SOLUTION: Three circles of radii 8.5, 11.9, and 15.7 cm are mutually tangent. Find the area bounded by the three circles. area =

Algebra ->  Trigonometry-basics -> SOLUTION: Three circles of radii 8.5, 11.9, and 15.7 cm are mutually tangent. Find the area bounded by the three circles. area =      Log On


   



Question 1043961: Three circles of radii 8.5, 11.9, and 15.7 cm are mutually tangent.
Find the area bounded by the three circles.
area =

Found 2 solutions by Alan3354, advanced_Learner:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Three circles of radii 8.5, 11.9, and 15.7 cm are mutually tangent.
Find the area bounded by the three circles.
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If you mean the area bounded by the tangent points:
Label the centers:
P - 8.5
Q - 11.9
R - 15.7
The lines between the centers form a triangle PQR.
Label the sides p, q & r, side p opposite angle QPR, etc.
---
Find the 3 angles:
Use the Cosine Law:
p^2 = q^2 + r^2 - 2qr*cos(P)
cos(P) = (8.5^2 - 11.9^2 - 15.7^2)/(-2*11.9*15.7) =~ 0.1856233
Angle P =~ 79.3 degs
---
Find a 2nd angle the same way.
The 3rd angle is 180 - sum of the 1st 2.
=================
Find the area of the triangle. Heron's Law will work, or any method.
Subtract the area of each of the 3 sectors.
eg, sector of P = Area of P * (79.3/360)

Answer by advanced_Learner(501) About Me  (Show Source):
You can put this solution on YOUR website!
this is indeed a classic challenging problem the decimals make it lenghy and hairy ,cant be solved with out drawing,i dont know how to draw three tangent circles but i uploaded an image for you to refer to.

please refer to picture called "circles"
.



first lets find semi-perimeter S of the triangle
semi-perimeter S of the triangle =%2827.6%2B24.2%2B20.4%29%2F%282%29
=36.1


area of the triangle=sqrt%28s%28s-a%29%28s-b%29%28s-c%29%29
area of the triangle=sqrt%2836.1%2836.1-27.6%29%2836.1-24.2%29%2836.1-20.4%29%29
area of the triangle=239.4343
find angle A ,B and C
NOTE it is not right triangle,use cosine law
Cos A=%28b%5E2%2Bc%5E2-a%5E2%29%2F%282bc%29
Cos A=%28%2827.6%29%5E2%2B%2820.4%29%5E2-%2824.2%29%5E2%29%2F%282%2A27.6%2A20.4%29
Cos A=%280.520069%29
A=58.66

Cos B=%28a%5E2%2Bc%5E2-b%5E2%29%2F%282ac%29
Cos B=%28%2827.6%5E2%29%2B%2824.2%5E2%29-%2820.4%5E2%29%29%2F%282%2A27.6%2A24.2%29
Cos B=0.697
B=45.81
cos C=%28180%29-%2858.66%2B45.81%29
cos C=75.53

or use the formula



Cos C=%28a%5E2%2Bb%5E2-c%5E2%29%2F%282ab%29
Cos C=%28%2824.2%5E2%29%2B%2820.4%5E2%29-%2827.6%5E2%29%29%2F%282%2A24.2%2A20.4%29
Cos C=0.2529
C=75.34 very very minor difference

area sector a=%28pi%2Ar%5E2%2Aangle+A%29%2F%28360%29
area sector a=%28pi%2A%2811.9%29%5E2%2A%2858.66%29%29%2F%28360%29
area sector a=%2872.45%29
area sector b=%28pi%2A%2815.7%29%5E2%2A%2845.8%29%29%2F%28360%29
area sector b=%2898.467%29
area sector c=%28pi%2A%288.5%29%5E2%2A%2875.34%29%29%2F%28360%29
area sector c=%2847.477%29

the area bounded by the three circles=%28%28Area+of+the+triangle%29+-%28sum+of+++sector+A+B+and+C%29%29%29
the area bounded by the three circles=%28239.43%29+-%2872.45%2B98.467%2B47.477%29
the area bounded by the three circles=~%2821cm%5E2%29 is the answer
your feedback is welcome