Using $2\pi\sqrt{\frac{a^2+b^2}{2}}$ to approximate the circumference of an ellipse

Question

I'm a 37 year old who hasn't had to think about the geometry of circles in 20 years, so I'm a bit stuck on my problem at hand. I am working on the design of a high tunnel greenhouse and trying to determine approximately (I don't need exact numbers as I can trim off extra) how much plastic I'll need to cover.

The width of my greenhouse is 8.5 ft and a height of 6 ft. Now, from what I've gathered online, the approx circumference of an ellipse can be determined with this formula: $2\pi\sqrt{\frac{a^2+b^2}{2}}$. If I plug in my number where a=12 (height x 2 since my greenhouse is only half of an ellipse) and b=8.5, I get a circumference of 65.33. Divide that by two (again, my greenhouse is only half of the ellipse) and I get 32.67 ft.

That looks right. But, just to check, I took a more simplistic look at a circle. The circumference of a circle is $\pi d$. If I used the width of my greenhouse, 8.5', which then means the height is also 8.5', which is higher than the actual 6' of my ellipse. $\pi 8.5 = 26.7$. Again, if I half that to get half the circumference, I get 13.35 ft.

Now, with the circle formula, I am assuming my height is greater than it actually is, yet the resulting number is significantly less than my result for the ellipse. So, I'm stumped here.

Answer 1

1. The $a, b$ for the ellipse are "half" of what you put in (see screenshot below), which works out to 32.44. Half of that is ~16.2
2. A width of 8.5 and a height of 8.5 for a "half arc" doesn't yield a semi circle. You need width = 8.5 and height = 4.25, which yield ~13.3.
3. Hence, your greenhouse is $~40\% = \frac{ 6-4.25}{4.25}$ higher than a semicircle, so the material used is more. $~22\% = \frac{ 16.2-13.3}{13.3}$ more materials doesn't seem that off.

Answer 2

Semi circumference of greenhouse arch =$\pi\sqrt{\dfrac{a^2+b^2}{2}}$

$ (a,b)$ are semi-major and minor axes. So plug in $$ a=\dfrac{8.5}{2} = 4.35,\, b=6 \,$$

calculates to $\approx 16.3336\,;$