Mathematics Question and Solution
a = ½(10+6)
a = 8 units.
a is the radius of the ascribed half circle.
b = a-6
b = 8-6
b = 2 units.
b is OB.
Observing similar triangles side lengths ratio.
b - (10-c)
(6-c) - b
Cross Multiply.
b² = (6-c)(10-c)
b² = 60-16c+c²
b = √(60-16c+c²) units.
d = (8-c) units.
It implies;
8² = d²+b²
64 = (8-c)²+√(60-16c+c²)²
64 = 64-16c+c²+60-16c+c²
0 = 60-32c+2c²
c²-16c+30 = 0
Resolving the above quadratic equation via completing the square approach to get c.
(c-8)² = -30+(8)²
(c-8)² = 34
c = 8±√(34)
It implies;
c ≠ 8+√(34)
c = 8-√(34)
c = 2.1690481052 units.
And;
b = √(60-16c+c²)
b = √(60-16*2.1690481052+2.16904810522)
b = 5.477225575 units.
e = 6-c
e = 6-2.1690481052
e = 3.8309518948 units.
f² = e²+b²
f² = 3.8309518948²+5.477225575²
f = 6.684025166 units.
f is BD.
g = 12-c
g = 10-2.1690481052
g = 7.8309518948 units.
h² = g²+b²
h² = 7.8309518948²+5.477225575²
h = 9.5563490716 units.
h is BC.
Therefore, area green is;
½*fh
= 0.5*6.684025166*9.5563490716
= 31.9374388448 square units.
