Mathematics Question and Solution
Calculating Area Pink.
a*b = 4*(4+1)
ab = 20
a = 20/b units.
c = a+b
c = (20/b)+b
c = (b²+20)/b units.
c is the diameter of the ascribed pink circle.
d = ½(c)
d = (b²+20)/(2b) units.
d is the radius of the pink circle.
e = b-d
e = b-((b²+20)/(2b))
e = (2b²-b²-20)/(2b)
e = (b²-20)/(2b) units.
f = ½(4+4+1)-4
f = 4.5-4
f = 0.5 units.
g = ½(4)
g = 2 units.
cosh = 2/b --- (1).
cosh = 0.5/((b²-20)/(2b))
cosh = b/(b²-20) --- (2).
Calculating b.
Equating (1) and (2).
2/b = b/(b²-20)
Cross Multiply.
2b²-40 = b²
b² = 40
b = √(40)
b = 2√(10) units.
b = 6.32455532034 units.
Calculating d.
Recall.
d = (b²+20)/(2b)
Where d is the radius of the pink circle.
And b = 2√(10) units.
d = ((2√(10))²+20)/(2*2√(10))
d = 60/(4√(10))
d = 15/√(10)
d = ½(3√(10)) units.
d = 4.74341649025 units.
Again, d is the radius of the pink circle.
Recall Again.
cosh = 2/b at (1).
And b = 2√(10) units.
cosh = 2/(2√(10))
h = acos(1/√(10))
h = 71.5650511771°
j = 90-h
j = 18.4349488229°
cosh = 2/k
k = 2/cos18.4349488229
k = ⅓(2√(10)) units.
k = 2.10818510678 units.
k is the radius of the inscribed white circle.
Therefore;
Area Pink is;
πd²-πk²
= π(½*3√(10))²-π(⅓*2√(10))²
= π((¼*9*10)-(⅑*4*10))
= π(½(45)-⅑(40))
= π((405-80)/18)
= (325/18)π square units.
