Sir Mike Ambrose is the author of the question.
Let the square side be x unit.
Calculating x.
x² = (x-4)² + ((x/2)+√((x²/4)-16)²
320 = 32x
x = 10 units.
Therefore;
Shaded Area is;...
a² = 3²+4²
a = √(25)
a = 5 cm.
a is the radius of the quarter circle.
Let r be the radius of the inscribed blue circle.
Calculating r.
b = (3+r) cm.
c = (5-r) cm.
It implies;
(...
Area red is;
Area semi circle with radius 5 cm - Area sector with radius 5 cm and angle (180-2atan(2))° - Area kite with diagonal lengths √(10²+5²) cm and √(200-200cos(2atan(½))) cm + Area secto...
Let the two equal lengths, AB and CD be 1 unit.
a = 180-50-15
a = 115°
a is angle ADB.
b = 180-a
b = 65°
b is angle ADC.
(1/sin115) = (c/sin50)
c = 0.8452365235 units.
c is AD.
d²...
Calculating x, length CD.
4² = 6²+6²-2*6*6cosa
72cosa = 72-16
cosa = 56/72
a = acos(7/9)
a = 38.942441269°
a is angle ABC.
b = ½(180-a)
b = 70.5287793655°
b is angle ADB = angle BAD....
Notice!
BD is the diameter of the circle.
a = ½(BD)
a = ½(14)
a = 7 cm.
a is OB = OD.
a is the radius of the circle.
b = 2(45)
b = 90°
b is angle AOB.
Calculating shaded area....
0.5*4*5sina = 8
Where a is the supplementary angle of angle ABC.
It implies;
10sina = 8
a = asin(8/10)
a = asin(4/5)°
or
a = acos(3/5)°
or
a = atan(4/3)°
a = 53.1301023542°
Therefo...
Notice!
The radius of the quarter circle is 4 units.
4² = 2²+a²
16-4 = a²
a = √(12)
a = 2√(3) units.
It implies, area green is;
Area triangle with height 2√(3) units and base 2 units...
Notice!
The side length of the small inscribed square is twice the side length of the big inscribed square.
Let the side length of the small inscribed square be a units
Therefore, the sid...
Area red is;
Area rectangle with length 4 units and width 3 units - Area triangle with height 4 units and base 1 unit - Area rectangle with length 2 units and width 1 unit - Area triangle with h...
