Mathematics Question and Solution
Calculating the required green area.
Notice.
OP = AP = r.
R = 2r.
a² = (2r)²+r²
a² = 4r²+r²
a = √(5r²)
a = √(5)r units.
a is CP = OQ.
r ~ a
b ~ 2r
It implies;
r ~ √(5)r
b ~ 2r
Cross Multiply.
2r² = √(5)rb
b = 2r/√(5) units.
b = ⅕(2√(5)r) units.
b is OM.
c²+b² = r²
c² = r²-(2r/√(5))²
c² = r²-(4r²/5)
c² = ⅕(r²)
c = √(⅕(r²))
c = ⅕√(5)r units.
c is MP.
Calculating r.
½*b*c = 6
½*⅕(2√(5)r)*⅕(√(5)r) = 6
⅕(r²) = 6
r² = 30
r = √(30) units.
r = 5.47722557505 units.
Recall.
R = 2r.
r = √(30) units.
R = 2√(30) units.
Let X be quadrilateral ANMP.
Therefore;
½*r*R = 6+X+8
½*√(30)*2√(30) = 14+X
30 = 14+X
X = 30-14
X = 16 square units.
Again, X is quadrilateral ANMP.
Notice.
Triangle AOQ Area = Triangle ACP Area.
Therefore, the required green area is;
6+8
= 14 square units.
