Calculating shaded area considering semi circle PQ.
Diameter of semi circle PQ will be 10cm, therefore radius is 5cm.
Shaded area will be;
Area semi circle PQ of radius 5cm and angle 180° -...
Let angle CAD = a
Let angle BAD = b
Let AD = c
tana = 3/c ----- (1).
tanb = 2/c ----- (2).
Adding (1) and (2).
Therefore;
tana+tanb = 5/c ----- (3).
Notice;
tan(a+b) = tana+tanb/(1-tanatanb)...
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Let the radius of one the the two congruent inscribed circles be a.
Therefore;
a = 2 cm.
tanb = (2/4)
b = atan(½)°
c = (180-2atan(½))°
c = 126.86989764584°
Shaded Region is;
Half area rectangl...
AC = √(12²+12²)
AC = 12√(2) cm.
AE = √(12²+6²)
AE = 6√(5) cm.
Angle BAE = atan(½)°
Angle CAE = 45-atan(1/2)
Angle CAE = 18.43494882292°
cos18.43494882292 = (AF)/12√(2)
AF = 16.099689438 cm.
EF...
Let AB = a.
Let BE = b.
It implies;
2a+b = 392
b = 392-2a
BC = ½(b-70)
BC = ½(392-2a-70)
BC = ½(322-2a)
Therefore;
BC = DE = 161-a
c = ½(70)+161-a
c = 196-a
Where c is half BE.
d² = a²-(196-a)²...
a² = 6²+12²
a = 6√(5) cm.
Where a is CE = EF.
b² = 2(a²)
b² = 2(6√(5))²
b² = 360
b = 6√(10) cm.
Where b is CF.
Angle DCE = atan(½)°
Angle ECF = 45°
Angle FCM = 90-45-atan(½)
Angle FCM = 18.43494...
Let CE = 1 unit.
(1/sin20) = (a/sin40)
a = 1.87938524157 units.
Where a is AE.
Notice;
b = a-1
b = 1.87938524157-1
b = 0.87938524157 units.
Where b is AD.
Angle AEC = 180-20+40
An...
a = ½(180-150)
a = 15°
b = 60-15
b = 45°
c² = 2(6²)
c = 6√(2) cm.
d² = 72+12²-144√(2)cos135
d= 6√(10) cm.
(6√(10)/sin135) = (12/sine)
e = 26.56505117708°
f = 180-135-26.56505117708
f = 18.4349...
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