a = ⅙*180(6-2)
a = 120°
a is the single interior angle of the regular hexagon.
b = a-90
b = 30°
tan30 = 6/c
c = 6/(1/√(3))
c = 6√(3) cm.
Therefore, area orange is;
½*6*c
= ½*6*6√(...
Let a be the radius of the quarter circle.
It implies;
a² = a²+4²-2*a*4cosb
8cosb = 16
acosb = 2
cosb = 2/a --- (1).
8² = a²+4²+2*a*4cosb
64 = a²+16+8acosb
48 = a²+8acosb --- (2)....
Notice!
The regular polygon has 12 sides (regular dodecagon).
a = (180(12-2))/12
a = 1800/12
a = 150°
a is the single interior angle of the regular dodecagon.
b = ½(a)
b = 75°
c = 90-b...
Let AB = AC = 1 unit.
It implies;
a² = 2-2cos(24+12)
a = 0.6180339887 units.
a is BC = AD.
b = ½(180-24-12)
b = ½(144)
b = 72°
b is angle ACB = angle ABC.
c² = 0.6180339887²+1-2*0....
30² = 18²+a²
a² = 900-324
a = √(576)
a = 24 units.
b = ½(a)
b = 12 units.
Therefore, the required length, x is;
x² = 18²+b²
x² = 18²+12²
x² = 324+144
x = √(468)
x = 6√(13) units....
tana = 7/4
a = atan(7/4)°
b = 180-a
b = (180-atan(7/4))°
c² = 7²+4²
c = √(49+16)
c = √(65) cm.
It implies, area blue triangle is;
Area triangle with height √(65) cm and base 4sin(18...
Let AB be x cm.
BC = 3x cm.
AC = AB+BC
AC = 4x cm.
sin60 = a/16
a = 8√(3) cm.
a is AC.
Therefore;
4x = 8√(3)
x = 2√(3) cm.
Again, x is AB.
BC = 6√(3) cm.
CE = CD = 8 cm....
tan30 = 8/a
a = 8√(3) cm.
b = ½(a)
b = 4√(3) cm.
Area Blue is;
½*8√(3)*10-½*10*4√(3)
= 40√(3)-20√(3)
= 20√(3) cm²
Area triangle ABC is;
½*10*8√(3)+½*8√(3)*8
= 40√(3)+32√(3)
= 72√(3) cm²
Theref...
a = (10-x) cm.
10 cm is the radius of the ascribed semi circle.
12² = a²+b²
b² = 144-(100-20x+x²)
b² = 44+20x-x²
b = √(44+20x-x²) cm.
10² = x²+b²
100 = x²+44+20x-x²
56 = 20x
14 = 5x
x...
Let a be the radius of the inscribed orange circle.
(3+a)² = a²+b²
b² = 9+6a+a²-a²
b = √(9+6a) cm.
c = c-3
c = (√(9+6a)-3) cm.
It implies;
(6-a)² = a²+c²
(6-a)² = a²+(√(9+6a)-3)²
36-12a+a² = a²...
