Let the side length of the regular pentagon be 1 unit.
x² = 2-2cos30
x = 0.5176380902 units.
y = 108-90
108° is the single interior angle of the regular pentagon.
y = 18°
c = y+½(180-3...
Let FC = x
Let CE = y
Therefore;
5+x = y+4
x = y-1 --- (1).
Notice;
Triangle AEF is similar to triangle EFC.
It implies;
5+x - y
4 - x
Cross Multiply.
x²+5x = 4y --- (2).
Su...
a² = 2(3+2√(3))²-2(3+2√(3))²cos120
a = 11.1961524227 units.
b = a-(3+2√(3))
b = 4.7320508076 units.
tan30 = c/(0.5(3+2√(3)))
c = 1.8660254038 units.
cos30 = (0.5(3+2√(3)))/d
d = 3.7320...
a² = 2b²
a = √(2)b cm.
(8+b)² = 2(4+a)²
(8+b)² = 2(4+√(2)b)²
64+16b+b² = 2(16+8√(2)b+2b²)
64+16b+b² = 32+16√(2)b+4b²
3b²+16√(2)b-16b-32 = 0
3b²+(16√(2)-16)b-32 = 0
It implies;
b = (8-4√(2...
Considering the quadrilateral cyclic.
a² = 2*10²-2*10²cosx
a² = 200-200cosx --- (1).
a² = 9²+7²-2*9*7cosx
a² = 130+126cosx --- (2).
Equating (1) and (2).
a² = a²
200-200cosx = 130+12...
Let the base of the ascribed triangle be 1 unit.
a = 180-12-42-12-36
a = 180-102
a = 78°
(1/sin(12+36)) = (b/sin78)
b = 1.3162274259 units.
c = 180-12-42
c = 180-54
c = 126°
(1.316...
Let AC be 2 units.
a = 180-15-30
a = 135°
b = 180-135
b = 45°
(1/sin15) = (c/sin30)
c = 1.9318516526 units.
d² = 1.9318516526²+1²-2*1.9318516526*1cos45
d = √(2) units.
d = 1.414213...
a² = 2²+1²
a = √(5) units.
b² = 2(a)²
b² = 2√(5)²
b² = 10
b = √(10) units.
b is the diagonal of the blue square.
tanc = (2/1)
c = atan(2)°
d = c-45
d = (atan(2)-45)°
Calculating...
a = ½(PQ)
a = ½(4)
a = 2 units.
b = 90-60
b = 30°
c = 180-45-30
c = 105°
d = 180-105-60
d = 15°
It implies;
(2/sin105) = (e/sin15)
e = 0.5358983849 units.
Therefore, Red Pai...
Calculating the area of the inscribed square.
Let the inscribed square length be a.
It implies;
b = 3-2
b = 1 unit.
Therefore;
The Area of the inscribed square (a²) is;
a² = 2²+1...
