an equilateral triangle is inscribed in a circle of radius 4 cm. find the area of the part of the circle other than the part covered by the triangle.

The area between the circle and the triangle = area of the
circle - area of the triangle.

Area of the circle = r^2*pi
= 4^2*22/7 = 50.29

Area of the
triangle:

Let ABC be a triangle, O is the center of the
circle.

Let us connect between OA, OB, and
OC

OA= OB= OC = r = 4

Now we
have divided the triangle into three equl triangles.

Then
the area of the triangle ABC = 3*area of the treiangle
AOB.

The tiangle AOB is an isoscele. and the angle AOB =
120 degree, then angle OAB = angle OBA = 30 degree.

Let OD
be perpindicular on AB ,

==> OA^2 = OD^2
+(AB/2)^2

==> 4^2 = OD^2 +
AB^2/4

But sinA = OD/
OA

==> sin30 = OD / 4 =
1/2

==> OD =
2

==> 4^2 = 2^2
+AB^2/4

==> 16 = 4
+AB^2/4

==> 12 =
AB^2/4

==> AB^2 =
48

==> AB = sqrt48 =
4sqrt3

Then area of the small triangle = (1/2)*4sqrt3*2 =
4sqrt3

The area of the big triangle = 3*4sqrt3 = 12sqrt3=
20.78

Then the area between the circle and the triangle
is:

a  = 50.29- 20-78= 29.51