Mathematics Question and Solution
a² = 12²+5²
a = √(144+25)
a = 13 cm.
a is AD.
12b+5b+13b = 12*5
30b = 60
b = 2 cm.
b is the radius of the inscribed circle.
tanc = 12/5
c = atan(12/5)°
tan(0.5c) = 2/d
⅔ = 2/d
d = 3 cm.
d is AB.
e = 12-2
e = 10 cm.
Let CD = f
f+3f = 13
4f = 13
f = ¼(13) cm.
Again, f is CD.
g = 3*¼(13)
g = ¼(39) cm.
g is AC.
h = g-3
h = ¼(39-12)
h = ¼(27) cm.
h is BC.
j = (180-2atan(12/5))°
k² = 2(¼(39))²-2(¼(39))²cos(180-2atan(12/5))
k = ½(15) cm.
Area green is;
½*½(15)*¼(27)sin(atan(12/5))
= 23.3653846154 cm²
Area blue is;
½*3*¼(39)sin(180-2atan(12/5))
= 10.3846153846 cm²
Therefore;
Area green ÷ Area blue in the term a/b where a and b are both less than 10 is;
23.3653846154÷10.3846153846
= 2.25
= 225/100
= 9/4
