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...
Let AC be 1 unit.
Area smaller pentagon is;
½(5)*(1/(2tan(36)))
= 1.72047740059 square units.
a² = 1²+0.5²-cos108
a = 1.24860602048 units.
Where a is CD.
(1.24860602048/sin108) =...
Sir Mike Ambrose is the author of the question.
Area R is;
Area triangle with height 4 cm and base (4tan52.5) cm - Area sector with angle 52.5° and radius 4 cm.
= ½*4(4tan52.5) - (52.5*4*4π)...
a² = 10²+16²-2*10*16cos120
a = 2√(129) units.
Where a is length AC.
(3√(129)/sin120) = (16/sinb)
b = 37.58908946897°
c = 90-b
c = 52.41091053103°
d = 60-b
d = 22.41091053103°
e...
Let the radius of the inscribed half circle be 1 unit.
Area rectangle is;
2*1
= 2 square units.
Calculating Shaded Area as a single fraction..
It is;
½(1*2) - 1² + ¼(1)²π - ((180-2a...
