Calculating blue area.
It is;
Area square with side length 3 cm + Area semi circle with radius 1 cm + Area semi circle with radius 2 cm.
= (3*3)+(π*½*1²)+(π*½*2²)
= 9+½(π)+2π
= ½(18+π+4π...
cos30 = a/16
a = 8√(3) cm.
cos30 = 10/b
b = ⅓(20√(3)) cm.
c = a-b
c = 8√(3)-⅓(20√(3))
c = ⅓(4√(3)) units.
d = 2c
d = ⅓(2√(3)) cm.
e = b+d
e = ⅓(20√(3))+⅓(2√(3))
e = ⅓(22√(3)) cm...
Notice!
Radius of the half circle is 9 units.
Let the three equal lengths be b.
5² = b²+a²-2bacos45
25 = b²+a²-√(2)ab --- (1).
9² = b²+a²-2abcos135
81 = b²+a²+√(2)ab --- (2).
Subtra...
Sir Mike Ambrose is the author of the question.
Calculating Yellow Area.
It is;
Area triangle with height 4sin(atan(⅔)) units and base 8√(13)/9 units - Area triangle with height 1 unit and bas...
a²+2² = 8²
a² = 64-4
a = √(60)
a= 2√(15) units.
a is BD
b = ½(a)
b = √(15) units.
c = ½(AB)
c = 4 units.
c is the radius of the half circle.
cosd = √(15)/4
d = acos(¼√(15))°
d is a...
Let the two equal lengths be a units each.
a² = 2b²
b² = a²/2
b = ½√(2)a unit.
It implies;
14+½√(2)a* ½√(2)a = 2a*2a
14+½(a)² = 4a²
14 = 4a²-½(a)²
14 = ½(7)a²
2 = ½(a)²
a² = 4
a =...
a² = 4²+2²
a = √(20)
a = 2√(5) units.
a is the diameter of the circle.
b = ½(a)
b = √(5) units.
b is the radius of the circle.
tanc = 4/2
c = atan(2)°
d = 180-c
d = (180-atan(2))°...
Let the small inscribed regular triangle side length be x.
It implies;
Calculating x.
2² = x²+(3x)²-2*x*3xcos60
4 = 10x²-3x²
4 = 7x²
x = ⅐(2√(7)) units.
Therefore;
3x is;
= 3*⅐(2...
tan60 = √(3)/a
a = 1 unit.
b = 8-a
b = 7 units.
tanc = √(3)/7
c = atan(√(3)/7)°
d = 2c
d = 2atan(√(3)/7)°
e = 180-60-d
e = (120-2atan(√(3)/7))°
f = ½(e)
f = (60-atan(√(3)/7))°...
a² = 2²+b²
a = √(4+b²)
a is AB.
b is the height of the yellow inscribed triangle.
0.5*b*c = 6
bc = 12
c = 12/b units.
c is the base of the inscribed yellow triangle.
d² = b²+(12/b)²
d²...
