AB = BC = AC = 6 cm.
cos60 = 2/BE
BE = 4 cm.
(DE)² = 4²-2²
DE = 2√(3) cm.
(CE)² = (2√(3))²+4²
CE = 12+16
CE = 2√(7) cm
EF = (2√(7)-4) cm.
EF = 1.29150262213 cm.
AE = 6-4
AE = 2...
a = ½(180-150)
a = 15°
b = 90-15
b = 75°
(c/sin45) = (12/sin60)
c = 9.79795897113 cm.
(12/sin60) = (d/sin45)
d = 9.79795897113 cm.
e² = 9.79795897113²+12²-24*9.79795897113cos135 ...
The radius of the circle is 6 cm.
a = atan(2)°
Where a is angle AOB = angle COE.
Angle BOE = 180-atan(2)
= 116.56505117708°
Angle OBE = ½(180-116.56505117708)
= 31.71747441146°
b =...
Let GD (inscribed square side) be a.
Calculating a.
2(a/(tan60))+a = 12
a = 12/((2/√(3))+1)
a = 5.56921938165 cm.
b = ½(12-5.56921938165)
b = 3.21539030918 cm.
Where b is BD = CE.
c = 3.21539030...
a² = 12²-6²
a = 6√(3) cm.
Where a is BD.
b = ½(a)
b = 3√(3) cm.
Where b is BG = DG.
c = atan(6/(3√(3))
c = 49.10660535087°
Where c is angle AGD = angle BGE.
d = 60-49.10660535087
d...
Let a be the radius of the inscribed circle.
Calculating a.
tan30 = a/6
a = 2√(3) cm.
b = inscribed circle diameter.
b = 4√(3) cm.
12² = c²+6²
c = 144-36
c = √(108)
c = 6√(3) cm.
d = 6√(3)-4√(3)...
18² = 14²+16²-2*14*16cosa
448cosa = 14²+16²-18²
a = acos((14²+16²-18²)/448)
a = 73.39845040098°
(18/sin73.39845040098) = (14/sinb)
b = 48.18968510422°
c = 180-48.18968510422-73.39845040098
c = 58....
a² = 11*11+10*10-2*11*10cos50
a = 8.92113926968 cm.
(8.92113926968/sin50) = (10/sinb)
b = 59.16920318624°
c = 180-50-59.16920318624
c = 70.83079681376°
d = 0.5a
d = 4.46056963484 cm.
e² = 4.460...
Sir Mike Ambrose is the author of the question.
Radius of the semicircle is 2 units.
Therefore;
Shaded area exactly in decimal square units is;
Area trapezoid with parallel sides 2 units...
Radius, r1 of the inscribed semicircle is;
sin60 = r1/6
r1 = 3√(3) cm.
Radius, r2 of the inscribed circle is;
tan60 = 3/r2
r2 = 3/√(3)
r2 = √(3) cm.
It implies;
Area of the coloured inscribe...
