Question
Solving it using GeoGebra
Area of the triange is $$0.04 x^2$$ where $$x$$ is the length of the side of the square.
Solving it without GeoGebra
Let the Quarter Circle ACD be $$C_1$$
Let the Semi Circle with Diameter along AB be $$C_2$$
Let the Semi Circle with Diameter along BC be $$C_3$$
$$ C_1: x^2 + y^2 = 1 $$
$$ C_2: (x-0.5)^2 + (y-1)^2 = 0.25 $$
$$ C_3: (x-1)^2 + (y-0.5)^2 = 0.25 $$
To Find $$C_1$$ intersection $$C_2$$
$$1: x^2 + y^2 = 1 $$
$$2: x^2 + 0.25 - x + y^2 + 1 - 2y = 0.25 $$
$$1-2: x + 2y - 1.25 = 0.75 $$
$$1-2: x + 2y = 2 $$
$$ L_1: x + 2y = 2 $$ is a line passing through A and G
Substituting $$ L_1 $$ in $$C_1$$ to find A and G
$$ 4 + 4y^2 - 8y + y^2 = 1 $$
$$ 5y^2 - 8y + 3 = 0 $$
$$ 5y^2 - 5y - 3y + 3 = 0 $$
$$ 5y(y-1) - 3(y-1) = 0 $$
$$ (5y-3)(y-1) = 0 $$
$$ A: (0, 1) $$
$$ G: (0.8, 0.6) $$
To Find $$C_1$$ intersection $$C_3$$
$$ 3: x^2 + y^2 = 1 $$
$$ 4: x^2 + 1 - 2x + y^2 + 0.25 - y = 0.25 $$
$$ 3-4: 2x + y - 1.25 = 0.75 $$
$$ 3-4: 2x + y = 2 $$
$$ L_2: 2x + y = 2 $$ is a line passing through F and C
Finding C and F by substituting $$L_2$$ in $$C_1$$
$$ C: (1, 0) $$
$$ F: (0.6, 0.8) $$
To Find $$C_2$$ intersection $$C_3$$
$$ 5: x^2 + 0.25 - x + y^2 + 1 - 2y = 0.25 $$
$$ 6: x^2 + 1 - 2x + y^2 + 0.25 - y = 0.25 = 0.25 $$
$$ 5-6: x - y = 0 $$
$$ L_3: x - y = 0 $$ is a line passing through E and B
Finding B and E by substituting $$L_3$$ in $$C_3$$
$$ B: (0, 0) $$
$$ E: (0.5, 0.5) $$
The triangle in the shaded region is EFG, where EF = EG
Base of EFG is FG = sqrt(0.08)
Height of EFG is = sqrt(0.08)
Area of EFG is 0.5 * base * height which is 0.04