Let's start by labeling the radius of ⊙O as r. So, CO = BO - BC = r - 2. Since AC = CO, AC = r - 2. Draw radius AO. △ACO is an isosceles right triangle. r = (r - 2)√2 r = r√2 - 2√2 r - r√2 = -(2√2) r(1 - √2) = -(2√2) r = [-(2√2)]/(1 - √2) Rationalize the denominator by multiplying the fraction by (1 + √2)/(1 + √2). = [-(2√2) * (1 + √2)]/(1 - 2) = [-(2√2) - 4]/(-1) = 2√2 + 4 Green Region Area = Sector AOB Area - △ACO Area Because △ACO is an isosceles right triangle, m∠AOB = 45°. A = πr² * (θ/360°) = π * (2√2 + 4)² * (45°/360°) = π * (8 + 16√2 + 16) * (1/8) = π * (24 + 16√2) * (1/8) = π(3 + 2√2) = 3π + (2π)√2 AC = CO = r - 2 = (4 + 2√2) - 2 = 2 + 2√2. A = (bh)/2 = 1/2 * (2 + 2√2) * (2 + 2√2) = 1/2 * (4 + 8√2 + 8) = 1/2 * (12 + 8√2) = 6 + 4√2 Green Region Area = [3π + (2π)√2] - (6 + 4√2) = 3π - 6 + (2π - 4)√2 So, the area of the green shaded region is π(3 + 2√2) - 6 - 4√2 square units, a. w. a. 3π - 6 + (2π - 4)√2 square units (exact), or about 6.65 square units (approximation). Spoiler (probably): Video rewrite of the answer: (3 + 2√2)(π - 2) square units
My way of solution ▶ we can find the radius of the circle a) according to the Pythagorean theorem: (r-2)²+(r-2)²= r² r²-4r+4+r²-4r+4= r² r²-8r+8=0 or: Intersecting chords theorem: (r-2)*(r-2)= 2*[(r-2)+r] (r-2)² = 2*(2r-2) r²-4r+4= 4r-4 r²-8r+8=0 both equations will give the radius of the circle, lets take one: r²-8r+8=0 Δ= 64-4*1*8 Δ= 32 √Δ= 4√2 r₁= (8-4√2)/2 r₁= 4-2√2 r₂= 4+√2 if r = 4-2√2, then r-2 would be equal to 2-2√2 < 0, that's why r= 4+2√2 Agreen= π*r²*(α/360°) - 1/2*(r-2)² r= 4+2√2 α= 45° ⇒ Agreen= π*(4+2√2)²*(45°/360°) - 1/2*(r-2)² = π*(4+2√2)²*(1/8) - 1/2*(r²-4r+4) = π*(16+16√2+8)*(1/8) - 1/2*[(4+2√2)²-4*(4+2√2) +4] = π*(24+16√2)*(1/8) - 1/2*[(16+16√2+8- 16 -8√2 +4] = π(3+2√2) - 1/2(8√2+12) = π(3+2√2) - 4√2-6 = π(3+2√2) - 2(2√2+3) Agreen= (3+2√2)[π-2] square units
Let R is Radius of circle AC=OC=a So R=a+2 ; a=R-2 a^2=(R+a)(2) (R-2)^2=2(R+R-2) So R=4+2√2 Connect O to A So right triangle isosrles AOC ∆ So AOC=45° a=4+2√2-2=2+2√2 Area of the sector =45/360(π)((4+2√2)^2=π(3+2√2) Area of right triangle=1/2(2+2√2)^2=6+4√2 Green area=π(3+2√2)-(6+4√2)=6.65 square units.❤❤❤v
Call the double-stroked length L. By the intersecting chords theorem we can immediately write L^2 = 2*(2*L+2) L^2 = 4*L + 4 L^2 - 4*L - 4 = 0 L = 2 + 2*sqrt(2), L = 2 - 2*sqrt(2) Looks from the problem like we need the larger one, so we will go with L = 2+2*sqrt(2). So the radius of the circle is R = 4+2*sqrt(2). Now for segment AO. The area of the sector AOB is (45/360)*pi*R^2, and the shaded area is that minus L^2/2. It comes out to 6.6537. Q.E.D.
I wish more people thought like engineers...... The ratio of the length BC to the radius is a constant (because AC=CO) AC represents a half of one side of the maximum square that will fit inside that circle. Had to do my own calculations to figure out the ratio, but we don't figure out π every time, could not for the life of me find the BC to radius ratio online, so simply applied random values (5) to lengths AC which equals CO, hence I know the length of two sides of right angle triangle so the radius is the square root of 50, so ratio is (6.8..... -5)/6.8..... =.2929... That's the ratio ( ✓50-5)/✓50 So knowing the radius of the circle and the dimensions of the square with insultingly simple maths, the shaded area is 1/8 of (the area of the circle minus the area of the square).
Draw OA. As OA and OB are both radii of circle O, OA = OB = r. Let OC = CA = x, so as CB = 2, OB = r = x+2. So x = r-2. Triangle ∆OCA: OC² + CA² = OA² x² + x² = r² = 2x² r² = 2(r-2)² = 2r² - 8r + 8 r² - 8r + 8 = 0 r = (-(-8)±√(-8)²-4(1)(8))/2(1) r = 4 ± √(64-32)/2 = 4 ± 2√2 r = 4 + 2√2 ---> r > 2 x = r - 2 = (4+2√2) - 2 = 2 + 2√2 As OC = CA, ∆OCA is an isosceles right triangle and ∠AOC = ∠CAO = (180°-90°)/2 = 45°. The green area is equal to the area of the sector covered by ∠AOC minus the area of the triangle ∆ OCA. Green area: Aɢ = (45°/360°)πr² - bh/2 Aɢ = π(4+2√2)²/8 - (2+2√2)²/2 Aɢ = (π/8)(16+16√2+8) - (4+8√2+8)/2 Aɢ = (3+2√2)π - (6+4√2) Aɢ = (3+2√2)(π-2) ≈ 6.65 sq units
In triangle AOC: (R -2)^2 + (R -2)^2 = R^2, with R the radius of the circle. So R^2 -8.R + 8 = 0 Deltaprime = 8, so R = 4 -2.sqrt(2) which is rejected as inferior to 2, or R = 4 + 2.sqrt(2) which is O.K. Then the area of triangle AOC is (1/2).(R -2)^2 =(1/2).(4 + 8 +8.sqrt(2)) = 6 +4.sqrt(2) The area of sector AOB is Pi.(R^2).(45°/360°) = Pi.(16 + 8 +16.sqrt(2)).(1/8) = (3 +2.sqrt(2)).Pi Finally, the green area is: (3 +2.sqrt(2).Pi - (6 +4.sqrt(2)). That was easy.
Let's find the area: . .. ... .... ..... First of all we apply the Pythagorean theorem to the right triangle OAC. With R being the radius of the circle we obtain: OA² = OC² + AC² = 2*OC² R² = 2*(R − 2)² R² = 2*R² − 8*R + 8 0 = R² − 8*R + 8 R = 4 ± √(4² − 8) = 4 ± √(16 − 8) = 4 ± √8 = 4 ± 2√2 Since R>2, the only useful solution is R=4+2√2. The triangle OAC is also an isosceles triangle. So we can conclude: ∠AOC = ∠OAC = (180° − ∠OCA)/2 = (180° − 90°)/2 = 90°/2 = 45° Now we are able to calculate the area of the green region: A(green) = A(circle sector OAB) − A(triangle OAC) = πR²*(∠AOC/360°) − (1/2)*OC*AC = πR²*(45°/360°) − (1/2)*(R − 2)*(R − 2) = πR²/8 − (1/2)*(R²/2) = πR²/8 − R²/4 = (π/2 − 1)*R²/4 = (π/2 − 1)*(4 + 2√2)²/4 = (π/2 − 1)*(16 + 16√2 + 8)/4 = (π/2 − 1)*(24 + 16√2)/4 = (π/2 − 1)*(6 + 4√2) ≈ 6.654 Best regards from Germany
Solution: r = radius of the circle. Pythagoras for the isosceles, right triangle AOC: (r-2)²*2 = r² ⟹ (r²-4r+4)*2 = r² ⟹ 2r²-8r+8 = r² |-r² r²-8r+8 = 0 |p-q-formula ⟹ r1/2 = 4±√(16-8) = 4±√8 ⟹ r1 = 4+√8 and r2 = 4-√8 < 2, but r must be larger than 2 ⟹ r1 = 4+√8 is correct. Angle BOA is 45° because of the isosceles, right triangle AOC and the section BOA is the 45°/360° = 1/8 part of the circle. ⟹ Green area = π*(4+√8)²/8-area of the isosceles, right triangle AOC = π*(16+8*√8+8)/8-(4+√8-2)²/2 = π*(24+8*√8)/8-(2+√8)²/2 = π*(3+√8)-(4+4*√8+8)/2 = π*(3+√8)-(6+2*√8) = π*(3+√8)-6-2*√8 ≈ 6,6537
CO is a perpendicular bisector. So you can extend AC to a point AD and you intersecting chords to solve for the unknown lengths CO and AC. OC and AC can make an isosceles right triangle with angles 45-45-90. Calculate the sector area for 45 degrees and subtract the isosceles triangle area to get the green shaded area.
π(2√2+3)-(4√2+6) I extend to the right the AC line to a colinear and mirrored straight line to a point D through the circle And extended down the BO line to a colinear and mirrored r line, to a point E AC = CO CO = r - 2 AC.CD = CE.BC (r-2)²=2(2r-2) r²-4r+4=4r-4 r²-8r+8=0 Completing squares, we have: r²-8r+16=8 (r-4)²=8 r-4=±√8 r=±√8+4 r=±2√2+4 r′=2√2+4 (ok) r″=-2√2+4 (not ok, since it is smaller than 2, i think) Then, triangle ACO area is: (r-2)².½ = 4√2+6 1/8 sector area is: (πr²)/8, then: [π(2√2+4)²]/8 = π(2√2+3)
STEP-BY-STEP RESOLUTION PROPOSAL : 01) OC = AC = X lin un 02) OB = OA = R = (X + 2) 03) X^2 + X^2 = R^2 ; 2X^2 = (X + 2)^2 ; 2X^2 = X^2 + 4X + 4 ; X^2 - 4X - 4 = 0 04) Two Solutions : X = (2 - 2*sqrt(2)) or X = (2 + 2*sqrt(2)). Let's choose the Positive Solution X = (2 + 2*sqrt(2)) lin un ; X ~ 4,83 lin un. 05) R = X + 2 ; R = (2 + 2*sqrt(2)) + 2 ; R ~ 6,83 lin un ; R^2 ~ 46,63 06) Sector [OAB] Area ~ 18,32 sq un 07) Triangle [AOC] Area ~ 11,66 sq un 08) Green Area = 18,32 - 11,66 ; Green Area = 6,66 Therefore, OUR BEST ANSWER (after correcting our mistakes inputing values in Wolfram Alpha) is : The Green Area is approx. equal to 6,66 Square Units. Best Regards from The Islamic Caliphate - Cordoba.
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Great explanation
🎉great video sir🙋🏻♂️thank you so much👍🏼
You bet😀
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Superb
(r- 2)^2 + (r- 2)^2 = r^2
r^2 - 8r + 8 = 0
r = 4 + 2✓2 or 4 - 2✓2 (not possible)
green area = (r^2)π/8 - [(r - 2)^2]/2 = π(4 + 2✓2)^2/8 - [(2 + 2✓2)^2]/2 = (3 + 2✓2)π - 6 - 4✓2 = (3 + 2✓2)(π - 2)
Let's start by labeling the radius of ⊙O as r.
So, CO = BO - BC = r - 2.
Since AC = CO, AC = r - 2.
Draw radius AO. △ACO is an isosceles right triangle.
r = (r - 2)√2
r = r√2 - 2√2
r - r√2 = -(2√2)
r(1 - √2) = -(2√2)
r = [-(2√2)]/(1 - √2)
Rationalize the denominator by multiplying the fraction by (1 + √2)/(1 + √2).
= [-(2√2) * (1 + √2)]/(1 - 2)
= [-(2√2) - 4]/(-1)
= 2√2 + 4
Green Region Area = Sector AOB Area - △ACO Area
Because △ACO is an isosceles right triangle, m∠AOB = 45°.
A = πr² * (θ/360°)
= π * (2√2 + 4)² * (45°/360°)
= π * (8 + 16√2 + 16) * (1/8)
= π * (24 + 16√2) * (1/8)
= π(3 + 2√2)
= 3π + (2π)√2
AC = CO = r - 2 = (4 + 2√2) - 2 = 2 + 2√2.
A = (bh)/2
= 1/2 * (2 + 2√2) * (2 + 2√2)
= 1/2 * (4 + 8√2 + 8)
= 1/2 * (12 + 8√2)
= 6 + 4√2
Green Region Area = [3π + (2π)√2] - (6 + 4√2)
= 3π - 6 + (2π - 4)√2
So, the area of the green shaded region is π(3 + 2√2) - 6 - 4√2 square units, a. w. a. 3π - 6 + (2π - 4)√2 square units (exact), or about 6.65 square units (approximation).
Spoiler (probably):
Video rewrite of the answer: (3 + 2√2)(π - 2) square units
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You are very welcome!
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❤️
AC=2(√2+1)..r=2(√2+2).. Agreen=Asett(45°)-AC^2/2=(π/8)r^2-2(3+2√2)=(π/8)4(6+4√2)-6-4√2=π(3+2√2)-2(3+2√2)=(π-2)(3+2√2)
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My way of solution ▶
we can find the radius of the circle
a) according to the Pythagorean theorem:
(r-2)²+(r-2)²= r²
r²-4r+4+r²-4r+4= r²
r²-8r+8=0
or: Intersecting chords theorem:
(r-2)*(r-2)= 2*[(r-2)+r]
(r-2)² = 2*(2r-2)
r²-4r+4= 4r-4
r²-8r+8=0
both equations will give the radius of the circle, lets take one:
r²-8r+8=0
Δ= 64-4*1*8
Δ= 32
√Δ= 4√2
r₁= (8-4√2)/2
r₁= 4-2√2
r₂= 4+√2
if r = 4-2√2, then r-2 would be equal to 2-2√2 < 0, that's why
r= 4+2√2
Agreen= π*r²*(α/360°) - 1/2*(r-2)²
r= 4+2√2
α= 45°
⇒
Agreen= π*(4+2√2)²*(45°/360°) - 1/2*(r-2)²
= π*(4+2√2)²*(1/8) - 1/2*(r²-4r+4)
= π*(16+16√2+8)*(1/8) - 1/2*[(4+2√2)²-4*(4+2√2) +4]
= π*(24+16√2)*(1/8) - 1/2*[(16+16√2+8- 16 -8√2 +4]
= π(3+2√2) - 1/2(8√2+12)
= π(3+2√2) - 4√2-6
= π(3+2√2) - 2(2√2+3)
Agreen= (3+2√2)[π-2] square units
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r²=2(r-2)²
-> r=2√2+4 -> a=2√2+2 -> a=4.8
-> θ=45°
sector area =
π/8(2√2+4)²=π(2√2+3)=5.8π
triangle area = a.a/2 = 11.7
green area = 5.8π-11.7 = 6.4u²
Let R is Radius of circle
AC=OC=a
So R=a+2 ; a=R-2
a^2=(R+a)(2)
(R-2)^2=2(R+R-2)
So R=4+2√2
Connect O to A
So right triangle isosrles AOC ∆
So AOC=45°
a=4+2√2-2=2+2√2
Area of the sector =45/360(π)((4+2√2)^2=π(3+2√2)
Area of right triangle=1/2(2+2√2)^2=6+4√2
Green area=π(3+2√2)-(6+4√2)=6.65 square units.❤❤❤v
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I calculated
r=a+2
r=4+2•Sqrt(2)
Area=r²((Pi-2)/8)
=(Pi-2)/8•[4+2•Sqrt(2)]²
=(Pi-2)/8•(24+16Sqrt(2))
=(Pi-2)•(3+2Sqrt(2))
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Hey I am confused with second step ????
..
And how u made tha formula
@@shalinisuryavanshi314
Area="Pizza slice"-triangle
Area=(45/360)•Pi•r²-1/2a²
Where a=Sqrt(2)/2•r, as shown.
Area=Pi/8r²-1/2(Sqrt(2)/2•r)²
=Pi•r²/8-2•r²/8
=r²((Pi-2)/8)
Call the double-stroked length L. By the intersecting chords theorem we can immediately write
L^2 = 2*(2*L+2)
L^2 = 4*L + 4
L^2 - 4*L - 4 = 0
L = 2 + 2*sqrt(2), L = 2 - 2*sqrt(2)
Looks from the problem like we need the larger one, so we will go with L = 2+2*sqrt(2). So the radius of the circle is R = 4+2*sqrt(2). Now for segment AO. The area of the sector AOB is (45/360)*pi*R^2, and the shaded area is that minus L^2/2. It comes out to 6.6537. Q.E.D.
AC=CO=x x * x = 2 * (x + x + 2) x² = 4x + 4 x² - 4x - 4 = (x - 2)² - 8 = 0
x - 2 = +-2√2 x > 0 , x = 2√2 + 2
radius is 2√2 + 4
Green area = (2√2 + 4)² * π * 45/360 - (2√2 + 2)² * 1/2
= (24 + 16√2)π/8 - (12 + 8√2)/2 = (3 + 2√2) π - (6 + 4√2) = (3 + 2√2) π - 2(3 + 2√2) = (3 + 2√2)(π - 2)
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I wish more people thought like engineers......
The ratio of the length BC to the radius is a constant (because AC=CO) AC represents a half of one side of the maximum square that will fit inside that circle. Had to do my own calculations to figure out the ratio, but we don't figure out π every time, could not for the life of me find the BC to radius ratio online, so simply applied random values (5) to lengths AC which equals CO, hence I know the length of two sides of right angle triangle so the radius is the square root of 50, so ratio is (6.8..... -5)/6.8..... =.2929...
That's the ratio ( ✓50-5)/✓50
So knowing the radius of the circle and the dimensions of the square with insultingly simple maths, the shaded area is 1/8 of (the area of the circle minus the area of the square).
Draw OA. As OA and OB are both radii of circle O, OA = OB = r. Let OC = CA = x, so as CB = 2, OB = r = x+2. So x = r-2.
Triangle ∆OCA:
OC² + CA² = OA²
x² + x² = r² = 2x²
r² = 2(r-2)² = 2r² - 8r + 8
r² - 8r + 8 = 0
r = (-(-8)±√(-8)²-4(1)(8))/2(1)
r = 4 ± √(64-32)/2 = 4 ± 2√2
r = 4 + 2√2 ---> r > 2
x = r - 2 = (4+2√2) - 2 = 2 + 2√2
As OC = CA, ∆OCA is an isosceles right triangle and ∠AOC = ∠CAO = (180°-90°)/2 = 45°. The green area is equal to the area of the sector covered by ∠AOC minus the area of the triangle ∆ OCA.
Green area:
Aɢ = (45°/360°)πr² - bh/2
Aɢ = π(4+2√2)²/8 - (2+2√2)²/2
Aɢ = (π/8)(16+16√2+8) - (4+8√2+8)/2
Aɢ = (3+2√2)π - (6+4√2)
Aɢ = (3+2√2)(π-2) ≈ 6.65 sq units
In triangle AOC: (R -2)^2 + (R -2)^2 = R^2, with R the radius of the circle. So R^2 -8.R + 8 = 0
Deltaprime = 8, so R = 4 -2.sqrt(2) which is rejected as inferior to 2, or R = 4 + 2.sqrt(2) which is O.K.
Then the area of triangle AOC is (1/2).(R -2)^2
=(1/2).(4 + 8 +8.sqrt(2)) = 6 +4.sqrt(2)
The area of sector AOB is Pi.(R^2).(45°/360°)
= Pi.(16 + 8 +16.sqrt(2)).(1/8) = (3 +2.sqrt(2)).Pi
Finally, the green area is: (3 +2.sqrt(2).Pi - (6 +4.sqrt(2)).
That was easy.
Thanks for the feedback ❤️
Let's find the area:
.
..
...
....
.....
First of all we apply the Pythagorean theorem to the right triangle OAC. With R being the radius of the circle we obtain:
OA² = OC² + AC² = 2*OC²
R² = 2*(R − 2)²
R² = 2*R² − 8*R + 8
0 = R² − 8*R + 8
R = 4 ± √(4² − 8) = 4 ± √(16 − 8) = 4 ± √8 = 4 ± 2√2
Since R>2, the only useful solution is R=4+2√2. The triangle OAC is also an isosceles triangle. So we can conclude:
∠AOC = ∠OAC = (180° − ∠OCA)/2 = (180° − 90°)/2 = 90°/2 = 45°
Now we are able to calculate the area of the green region:
A(green)
= A(circle sector OAB) − A(triangle OAC)
= πR²*(∠AOC/360°) − (1/2)*OC*AC
= πR²*(45°/360°) − (1/2)*(R − 2)*(R − 2)
= πR²/8 − (1/2)*(R²/2)
= πR²/8 − R²/4
= (π/2 − 1)*R²/4
= (π/2 − 1)*(4 + 2√2)²/4
= (π/2 − 1)*(16 + 16√2 + 8)/4
= (π/2 − 1)*(24 + 16√2)/4
= (π/2 − 1)*(6 + 4√2)
≈ 6.654
Best regards from Germany
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Solution:
r = radius of the circle.
Pythagoras for the isosceles, right triangle AOC: (r-2)²*2 = r² ⟹
(r²-4r+4)*2 = r² ⟹
2r²-8r+8 = r² |-r²
r²-8r+8 = 0 |p-q-formula ⟹
r1/2 = 4±√(16-8) = 4±√8 ⟹
r1 = 4+√8 and r2 = 4-√8 < 2, but r must be larger than 2 ⟹ r1 = 4+√8 is correct.
Angle BOA is 45° because of the isosceles, right triangle AOC and the section BOA is the 45°/360° = 1/8 part of the circle. ⟹
Green area = π*(4+√8)²/8-area of the isosceles, right triangle AOC
= π*(16+8*√8+8)/8-(4+√8-2)²/2 = π*(24+8*√8)/8-(2+√8)²/2
= π*(3+√8)-(4+4*√8+8)/2 = π*(3+√8)-(6+2*√8) = π*(3+√8)-6-2*√8 ≈ 6,6537
Seems not difficult😮, 2(r-2)^2=r^2, r^2-8r+8=0, the area is 1/8 r^2 pi-1/4r^2=1/8 r^2(pi-2)... 😊
S=(3+2√2)(π-2)=(√2+1)²(π-2)≈6,66
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Thanks for sharing ❤️
CO is a perpendicular bisector. So you can extend AC to a point AD and you intersecting chords to solve for the unknown lengths CO and AC. OC and AC can make an isosceles right triangle with angles 45-45-90. Calculate the sector area for 45 degrees and subtract the isosceles triangle area to get the green shaded area.
Thanks for the feedback ❤️
@@PreMath Even though I am well past my geometry years, I enjoy your math puzzles each morning. Thanks.
@@bakrantz
Thanks my dear friend. Stay blessed🙏
Nice alternate method. 👍
π(2√2+3)-(4√2+6)
I extend to the right the AC line to a colinear and mirrored straight line to a point D through the circle
And extended down the BO line to a colinear and mirrored r line, to a point E
AC = CO
CO = r - 2
AC.CD = CE.BC
(r-2)²=2(2r-2)
r²-4r+4=4r-4
r²-8r+8=0
Completing squares, we have:
r²-8r+16=8
(r-4)²=8
r-4=±√8
r=±√8+4
r=±2√2+4
r′=2√2+4 (ok)
r″=-2√2+4 (not ok, since it is smaller than 2, i think)
Then, triangle ACO area is:
(r-2)².½ = 4√2+6
1/8 sector area is:
(πr²)/8, then:
[π(2√2+4)²]/8 = π(2√2+3)
Solution:
By Chords Theorem
(r - 2) . (r - 2) = 2 . (2r - 2)
r² - 4r + 4 = 4r - 4
r² - 8r + 8 = 0
r = 8 ± √32/2
r = 8 ± 4√2/2
r = 4 ± 2√2
r = 4 + 2√2 Accepted
r = 4 - 2√2 Rejected
Green Area (GA) = Sector Area - Triangle Area ... ¹
Sector Area = 45°/360° π (4 + 2√2)²
Sector Area = ⅛ π (16 + 16√2 + 8)
Sector Area = ⅛ π (24 + 16√2)
Sector Area = π (3 + 2√2)
a = r - 2 = 4 + 2√2 - 2
a = 2 + 2√2
Triangule Area = ½ a²
Triangule Area = ½ (2 + 2√2)²
Triangule Area = ½ (4 + 8√2 + 8)
Triangule Area = ½ (12 + 8√2)
Triangule Area = 6 + 4√2
GA = π (3 + 2√2) - (6 + 4√2)
GA = π (3 + 2√2) - 2 (3 + 2√2)
GA = (π - 2) . (3 + 2√2) Square Units
GA ~= 6,654 Square Units
Guess I'll have too bake an upside down cake today. 🙂
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STEP-BY-STEP RESOLUTION PROPOSAL :
01) OC = AC = X lin un
02) OB = OA = R = (X + 2)
03) X^2 + X^2 = R^2 ; 2X^2 = (X + 2)^2 ; 2X^2 = X^2 + 4X + 4 ; X^2 - 4X - 4 = 0
04) Two Solutions : X = (2 - 2*sqrt(2)) or X = (2 + 2*sqrt(2)). Let's choose the Positive Solution X = (2 + 2*sqrt(2)) lin un ; X ~ 4,83 lin un.
05) R = X + 2 ; R = (2 + 2*sqrt(2)) + 2 ; R ~ 6,83 lin un ; R^2 ~ 46,63
06) Sector [OAB] Area ~ 18,32 sq un
07) Triangle [AOC] Area ~ 11,66 sq un
08) Green Area = 18,32 - 11,66 ; Green Area = 6,66
Therefore,
OUR BEST ANSWER (after correcting our mistakes inputing values in Wolfram Alpha) is :
The Green Area is approx. equal to 6,66 Square Units.
Best Regards from The Islamic Caliphate - Cordoba.
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Thanks for sharing ❤️