a² = 12²-6²
a = 6√(3) cm.
a = 10.39230484541 cm.
a = circle's radius = AD = AG.
b² = 2*10.39230484541²-2*10.39230484541²cos120
b = 18 cm.
b = DG.
c = 12-10.39230484541
c = 1.60769515459 cm.
c = CD...
Let the side of the regular pentagon be 1 unit.
a = 360-2(108)-90
a = 54°
tan72 = b/0.5
b = 1.53884176859 units.
tan54 = 1.53884176859/c
c = 1.11803398875 units.
d = c-0.5
d = 0.61803398875 unit...
a² = 6²+8²-2*6*8cos30
a = 4.10628314132 cm.
a = DE.
(4.10628314132/sin30) = (8/sinb)
b = 103.06431342951°
b = Angle ADE.
Angle AED = 180-30-103.06431342951
= 46.93568657049°
tan30 = c/6
c = 3.46...
Calculating r, radius of the circle.
a = 2+½(1)
a = (2½)
a = (5/2) units.
r² = (½)²+b²
b = √(r²-(¼)) units.
c = 3-b
c = (3-√(r²-(¼))) units.
It implies;
r² = (5/2)²+(3-√(r²-¼))²
r² = (25/4)+9-6...
Let a side length of the three congruent regular pentagon be 1 unit.
Single interior angle of the three congruent regular pentagon is;
⅕(180*3)
= 108°
a² = 1²+0.5²-2*1*0.5cos108
a = 1.24860602048...
a² = 8²-2²
a = √(60)
a = 2√(15) units.
Let the radius of the inscribed circle be r.
Considering similar triangle ratios, calculating r.
r = 2√(15)
1 = 2
Cross Multiply.
r = √(15) units.
b² = 1+r²...
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Notice;
Angle DCE = -(Angle BCG)
Let Angle DCE be x.
Calculating K.
4k² = 4k²+k²-4k*kcosx
cosx = (¼) --- (1).
144 = 4k²+k²-2k²(-cosx)
144 = 5k²-2k²(-cosx)
Substituting (1) in (2).
144 = 5k²-2...
Let a be the side of the inscribed square.
24² = a²+(2a)²
576 = a²+4a²
576 = 5a²
a = √(576/5)
a = 10.733126292 cm.
b = ½(24-10.733126292)
b = 6.633436854 cm.
b = AM = DR.
c = atan(6.633436854/(...
AB = BC = AC = 6 cm.
cos60 = 2/BE
BE = 4 cm.
(DE)² = 4²-2²
DE = 2√(3) cm.
(CE)² = (2√(3))²+4²
CE = 12+16
CE = 2√(7) cm
EF = (2√(7)-4) cm.
EF = 1.29150262213 cm.
AE = 6-4
AE = 2 cm.
Angle AEF =...
