a = 8+9
a = 17 units.
a is the radius of the half circle.
b²+8² = 17²
b = √(225)
b = 15 units.
c²+d² = 17²
c = √(289-d²) units.
e = (17-d) units.
f = ½(c)
f = ½√(289-d²) unit.
g = 17+17-9
g = 3...
Let the equal side lengths be a.
14² = a²+16²-2*16acosb
196 = a²+256-32acosb --- (1).
21² = 2a²+2a²cosb
cosb = (441-2a²)/2a² --- (2).
Substituting (2) in (1).
196 = a²+256-32a(441-2a²...
Notice!
r, radius of the inscribed quarter circle is also the side length of the ascribed square.
Let alpha be a.
Notice again.
2a = 60°
a = 30°
Alpha = 30°
It implies;
sin30 =...
Let the square side length be a.
Observing similar plane shape (right-angled) side length ratios.
8 - ½(a)
b - a
Cross Multiply.
½(ab) = 8a
b = 16 units.
c² = 8²+16²
c = √(64+256)
c = 8√(5) units...
let theta be a.
tana = k/3k
a = atan(⅓)°
Notice!
3k = 4.
b = 2a
b = 2atan(⅓)°
It implies;
tan(2atan(⅓)) = c/4
c = 3 units.
Therefore, ? the required length is;
Let if be d.
d = 4-c
d = 4-3
d...
Calculating angle x (angle CBD).
Let AC = 1 unit.
a = 180-12-18
a = 150°
a is angle ADC.
(b/sin12) = (1/sin150)
b = 0.4158233816 units.
b is CD.
c = 180-(54+12)-(48+18)
c = 180-66-66
c = 48°
c i...
a = 180-54-48
a = 180+102
a = 78°
b = 30+48
b = 78°
c = 180-2*72
c = 36°
Let the two congruent side lengths be 1 unit each.
d² = 2-2cos36
d = 0.6180339887 units.
e = 180-30-24...
Let AD be x.
tana = 4/x --- (1).
tana = x/9 --- (2).
Equating (1) and (2).
4/x = x/9
Cross Multiply.
x² = 36
x = 6 units.
Again, x is AD, the height of the ascribed trapezoid AB...
a = ½(24)
a = 12 units.
a is the radius of the half by circle.
sinb = 8/24
b = asin(⅓)°
tan(asin(⅓)) = c/12
c = 4.2426406871 units.
tand = 12/4.2426406871
d = 70.5287793656°
e = ½(d)
e = 35.2643...
a = (r-7) units.
b = (r-14) units.
Therefore, r, radius of the inscribed circle is;
r² = a²+b²
r² = (r-7)²+(r-14)²
r² = r²-14r+49+r²-28r+196
r²-42r+245 = 0
Resolving the above quadratic equation...
