Mathematics Question and Solution
c² = 5³+12²
c = √(169)
c = 13 units.
Calculating a, radius of the inscribed yellow circle.
5a+12a+13a = (12*5)
30a = 60
a = 2 units.
Calculating b, radius of the blue inscribed circle.
Let d be the radius of the big ascribed circle.
d+x = 2b
b = ½(d+x) --- (1).
d² = (½(13))²+x²
d² = 42.25+x²
x = √(d²-42.25) --- (2).
e = (d-2) units.
f = 6.5-3
f = 3.5 units.
g = (x+2) units.
It implies;
e² = f²+g²
(d-2)² = 3.5²+(x+2)²
d²-4d+4 = 12.25+x²+4x+4
d²-4d = 12.25+x²+4x --- (3).
Substituting (2) in (3).
d²-4d = 12.25+(d²-42.25)+4√(d²-42.25)
30-4d = 4√(d²-42.25)
15-2d = 2√(d²-42.25)
(15-2d)² = 4d²-169
225-60d+4d² = 4d²-169
394 = 60d
d = 394/60
d = (197/30) units.
Again, d is the radius of the ascribed bigger circle.
Recall.
x = √(d²-42.25)
And d = (197/30) units.
x = √((197/30)²-42.25)
x = (28/30) units.
Recall Again.
b = ½(d+x) --- (1).
Where b is the radius of the inscribed blue circle.
d = (197/30) units.
x = (28/30) units.
Therefore;
b = ½((197/30)+(28/30))
b = ½(225/30)
b = ½(15/2)
b = ¼(15) units.
It implies;
b÷a is;
= (¼(15))÷2
= ⅛(15)
