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?

help me

Answer 1

Draw a circle, with the equilateral triangle inscribed in it. They share the same center ; mark the center point as "O". Mark the triangle vertices "A", "B", "C" .
Draw a line from "O" (the center) to vertex "A".
Draw a line from "O" (the center) to vertex "B".
Draw a line from "O" (the center) to the center of the line "AB" (let's call that point "D") .
Since "AB" is a side of the triangle, and it's length is 2sqrt3 , then the length of "AD" is half of "AB" .

So now you have a new right triangle, with vertices "O" "D" "A" , on which we will focus.
The angle CAB = 60 degrees (equalateral triangle) , so the angle OAD is 30 degrees.
The angle ADO is 90 degrees.

we know that cos( 30 degrees) = 0.5 * sqrt3

We also know that for a right triagnle ,
cos (angle) = ( length of side adjacent the angle ) / ( length of hypotenuse )

In our case:

cos(30 degrees) = "AD" / "OA" (adj / hyp)
==> 0.5 * sqrt3 = sqrt3 / "OA"
==> "OA" = 2

Now, "OA" is also the radius of the circle.

So the area of the entire circle is
pi * 2^2 = 4 * pi

but we need the area of only one sector , which is a third of the entire circle -

So the area of one sector is (4/3) * pi (if I didn't mess something up :) )