Calculating angle x.
Let the side length of the regular pentagon be 1 unit.
Calculating r, radius of the inscribed two equal circles.
a = ⅕*180(5-2)
a = 108°
b = ½(a)
b = ½*108
b = 54...
a = 2+2+R+R
a = (4+2R)
a is the diameter of the ascribed circle.
b = ½(a)
b = ½(4+2R)
b = (2+R) units.
b is the radius of the ascribed circle.
c = b-R
c = (2+R)-R
c = 2 units.
d = b...
a² = 13²-5²
a² = 169-25
a = √(144)
a = 12 units.
a is AE.
tanb = 5/12
b = atan(5/12)°
b is angle BAE.
c = x+a
c = (x+12) units.
c is AG.
Notice!
Triangle ABG is isosceles.
AB = AG
Therefore;
13...
Calculating R, radius of the inscribed circle.
Observing similar plane shape (right-angled triangle) side length ratios.
(12+x) - 6
6 - x
Cross Multiply.
36 = 12x+x²
x²+12x-36 = 0
(x+6)² =...
a² = 16+9-2*3*4cos150
a = 6.7664325675 cm.
a is the side length of the regular triangle.
(6.7664325675/sin150) = (4/sinb)
b = 17.192123734°
c = 60-b
c = 60-17.192123734
c = 42.807876266...
Calculating lengths x plus y.
Calculating y.
y² = 2(10)²
y = √(2(10)²)
y = 10√(2) cm.
y = 14.1421356237 cm.
Calculating x.
sina = 12/20
a = asin(3/5)°
b = ½(a)
b = (0.5asin(3/5))°
Therefore, l...
Let R be 1 unit (radius of the inscribed big circle).
Let a be the ascribed square side length.
Calculating a.
(a-1)² = 1²+(0.5a)²
a²-2a+1 = 1+¼(a²)
4a²-a² = 8a
3a = 8
a = ⅓(8) units.
Again, a is...
(x³)² = (x²)²+x²
x⁶ = x⁴+x²
Dividing through by x²
x⁴ = x²+1
Let p be x²
p² = p+1
p²-p-1 = 0
Resolving the above quadratic equation via completing the square approach to get p.
(p-½)² = 1+(-½)²
(p...
Calculating angle EHB.
Let the side length of the regular pentagon be 1 unit.
Therefore;
AB = BC = CD = DE = AE = 1 unit.
a² = 2-2cos108
a = 1.6180339887 units.
a is BE, the square side...
Let the radius of the circle be r.
Calculating r.
2(½*12r)+(r*r)+2(½*8r) = ½(12+r)(8+r)
(12r)+r²+(8r) = ½(12+r)(8+r)
24r+2r²+16r = 96+20r+r²
40r+2r² = 96+20r+r²
r²+20r-96 = 0
(r+10)² = 96+100
r = -...
