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Calculating R, radius of the four inscribed equal half circles.
Notice!
Radius of the four equal inscribed circles is 1 unit.
Therefore;
(2R-1)² = 1²+a²
4R²-4R+1 = 1+a²
a = √(4R²-4R) uni...
55+2(90)+270+a = 180(5-2)
a+505 = 540
540° is the sum of the interior angles of a pentagon.
It implies;
a = 540-505
a = 35°
tan55 = b/4
b = 5.712592027 cm.
c = b-4
c = 1.712592027 cm....
a² = 4²+6²-2*4*6cos60
a = 5.2915026221 units.
(5.2915026221/sin60) = (4/sinb)
b = 40.8933946491°
c² = 6²+2²-2*6*2cos60
c = 5.2915026221 units.
(5.2915026221/sin60) = (2/sind)
d = 19.1066053509°...
Let the inscribed blue equilateral triangle side length be 1 unit.
Area Blue is;
0.5*1*1sin60
= ¼√(3) square units.
= 0.4330127019 square units.
0.5² = 2-2cosa
2cosa = 2-0.25
cosa = (1....
Let the radius of the ascribed circle be 1 unit.
Area ascribed circle is;
πr²
= π(1)²
= π square units.
Calculating area inscribed red circle.
Let it's radius be r.
sin30 = a/1
a...
Method A.
15² = (0.5*24)²+a²
a² = 225-144
a = √(81)
a = 9 units.
a is the side length of the two equal lengths each.
tanb = 9/12
b = atan(¾)°
c = 90-b
c = atan(4/3)°
Therefore, r,...
a*x = 15*15
ax = 225
a = 225/x --- (1).
(30+a)*x = (30+15)*(30+15)
ax+30x = 2025 --- (2).
Substituting (1) in (2).
(225/x)x+30x = 2025
225+30x = 2025
30x = 2025-225
30x = 1800
x = 1...
Let R=the bigger inscribed circle radius.
Let r=the smaller inscribed circle radius.
It implies;
R+r=6√(3)-6=4.4
R+r=4.4 -------(1).
And the radius of the ascribed semi circle is 6 un...
a² = 2-2cos120
a = √(3) units.
(2/sin30) = (√(3)/sinb)
b = 25.6589062733°
c = 180-30-b
c = 124.3410937267°
d² = √(3)²+2²-2*2√(3)cos124.3410937267
d = 3.3027756377 units.
e² = 3.3027...
