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√...
AB = BC = AC = 6 cm.
cos60 = 2/BE
BE = 4 cm.
(DE)² = 4²-2²
DE = 2√(3) cm.
(CE)² = (2√(3))²+4²
CE = 12+16
CE = 2√(7) cm
EF = (2√(7)-4) cm.
EF = 1.29150262213 cm.
AE = 6-4
AE = 2...
a = ½(180-150)
a = 15°
b = 90-15
b = 75°
(c/sin45) = (12/sin60)
c = 9.79795897113 cm.
(12/sin60) = (d/sin45)
d = 9.79795897113 cm.
e² = 9.79795897113²+12²-24*9.79795897113cos135 ...
The radius of the circle is 6 cm.
a = atan(2)°
Where a is angle AOB = angle COE.
Angle BOE = 180-atan(2)
= 116.56505117708°
Angle OBE = ½(180-116.56505117708)
= 31.71747441146°
b =...
Let GD (inscribed square side) be a.
Calculating a.
2(a/(tan60))+a = 12
a = 12/((2/√(3))+1)
a = 5.56921938165 cm.
b = ½(12-5.56921938165)
b = 3.21539030918 cm.
Where b is BD = CE.
c = 3.21539030...
a² = 12²-6²
a = 6√(3) cm.
Where a is BD.
b = ½(a)
b = 3√(3) cm.
Where b is BG = DG.
c = atan(6/(3√(3))
c = 49.10660535087°
Where c is angle AGD = angle BGE.
d = 60-49.10660535087
d...
Let a be the radius of the inscribed circle.
Calculating a.
tan30 = a/6
a = 2√(3) cm.
b = inscribed circle diameter.
b = 4√(3) cm.
12² = c²+6²
c = 144-36
c = √(108)
c = 6√(3) cm.
d = 6√(3)-4√(3)...
