SOLUTION: An equilateral triangle is inscribed in a circle with a radius of 6". Find the area of the segment cut off by one side of the triangle

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Question 254908: An equilateral triangle is inscribed in a circle with a radius of 6". Find the area of the segment cut off by one side of the triangle
Answer by edjones(8007) About Me  (Show Source):
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Solved by pluggable solver: Calculate side length of equilateral triangle inscribed in the circle

The circle of radius 6 with center O with an equilateral triangle ABC inscribed in it.


Join edges A,B and C with center O as shown in figure.





Consider Triangle AOB and AOC,


1.> AB=AC .............(Sides of Equilateral triangle are equal)


2.>OC=OB .............(These are radius of a circle)


3.>OA is common side to both Triangles.


From conditions 1,2 and 3


Triangle AOB and AOC are congruent to each other. (SSS congruency condition)


Now since triangles are congruent,


Therefore, Angle+OAB=Angle+OAC ...............(4)


Similarly, Angle+OBC=+Angle+OBA ...............(5)


As, Angle CAB is an angle in a equilateral triangle


Hence, Angle+CAB=60 degrees ...............(6)


Now, Angle+CAB=Angle+OAB+%2B+Angle+OAC


Angle+CAB=Angle+OAB+%2B+Angle+OAB ..........(From (4))


60=2+Angle+OAB


Angle+OAB+=+60%2F2 degrees


Angle+OAB=Angle+OAC=30 degrees ...........(6)


Similarly, Angle+CBA=60


and from condition 5, Angle+OBA=+Angle+OBC=30 degrees ...........(7)


Now consider Triangle AOB,


Sum of angles in a Triangle is 180 degrees.


Hence, Angle+OAB+%2B+Angle+OBA+%2B+Angle+AOB+=+180


Angle+AOB+%2B+30+%2B+30+=+180 ........(From 6 and 7)


Angle+AOB+=+180-%2830%2B30%29


Angle+AOB+=+180-60+=+120 ...........(8)


In Triangle AOB using sine rule,


%28sin+AOB%2FAB%29=%28sin+OAB%2FOB%29+=%28sin+OBA%2FOA%29


%28sin+120%2FAB%29=%28sin+30%2FOB%29+=%28sin+30%2FOA%29


%28sin+120%2FAB%29=%28sin+30%2FOB%29


AB=%28sin+120+%2A+OB%29%2Fsin+30


AB=%28%28sqrt%283%29%2F2%29%2A6%29%2F%281%2F2%29


AB=10.3923048


Hence the side of an equilateral triangle inscribed in a circle of radius 6 is 10.3923048.


.
A(equilateral triangle)=s^2sqrt(3)/4
(10.39^2*sqrt(3))/4
=46.74 sq in
.
A(circle)=pi*r^2
=pi*6^2
=113.1 sq in
.
113.1-46.74=66.35 area of all 3 segments cut off by a side of the triangle.
66.35/3
=22.12 sq in, area of the segment cut off by one side of the triangle.
.
Ed