SOLUTION: Three identical circles are tangent to each other externally. If the area of the curvilinear triangle enclosed between the points of tangency of the 3 circles is 16.13, compute the

Algebra ->  Circles -> SOLUTION: Three identical circles are tangent to each other externally. If the area of the curvilinear triangle enclosed between the points of tangency of the 3 circles is 16.13, compute the      Log On


   

Question 891679: Three identical circles are tangent to each other externally. If the area of the curvilinear triangle enclosed between the points of tangency of the 3 circles is 16.13, compute the radius of each circle.
Found 2 solutions by Fombitz, robertb:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!

The diagram helps to solve the problem.
Calculate the area of the large blue triangle.
Calculate the circular sector areas.
Subtract to get the area of the curvilinear triangle.
Blue triangle - Equilateral triangle with side 2R.
A%5BET%5D=%28sqrt%283%29%2F4%29s%5E2
A%5BET%5D=%28sqrt%283%29%2F4%29%282R%29%5E2
A%5BET%5D=sqrt%283%29R%5E2
.
.
.
Circular sector : Each sector is 60 degrees. The whole circle is 360 degrees.
So the area of each circular sector is 1/6 of the whole circle area.
A%5BCS%5D=%28pi%2F6%29R%5E2
Since there are 3 circular sector within the equilateral triangle,
A%5BCST%5D=%28pi%2F2%29R%5E2
So then the area of the curvilinear triangle is,
A%5BCT%5D=A%5BET%5D-A%5BCST%5D
A%5BCT%5D=sqrt%283%29R%5E2-%28pi%2F2%29R%5E2=16.13
%28sqrt%283%29-pi%2F2%29R%5E2=16.13
R%5E2=16.13%2F%281.73-1.57%29
R%5E2=16.13%2F0.1613
R%5E2=100.03
R=10

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Note that the figure formed is an equilateral, equiangular, triangle with side 2r and interior angles of 60 degrees. For each circle, the area of the intercepted sector is A%5Bs%5D+=+%28pi%2F6%29%2Ar%5E2
==> the combined area of the sectors is 3%2AA%5Bs%5D+=+%28pi%2F2%29%2Ar%5E2.
But the area of the equilateral triangle with side 2r is
sqrt%283%29%2Ar%5E2, hence
16.13+=+sqrt%283%29%2Ar%5E2+-+%28pi%2F2%29%2Ar%5E2
<==> 16.13%2F%28sqrt%283%29+-+pi%2F2%29+=+r%5E2
==> sqrt%2816.13%2F%28sqrt%283%29+-+pi%2F2%29%29+=+r
Then r = 10.00141131, or practically, r = 10 units.