Mathematics Question and Solution
Let the radius of the red inscribed circle be r.
Calculating r.
2a² = 6²
a = √(18)
a = 3√(2) units.
2b² = 8²
b = √(32)
b = 4√(2) units.
c² = 6²+8²
c = √(36+64)
c = 10 units.
10² = 2d²
d² = 50
d = √(50)
d = 5√(2) units.
d is the radius of the yellow ascribed semi circle.
(5√(2))² = e²+(4√(2))²
e² = 50-32
e = √(18)
e = a = 3√(2) units.
f = b-e
f = 4√(2)-3√(2)
f = √(2) units.
f is length OD.
g = d-r
g = (5√(2)-r) units.
h² = 2r²
h = √(2)r units.
It implies;
g² = h²+f²
(5√(2)-r)² = (√(2)r)²+(√(2))²
50-10√(2)r+r² = 2r²+2
r²+10√(2)r-48 = 0
Resolving the quadratic equation via completing the square approach.
(r+5√(2))² = 48+(5√(2))²
(r+5√(2))² = 98
r = -5√(2)±√(98)
r = -5√(2)±7√(2)
Therefore;
r ≠ -5√(2)-7√(2)
r = -5√(2)+7√(2)
r = 2√(2) units.
Again, r is the radius of the red inscribed circle.
Therefore, area red inscribed circle is;
πr²
= π(2√(2))²
= 8π square units.
= 25.1327412287 square units.
