Let the side of the regular pentagon be 1 unit.
Notice;
The regular pentagon side is equal the regular heptagon side.
a = ⅐(180*5)
a = ⅐(900)°, the single interior angle of the regular heptagon....
It implies;
tan(0.5acos(3/5)) = r/3
Where r is the radius of the inscribed semi circle.
r = 3tan(0.5acos(3/5))
r = 1.5 units.
r = 1½ units.
Let the side length of the square be 2 unit.
Therefore area ABD (equilateral triangle) is;
Area triangle with height 2 unit and base (2sin60) unit.
= ½*2*2sin60
= √(3) square unit.
Area purple is...
Let BD = y units.
Let angle BAD = a
It implies;
tana = y/6 --- (1).
tan(a+45) = (y+10)/6
(tana+tan45)/(1-tanatan45) = (y+10)/6 --- (2)
Therefore, Substituting (1) in (2).
Notice.
tan45 = 1 uni...
a = 3x+x
a = 4x
a is the square side.
it implies;
Area Square is;
a²
= (4x)²
= 16x² square units.
Calculating Shaded Area.
b = 45°
tanc = (4x/3x)
c = atan(4/3)°
d = 180-b-c
d = (180-45-atan(4/...
Let the inscribed purple regular heptagon side be 1 unit.
a = ⅐(180*5)
a = ⅐(900)°
b = 180-⅐(900))
b = ⅐(360)°
c = 180-2(⅐(360))
c = ⅐(540)°
(1/sin⅐(540)) = (d/sin⅐(360))
d = 0.8019377358 unit....
Calculating x.
(6+12)² = (6+x)²+(12+x)²
18² = 36+12x+x²+144+24x+x²
324 = 180+36x+2x²
2x²+36x+180-324 = 0
2x²+36x-144 = 0
x²+18x-72 = 0
Resolving the quadratic equation.
(x+9)² = 72+81
x...
Let the square side be a.
b² = 3²+4²
b = √(25)
b = 5 units.
tanc = 3/4
c = atan(¾)°
d = 90-c
d = 90-atan(¾)
d = atan(4/3)°
It implies;
Calculating a, the square side.
tan(d) = a/(5-a)
tan(atan(...
Let a be the diameter of the semi circle.
a² = b²+14²
b² = a²-196
b = √(a²-196)
It implies;
b² = 6²+6²-2*6*6(-cosc)
a²-196 = 72+72cosc
a²-196-72 = 72cosc
72cosc = a²-268
cosc = (a²-26...
Let BD = 1 unit.
It implies;
CD = 2BD = 2 units.
a = 180-60
a = 120°
a is angle ADB.
b = 180-120-45
b = 15°
b is angle BAD
(1/sin15) = (c/sin120)
c = 3.346065215 units.
c is AB.
d² = 3.34606521...
