Mathematics Question and Solution
Let a be the side length of the two equal inscribed lengths.
r = 2+1
r = 3 units.
r is the radius of the ascribed half circle.
b²+(a/2)² = 3²
b²= 9-¼(a²)
b = √(9-¼(a²))
b = ½√(36-a²) units.
c = a-b
c = a-½√(36-a²)
c = ½(2a-√(36-a²)) units.
Calculating a.
2² = c²+(a/2)²
4 = (½(2a-√(36-a²)))²+¼(a²)
4 = ¼(4a²-4a√(36-a²)+(36-a²))+¼(a²)
16 = 4a²-4a√(36-a²)+36-a²+a²
16 = 4a²-4a√(36-a²)+36
4 = a²-a√(36-a²)+9
a√(36-a²) = a²+5
a²(36-a²) = (a²+5)²
36a²-a⁴ = a⁴+10a²+25
2a⁴-26a²+25 = 0
Let p = a²
It implies;
2p²-26p+25 = 0
(p-½(13))² = (-½(13))²-(25/2)
(p-½(13))² = ¼(169)-½(25)
(p-½(13))² = ¼(169-50)
(p-½(13))² = ¼(119)
p = ½(13)±√(¼(119))
p = ½(13)±½√(119)
Therefore;
p = ½(13-√(119)) units.
Or
p = ½(13+√(119)) units.
Recall.
p = a²
It implies;
For p = ½(13-√(119)) units.
a = √(½(13-√(119)))
a = 1.02256732917 units.
For p = ½(13+√(119)) units.
a = √(½(13+√(119)))
a = 3.45750720279 units.
Notice.
a ≠ 1.02256732917 units.
a = 3.45750720279 units.
sind = 0.5a/3
d = asin(0.5*3.45750720279/3)
d = 35.1873016959°
e = 2d
e = 70.3746033917°
(3.45750720279/sinf) = (2/sin35.1873016959)
f = 85.0010782563°
g = e+f
g = 70.3746033917+85.0010782563
g = 155.375681648°
h = ½(180-g)
h = ½(180-155.375681648)
h = 12.312159176°
j = 90-h
j = 77.687840824°
It implies;
cos77.687840824 = k/(1+2+3)
k = 6cos77.687840824
k = 1.27942636679 units.
Therefore, the required inscribed area, area pink shaded triangle is;
Area triangle with height 1.27942636679 units and base 6sin77.687840824
= 0.5*1.27942636679*6sin77.687840824
= 3.75 square units.
