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I'm asking this question again since now I have a bit more insight into how to solve the problem

my current method which is wrong was constructing a smaller trapezium inside the larger one using the centers of the circles you could then determine the height of the triangle by using Pythagoras' theorem and adding 2r to find the height of the larger trapezium which could be divided by the height of the smaller trapezium and get the scalability factor to then scale the smaller trapezium up to a larger size to get the rest of the lengths but as you can see in the image below this does not work as the length of the trapezium is not the same scalability as the top and bottom lengths have a different scalability as you can see in the image current formula how we tried to solve it

image not to scale

the radius of the circles is 4.5cm

if you have any suggestions for how I could improve this question or extra tags I want to add feel free to leave a comment as this is only my second post

Jim Bellette
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  • Firstly, welcome. Secondly, algebraic geometry is the wrong tag. Thirdly, add your own work. Fourthly, enjoy the benefits of doing all three. Fifthly, accept an amazing answer written by the amazing people on this site. – Bumblebee Apr 24 '23 at 04:27
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    The centers of the circles in the problem form several equilateral triangles. The centers in the second picture do not, which is why it doesn't help. Focus on the first drawing. For example, the height of the trapezoid is $2R ;+$ the height of an equilateral triangle of side $2R$. You can find more such useful relations to help solve the problem. – dxiv Apr 24 '23 at 04:49
  • Hint: The angle between a vertical line and the line joining two adjacent circles on the first and second row is $\frac{\pi}{6}$. – Gribouillis Apr 24 '23 at 04:55

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