Let AD be x.
a² = x²+4²-2*4*xcos30
a = √(x²+16-4√(3)x) units.
a is BD.
b² = x²+3²-2*3*xcos30
b = √(x²+9-3√(3)x) units.
b is CD.
c² = 4²+3²-2*3*4cos60
c² = 25-12
c = √(13) units.
c is BC.
Calcula...
Calculating R, radius of the circle.
a²+4² = 6²
a = √(36-16)
a = √(20)
a = 2√(5) units.
b²+4² = 10²
b = √(100-16)
b = √(84)
b = 2√(21) units.
Therefore;
2√(5)*2√(21) = 4c
c = √(1...
Let the two equal lengths be 1 unit each.
(1/sin20) = (a/sin40)
a = 1.87938524157 units.
b = a-1
b = 0.87938524157 units.
(1/sin20) = (c/sin(180-40-20))
c = 2.53208888624 units.
d² = 2(2.5320888...
Green Area is;
Area square with length 4 units - Area quarter circle with radius 2 units - Area square with length 2 units - 3(Area square with length 1 unit) - Area square with length 3 units +...
Calculating Length x.
7² = x²+(x+5)²-2*(5+x)xcosa
49 = 2x²+10x+25-(10x+2x²)cosa
(10x+2x²)cosa = 2x²+10x-24
cosa = (2x²+10x-24)/(10x+2x²) --- (1).
7² = x²+5²+2*x*5cosa
49-25-x² = 10xcosa
24-x² = 10...
Calculating length c.
cosd = 9/12
d = acos(3/4)°
e = 2d
e = 2acos(¾)°
Therefore, length c is;
cose = c/a
cos(2acos(¾)) = c/12
c = 12cos(2acos(¾))
c = 12*⅛
c = ½(3) units.
c = 1.5 units.
Let a be the diameter of the circle.
x² = 6²+6²+2*6*6cosb
x² = 72+72cosb --- (1).
x² = 14²+a²-2*14acosb
x² = 196+a²-28acosb --- (2).
Notice.
x is length AC
b is angle ADC.
cosb =...
Let a = AB = AD = CD.
b² = 2a²
b = √(2)a units.
b is BD.
Notice.
Triangle BCD is a right-angled triangle, where BC is the hypotenuse.
c² =(√(2)a)²+a²
c = 3a²
c = √(3)a units.
c is BC.
cosd = (0....
Sir Mike Ambrose is the author of the question.
Calculating Exactly Green Area.
Let x be the side length of each of the three inscribed congruent squares.
a² = 2x²
a = √(2)x
a is the diago...
Let a be the radius of the quarter circle.
b = (a-6) units.
c = (a-8) units.
Let x be the side length of the square.
x² = (a-8)²+(a-6)²
x² = a²-16a+64+a²-12a+36
x² = 2a²-28a+100
x =...
