Mathematics Question and Solution
Calculating Blue Area.
Let 2x units be the radius of the ascribed quarter circle.
½(2x) = x units.
x is the radius of the inscribed yellow (20 square units) and white half circle.
Calculating x.
a = ½(90)
a = 45°
b² = 2x²
b = √(2)x units.
b is the radius of the inscribed yellow (20 square units) and white quarter circle.
It implies;
Area quarter circle with radius x units + Area sector with radius √(2)x units and angle 45° - Area isosceles right-angled triangle with equal side lengths x units = 20
(¼(x²)π)+(45π(√(2)x)²÷360)-(½(x)²) = 20
¼(πx²)+⅛(2πx²)-½(x²) = 20
¼(πx²)+¼(πx²)-½(x²) = 20
πx²+πx²-2x² = 80
2(πx²-x²) = 80
πx²-x² = 40
x²(π-1) = 40
x = √(40/(π-1)) units.
Again, x is the radius of the inscribed yellow (20 square units) and white half circle.
It implies 2x is;
2√(40/(π-1)) units.
It is the radius of the ascribed quarter circle.
Recall.
b = √(2)x units.
And x = √(40/(π-1)) units.
b = √(80/(π-1)) units.
Again, b is the radius of the inscribed yellow (20 square units) and white quarter circle.
It implies;
Area Blue is;
Area quarter circle with radius 2√(40/(π-1)) units - Area quarter circle with radius √(40/(π-1)) units - Area sector radius √(80/(π-1)) units and angle 45° - Area isosceles right-angled triangle with equal sides √(40/(π-1)) units.
= (¼(2√(40/(π-1)))²π)-(¼(√(40/(π-1)))²π)-(⅛(√(80/(π-1)))²π)-(½(√(40/(π-1)))²)
= 20 square units.
Therefore;
Inscribed yellow area = Inscribed blue area.
