Let a be the radius of the inscribed semi circle.
b = 2+4+a
b = (6+a) cm.
b is the diameter of the ascribed circle.
c = ½(b)
c = ½(6+a) cm.
c is the radius of the ascribed circle.
e = c-2
e = ½(6...
Calculating R, radius of the circle.
It implies;
10²+11²-2*10*11cosa = 7²+8²+2*7*8cosa
221-220cosa = 113+112cosa
221-113 = 112cosa+220cosa
108 = 332cosa
cosa = 108/332
a = acos(108/332)
a = 71.0161...
a = ⅙(180(6-2))
a = ⅙(180*4)
a = 120°
a is the single interior angle of the regular hexagon.
b² = 2²+4²-2*2*4cos120
b = √(20+7)
b = 2√(7) units.
(2√(7)/sin120) = (2/sinc)
c = 19.1066053...
Calculating x.
a = 9+3
a = 12 units.
Therefore the ascribed plane shape (quadrilateral or irregular quadrilateral) is an equilateral triangle of side length 12 units with a subtraction of a s...
Notice.
The ascribed triangle is equilateral.
Let r be the radius of the inscribed half circle.
Calculating r.
sin60 = r/6
√(3)/2 = r/6
r = 3√(3) cm.
It implies, area orange (inscribed semi circl...
Let the side length of the regular octagon be 1 unit.
a = ⅛(180(8-2))
a= ⅛(180*6)
a = 135°
a is the single interior angle of the regular octagon.
2b² = 1²
b = √(1/2)
b = ½√(2) units.
c = 2b+1
c =...
Calculating red length.
Let r be the radius of the circle.
sin30 = a/2
a = 1 unit.
cos30 = b/2
b = √(3) units.
Or
b = ½√(2²+2²-2*2*2cos120)
b = ½√(8+4)
b = ½√(12)
b = ½*2√(3)
b = √(3) units.
c =...
8 = 0.5b²sin113.7
16 = b²sin113.7
b = √(16/sin113.7)
b = 4.1801537416 units.
Therefore;
x² = 2*4.1801537416²-2*4.1801537416²cos113.7
x = 6.9995991687 units.
x = 7 units.
Notice.
Since x+y = 1 cm.
It explains that the pentagon or irregular pentagon can be presented in form of a regular quadrilateral (square) with side length 1 unit.
Therefore;
Area red is;
1²
=...
a = 180-72-36
a = 180-108
a = 72°
Notice.
Triangle BCD is isosceles.
Let AD = BC = 1 unit.
Therefore, calculate b.
((b+1)/sin72) = (1/sin36)
bsin36+sin36 = sin72
bsin36 = sin72-sin36
b = (sin72-s...
