Mathematics Question and Solution
Calculating area of the inscribed red circle.
Let the bigger circle radius be a.
Let the big circle radius be b.
It implies;
a+b = 6 --- (1).
½(b²)+½(a²)+14 = 6*a
b²+a²+28 = 12a --- (2).
At (1).
b = 6-a --- (3).
Therefore, substituting (3) in (2) to get a.
(6-a)²+a²+28 = 12a
36-12a+a²+a²+28 = 12a
2a²-24a+64 = 0
a²-12a+32 = 0
(a-6)² = -32+(-6)²
(a-6)² = 4
a = 6±2
Therefore;
a = 6-2
a = 4 units.
Again, a is the radius of the bigger circle.
At (3).
b = 6-a
And a = 4 units.
b = 6-4
b = 2 units.
Again, b is the radius of the big circle.
c = a+2b
c = 8 units.
d = c+a
d = 12 units.
e² = 2a²
e² = 2(4²)
e = 4√(2) units.
f² = 4²+8²
f² = 80
f = √(80)
f = 4√(5) units.
It implies;
12g+4√(5)g+4√(2)g = 12*4
(12+4√(5)+4√(2))g = 12*4
(3+√(2)+√(5))g = 12
g = 12/(3+√(2)+√(5)) units.
g = 1.8044348842 units.
g is the radius of the red inscribed circle.
Area inscribed red circle is;
πg²
= π(12/(3+√(2)+√(5)))²
= 10.2289793452 square units.
