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/sci/ - Science & Math

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4296026 No.4296026 [Reply] [Original]

so i came up with a babby's first geometry problem to have my highschooler math nerd friends have a crack at. so far the only person to have solved it in less than an hour was my dad, who took about three minutes, if when i was asking for only the conceptual method and not the actual expression.

how difficult is it really, /sci/? i don't think this is that big of a conceptual leap for a high schooler, especially my circle of math nerd friends.

picture related: PROBLEM --> find the shaded area if square ABCD has side length R. (yes, those are quarter circles.) answer: <span class="math">r^2\left(\frac{\pi}{6} + \frac{\sqrt{3}}{4}\right)[/spoiler]

>> No.4296054

It's not very difficult.

It just involves cutting the region into manageable chunks. Choosing how to best cut it up may take a few mental tries, though.

I'm opting for the equilateral and two 30' sectors.

>> No.4296055

you are an idiot and so are your friends

>> No.4296086

>>4296054
that was my approach, too, though i distinctly remember one of them suggesting integrating the curve of the top of the circle and doubling that, which seems to me a little more difficult and more complex than it needed to be.

how would you even integrate y = sqrt(r^2 - x^2) ?

>> No.4296111
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4296111

>>4296026

so fucking easy, if R is any segment of the square then just take what would be the area of the circle and divided by 4 if it is a quarter, R is also the radio of the circle if it is the segment then you get the shaded area and multiply it by 2, or easier, divide it by 2 instead of 4, this leaving a (PiR^2)/2

>> No.4296144

>>4296086
I did the integration method and got your answer, but with only "r" as opposed to "r²". Close enough.

To integrate sqrt(r^2 - x^2), you first factor out the r. You can then substitute x/r for sin(t), which works out nicely since sqrt(1 - sin²t) is just cos²(t). You then use the half angle identity (google it) to get something integrateble.

>> No.4296279
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4296279

>>4296111

>> No.4296404
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4296404

the area is 2 times the integral of r sin(theta) from pi/3 to pi/2.

this is - 2r cos(pi/2) + 2r cos(pi/3)

which equals 2r sqrt(3/2)

I balls-d this up.