Sir Mike Ambrose is the author of the question.
Let the side length of one of the three congruent squares be 2 units.
Therefore;
Area square is;
2²
= 4 square units.
Area Green is;
Area triangle...
Let the side of the regular heptagon be 2 units.
Calculating PH.
AH is;
AH = 2sin(180/7)÷sin(720/7)
AH = 0.89008373583 unit.
AP is;
AP = √(2-2cos(540/7))
AP = 1.24697960372 unit...
Let a be the side length of the blue inscribed square.
Calculating Area of the blue inscribed square.
tan60 = a/b
b = ⅓(√(3)a) cm.
b = 0.5773502692a cm.
c = a+b
c = 1.5773502692a cm.
Calculating...
Let the square side length be 1 unit.
a = 1+1
a = 2 units.
a is twice the side length of the square which is also the height of the equilateral triangle.
Therefore;
tan60 = 2/b
b = ⅓(2√(3)) units...
a² = 3²+4²-2*3*4cos60
a² = 25-12
a = √(13) units.
a = 3.6055512755 units.
a is the side length of the square.
(√(13)/sin60) = (4/sinb)
b = 73.897886248°
sin73.897886248 = c/3
c = 2.8823067685 unit...
Let the radius of the ascribed half circle be 2 units.
Therefore;
Area Blue is;
Area triangle with height 4 units and base √(2)sin45 units.
= ½*4*√(2)sin45
= 2 square units.
Or
Area...
Calculating the area of the blue inscribed circle.
a = 8-5
a = 3 cm.
b²+3² = c²
c is the side length of the regular triangle.
b = √(c²-9) cm.
5²+d² = c²
d = √(c²-25) cm.
8²+e² = c²...
Let a be the side length of the inscribed square.
b² = 2a²
b = √(2)a units.
b is the radius of the half circle.
tanc = a/√(2)a
c = atan(1/√(2))°
d² = a²+(√(2)a)²
d² = a²+2a²
d = √(3)a units.
e =...
Let a be the radius of the inscribed semi circle.
Therefore;
Calculating a.
(2a)²+a² = 25²
5a² = 25*25
a = 5√(5) units.
2a is;
2*5√(5)
= 10√(5) units.
Therefore, red shaded region is;
Area tri...
a² = 3²+4²
a = 5 units.
a is the diagonal of the ascribed rectangle.
3b+5b+4b = 12
b = 1 unit.
b is the radius of the inscribed circle.
c = 3-1
c = 2 units.
tand = 2/1
d = atan(2)°
e = 360-90-2d...
