Mathematics Question and Solution
Let a be the diameter of the big circle.
Radius = ½(a).
Let b be the diameter of the small inscribed blue circle.
Radius = ½(b).
Diameter of the semi circle is;
2+16+6 = 24 units.
It radius = ½(24) = 12 units.
It implies;
12² = (12-b)²+(½(a))²
144 = 144-24b+b²+¼(a²)
24b-b² = ¼(a²)
96b-4b² = a² --- (1).
(½(a))² = (½(16))²+c²
¼(a²) = 64+c²
4c² = a²-256
c = ½(√(a²-256)) --- (2).
d = 12-8-2
d = 2 units.
(12-b)² = 2²+c²
144-24b+b² = 4+(½(√(a²-256)))²
140-24b+b² = ¼(a²-256)
560-96b+4b² = a²-256
a² = 816-96b+4b² --- (3).
Calculating b.
Substituting (3) in (1) or equating (3) and (1).
a² = a²
96b-4b² = 816-96b+4b²
0 = 816-192b+8b²
8b²-192b+816 = 0
Resolving the above quadratic equation via completing the square approach to get b.
8b²-192b+816 = 0 (dividing through by 8, the coefficient of b²).
b²-24b+102 = 0
b²-24b= -102
b²-12b = -102
(b-12)² = -102+(-12)²
(b-12)² = -102+144
b-12 = ±√(42)
b = 12±√(42)
It implies;
b ≠ 12+√(42)
b = 12-√(42) units.
b is the diameter of the small inscribed blue circle.
Therefore radius, r of the small inscribed blue circle is;
r = ½(b)
r = ½(12-√(42)) units.
r = 2.7596296508 units.
It implies;
Area small inscribed blue circle is;
πr²
= π(2.7596296508)²
= 23.9249741843 square units.
= 23.92 square units to 2 decimal place.
