# Geometry question

Tags:
1. Feb 28, 2015

1. The problem statement, all variables and given/known data
In the attached drawing, find R in terms of L and c. Also, at the bottom of the picture I wrote something wrong. I said c, which equals 0.5, is the arc-length of each semi-circle, but I really meant to say each quarter circle. My bad. I'm not given a number for L so that can just remain a variable.

Can anyone see the solution? I'd appreciate any help.

2. Relevant equations

3. The attempt at a solution
I think I need to take advantage of c somehow, but I wasn't able to figure it out. Obviously, R=(a^2 + L^2)^(1/2), so I just need to figure out a in terms of L and c (or something else)?

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2. Feb 28, 2015

### Svein

From your figure a=R⋅sin(c) and L=R⋅cos(c). Then...

3. Feb 28, 2015

### titasB

First,

a / R = sin (c) => a = R sin (c) --- (i)

Next,

R^2 = a^2 + L^2 => a = (R^2-L^2)^1/2 ---(ii)

Equating (i) and (ii) gives: R sin (c) = (R^2-L^2)^1/2 => R = L / cos(c)

4. Feb 28, 2015

Can I ask, are you treating c like an angle? Also, after speaking to a friend, I was mistaken in assuming that those arcs belonged to quarter circles. In fact, they belong to an ellipse (so the entire big arc you see is actually half an ellipse). How does that change your answer?
Thanks though.

5. Feb 28, 2015

Hi titasB,
I apologize, but as in my other reply, does this answer change if the big arc is actually half an ellipse, and not have a circle as I first assumed?

6. Mar 1, 2015

### Svein

Yes. If not, we need to find some other way. If it is part of a circle (as you mentioned in your original post), then the angle is c/R. Then you get a slightly more complicated expression. From the note on your figure, c=0.5, so the angle is 0.5/R. Inserting this, you get L=R⋅cos(0.5/R) - which is fine for finding L, but complicated for finding R. On the other hand, tg(0.5/R) = a/L - which gives 0.5/R = arctg(a/L).

7. Mar 1, 2015