Mathematics Question and Solution
Let the blue square side length be x.
2² = a²+b²
b² = 4-a²
b = √(4-a²)
4² = a²+(√(4-a²)+x)²
16 = a²+((4-a²)+2x√(4-a²)+x²
12 = 2x√(4-a²)+x² --- (1).
5² = (x+a)²+(4-a²)
25 = x²+2xa+a²+4-a²
21 = x²+2xa
a = (21-x²)/(2x) --- (2).
Calculating x. Substituting (2) and (1)
12 = 2x√(4-((21-x²)/(2x))²)+x²
12 = 2x√(4-((441-42x²+x⁴)/(4x²))+x²
(12-x²)² = 4x²((58x²-x⁴-441)/4x²)
144-24x² +x⁴ = 58x²-x⁴-441
2x⁴-82x²+585 = 0
Let x² = p.
It implies;
2p²-82p+585 = 0
Resolving the above quadratic equation.
(p-(41/2))² = (-585/2)+(1681/4)
(p-(41/2))² = (511/4)
p-(41/2) = ±½√(511)
p = ½(41±√(511))
Notice.
p ≠ ½(41+√(511))
p = ½(41-√(511))
And p = x², where x² is the area of the blue square.
x = √(p)
x = √(½(41-√(511))) units.
It implies;
Area of the blue square (x²) is;
p = x²
x² = ½(41-√(511)) square units.
x² = 9.1973454445 square units exactly in decimal.
