Sir Mike Ambrose is the author of the question.
Area purple is;
Area trapezium with two parallel side length (58/15) cm and (46/5) cm, and height 4 cm.
= ½((58/15)+(46/5))*4
= (392/15) cm²...
Calculating area shaded blue.
a² = (2+√(3))²+(3+2√(3))²
a² = 4+4√(3)+3+9+12√(3)+12
a² = 28+16√(3)
a = √(28+16√(3)) units.
a = 7.4641016151 units.
a is the hypotenuse of the ascribed right-a...
Calculating red angle.
Let the equal two red angles be x each.
An irregular hexagon is drawn/derived.
It implies;
90+90+109+(360-15)+2x = 180(6-2)
289+345+2x = 720
634+2x = 720
2x =...
Blue area is;
8(area quarter circle with radius 4 cm - area triangle with height and base 4 cm each) -
8(area quarter circle with radius 2 cm - area triangle with height and base 2 cm each)...
a = ½(12)
a = 6 cm.
a is the radius of the inscribed half circle.
c = (6+b) cm.
b is the radius of the inscribed circle.
d = (12-b) cm.
Calculating b.
(6+b)²+6² = (12-b)²
36+12b+b²+...
tan60 = a/2
a = 2√(3) units.
a is the height of the inscribed red regular triangle.
b = 2+a
b = (2+2√(3)) units.
Area green is;
(½*4*(2+2√(3)))-(½*2√(3)*4)
= 4+4√(2)-4√(3)
= 4 square...
Notice.
OE = AE = ½(6)
OE = AE = 3 cm.
a²+3² = 6²
a² = 36-9
a = √(27)
a = 3√(3) cm.
tanb = a/3
b = atan(√(3))
a = 60°
Shaded Area is;
(0.5*3*3√(3))+(60π*6*6/360)-(0.5*6*6sin60...
a = (2+x) units.
a is the radius of the ascribed sector.
b = 2x units.
x is the radius of the inscribed quarter circle.
Calculating x.
(2+x)² = x²+(2x)²
4+4x+x² = 5x²
x²-x-1 = 0
(x-½)...
Calculating x.
2*2^(x)-9^(x) = 0
2*2^(x) = 9^(x)
Dividing through by 2^(x).
2 = (4.5)^(x)
Taking log base 10 at both sides of the equation.
log2 = log(4.5)^(x)
x = log2/log(4.5)
x =...
Area red is;
½(area square with single side length 24 units - 4(area circle with radius 6 units))
= ½(24²-4(36π))
= ½(576-144π)
= 288-72π
= 72(4-π) square units.
Or
2(area square wit...
