You are really a genius❤❤❤

Love your channel please never stop posting videos

I'm impressed how the math worked out!

Let R is Radius of semicircle and x is side length of the square R^2=x^2+x^2 so R=√2x √2x/√3x=(2+√3x)/2√2x So x=2√3 Green area=(2√3)^2=12 square units.❤

Si "a" es el lado del cuadrado, "P" es el vértice superior derecho del cuadrado y MD=r=Radio del semicírculo =Diagonal del cuadrado =a√2=r → PD=√[a²+(a√2)²] =a√3 → Potencia de P respecto a la circunferencia =2*a√3 =(r-a)(r+a)=r²-a²=(a√2)²-a²=a²→ a²-2a√3=0→ a=2√3→ Área verde =a²=4*3=12 ud². Gracias y saludos.

This is really an amazing question, but soluble 🤭 Let's check the answer ▶ The center of this semicircle is M, AM= MB= r We draw the radius to the line where it couches the circumsference of the semicircle, this point is D MD= AM= r if we combine point A und D we get an isosceles triangle: ΔAMD ∠ ADM= α ∠ DAM= α the diagonal of the square is equal to the radius of the semicircle: a²+a²= r² 2a²= r² a²= r²/2 a= r/√2 a= √2r/2 this is the relation between the side of the square (a) and the radius of the semicircle ! ∠ ADB will be equal to 90° according to the Thales's theorem, while all triangles circumscribed by a semicircle are right-angled ❗ ∠ ADB= 90° In the triangle ΔMBD ∠ BDM= 90-α = β while DM= MB= r ∠ MBD= β in the triangle ΔMBC: ∠ CMB= 90° ∠ MBC= β α+β=90° ⇒ ∠ BCM= α, and: MB= r CM= a ⇒ tan(β)= CM/MB tan(β)= a/r a= √2r/2 ⇒ tan(β)= √2r/2 / r tan(β)= √2/2 β= arctan(√2/2) β≈ 35,2644° α≈ 54,7356° while ∠ BCM= α and the angle ∠ MCE= 90° (in the square) ∠ FCD= 180°-90°-α ∠ FCD= 90° - α ∠ FCD= β ⇒ ΔFCD is an obtuse-angled isosceles triangle: CD= 2 FC= DF a) Let's consider the triangle ΔDMC: ∠ MCD= 90°+ β ∠ MCD= 125,2644° ∠ CDM= β ∠ CDM= 35,2644° ∠ DMC= 180°-(125,2644°+35,2644°) ∠ DMC= 19,4712° according to the sine theorem: sin(∠ DMC)/CD = sin(β)/MC CD= 2 MC= a sin(19,4712°)= 0,3333 sin(35,2644°)= 0,5774 ⇒ 0,3333/2= 0,5774/a a= 3,4644 length units Agreen= a² Agreen= 3,4644² Agreen= 12,00 square units ✅ b) if we consider the triangle ΔFCD : CD= 2 FC= DF ∠ CDF= β = 35,2644° ∠ FCD= β= 35,2644° ∠ DFC= 180°- 2β ∠ DFC= 109,4712° according to the sine theorem: sin(β)/FC = sin(DFC)/CD CD= 2 sin(β)= 0,5774 sin(109,4712°)= 0,9428 ⇒ 0,5774/FC= 0,9428/2 FC= 1,2249 DF= FC ⇒ DF= 1,2249 MD=r MF= r-DF MF= r-1,2249 Let's consider the triangle ΔMCF and apply the Pythagorean theorem: MC²+CF²= MF² MF= r-1,2249 CF= 1,2249 MC= a a= √2r/2 ⬆ ⇒ (r-1,2249)²= (√2r/2)²+1,2249² r²-2,4498r+1,50038 = r²/2 + 1,50038 r²/2 - 2,4498r = 0 r² - 4,8996 r = 0 r(r-4,8996)= 0 r₁ = 0 which can't be the case ❗ r-4,8996=0 r= 4,8996 ✅ a= √2r/2 a= (√2/2)* 4,8996 a= 3,46454 Agreen= a² Agreen= 3,46454² Agreen= 12,00 square units ✅

a = r/√2 = r√2/2; x = a√3 = r√6/2; r + r√2/2 = (r/2)(2 + √2) → r - r√2/2 = (r/2)(2 - √2) → (r/2)(2 + √2)(r/2)(2 - √2) = r√6 → r = 2√6 → a = 2√3 → a^2 = 12