How do I find the area of the blue section in this trig circle problem?

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To find the area of the blue section in the trig circle problem, first, form a triangle by joining the centers of the circles, which have sides measuring 11, 9, and 10 units. The area of this triangle includes both the blue section and a quarter of each circle's area. To isolate the blue area, subtract one-fourth of the total area of the circles from the area of the triangle. This approach provides a clear method to calculate the desired area. The solution effectively combines geometric and trigonometric principles.
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Homework Statement


This is my own sketch of the problem. The radii are listed, and the circles are all tangent. I need to find the area of the blue part.

Homework Equations


The Attempt at a Solution


I've been staring at this for hours. I really have no idea where to start. I'm just looking for a push in the right direction.

Thanks for any help.

Homework Statement


Homework Equations


The Attempt at a Solution

 
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Join centers of the circles to form a triangle with sides of 11,9 and 10 units. As you can see this triangle contains blue part+ quarter of each circle.
thus area of the blue part is:
area of the triangle joining centers - 1/4*(area of each circle)
 
santapuri.1 said:
Join centers of the circles to form a triangle with sides of 11,9 and 10 units. As you can see this triangle contains blue part+ quarter of each circle.
thus area of the blue part is:
area of the triangle joining centers - 1/4*(area of each circle)

Thanks a ton!
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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