sin30 = a/12
a = 6 cm.
Where a is the height of the green regular triangle.
sin60 = 6/b
½√(3) = 6/b
b = 12/√(3)
b = 4√(3) cm.
b is the side length of the green regular triangle.
tan30 =...
Let the side length of the square be a.
b = ½(a) cm.
b is the radius of the inscribed half circle.
c = (a-2) cm.
d²+2² = (a-2)²
d² = a²-4a+4-4
d = √(a²-4a) cm.
d is the radius of the b...
Let the side length of the regular pentagon be 2 units.
a = ⅕*180(5-2)
a = 108°
a is the single interior angle of the regular pentagon.
b² = 1²+2²-2*1*2cos108
b = 2.497212041 units.
(2....
Let the side length of the ascribed regular hexagon be 1 unit.
a = ⅙*180(6-2)
a = 120°
a is the single interior angle of the ascribed regular hexagon.
b = ¼(a)
b = 30°
d = (1-c)
c is t...
Calculating r, the radius of the blue inscribed circle.
Therefore;
Consider triangle OBC.
(6-r)² = r² + x², and x = √(9+6r) - 3
It implies;
(6-r)² = r² + (√(9+6r)-3)²
36-12r+r²=r²+9-...
Let the radius of the quarter circle be 2 units.
Therefore;
OD = BD = ½(2)
OD = BD = 1 units.
a² = 1²+2²-2*1*2cos45
a = 1.4736257582 units.
a is CD.
(1.4736257582/sin45) = (2/sinb)...
Let a be the side length of the square.
b² = 13²+a²
b = √(169+a²) units.
c² = 2a²
c = √(2)a units.
d²+5² = a²
d = √(a²-25) units.
e = b-d
e = (√(169+a²)-√(a²-25)) units.
Calculat...
b = √(625-a²) units.
c = √(a²-256) units.
Where a is the diameter of the blue circle.
Calculating a.
b² = c²+9²
√(625-a²)² = √(a²-256)²+9²
625-a² = a²-256+81
625-a² = a²-175
2a² = 8...
Calculating angle x.
x = 180-½(360-100-160)
x = 180-½(360-260)
x = 180-½(100)
x = 180-50
x = 130°
a² = 4²+1²
a = √(17) units.
a is the diameter of the circle.
b = ½(a)
b = 0.5√(17) units.
b is the radius of the circle.
Area circle is;
π(b²)
= π(0.5√(17))²
= ¼(17π) square units....
