Notice.
ABCD is a square.
Let AB = 1 unit.
cos50 = a/1
a = 0.6427876097 units.
a is CE.
b = ½√(1²+1²)
b = ½√(2) units.
b = 0.7071067812 units.
b is OC.
c² = 0.7071067812²+0.6427876097²-2*0.70710...
tana = (10/5)
a = atan(2)°
b = ½(180-a)
b = ½(180-atan(2))°
cosb = c/10
cos(½(180-atan(2))) = c/10
c = 5.2573111212 cm.
sinb = d/10
sin(½(180-atan(2))) = d/10
d = 8.5065080835 cm.
It implies, bl...
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(...
