Let a be the diameter of the smaller inscribed circle.
2a is the radius of the ascribed semi circle.
(2a)² = (0.5*3)²+b²
b² = 4a²-2.25
b = √(4a²-2.25) units.
(2a)² = (0.5*5)²+c²
c² = 4a²-6.25
c =...
a² = 2√(6)²
a² = 12
a = √(12)
a = 2√(3) units.
b = ½(a)
b = ½(2√(3))
b = √(3) units.
2c² = √(3)²
2c² = 3
c = √(3/2) units.
c =½√(6) units.
d = 2√(6)-½√(6)
d = ½(4√(6)-√(6))
d = ½(3√(6)) units.
d...
a = 90+20
a = 110°
sin20 = b/1
b = sin20 units.
b = 0.3420201433 units.
cos20 = c/1
c = cos20 units.
c = 0.9396926208 units.
Calculating d, the square side length observing similar plane shape si...
Let a be 1 unit.
c² = 1²+1²-2*1*1cos120
c² = 2-2cos120
c = √(3) units.
c = 1.7320508076 units.
It implies;
1+b = c
1+b = √(3)
b = (√(3)-1) units.
b = (1.7320508076-1) units.
b = 0.7320508076 unit...
a² = 30²+10²-2*30*10cos120
a = 36.0555127546 cm.
(36.0555127546sin120) = (30/sinb)
b = 46.1021137521°
cosb = 10/d
Where d is PQ.
cos46.1021137521 = 10/d
d = 10/cos46.1021137521
d = 14.4222051019 c...
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Notice, AB = 4 units is the height of the inscribed regular pentagon.
Calculating Area Blue.
a = ⅕(180(5-2))
a = 108°
a is the single interior angle of the inscribed regular pentagon.
b = ½(a)
b...
Calculating x, the diameter of the semi circle.
Notice, we have a cyclic quadrilateral in the complete circle.
Therefore;
x² =11²+y²
y² = x²-121 --- (1).
cosa = (11/x) --- (2).
y² = 7²+2²+2*2*7c...
Notice.
Area Blue is an inscribed half circle.
Let r be the radius of area blue.
r² = (r-1)²+(r-2)²
r² = r²-2r+1+r²-4r+4
0 = r²-6r+5
Resolving the above quadratic equation via completing the squa...
Calculating r, radius of the two congruent circles.
r² = 2²+a²
a = √(r²-4) units.
r² = 1²+b²
b = √(r²-1) units.
c = 2+1+1
c = 4 units.
d = b-a
d = (√(r²-1)-√(r²-4)) units.
Therefore, r, radius...
