7² = 3²+8²-2*3*8cosa
48cosa = 9+64-49
a = acos((9+64-49)/48)
a = 60°
Calculating b, radius of the inscribed circle.
3b+8b+7xb = 3*8sin60
18b = 12√(3)
3b = 2√(3)
b = ⅔√(3) units.
b = 1....
Sir Mike Ambrose is the author of the question.
Let the side of the regular pentagon be 1 unit.
Therefore;
Area purple is;
Area trapezium with two parallel side 0.07294901687 units and 0....
Notice;
Green regular hexagon's side is 8 cm.
sin60 = a/8
a = 4√(3) cm
Where a is the the height of the white triangle enclosed by the two green regular hexagon and the regular pentagon....
Radius of the circle, r is;
√(2-2cos120)
r = √(3) units.
Therefore!
Area yellow is;
Area regular hexagon with side 2 units - 2(area triangle with height ½ units and base √(3) units) -...
Let the base of the triangle ascribing the green and yellow area b 1 unit.
tan55 = a/1
a = 1.42814800674 units.
tan55 = b/(1-b)
b = 0.58816349035 units.
Where b is the side of the white...
Sir Mike Ambrose is the author of the question.
Notice;
Square side is 4 units.
Small regular hexagon side is 2 units.
Big regular hexagon side is 4 units.
Therefore;
Area shaded, exa...
Let the regular octagon side be 2 units.
Area Octagon is;
½(8*2)*(2/(2tan(180/8)))
= 8/(tan(180/8))
= 19.31370849898 square units.
Calculating Area Red.
a = ⅛(180*6)
a = 135°
Le...
a² = R²+6²
a = √(R²+36) units.
a is OB.
b = a-4
b = (√(R²+36)-4) units.
c = 6+4
c = 10 units.
It implies;
10² = R²+(√(R²+36)-4)²
100 = R²+R²+36-8√(R²+36)+16
48 = 2R²-8√(R²+36)
8√...
Let the single side length of the two congruent inscribed regular pentagon be 1 unit.
Area Green is;
0.5*5(1/(2tan(180/5)))
= 1.72047740059 square units.
Calculating Area Shaded.
a = ⅕...
Let a be the big inscribed circle radius.
Let b be the bigger inscribed circle radius.
Let c be the biggest inscribed circle radius.
It implies;
a+b = 12 --- (1).
a+c = 22 --- (2)....
