Let the single side length of the square be x.
Therefore;
½(x*(3x/4))+½(4√(x²+(3x/4)²)+½(8x)+½(¼(x)(x-4))=x²
(3x²)/8+2√(x²+(9x²/16))+4x+(x²/8)-(x/2)=x²
3x²+16√(x²+(9x²/16))+32x+x²-4x=8x²
1...
Exact enclosed area R will be;
Area sector with radius 2 unit and angle (90-2tan–¹(½)) - Area sector with radius 1 unit and angle (180-4tan–¹(½)) + Area isosceles triangle with two equal lengths...
tan30 = 6/a
a = 6√(3) units.
a is AB.
cos30 = 6/b
b = 4√(3) units.
b is CR.
c = a-b
c = 2√(3) units.
c is DR.
d = 180-30
d = 150°
d is angle DRQ.
It implies, shaded area is;...
a² = 6²+6²-2*6*6cos120
a = 6√(3) units.
tanb = 3/(6√(3))
b = atan(2/(2√(3)))
b = 16.102113752°
c = 90-16.102113752
c = 73.897886248°
d = 30+16.102113752
d = 46.102113752°
It implie...
Let the radius of the inscribed circle be a.
tanb = 4/3
b = atan(4/3)°
c = ½(b)
c = ½(atan(4/3))°
d = (3-a) units.
It implies;
tan(½(atan(4/3))) = a/(3-a)
½ = a/(3-a)
2a = 3-a
a...
Let the inscribed square side length be a.
2² = 1²+b²
b² = 4-1
b = √(3) units.
c = 2b+a
c = (a+2√(3)) units.
c is BC.
d = (a+1) units.
sin60 = (a+1)/e
½√(3) = (a+1)/e
√(3)e = 2a+2...
Let x be 1 unit.
a - 1
1 - √(2)
Cross Multiply.
√(2)a = 1
a = ½√(2) units.
It implies;
Area yellow is;
½*1*½√(2)
= ¼√(2) square units.
Area Blue is;
½(√(2)+½√(2))*1
= ½(½(3√...
Let the side length of the two congruent inscribed regular hexagon be 1 unit each.
Calculating the radius, a, of one of the two congruent inscribed yellow circles.
tan60 = a/(0.5)
a = 0.8660...
a = 180-60
a = 120°
b² = 4²+6²-2*4*6cos120
b = 2√(19) units.
b = 8.7177978871 units.
b is the side length of the inscribed regular triangle ABC (equilateral triangle).
(8.7177978871/sin12...
a² = 3²+4²
a = √(25)
a = 5 units.
3b+4b+5b = 3*4
12b = 12
b = 1 unit.
b is the radius of the inscribed circle.
tanc = 3/4
c = atan(¾)°
d = ½(180-atan(¾))
d = 71.5650511771°
tane...
