Mathematics Question and Solution
a = ½(10)
a = 5 cm.
a is the radius of the ascribed semi circle.
b = (5-c) cm.
c is the radius of the inscribed blue circle.
It implies;
(5-c)² = c²+d²
25-10c+c² = c²+d²
d² = 25-10c
d = √(25-10c) cm.
e = 5+d
e = (5+√(25-10c)) cm.
Therefore;
calculation c, radius of the blue inscribed circle.
tan15 = c/(5+√(25-10c))
0.2679491924(5+√CT(25-10c)) = c
c-1.3397459622 = 0.2679491924√(25-10c)
(3.732050808c-5)² = 25-10c
13.9282032335c²-37.32050808c+25 = 25-10c
13.9282032335c² = 27.32050808c
13.9282032335c = 27.32050808
c = 1.9615242269 cm.
Again, c is the radius of the blue inscribed circle.
Area Blue Circle is;
πr²
= π(1.9615242269)²
= 12.0875205572 cm²
