Notice!
The square side length is;
4+4+4
= 12 units.
Calculating Area Red.
Area quarter circle with radius 12 units - Area triangle with height and base 12 units respectively - Area quarte...
Let the diameter of the big inscribed half circle be a units.
It's radius is ½(a) units.
Let the diameter of the small inscribed half circle be b units.
It's radius is ½(b) units.
It im...
Equation of the curve is;
y = ½(x^(3/2))
It implies;
PR = √(4²+6²)
PR = 2√(13) units
QR = (2√(13))/(cos(90-atan(⅔)))
= 13 units.
Therefore;
Shaded Area is;
(Area under the cu...
sin60 = 5/a
a = 5/(√(3)/2)
a = ⅓(10√(3)) units.
a is the side length of the inscribed regular triangle.
b = ½(a)
b = ⅓(5√(3)) units.
sin60 = c/⅓(5√(3))
√(3)/2 = 3c/(5√(3))
6c = 15
c =...
Notice!
The ascribed triangle is equilateral.
Calculating the required angle x.
a² = 1+3²-2*3cos60
a = √(7) units.
(√(7)/sin60) = (1/sinb)
b = 19.1066053509°
c = 60-b
c = 40.8933...
Let CE be 3 units.
Therefore:
Area Orange is;
Area triangle with height √(19) units and base 6sin(60-acos(4/√(19))) units - Area triangle with height (12√(19)/19) units and base (36√(57)/...
Let x be the single side length of the square.
Therefore;
x²=28-4√(x²-16) via similar triangle.
16(x²-16)=(28-x²)²
16x²-256=784-56x²+x⁴
x⁴-72x²+1040=0
Let x²(blue area) = P
Therefore;
P...
Let a be the radius of the half circle.
b² = (3√(3))²+a²
b = √(27+a²) units.
cosc = 3√(3)/√(27+a²) --- (1).
a² = 3²+(3√(3))²-2*3*3√(3)cosc
a² = 36+18√(3)cosc --- (2).
Substituting (1)...
Let the square side length be 3 units.
tana = (3/2)
a = atan(3/2)°
tanb = 1.5/1
b = atan(3/2)°
c = 180-45-a
c = (135-atan(3/2))°
It implies;
(2/sin(135-atan(3/2))) = (d/sin(atan(3/2)))
d = 1.69...
a = (8-2r) units.
Observing similar plane shape (right-angled) side length ratios.
r - 2r
b - (8-2r)
Cross Multiply.
2b = (8-2r)
b = (4-r) units.
c = r+b
c = r+(4-r)
c = 4 units.
tand = 8/4
d =...
