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an equilateral triangle is inscribed in a circle of radius 4 cm. find the area of the...

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utkr940 | Student, Grade 11 | Honors

Posted August 20, 2010 at 4:40 PM via web

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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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neela | High School Teacher | Valedictorian

Posted August 20, 2010 at 6:33 PM (Answer #1)

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We know that  area of the triangle = (1/2)bcsinA

We know that  a/sinA = 2R, or a =  2RsinA. where R is circum radius. a,b,c are ths sides and A = angle between the sides b and c of the triangle.

In case of equilateral triangle a = b =c  and angles  A=B= C = 60 deg, each.  The circumradius R = 4cm  is given.

So  the area of the equilateral triangle = (1/2) a^2* sin60  = (1/2) (2R*sinA)^2 * sinA = 2R^2*(sin60)^3 = 2*4^2*(sqrt3/2)^3 =

=32*(sqrt3/2)^3 =20.7846 sq cm

Area of the circum circle = pi*R^2 = pi*4^2 =  50.2655sq cm

Therefore the area of the circumcircle uncovered by the inscribed equilateral triangle = (50.2655-20.7846)sq cm = 29.4809sq cm.

 

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thewriter | College Teacher | Valedictorian

Posted August 20, 2010 at 9:34 PM (Answer #2)

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For an equilateral triangle with side a, the radius of the circumcircle is (1/2)*a*sec(pi/6) or (1/2)(2/sqrt3)a or a/sqrt3.

The area of an equilateral triangle with side a is (sqrt3/4)a^2.

As the radius of the circumcircle is given as 4:

a/sqrt3=4 or a =4sqrt3

The area of the triangle works out to (sqrt3/4)a^2=(sqrt3/4)a^2=(sqrt3/4)(4sqrt3)^2= 12sqrt3

As the area of the circle is pi*r^2=pi*4^2=pi*16, the area left over after the area of the triangle is eliminated is 16*pi-12sqrt3= 50.2654-20.7846=29.4808

 

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted August 21, 2010 at 12:32 AM (Answer #3)

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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

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