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[gervasioraffo]

I just calculated AO^2=r^2 + r^2 = 2r^2, then AO= r√2 and then 1=r+r√2 .... Etc.

[LuisdeBritoCamacho]

Ok, let me give you my Resolution Proposal.
1) Let the Radius of the Green Circle be equal to "R"
2) Mentally Draw two Straight Lines:
a) From the Center of Green Circle to the Point of Tangency in the bottom; xx Axis.
b) From the Center of Green Circle to the Point of Tangency in the left; yy Axis.
3) Now we have a Square, and the Diagonal of any Square is always equal to Side*sqrt(2).
4) Our Diagonal = R * sqrt(2)
5) From the Origin of the Quarter of a Circle until the Point pf Tangency with the Green Circle is a distance of 1
6) So, R*sqrt(2) + R = 1 ; R * (sqrt(2) + 1) = 1 ; R = 1 / (sqrt(2) + 1) ; R = (sqrt(2) - 1) after rationalize the denominator ; R ~ 0, 414
7) Green Area = Pi * [(sqrt(2) - 1)]^2 =
8) Answer: Green Area equal to (sqrt(2) - 1)^2 * Pi Square Units or approx. equal to 0, 539 Square Units.

[HoSza]

Found it. It's right inside the white circle sector. EZ.

[KipIngram]

Well, I like to try to solve these before watching the video, looking only at the initial diagram. And in this case it's hard to tell whether that 1 is supposed to be the radius of the quarter circle, or the vertical coordinate of the center of the green circle. Obviously that makes a difference. I recognize you verbally clarified, but it would be nice if such things were apparent from the drawing itself.

Ok, so let (r, r) be the coordinates of the green circle's center. Then the distance of that center from the origin is r*sqrt(2), and we have to go out a further distance r to get to the tangent point. So the radius of the quarter circle is r*(1+sqrt(2)). We are given that that equals 1:

r*(1+sqrt(2)) = 1 --> r = 1 / (1+sqrt(2)) --> r^2 = 1/(1+sqrt(2))^2

So

Area = pi/(1+sqrt(2))^2

This seems a lot more simple and direct than the process you went through - notice I don't have to solve a quadratic. To get from my solution to yours, one notes that 1/(1+sqrt(2)) = sqrt(2) - 1. This is easy to prove:

1 / (sqrt(2)+1) = sqrt(2) - 1
1 = (sqrt(2)+1)*(sqrt(2)-1)
1 = 2 - 1. Q.E.D.

[prossvay]

Green area=(√2-1)^2(π)=3π-2√2π=0.54 square units.

[dtikvxcdgjbv]

Another dimensional solution is with negative r 😊

Surface is 3 + 2 * sqrt(2)

[darkspel]

To get the radius of the inner circle: sin(pi/4)/(1+sin(pi/4))

[tedn]

No need for a binomial to solve this. Instead of 1 - r you know that is sqrt2*r

[Birol]

My way of solution is ▶
the radius of the large circle is R,
R=1
the radius of the green circle is r
When we draw a line from the center of the large circle to the tangent line of the circle, we obtain the equation:
R= 2r+x
R=1
⇒
1= 2r+x equation-1
where x represents the remaining distance (white area) between 1 and 2r.

Now, if we draw a square within the green circle with side length a= r, the diagonal of this square would be equal to √2r
Therefore, we have the equation:
d= r+x
d= √2r
⇒
r+x= √2r equation-2

Solving for x, we find:
x= √2r -r

Substituting the expression for x from equation-2 into equation-1, we obtain:
1= 2r+x ⬆
1= 2r+√2r -r
1= r+√2r
1= r(1+√2)
r= 1/(1+√2)
r= (1-√2)/(1-√2²)
r= √2-1 length units

Agreen= π*r²
= π*(√2-1)²
= π*(2-2√2+1)
Agreen= (3-2√2)π
Agreen ≈ 0, 54 square units ✅

[IvanIvanov-trxs]

Оne should proof that A, O and tangent are on one and the same line….

[GentlemanH]

This problem I managed to solve myself although using a slightly different method than the one which is shown in the video. 🙂