# Deriving diameter from given parallel chord

I'm trying to design a toddler bed (crescent moon). I should be fine once I get some basic measurements but math is proving to be a weak spot for me.

If I have circle, and there is a chord (which is parallel to a diameter ((perfectly horizontal))) that is 4' long, how long would that diameter be?

Additionally, how long would a another parallel chord be at 10% into the diameter?

For the purpose of an explanation, the chords and would be parallel to a WE diameter (West East) and a perpendicular diameter can be identified as NS (North South). So the bigger chord would be 33% from S and the smaller unknown chord would be 10% from S.

Thanks for any help.

• Just to be clear - you're saying the longer chord is horizontal, and it is 33% of the way from the bottom of the circle to the top? And the smaller chord is also horizontal and 10% of the way from the bottom of the circle to the top?
– kccu
Commented Apr 26, 2017 at 18:34
• yes, that would be correct Commented Apr 26, 2017 at 18:39
• i'm planning to put the bed platform 33% up from the floor, and the bottom of the circle is (10%) would be flat for stability Commented Apr 26, 2017 at 18:40
• Give me a couple minutes to type up an answer for you.
– kccu
Commented Apr 26, 2017 at 18:42

Sorry for the crappy microsoft paint pictures.

The longer chord is 1/3 of the way from the bottom of the circle, so it's 2/3 of a radius up from the bottom. Let $R$ be the radius of the circle. Then we can label the vertical segments $2R/3$ and $R/3$, and we can label the diagonal radius $R$. The chord has length $4$ and is bisected by the vertical radius, so we can label each half with length $2$. Now apply the Pythagorean theorem to the right triangle with sides $2$, $R/3$ and $R$:

\begin{align*} \left(\frac{R}{3}\right)^2 + 2^2 &= R^2 \\ \frac{R^2}{9}+4 &= R^2 \\ R^2 + 36 &= 9R^2 \quad \text{(multiply through by 9)}\\ 36 &= 8 R^2 \\ \frac{9}{2} &= R^2\\ \frac{3}{\sqrt{2}} &= R \end{align*} so finally the diameter is $D=2R=2\frac{3}{\sqrt{2}}= 3\sqrt{2} \approx4.2426' \approx 4' 2 \frac{7}{8}''.$

For the second chord, it is $10\%$ of the way from the bottom, so $1/5$ of a radius from the bottom. So we get the vertical segment labeled $4R/5$. Again there is a diagonal radius of length $R$, and we have $L/2$, half the length of the shorter chord. Applying the Pythagorean theorem:

\begin{align*} \left(\frac{4R}{5}\right)^2 + \left(\frac{L}{2}\right)^2 &= R^2 \\ \frac{16R^2}{25}+ \frac{L^2}{4} &= \frac{25 R^2}{25} \\ \frac{L^2}{4} &= \frac{25R^2-16R^2}{25}\\ \frac{L^2}{4}&= \frac{9R^2}{25}\\ \frac{L}{2} &= \frac{3R}{5}\\ L&= \frac{6R}{5} = \frac{9\sqrt{2}}{5} \end{align*} i.e., $L \approx 2.5456' \approx 2' 6 \frac{1}{2}''.$

• No problem. I just corrected a mistake in my calculation for $L$, but it should be correct now.
– kccu
Commented Apr 26, 2017 at 18:59