MHB Solve Circle Radius Given Trapezoid Height & Length

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Determining the radius of a circle given the height and length of a trapezoid is complex and requires additional information. The discussion highlights the use of right-angled triangles and Pythagorean theorem to derive equations relating the radius and lengths. By manipulating these equations, a quadratic formula for the radius can be established in terms of the trapezoid's dimensions. The configuration can be clarified if a horizontal line is tangential to the circle, which helps in solving for the unknowns. The participants express gratitude for the assistance and encouragement in tackling the problem.
Jeppe1
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I don't know if this can be calculated.
I have tried for hours and days to isolate/calculate the radius and angles of the circle in order to be able to calculate length 1. I have tried using cos/sin-relation formulas and triangle formas - but Iam stuck. Any hints would be greatly appreciated. The task is one i have put on my self for cutting out a wooden plate. I have made the cut - but by approximation :-)

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Hi Jeppe,

Unfortunately, without more information, the four unknowns cannot be determined.
 
Draw horizontal lines through the points where the lines $h_1$ and $h_2$ meet the circle. That will give you a pair of right-angled triangles. You can then use Pythagoras to get the equations $$(r-h_1)^2 + l_1^2 = r^2,$$ $$(r-h_2)^2 + (l_1+l_2)^2 = r^2.$$ After a bit of algebra (expanding those brackets), the equations reduce to $$h_1^2 - 2rh_1 + l_1^2 = 0,\qquad (*)$$ $$h_2^2 - 2rh_2 + l_1^2 + 2l_1l_2 + l_2^2.$$ Subtract the first of those equations from the second: $$h_2^2 - h_1^2 - 2r(h_2 - h_1) + 2l_1l_2 + l_2^2 = 0.$$ Solve that for $l_1$: $$l_1 = \frac{(h_2 - h_1)(2r - h_1 - h_2) - l_2^2}{2l_2}.$$ Substitute that expression for $l_1$ into the equation labelled (*) and you will have a quadratic equation (admittedly quite a messy one) for $r$ in terms of $h_1$, $h_2$ and $l_2$.

Edit (@Euge): I am assuming that the horizontal blue line is meant to be tangential to the circle. That should determine the configuration, shouldn't it?
 
Opalg said:
Edit (@Euge): I am assuming that the horizontal blue line is meant to be tangential to the circle. That should determine the configuration, shouldn't it?

Yes, in that case it does. We would then have $\tan \phi_1 = \frac{\ell_1}{r-h_1}$ and $\tan (\phi_1 + \phi_2) = \frac{\ell_1 + \ell_2}{r-h_2}$, so then

$$\tan \phi_2 = \frac{\tan (\phi_1 + \phi_2) - \tan \phi_1}{1 + \tan (\phi_1 + \phi_2) \tan \phi_1} = \frac{(r - h_2)\ell_1 - (r - h_1)(\ell_1 + \ell_2)}{(r - h_1)(r - h_2) + \ell_1(\ell_1 + \ell_2)}.$$

Since $r$ and $\ell_1$ have been determined, it now follows that the entire configuration is determined.
 
Wow - that was fast ! - I will get working on the quadratic!
Thanks and thanks again! - best forum and page ever :)
 
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