Mathematics Question and Solution
Calculating the circumference of the blue inscribed circle.
Let x be the base of the inscribed right-angled triangle with height 10 units.
x²+10² = a²
a = √(x²+100) units.
b = ½(a)
b = ½√(x²+100) units.
Calculating x.
10² = 10²+(½√(x²+100))²-10√(x²+100)cosy
10√(x²+100)cosy = ¼(x²+100) --- (1).
cosy = x/√(x²+100) --- (2).
Substituting (2) in (1).
10√(x²+100)*x/√(x²+100) = ¼(x²+100)
10x = ¼(x²+100)
40x = x²+100
x²-40x+100 = 0
(x-20)² = -100+(-20)²
(x-20)² = 300
x = 20±√(300)
x = 20±10√(3)
It implies;
x ≠ 20+10√(3) units.
x = 20-10√(3) units.
x = 2.67949192431 units.
Recall.
b = ½√(x²+100) units.
And x = 2.67949192431 units.
b = and 0.5√(2.67949192431²+100)
b = 5.17638090205 units.
sinc = (0.5*5.17638090205/10)
c = 15°
d = 90-2c
d = 90-2(15)
d = 60°
e = (10-f) units.
Where f is the radius of the blue inscribed circle.
sin(0.5*60) = f/(10-f)
½ = f/(10-f)
2f = 10 -f
3f = 10
f = ⅓(10) units.
Again, f is the radius of the blue inscribed circle.
It implies;
Circumference of the blue inscribed circle is;
2πf
= 2*π(10/3)
= ⅓(20π) units.
