Question

Express answers in simplest exact form.

An equilateral triangle with side of 2sqrt3 is inscribed in a circle. What is the area of one of the sectors formed by the radii to the vertices of the triangle?

Answer 1

angle should be 360/3 =120
then there is some formula which i don't remember sorry : (

Answer 2

If you draw this as a picture you will note that a line from the center of the circle to any of the corners of the trianngle is the radius. This makes an obtuse triangle. If you split the 3 triangles in 2 you get some right triangles to work with. The angles of these triangles are 30-60-90. The adjacent side to the 30 degree angle is sqrt3. This allows us to solve the hypotenuse or radius:
\[cos(30) = \sqrt {3}/x \]
\[x= \sqrt{3}/cos(30)\]
\[x=2\]
Area of circle = 2 pi r
Area of sector = 2/3 pi r
Area of sector = \[(2/3) \pi (2) = (4/3) \pi\]

Answer 3

so it 4/3?

Answer 4

It's 4/3 pi
The pi matters :)
Sorry been working an integral and was preoccupied :P