Let the side length of the ascribed regular hexagon be 2 units.
Calculating Area B.
½*1*1sin60
= ¼√(3) square units.
a = ⅙*180(6-2)
a = 120°
a is the single interior angle of the ascrib...
Notice.
ABCD is a square.
Let AB equal 1 unit.
sin70 = 1/a
a = 1.0641777725 units.
a is AE.
b = 90-70
b = 20°
b is angle DAE.
c = 90-45-20
c = 25°
c is angle BAF.
cos25 = 1/...
Let the radius of the quarter circle be a.
b = (a-4) units.
c = (a-2) units.
It implies;
a² = (a-4)²+(a-2)²
a² = a²-8a+16+a²-4a+4
a²-12a+20 = 0
a²-2a-10a+20 = 0
a(a-2)-10(a-2) = 0...
Sir Mike Ambrose is the author of the question.
Area red is;
Area triangle with height (18/7) units and base (24/7) units.
= ½((18/7)*(24/7))
= (216/49) square unit.
Area orange is;
Ar...
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...
