Mathematics Question and Solution
Let the side length of the square be a.
b = ½(a) cm.
b is the radius of the inscribed half circle.
c = (a-2) cm.
d²+2² = (a-2)²
d² = a²-4a+4-4
d = √(a²-4a) cm.
d is the radius of the bigger inscribed quarter circle.
e = b+d
e = (½(a)+√(a²-4a)) cm.
Calculating a.
a²+(½(a))² = (½(a)+√(a²-4a))²
¼(5a²) = ¼(a²)+a√(a²-4a)+a²-4a
4a = a√(a²-4a)
4 = √(a²-4a)
a²-4a-16 = 0
(a-2)² = 16+(-2)²
(a-2)² = 20
a = 2±√(20)
a = 2±2√(5)
It implies;
a ≠ 2-2√(5)
a = 2+2√(5) cm.
a = 2(1+√(5)) cm
a = 6.472135955 cm.
Again, a is the side length of the ascribed square.
Recall.
b = ½(a)
And a = 2(1+√(5)) cm.
b = ½(2(1+√(5)))
b = (1+√(5)) cm.
b = 3.2360679775 cm.
Again, b is the radius of the inscribed half circle.
f = a-2
f = 2+2√(5)-2
f = 2√(5) cm.
g²+2² = (2√(5))²
g² = 20-4
g = √(16)
g = 4 cm.
g = d, the radius of the bigger inscribed quarter circle.
h = a-d
h = 2+2√(5)-4
h = (2√(5)-2) cm.
h = 2.472135955 cm.
h is the radius of the big inscribed quarter circle.
It implies;
Area Painted is;
Area square with side length (2+2√(5)) cm - Area half circle with radius (1+√(5)) cm - Area quarter circle with radius 4 cm - Area quarter circle with radius (2√(5)-2) cm.
= (2+2√(5))²-(½*π(1+√(5))²)-(¼*π(4)²)-(¼*π(2√(5)-2)²)
= 4+8√(5)+20-(½*π(1+2√(5)+5)-4π-(¼*π(20-8√(5)+4)
= 24+8√(5)-(½*π(6+2√(5))-4π-(¼*π(24-8√(5))
= 24+8√(5)-(π(3+√(5))-4π-(π(6-2√(5))
= 24+8√(5)-3π-√(5)π-4π-6π+2√(5)π
= (24+8√(5)-13π+√(5)π) cm²
= (8(3+√(5))-π(13-√(5))) cm²
= 8.0726540544 cm²
