White area is;
Area sector with radius 6 cm and angle 120° + Area sector with radius 2 cm and angle 60°
= (120*36π÷360) + (60*4π÷360)
= 12π + ⅔(π)
= ⅓(38π) cm²
Green area is;
Area circle with radius 5 units + 2(area sector with radius 20 units and angle 60°) - Area equilateral triangle with side 20 units.
= π(5²) + 2(60÷360*20²π) - ½(20²)sin60
= 25...
Let the radius of the inscribed circle be r.
The radius of the semi circle is 5 units.
Therefore;
(2r)² + r² = 5²
4r² + r² = 25
r = √(5) units.
Therefore shaded area/area blue is;...
Sir Mike Ambrose is the author of the question.
AB is 2 cm.
BC is 6 cm.
The single side length of equilateral triangle BDE is;
= ⅓(16√(3)) cm.
Therefore;
Area Orange is;
Area trian...
sin30 = a/12
a = 6 cm.
Where a is the height of the green regular triangle.
sin60 = 6/b
½√(3) = 6/b
b = 12/√(3)
b = 4√(3) cm.
b is the side length of the green regular triangle.
tan30 =...
Let the side length of the square be a.
b = ½(a) cm.
b is the radius of the inscribed half circle.
c = (a-2) cm.
d²+2² = (a-2)²
d² = a²-4a+4-4
d = √(a²-4a) cm.
d is the radius of the b...
Let the side length of the regular pentagon be 2 units.
a = ⅕*180(5-2)
a = 108°
a is the single interior angle of the regular pentagon.
b² = 1²+2²-2*1*2cos108
b = 2.497212041 units.
(2....
Let the side length of the ascribed regular hexagon be 1 unit.
a = ⅙*180(6-2)
a = 120°
a is the single interior angle of the ascribed regular hexagon.
b = ¼(a)
b = 30°
d = (1-c)
c is t...
Calculating r, the radius of the blue inscribed circle.
Therefore;
Consider triangle OBC.
(6-r)² = r² + x², and x = √(9+6r) - 3
It implies;
(6-r)² = r² + (√(9+6r)-3)²
36-12r+r²=r²+9-...
Let the radius of the quarter circle be 2 units.
Therefore;
OD = BD = ½(2)
OD = BD = 1 units.
a² = 1²+2²-2*1*2cos45
a = 1.4736257582 units.
a is CD.
(1.4736257582/sin45) = (2/sinb)...
