circle shaded area

[franklin91]

hi geometry is not really my strong suit and was wondering if someone could help me figure this problem out.....thank you

three circles of radii 4,5,6 cm are mutually tangent. Find the shaded area enclosed between the circles.

 

[mmm4444bot]

By connecting the centers, a triangle is formed.

Subtract away the area of the sectors from this triangle; what's left is the area between the tangent circles.

See discussions here:

http://answers.yahoo.com/question/index ... 449AAs1nDQ

http://answers.yahoo.com/question/index ... 003AAmi4pl

Here's an image, to look at. The labels on the image might not match the discussions above, but this is what the triangle looks like.

(Double-click image to expand, if needed.)

[attachment=0:3rgifeup]Sectors.JPG[/attachment:3rgifeup]

 

[soroban]

Hello, franklin911

mmm444bot has an excellent game plan . . .

Three circles of radii 4,5,6 cm are mutually tangent.
Find the shaded area enclosed between the circles.

$\displaystyle \text{Let the centers of the circles be }A, B, C\text{, respectively.}$
$\displaystyle \text{We have }\Delta ABC\text{ with sides: }\:AB = 9,\:AC = 10,\:BC = 11.$


$\displaystyle \text{Using the Law of Cosines, we have:}$

. . $\displaystyle \cos A \:=\:\frac{10^2+9^2-11^2}{2(10)(9)} \;=\;\tfrac{60}{180} \quad\Rightarrow\quad A \:=\:\cos^{-1}(\tfrac{1}{3}) \:\approx\:1.23 \quad\Rightarrow\quad \sin A \:=\:\tfrac{2\sqrt{2}}{3}$

. . $\displaystyle \cos B \:=\:\frac{9^2+11^2-10^2}{2(9)(11)} \;=\;\tfrac{102}{198} \quad\Rightarrow\quad B \:=\:\cos^{-1}(\tfrac{17}{33}) \;\approx\;1.03$

. . $\displaystyle \cos C \;=\;\frac{10^2+11^2-9^2}{2(10)(11)} \;=\;\tfrac{140}{220} \quad\Rightarrow\quad C \:=\:\cos^{-1}(\tfrac{7}{11}) \;\approx\;0.88$


$\displaystyle \text{Area }(\Delta ABC) \;=\;\tfrac{1}{2}bc\sin A \;=\;\tfrac{1}{2}(10)(9)\tfrac{2\sqrt{2}}{3} \:=\:30\sqrt{2} \;\approx\;42.43$

. . \(\displaystyle \text{Area (sector }A)} \;=\;\tfrac{1}{2}(4^2)(1.23) \;=\;9.85\)

. . $\displaystyle \text{Area (sector B)} \;=\;\tfrac{1}{2}(5^2)(1.03) \:=\;12.87$

. . $\displaystyle \text{Area (sector C)} \;=\;\tfrac{1}{2}(6^2)(0.88) \;=\;15.86$


$\displaystyle \text{Therefore: }\:\text{Area} \;=\;42.43 - 9.85 - 12.87 - 15.86 \;=\;3.85\text{ cm}^2$