Mathematics Question and Solution
Let the radius of the small yellow inscribed circle be a.
Let the radius of the big green inscribed circle be b.
Notice!
PQ = a+b
Calculating radius of the ascribed semi circle.
c² = 12²+16²
c = √(144+256)
c = √(400)
c = 20 units.
c is the diameter of the ascribed semi circle.
d = ½(c)
d = ½(20)
d = 10 units.
d is the radius of the ascribed semi circle.
tane = 16/12
e = atan(4/3)°
e is angle BAC.
sin(atan(4/3)) = f/12
4/5 =f/12
f =48/5
f = 9.6 units.
cos(atan(4/3)) = g/12
3/5 = g/12
g = 36/5
g = 7.2 units.
h² = a²+9.6²
h = √(a²+92.16) units.
h is CP.
tani = 12/16
i = atan(3/4)
I is angle ABC.
10² = 9.6²+j²
j = √(10²-9.6²)
j = 2.8 units.
Calculating a.
(10-a)² = a²+(a+2.8)²
100-20a+a² = a²+a²+5.6a+7.84
100 = a²+25.6a+7.84
a²+25.6a-92.16 = 0
Resolving the above quadratic equation via completing the square approach.
(a+12.8)² = 92.16+(14.8)²
(a+12.8)² = 256
a+12.8 = ±16
a = -12.8±16
It implies;
a ≠ -12.8-16
a = -12.8+16
a = 3.2 units.
a is the radius of the small yellow inscribed circle.
Calculating b.
(10-b)² = b²+(b-2.8)²
100-20b+b² = b²+b²-5.6b+7.84
100 = b²+14.4b+7.84
b²+14.4b-92.16 = 0
Resolving the above quadratic equation via completing the square approach.
(b+7.2)² = 92.16+(7.2)²
(b+7.2)² = 144
b+7.2 = ±12
b = -7.2±12
It implies;
b ≠ -7.2-12
b = -7.2+12
b = 4.8 units.
b is the radius of the big green inscribed circle.
Therefore;
PQ is;
PQ = a+b
PQ = 3.2+4.8
PQ = 8 units.
