Let a be the radius of the inscribed circle.
Calculating a.
(1+a)² = a²+b²
b² = (1+a)²-a²
b² = 1+2a
b = √(1+2a) cm.
c = b-1
c = (√(1+2a)-1) cm.
It implies;
(2-a)² = a²+c²
(2-a)² = a²+(√(1+2a)-1...
a = 4+4+x
a = (8+x) cm.
a is the diameter of the bigger inscribed circle and also the side length of the square.
b = ½(a)
b = ½(8+x) cm.
6+y = b
6+y = ½(8+x)
12+2y = 8+x
x = 2y+4 --- (1...
a = (180(12-2))/12
a = 1800/12
a = 300/2
a = 150°
a is the single interior angle of the regular dodecagon.
b = 360-150-90
b = 360-240
b = 120°
(6√(3))² = 2c²-2c²cos120
108 = 3c²
c² = 36
c = 6 unit...
Calculating x, side length of the square.
a² = 2x²
a = √(2)x units.
a is BD, diagonal of the square.
tanb = x/(0.5x)
b = atan(2)°
c = 2b
c = 2atan(2)°
c is angle DEF.
d² = 2(0.5x)²-2(0.5x)²cos(2...
a = 180-60-40
a = 80°
(6/sin80) = (b/sin40)
b = 3.916221868 units.
(6/sin80) = (c/sin60)
c = 5.2763114494 units.
Calculating r, radius of the inscribed circle.
5.2763114494r+3.916221...
Let the two congruent blue semi circle's radius be a.
1 = πb²
b² = 1/π
b = √(1/π) units.
b is the radius of the yellow circle.
tanc = a/2a
c = atan(½)°
d = 2c
d = 2atan(½)°
sin(atan(½)).= √(1/π)...
Calculating r, radius of the inscribed circle.
4r+5r+3r = (4*3)
12r = 12
r = 1 unit.
tana = 4/3
a = atan(4/3)°
b = 180-a
b = (180-atan(4/3))°
c² = 1+1-2cos(180-atan(4/3))
c = 1.788854382 units....
Let the ascribed regular pentagon side length be 1 unit.
a = ⅕(180(5-2))
a = 108°
a is the single interior angle of the ascribed and the smaller inscribed regular pentagon.
b = ½(180-108)
b = 36°...
Let the side length of the regular hexagon be 2 units.
a = ⅙(180(6-2))
a = 120°
a is the single interior angle of the regular hexagon.
b² = 2²+1²-2*2*1cos120
b = √(7) units.
b = 2.6457513111 units...
a² = 10²+5²
a = √(125)
a = 5√(5) units.
a = 11.1803398875 units.
a is the side length of the square.
tanb = 10/5
b = atan(2)°
c = atan(1/2)°
d = 45-c
d = (45-atan(1/2))°
d = 18.4349488229°
e = 4...
