Let a be the radius of the inscribed pink circle.
b = (a+7) units.
c = (7-a) units.
d = (a-2) units.
Calculating a.
It implies;
a ~ (a+7)
(a-2) ~ (7-a)
Cross Multiply.
a(7-a) = (a-2)(a+7)
7a...
Let 2 units be the side length of the ascribed square.
Therefore;
a = 1 units.
a is the radius of the inscribed circle.
Calculating angle x.
b² = 2(1)²
b = √(2) units.
c = ½(b)
c...
Calculating shaded area.
tana = 10/5
a = atan(2)°
b = 2a
b = 2atan(2)°
c = 180-b
c = (180-2atan(2))°
d = ½(c)
d = (90-atan(2))
d = atan(0.5)°
It implies;
tan(atan(0.5)) = e/5...
a² = r²+(½(r))²
a = ½√(5)r units.
2r ~ ½√(5)r
b ~ ½(r)
Cross Multiply.
½√(5)rb = r²
√(5)rb = 2r²
b = 2r/√(5)
b = ⅕(2√(5))r units.
b is the height of the inscribed triangle.
c²+b²...
Let a be the radius of the circ.
Calculating a.
b = ½(12+2)
b = 7 units.
c = ½(4+6)
c = 5 units.
d = b-2
d = 7-2
d = 5 units.
It implies;
a² = c²+d²
a = √(5²+5²)
a = √(50)...
Let a be the radius of the white inscribed circle.
b = (10-a) units.
Calculating a.
b² = a²+5²
(10-a)² = a²+25
100-20a+a² = a²+25
75 = 20a
a = ¼(15) units.
a = 3.75 units.
c = ½(10)
c = 5 units...
Let a be the radius of the half circle.
b = (a-1) units.
c²+1² = b²
c² = a²-2a+1-1
c = √(a²-2a) units.
d = a+c
d = (a+√(a²-2a)) units.
d is the required length.
e = (a+1) units....
x+a = y+2a
4+a = 3+2a
a = 1 unit.
It could be;
a = √(4²+3²)-4
a = 5-4
a = 1 unit.
Or
2a = √(4²+3²)-3
2a = 5-3
a = 1 unit.
Which makes;
b = y+a
b = 4+a
And a = 1 unit.
b = 5 units.
Or
b = y+2a
b =...
c² = 5³+12²
c = √(169)
c = 13 units.
Calculating a, radius of the inscribed yellow circle.
5a+12a+13a = (12*5)
30a = 60
a = 2 units.
Calculating b, radius of the blue inscribed circle....
a²+m² = 12²
a = √(144-m²) units.
Calculating m.
(4m)² = (3m)²+√(144-m²)²
16m² = 9m²+144-m²
8m² = 144
2m² = 36
m² = 18
m = 3√(2) units.
Recall.
a = √(144-m²)
And m = 3√(2) units.
a = √(144-(3√(2)...
