Notice!
Length 3 units and length 6 units are both tangent to the green inscribed circle.
It implies;
Radius r, of the ascribed semi circle is;
r = ½(3+6)
r = ½(9)
r = 4.5 units.
Where 9 units is...
Let the side length of the regular pentagon be a.
b = ⅕(180(5-2))
b = 108°
b is the single interior angle of the regular pentagon.
c = 108-90
c = 18°
d = (180-2(18))
d = (180-36)
d = 144°
e = ½(...
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Let AE = CE = BE = b.
Let OE = a.
4² = a²+b²-2abcos45
16 = a²+b²-√(2)ab --- (1).
6² = a²+b²-2abcos135
36 = a²+b²+√(2)ab --- (2).
Solving (1) and (2) simultaneously.
Subtracting (1) from...
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Area of coloured region is;
Area triangle with height 53.3479501083 m and base 60sin(102.042575143) m + Area triangle with height 100.027406452 m and base √(5924)sin(12.5245961777) m
= (½*53.34795...
Observing Pythagoras Rule.
(n³)² = (n²)²+n²
n⁶ = n⁴+n²
n⁶-n⁴-n² = 0
n²(n⁴-n²-1) = 0
Let n² = p.
It implies;
p(p²-p-1) = 0
p²-p-1 = 0
Resolving the above quadratic equation, applying c...
Sir Mike Ambrose is the author of the question.
Let beta be x.
x = (½*atan(5/4))
Let alpha be y.
y = (½*atan(4/5))°
Therefore;
Area blue exactly in cm² decimal is;
Area triangle with hei...
Let the two equal red lengths of the triangle, ABC be 1 unit.
Therefore;
AB = CE = 1 unit.
a = 180-100-20
a = 60°
(b/sin100) = (1/sin20)
b = 2.8793852416 units.
b is BC.
c = 2.8793852416-1
c = 1...
Sir Mike Ambrose is the author of the question.
Calculating a.
½((0, 6√(3)) (-3, 3√(3)) (a, (16a/3)) (0, 6√(3))) = ¼(9+20√(3))
Therefore;
a = x = ½√(3) unit
y = 16a/3 = (16/3)(√(3)/2)
y = ⅓(8√(3))...
