Solution: a = side of the square inside the semicircle, r = radius of the semicircle = 10/2 = 5, x = side of the blue square. Pythagoras: (a/2)²+a² = r² = 5² ⟹ a²/4+a² = 5/4*a² = 25 |*4/5 ⟹ a² = 20 |√() ⟹ a = √20 ⟹ Pythagoras: (r-a/2)²+a² = x² ⟹ (5-√20/2)²+√20² = x² ⟹ 25-5*√20+20/4+20 = x² ⟹ 50-5*√20 = x² = area of the blue square ≈ 27.6393[cm²]
Great vid
Solution:
a = side of the square inside the semicircle,
r = radius of the semicircle = 10/2 = 5,
x = side of the blue square.
Pythagoras:
(a/2)²+a² = r² = 5² ⟹
a²/4+a² = 5/4*a² = 25 |*4/5 ⟹
a² = 20 |√() ⟹
a = √20 ⟹
Pythagoras:
(r-a/2)²+a² = x² ⟹
(5-√20/2)²+√20² = x² ⟹
25-5*√20+20/4+20 = x² ⟹
50-5*√20 = x² = area of the blue square ≈ 27.6393[cm²]
Did it after bottle of prosecco 🤪
Let the side of the blue square be S. Let FG = x.
(5-x)^2 + 2*(5 - x)^2 = 5^2
(5 - x)^2 = 25/3
x = 5 - (5/√3)
x^2 + 2*(5 - x)^2 = S^2