Let the side length of square ABCD be a.
Area Square ABCD will be a²
Calculating a²
Notice, he three inscribed plane shapes, two congruent right-angled triangle and a parallelogram has equal area....
Let the regular pentagon side be 1 unit.
a = ⅕(180*3)
a = 108°
a is an angle of the regular pentagon.
b = 108-36
b = 72°
c = 180-45-72
c = 180-117
c = 63°
(d/sin63) = (1/sin45)
d = 1.26007351067...
Let the inscribed regular pentagon side be 1 unit.
Area small/inscribed pentagon is;
0.5*5(1/(2tan(180/5)))
= 1.72047740059 square units.
Calculating Shaded Area.
a² = 2-2cos(60+108)
a = 1.989043...
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Let BD be 1 unit.
Therefore;
DC = 2 units.
a = 180-60
a = 120°
b = 180-120-45
b = 15°
(1/sin15) = (c/sin120)
c = 3.34606521495 units.
d² = 3.34606521495²+3²-2*3.34606521495*3cos45
d = 2.449489...
a² = 2*12²
a = 12√(2) cm.
Where a is AC = CF.
b = 90+45
b = 135°
Where b is angle BCE.
c² = (12√(2))²+12²-24*12√(2)cos135
c = 12√(5) cm.
Where c is BE.
(12√(5)/sin135) = (12/sind)
d = 18.43494882...
Let the single side length of the two congruent inscribed regular pentagon be 1 unit.
Area Green is;
0.5*5(1/(2tan(180/5)))
= 1.72047740059 square units.
Calculating Area Shaded.
a = ⅕(180*3)
=...
(1) 2x²+5x+3 = 0
Using smart completing the square approach.
(x+(5/4))²+(3/2)-(5/4)²
= 0
(x+(5/4))² = (25/16)-(3/2)
(x+(5/4))² = (1/16)
x = -(5/4)±(1/4)
x = -1
Or
x = -1.5
Using factorization appro...
a² = 3²+4²
a = 5cm.
Where a is BC.
b² = 2.5²+4²-8*2.5cos(atan(3/4))
b = 2.5 cm.
c = 90-atan(3/4)
c = atan(4/3)
sin(atan(4/3)) = d/4
d = 3.2cm.
Where d is CN.
e = 5-3.2
e = 1.8cm.
Where e is BN....
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