In the same vein as yesterday's problem, find the radius $r$ such that the area of the semicircle is the same as the area of the rectangle. The green region has an area of 10 (a.u.).
Previous iteration: #717118
pull down to refresh
pull down to refresh
In the same vein as yesterday's problem, find the radius $r$ such that the area of the semicircle is the same as the area of the rectangle. The green region has an area of 10 (a.u.).
Previous iteration: #717118
Ok, for real this time. Let ℓ be the length of the rectangle.
We know rℓ=0.5πr2 and so ℓ=0.5πr
Moreover, it'll be useful to write: ℓ−r=(0.5π−1)r
The angle of the "pizza slice" that the green region covers is:
The area of the pizza slice is 0.5θr2 and the "height" of the green region is rsinθ.
Thus:
Or:
Which gives r=6.3587
I'm pretty sure I had the same answer, but I was trying to work it out analytically.
I'm not as much of a fan of these brute force calculations, but once I start on something I find it very hard to stop...
I gave myself 15 minutes to do this, and I thought I had it, but it turns out I made a mistake. Ended up spending maybe 30-40 minutes on this problem... not great for my productivity today lol
I don't think it simplified nicely, which isn't surprising when you put stuff like that into an inverse cosine.
Yeah, i also usually prefer the ones that simplify nicely. Also because it's an indirect confirmation one is on the right track.
It feels like a better puzzle when the answer comes out nice and simple.
The green area is half a circle segment with height h.
The area of the rectangle is r*(r + r - h).
The area of the semicircle is (pi * r^2) / 2.
Since they are equal, you get h = 2*r - (pi * r)/2
Insert that h into the circular segment area based on height h formula and you get
r ~ 6.3587
the reason this would take a while for me is that i like to derive the formulas from scratch for such problems... otherwise i trust not verify, haha.
sometimes being good at math is knowing that someone else derived the formula you need, and knowing where to find that formula.
yes, it's a process... i have the stamina and interest to relearn geometry, because i have a dream to build in stone, make true masonry great again.
i actually ran into a real mason recently. he retired and started making canvas art instead, based on his work as a mason. very intricate, never seen anything like it.
That's a very cool reason to want to learn geometry. I wish you luck in your pursuits!
Nice find, didn't even though there was a formula available for this on Wikipedia
This is another level. I'm not even going to try!
:)
I'm just not going to try it because I've got a headache and I don't want to make it worse! Today is Friday ... Ahahaha
We say 불금 here, fire-friday... people getting drunk and letting free after a hard week of work. No alcohol for me today, though...
this? 🤠
I don't need a lot of alcohol to have fun. Two whiskies is enough for the whole night! Ahhahaha
Social media campaign from Hana Bank it seems.
A modern take on 불금 where one prefers staying in with some fried chicken and beer (chimaek, 치맥) to watch a Korean drama on the laptop~~
Ah ok! Or read Han Kang's works. #717216
Missed that post. Tnx for linking. Enjoy, time for me to sleep here.
Should there be an elegant solution or is it a big mess of radicals and inverse trig functions?
i have a bet with an older friend that i will pass the differential calculus exam when i reach the age at which he failed it (about 40yo), after having the same amount of study time. i shall also pass the integral calculus exam, because it's not about the age, but the amount of cobwebs in the brain.
We should have a button for MathJax syntax like the markdown instructions button in the upper right of the text area input. I am familiar with markdown, but not MathJax.
deleted by author
wait I made a mistake... urk
deleted by author