 triangles, a well known configuration. The height is equal to squrt(3) times the base, so the base is 4.5 / sqrt(3). Remember, that's the base of half the triangle only. The whole base is therefore 9 / sqrt(3) = 3*sqrt(3). The area of the triangle is half base times height, so A = 0.5*3*sqrt(3)*4.5 = 6.75*sqrt(3) = 11.69.

Answer 3

Its 3*(2*20*sin(360/(3*2))) = 103.92 I think. That is perimeter of triangle. The area of 3 segments is pi*20^2-(103.92*sqrt(,75)* .5*(1/3)) = 737

Answer 4

for an inscribed equilateral triangle
A=3/2 * r^2 * sqrt(3)/2
27/4sqrt(3)=5.846

Answer 5

I think it's 11.69