Mathematics Question and Solution
a = (10-x) cm.
10 cm is the radius of the ascribed semi circle.
12² = a²+b²
b² = 144-(100-20x+x²)
b² = 44+20x-x²
b = √(44+20x-x²) cm.
10² = x²+b²
100 = x²+44+20x-x²
56 = 20x
14 = 5x
x = (14/5) cm.
x = 2.8 cm.
Let r be the radius of the inscribed orange circle.
c = r-x
c = (r-(14/5)) cm.
d = (10-r) cm.
It implies;
d² = r²+c²
(10-r)² = r²+(r-(14/5))²
100-20r+r² = r²+r²-(28/5)r+(196/25)
100-20r = r²-(28/5)r+(196/25)
2500-500r = 25r²-140r+196
25r²+360r-2304 = 0
Resolving the above quadratic equation via completing the square approach to get r, radius of the inscribed orange circle.
r²+(180/25)r = (2304/25)+(180/25)²
(r+(180/25))² = 144
r+(180/25) = ±√(144)
r = -(180/25)±12
It implies;
r = 12-(180/25)
r = (120/25)
r = ⅕(24) cm.
r = 4.8 cm.
Again, r is the radius of the inscribed orange circle.
Area inscribed orange circle is;
πr²
= π(24/5)²
= (576π/25) cm²
= 72.3822947387 cm²
