Mathematics Question and Solution
Let the radius of the ascribed quarter circle be a.
It implies;
a² = 5²+b²
b = √(a²-25) units.
6² = b²+c²
c² = 36-b²
c = √(36-√(a²-25)²)
c = √(36-(a²-25))
c = √(61-a²) units.
d = 5+c
d = (5+√(61-a²)) units.
Calculating a.
(5+√(61-a²))² = a²+6²
25+10√(61-a²)+61-a² = a²+36
10√(61-a²)-a² = a²+36-86
100(61-a²) = (2a²-50)²
100(61-a²) = 4a⁴-200a²+2500
6100-100a² = 4a⁴-200a²+2500
4a⁴-100a²-3600 = 0
a⁴-25a²-900 = 0
Let a² = p.
p²-25p-900 = 0
Resolving the above quadratic equation via completing the square approach to get p.
p²-½(25)p = 900
(p-½(25))² = 900+(625/4)
(p-½(25))² = ¼(3600+625)
(p-½(25))² = ¼(4225)
p-½(25) = ±√(¼(4225))
p = ½(25)±½(65)
Therefore;
p ≠ ½(25-65)
p = ½(25+65)
p = ½(90)
p = 45 units.
And p = a²
Therefore;
a² = 45
a = √(45)
a = 3√(5) units.
Again, a is the diameter of the ascribed quarter circle.
e = ½(a)
e = ½(3√(5)) units.
e is the radius of the ascribed quarter circle.
tanf = 6/(3√(5))
f = atan(2/√(5))°
tang = (3√(5))/6
g = atan(½√(5))°
Area yellow is;
Area quarter circle with radius 3√(5) units - Area triangle with height and base 3√(5) units and 6 units respectively - Area sector with radius 1.5√(5) and angle (180-2atan(2/√(5))) + Area triangle with height 1.5√(5) units and base 1.5√(5)sin(180-2atan(2/√(5))) - Area sector with radius 3 units and angle (180-2atan(½√(5))) + Area triangle with height 3 units and base 3sin(180-2atan(½√(5)))
= ¼(√(45)²π)-½(6√(45))-((180-2atan(2/√(5)))π*(1.5√(5))²/360)+(0.5*(1.5√(5))²sin(180-2atan(2/√(5))))-((180-2atan(0.5√(5)))π*3²/360)+(0.5*3²sin(180-2atan(0.5√(5))))
= 35.3429173529-20.1246117975-9.4620225439+5.5901699437-6.567548906+4.472135955
= 9.2510400042 square units.
