# Calculate the radius of a circle given the chord length and height of a segment

I have a (circular) segment of known height and known chord length. Is is possible to determine the radius of the circle?

Any help much appreciated.

• The formula I derived is simple: radius is equal to the added square of the chord straight length and the fourth multiple of the perpendicular height squared (as measured from midpoints of arc and chord) all divided by the eighth multiple of of that perpendicular height. I would have posted a picture of the calculation but I'm not allowed to answer any questions here. – Rhodie Sep 2 '18 at 1:49

We can apply the Intersecting Chords Theorem.

You chord length is the length $UV$ and the segment height is the length $PX$.

The intersecting chords theorem tells us that $XP \times XQ = XU \times XV$.

Let $\ell = UV$ and $h=XP$. It follows that $UX = XV = \tfrac{1}{2}\ell$. The ICT then tells us that

$$\tfrac{1}{2}\ell \times \tfrac{1}{2}\ell = h \times XQ \, ,$$ i.e. $XQ = \tfrac{1}{4h}\ell^2$. The diameter $PQ=PX+XQ$ and $$PX + XQ = h + \frac{\ell^2}{4h}=\frac{4h^2+\ell^2}{4h}$$

The radius is then one half of this, i.e.

$$CQ = \frac{4h^2+\ell^2}{8h} \, .$$

Here is the solution using trigonometry

Consider given the height $h=DE$ and the half length $a=DB$. The triangle ADB produces the following two equations

$$a = r \sin \theta \\ r-h = r \cos \theta$$

where $r$ is the unknown radius, and $\theta$ is the included half angle (denoted on point A above)

If these two equations are squared and added together they make

$$a^2 + (r-h)^2 = (r \sin\theta)^2 + (r \cos \theta)^2 = r^2$$

The solution to this is $$\boxed{ r = \dfrac{a^2+h^2}{2 h} }$$

This problem can be solved as follows. There are two knowns and two unknowns. The two knowns are the chord length $UV$ ($l$) and the chord height $XP$ ($h$). The two unknowns are the radius ($PC$ or $r$) and $XC$ which is part of the radius. Lets call this $q$. I used the drawing and symbols from the first answer to the problem.

We can write two distinct equations using the two knowns and two unknowns, so we can solve this problem:

(1) $h + q = r$

(2) Using the Pythagorean theorem, $(\frac{l}{2})^2 + q^2 = r^2$. To see this draw a line from $C$ to $V$ (this distance is $r$ of course). Since $PC$ bisects $UV$, the angle $CXV$ must be a right angle, so the Pythagorean theorem applies

By rearranging (1) as $q = r - h$, substitute $r - h$ for $q$ in equation (2)

Without going through all the math, you can solve equation (2) for $r$

The solution is: $$r = \dfrac{\left(\frac{l}{2}\right)^2 + h^2}{2h}$$

or

$$r = \dfrac{\frac{l^2}{4} + h^2}{2h}$$

Multiply numerator and denominator by $4$ and you get the same answer as previously submitted

• Please see this tutorial on how to format mathematics on this site. – N. F. Taussig Mar 5 '15 at 14:53

Any two cords that intersect within the same circle will create two equal area rectangles if you multiply the two segments of the same cord together, the segments created by the intersection of the other second cord.

Given one cord and creating a second imaginary cord that intersects the given cord at the mid point and also intersects the circles center, we can find the diameter of the circle. The height of the circular segment becomes one of the segments of the second imaginary cord. We can solve the second segment by dividing the square of the given two segments by the height of the circular segment. The height of the circular segment is one of the segments of our imaginary created cord. If we add them both together they create the diameter length of the circle.

(1/2 cord)^2 / circular segment height, equals the diameter if you add the height of the circular segment to it. If you want the radius just divide the diameter by 2.

Sincerely,

William McCormick

I have been exploring the design of certain church doorways from the mid 1700's and this problem came up. I tried to think of how they might have solved this problem back then and came up with the following solution:

Draw a long vertical line. Measure and mark h from the top. Draw a horizontal line equal to l centered on that point. Take your compass with the sharp end moving down the vertical line and expand it until it scribes an arc that includes the top of the horizontal line and the end of the vertical lines. Measure that length and you get the radius.

Math is a powerful interpretive language to describe certain things, but it is not those things. It's interesting to speculate how the masons of those days solved these complex problems with little or no knowledge of the calculations that we would use today.

• Masons were not obligated to use compass and straight edge according to the rules of Euclidian geometry. They could, for example, easily trisect an angle. – richard1941 Aug 3 '17 at 19:52

I would suggest that you read a book on telescope mirror making. The telescope maker's approximate formula for the saggita of a mirror is s = r^2/2R, where r is the radius of the mirror surface, R is the big radius (half the focal length). That means R = r^2/2s. If the mirror is a parabola instead of a sphere, this formula is exact.

For an exact formula, use the Pythagorean theorem to get (after some algebra) R=(r^2+s^2)/2s.

In real world telescope mirror making, no mirror comes out exactly spherical, not even for John Dobson (PBUH), so the approximation is good enough.

Contact me richard1941@gmail.com if you need help building a Foucault tester or interpreting the test results. I have plenty of glass from the estate of Bill Kelly waiting to made into mirrors. You would honor a great inventor.

Let $h$ be the segment height, $2c$ the chord and $r$ radius. For Pythagoras: $$r^2=c^2+(r-h)^2\rightarrow r^2=c^2+r^2-2hr+h^2\rightarrow h^2-2rh+c^2=0\rightarrow$$ $$2rh=h^2+c^2\rightarrow r=\dfrac{1}{2h}(h^2+c^2)$$

Consider one half of the segment. Slant side length = $$\sqrt{c^2+h^2}$$.

Note similarity of this right triangle of smaller sides $$(c,h)$$ at left with main triangle diameter as hypotenuse $$2r$$

Ratio of corresponding sides should be same:

$$\dfrac{h}{\sqrt{c^2+h^2}}=\dfrac{\sqrt{c^2+h^2}}{2r} \rightarrow 2rh ={c^2+h^2}$$

Note that the result is true either if $$h$$ is the major segment height or the minor segment height.

## protected by Community♦May 5 '17 at 4:18

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