Sir Mike Ambrose is the author of the question.
Let CD = 2 units.
Therefore;
Calculating AB.
AB = 1 + (2sin18*tan72 + 2sin36)tan54
AB = 1 + 3.07768353717tan54
AB = 1 + 4.23606797749...
x² = (r+1)² - (r-1)²
Therefore x = 2√(r) unit.
y² = (r+1)² - (2√(r))²
Therefore y = √(r²-2r+1)
y = √(r-1)²
y = (r-1) unit
z = (r+1) - y
z = (r+1) - (r-1)
Therefore z = 2 units
a = r+...
Area R is;
Area sector with radius 8 cm and angle 20.16632290743° - Area triangle with height 8 cm and base (4.2sin 20.16632290743) cm
= ( 20.16632290743 *64π÷360) - (½*8*4.2sin 20.166322907...
Area coloured is;
Area sector with radius 3 units and angle 120° - Area triangle with height 3 units and base 3sin120 units + Area triangle with height 3 units and base 3sin60
= (120*9π÷360)...
Shaded area is;
Area sector with radius 10 cm and angle 60°
(100π*60)/360
= (100π)/6
= ⅓(50π) cm²
Let the side length of the regular heptagon be 1 unit.
a = ⅐(180*5)
a = ⅐(900)°
b = 0.5(360-2(900/7))
b = ⅐(360)°
c = 2cos(360/7)+1
c = 2.24697960372 units.
d² = 2-2cos(90/7)
d = 1....
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 = 180-48.18968510422-73.3984504009...
8 = 4
a = (8-a) Cross multiply.
64-8a = 4a
64 = 12a
a = ⅓(16) cm.
b = 8-a
b = ⅓(8) cm.
c = atan(⅔)°
d = atan(3/2)°
⅔ = 8/e
e = 12 cm
⅔ = f/8
f = ⅓(16) cm.
g = 8-f
g = ⅓(...
Area coloured regions is;
Half the area of the ascribed square.
= ½(2*2)
= 2 square units.
Sir Mike Ambrose is the author of the question.
Let AC = 2 units.
Therefore;
AB = ⅖ unit.
BC = (8/5) unit.
It implies;
Area Blue is;
Area triangle with height ⅖ unit and base 2si...
