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 June 3rd, 2019, 12:31 AM #1 Newbie   Joined: May 2019 From: Australia Posts: 8 Thanks: 0 The height of a section of overlapping circles. Say I have two identical circles, both of radii of one, overlapping, as shown in the diagram below: In this diagram, x is the circumference of the circles, and the bit of the bottom circle which is drawn blue (the overlapping bit) is 1/6th of the whole circumference. What I'm looking for is y, which is this: Now, working out x is easy - it's 2 \pi r, thus the overlapping bit is 1/3 \pi r. But how do I proceed in finding y from here? Help is much appreciated! Thanks!
 June 3rd, 2019, 03:48 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Imagine a right triangle from the centre of a circle, the midpoint of the segment y and one of the points of intersection of the circles. This triangle has hypotenuse 1 and side opposite the centre of the circle 1/2. Also, this triangle is 30-60-90. Can you use some basic trig and the Pythagorean theorem to find y?
June 3rd, 2019, 04:11 AM   #3
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Quote:
 Originally Posted by greg1313 Imagine a right triangle from the centre of a circle, the midpoint of the segment y and one of the points of intersection of the circles. This triangle has hypotenuse 1 and side opposite the centre of the circle 1/2. Also, this triangle is 30-60-90. Can you use some basic trig and the Pythagorean theorem to find y?
Although I know of trigonometry and Pythagoras, I don't see how having a point of the triangle at the midpoint of y can help me find the whole of y.

 June 3rd, 2019, 11:00 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2219 Can you post a diagram with the suggested triangle shown? Certain lines are radii, so they have length 1.
 June 4th, 2019, 08:40 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond $\displaystyle \frac{y}{2}=1-\cos30=1-\frac{\sqrt3}{2}=\frac{2-\sqrt3}{2}\implies y=2-\sqrt3$

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