Three circles of radii 6.7, 7.9, and 10.3 cm are mutually tangent. Find the area bounded by the three circles.
+130516 I assume you want the "middle area" "......
By Euclid, if circles are tangent, the radial lines joining their centers go through the point of tangency.
So......we have a triangle formed by these lines that has sides of (6.7 + 7.9), (7.9 + 10.3) and (10.3 + 6.7) = 14.6, 18.2 and 17
We need to find the area of this triangle.....let's find its angles
We can find two of the angles between the sides using the Law of Cosines
So we have
14.6^2 = 18.2^2 + 17^2-2(18.2)(17)cos(Theta 1)
cos-1 [( 14.6^2 - 18.2^2 - 17^2) / (-2)(18.2)(17)] = about 48.86° = Theta 1 = about .853 rads
And to find the second angle we have
17^2 = 18.2^2 + 14.6^2-2(18.2)(14.6)cos(Theta 2)
cos-1 [( 17^2 - 18.2^2 - 14.6^2) / (-2)(18.2)(14.6)] = about 61.28° = Theta 2 = about 1.07 rads
And the remaining angle = 180 - 48.86 - 61.28 = 69.86° = Theta 3 = about 1.22 rads
So...the area of this triangle = (1/2)(18.2)(14.6)sin(61.28) = about 116.52 cm^2
From this, we need to subtract the area of the sectors formed in the circles by their respective radiuses and central angles....we need to be careful here to "line" things up correctly!!!!
For the circle with the radius of 6.7, we have ..... (1/2)(6.7)^2(1.22) = about 27.38cm^2
For the dircle with the 7.9 radius, we have (1/2)(7.9)^2(1.07) = about 24.02cm^2
And for the circle with the radius of 10.3, we have (1/2)(10.3)^2(.853) = about 42.25cm^2
So, finally, the "middle area" = 116.52 - 27.38 - 24.02 - 42.25 = about 22.87cm^2
{If I haven't made any egregious errros, I believe this is correct }
+130516 I assume you want the "middle area" "......
By Euclid, if circles are tangent, the radial lines joining their centers go through the point of tangency.
So......we have a triangle formed by these lines that has sides of (6.7 + 7.9), (7.9 + 10.3) and (10.3 + 6.7) = 14.6, 18.2 and 17
We need to find the area of this triangle.....let's find its angles
We can find two of the angles between the sides using the Law of Cosines
So we have
14.6^2 = 18.2^2 + 17^2-2(18.2)(17)cos(Theta 1)
cos-1 [( 14.6^2 - 18.2^2 - 17^2) / (-2)(18.2)(17)] = about 48.86° = Theta 1 = about .853 rads
And to find the second angle we have
17^2 = 18.2^2 + 14.6^2-2(18.2)(14.6)cos(Theta 2)
cos-1 [( 17^2 - 18.2^2 - 14.6^2) / (-2)(18.2)(14.6)] = about 61.28° = Theta 2 = about 1.07 rads
And the remaining angle = 180 - 48.86 - 61.28 = 69.86° = Theta 3 = about 1.22 rads
So...the area of this triangle = (1/2)(18.2)(14.6)sin(61.28) = about 116.52 cm^2
From this, we need to subtract the area of the sectors formed in the circles by their respective radiuses and central angles....we need to be careful here to "line" things up correctly!!!!
For the circle with the radius of 6.7, we have ..... (1/2)(6.7)^2(1.22) = about 27.38cm^2
For the dircle with the 7.9 radius, we have (1/2)(7.9)^2(1.07) = about 24.02cm^2
And for the circle with the radius of 10.3, we have (1/2)(10.3)^2(.853) = about 42.25cm^2
So, finally, the "middle area" = 116.52 - 27.38 - 24.02 - 42.25 = about 22.87cm^2
{If I haven't made any egregious errros, I believe this is correct }
