Let a be the radius of the half circle.
a²+8² = b²
b = √(a²+64) units.
b is OC.
Observing similar plane shape (right-angled) side length ratios.
8 - c
√(a²+64) - a
Cross Multiply.
c =...
a²+1.5² = 4²
a² = 16-2.25
a = 3.7080992435 units.
tanb = a/1.5
b = atan(3.7080992435/1.5)
b = 67.9756871627°
b is angle ODC.
c = 90-b
c = 22.0243128373°
c is angle ODT.
It implies;...
Sir Mike Ambrose is the author of the question.
The base/length of the rectangle with height or width 4 units is 3√(3) units.
Therefore calculating red area (right-angled triangle with hypotenu...
a = (4-x) units.
It implies;
4² = (4-x)²+2²
12 = 16-8x+x²
x²-8x+4 = 0
(x-4)² = -4+(-4)²
(x-4)² = -4+16
(x-4)² = 12
x = 4±√(12)
x = 4±2√(3)
It implies;
x = (4-2√(3)) units.
a = (4-x)...
a = (b+1) units.
a is the diameter of the circle.
c = ½(a)
c = ½(b+1) units.
c is the radius of the circle.
d² = 1+1
d = √(2) units.
d is the diagonal of a square.
e = d-c
e = √(2)-½(...
tana = 4/7
a = atan(7/4)°
b = 180-a
b = (90+atan(4/7))°
c² = 7²+4²
c = √(49+16)
c = √(65) units.
Therefore, Blue Area is;
Area triangle with height √(65) units and base 4sin(90+atan...
a = 180-44
a = 136°
b = ½(180+44)
b = ½(224)
b = 112=
c = ½(a)
c = 68°
d = ½(b)
d = 56
e = 180-c-d
e = 180-68-56
e = 180-124
e = 56°
Where e is the required angle, angle CBO....
7² = 5²+6²-2*5*6cosa
60cosa = 25+36-49
cosa = (61-49)/60
a = acos(12/60)
a = 78.4630409672°
Calculating r, radius of the inscribed half circle.
5r+6r = 5*6sin78.4630409672
11r = 30sin78....
tan30 = a/h
a = htan30 units.
tan60 = b/h
b = htan60 units.
It implies;
a+b = 100
htan30+htan60 = 100
⅓(√(3)h)+√(3)h = 100
⅓(4√(3)h) = 100
4√(3)h = 300
h = 300/(4√(3))
h = ¼*⅓*300√...
a²+R² = 7²
a = √(49-R²) units.
b = (3+R) units.
c²+R² = 3²
c = √(9-R²) units.
d = a+c
d = √(49-R²)+√(9-R²) units.
R is;
(√(49-R²)+√(9-R²))² = (3+R)²+√(49-R²)²
49-R²+2√((49-R²)(9-...
