Let the single side length of the two congruent inscribed regular pentagon be 1 unit.
Area Green is;
0.5*5(1/(2tan(180/5)))
= 1.72047740059 square units.
Calculating Area Shaded.
a =...
Sir Mike Ambrose is the author of the question.
Area Orange exactly in square cm decimal is;
Area triangle with height 1.64419338427 cm and base (4.95325421888sin129.5411167921) cm
= ½*1.644...
a = atan(3/2)°
b = atan(2/3)°
b = Angle BEG.
tan(atan(2/3)) = c/2
⅔ = c/2
c = (4/3) cm.
c = BG.
OG = 2-(4/3)
OG = ⅔ cm.
Notice;
Radius of the circle is 2 cm.
Angle OGH = 180-...
Sir Mike Ambrose is the author of the question
Let the side of the regular Pentagon be 2 units.
Therefore;
Area orange is;
Area triangle with height 1.7013016167 units and base 1.8345763...
a² = 2*12²-2*12²cos120
a = 12√(3) cm.
a = 20.78460969083 cm.
a = AB.
b = 2a
b = 24√(3) cm.
b = 41.56921938165 cm.
b = AC.
c² = 12²+41.56921938165²-24*41.56921938165cos30
c = 31.7490157...
a² =2*12²
a = 12√(2) cm.
a = AC.
12b+12b+12√(2)b = 12*12
b = 144/(24+12√(2))
b = 12-6√(2) cm.
b = 3.51471862576 cm.
b = radius of the inscribed circle.
BF = ½(12√(2))
BF = 6√(2) cm....
a² = 15²+12²-2*12*15cos120
a = 23.43074902772 cm.
Where a is the side length of the equilateral triangle ABC.
(23.43074902772/sin120) = (12/sinb)
b = 26.32950349168°
c = 60-b
c = 33.67049...
Sir Mike Ambrose is the author of the question.
Area Green exactly is;
Area trapezoid with parallel sides 3 cm and (6-2√(3)) cm, and height (6-3√(3)) cm.
= ½(6-3√(3))(3+(6-2√(3)))
= ½(6-...
a² = 12²-6²
a = 6√(3) cm.
a = 10.39230484541 cm.
a = circle's radius = AD = AG.
b² = 2*10.39230484541²-2*10.39230484541²cos120
b = 18 cm.
b = DG.
c = 12-10.39230484541
c = 1.60769515459...
Let AB be 5 units.
Radius, r of the inscribed circle is;
5*(2/5)
r = 2 units.
a = ⅛(180*6)
a = 135°
2b² = 25
b = ½(5√(2)) units.
c = 180-0.5(135)-45
c = 67.5°
sin22.5 = ½(5√...
