Can you find area of the Blue portion? | (Fun Geometry Problem) |

[@marioalb9726]

Radius of semicircle :
R = 9/2 = 4,5 cm
Internal angles :
cos2α=7/9 --> α=19,47° --> sinα=1/3
Chord QR and RS:
n = 2R sinα = 9/3 = 3 cm
Circular segment QR :
A₁ = ½R²(β-sinβ) = ½ 4,5² (2α-sin2α)
A₁ = 0,517736 cm²
Triangle PQR:
A₂ = ½ m.n.sin(90°-α) = ½*7*3*cosα
A₂ = 9,8995 cm²
Blue shaded area:
A = A₁+ A₂ = 10,4172 cm² ( Solved √ )

[@MathandEngineering]

@@marioalb9726 this is Excellent friend. Thanks

[@santiagoarosam430]

QS=4√2---> sen(2a)=4√2/9---> 2a=38,94º.
Área azul = (πr²/2)-[(πr²/2)-2*Sector(2a)]-[Sector(2a)]+[r²*sen(4a)/2]-[r²*sen(2a)/2] =[Sector(2a)]+[(r²/2)*(sen(4a}-sen(2a)] =10,4156 cm².
Gracias y saludos

[@MathandEngineering]

@@santiagoarosam430 ok friend I see you found. Shorter solution, thanks for sharing, it's simple and perfect

[@murdock5537]

φ = 30° → sin⁡(3φ) = 1; ∆ PSQ → sin⁡(SQP) = 1; PQ = 7; PS = PO + SO = r + r = 9 → QS = 4√2
∆ PSR → PS = 9; sin⁡(SRP) = 1; QPR = RPS = δ → QPS = 2δ →
sin⁡(2δ) = 4√2/9 → cos⁡(2δ) = 7/9 → sin⁡(δ) = √((1/2)(1 - cos⁡(2δ))) = 1/3 →
∆ QOR → QO = RO = r = 9/2 → (1/2)sin⁡(2δ)(81/4) = 9√2/2 = area ∆ QOR
area ∆ PRQ = (1/2)sin⁡(δ)7(6√2) = 7√2
cos⁡(2δ) = 7/9 > √2/2 = sin⁡(3φ/2) → θ = 3φ/2 - 2δ →
sin⁡(θ) = sin⁡(3φ/2 - 2δ) = sin⁡(3φ/2)cos⁡(2δ) - sin⁡(2δ)cos⁡(3φ/2) = (√2/18)(7 - 4√2) ≈ θ = 0,1055275 →
θ(180°/π) = 6,0462° → 2δ = 38,9538° → blue area = 5√2/2 + (δ/6φ)(81π/4) ≈ 10.42
or with π ≈ 22/7; 3φ/2 - θ = 2δ →
3φ/2 - (√2/18)(7 - 4√2)(6φ/π) = 3φ/2 - (√2/18)(7 - 4√2)(42φ/7) = (1/66)(155 - 49√2)φ ∶= k = 2δ
πr^2 = (22/7)(81/4) = (81/2)(11/7) → πr^2(k/12φ) = (81/2)(11/7)(155 - 49√2)(1/66)(1/12) →
blue shaded area = πr^2(k/12φ) + 7√2 - 9√2/2 = (1/112)(1395 - 161√2) ≈ 10,422

[@MathandEngineering]

@murdock5537 friend this is perfect, thanks for sharing

[@juanalfaro7522]

I obtained Area=8.90619. Probably I made an arithmetic error somewhere.

[@MathandEngineering]

@@juanalfaro7522 ok, did you find it?