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.

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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,...

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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

Approved by eNotes Editorial Team