Mathematics Question and Solution
Let a be the diameter of the smaller inscribed circle.
2a is the radius of the ascribed semi circle.
(2a)² = (0.5*3)²+b²
b² = 4a²-2.25
b = √(4a²-2.25) units.
(2a)² = (0.5*5)²+c²
c² = 4a²-6.25
c = √(4a²-6.25) units.
c = a
a = √(4a²-6.25)
a² = 4a²-6.25
6.25 = 3a²
a² = 6.25/3
a = √(25/12) units.
a = 5/2√(3)
a = ⅙(5√(3)) units.
a = 1.443375673 units.
Again, a is the diameter of the smaller inscribed circle.
c = ½(a)
c = ½*⅙(5√(3))
c = (5√(3))/12 units.
c = 0.7216878365 units.
c is the radius of the smaller inscribed circle.
Notice.
e = a+d
e is the diameter of the bigger inscribed circle.
Therefore;
a+d = √(4a²-2.25)
And a = ⅙(5√(3)) units.
⅙(5√(3))+d = √(4(⅙(5√(3)))²-2.25)
(⅙(5√(3))+d)² = 4(⅙(5√(3)))²-2.25
(25/12)+⅓(5√(3))d+d² = 4(25/12)-(9/4)
d²+⅓(5√(3)d)-(75/12)+(9/4) = 0
d²+⅓(5√(3))d-(48/12) = 0
12d²+20√(3)d-48 = 0
3d²+5√(3)d-12 = 0
Resolving the above quadratic equation via completing the square approach to get d.
d²+⅓*5√(3)d = 4
d²+⅙*5√(3)d = 4
(d+(5√(3)/6))² = 4+(5√(3)/6)²
(d+(5√(3)/6))² = 4+(25/12)
(d+(5√(3)/6))² = (48+25)/12
(d+(5√(3)/6))² = (73/12)
d+(5√(3)/6) = ±√(73/12)
d+(5√(3)/6) = ±⅙√(219)
d = -(5√(3)/6)±(⅙√(219))
It implies;
d ≠ -(5√(3)/6)-(⅙√(219))
d = -(5√(3)/6)+(⅙√(219))
d = ⅙(√(219)-5√(3)) units.
d = 1.0230657582 units.
Therefore;
e = a+d
e = ⅙(5√(3))+⅙(√(219)-5√(3))
e = ⅙√(219) units.
Again, e is the diameter of the bigger inscribed circle.
f = ½(e)
f = ½(⅙√(219))
f = √(219)/12 units.
f = 1.2332207156 units.
f is the radius of the bigger inscribed circle.
Therefore, area yellow is;
Area bigger inscribed circle with radius √(219)/12 units - Area smaller inscribed circle with radius (5√(3))/12 units.
= π(√(219)/12)²-π((5√(3))/12)²
= π((219/144)-(75/144))
= π(144/144)
= π square units.
