Let the base of the ascribed triangle be 1 unit.
a = 180-20-20-10-10
a = 120°
(1/sin120) = (b/sin20)
b = 0.3949308436 units.
c = 180-120-20
c = 40°
(0.3949308436/sin40) = (d/sin120)
d = 0.532088...
a = 180-20-50-40
a = 70°
It implies the triangle is isosceles.
Let the base of the isosceles triangle be 1 unit.
1² = 2b²-2b²cos40
1 = 0.4679111138b²
b = √(1/0.4679111138)
b = 1.4619022...
Notice!
R is the radius of the ascribed semi circle.
Calculating R.
a² = 2*√(2)²
a² = 4
a = 2 units.
a is the diagonal of the inscribed red and white square.
√(2)² = 2b²
b² = 2/2
b = 1 unit.
It...
a² = 2*5²
a = √(50)
a = 5√(2) cm.
tanb = 10/5
b = atan(2)°
c = b-45
c = (atan(2)-45)°
c = 18.4349488229°
d = 90-b
d = (90-atan(2))°
d = 26.5650511771°
e = 180-2d
e = 180-2*26.56...
Calculating x.
Notice!
AD = 100 cm.
Therefore;
Area triangle ABD + Area triangle ACD = Area triangle ABC
0.5*2x*100sin60+0.5*x*100sin60 = 0.5*2x*xsin120
100x+50x = x²
x = 100+50
x = 150 cm.
It i...
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...
