Sir Mike Ambrose is the author of the question.
Let the side of the square be 2 units.
Area square is;
2²
= 4 square units.
Area orange is;
Area triangle with two side 0.70710678119 u...
Sir Mike Ambrose is the author of the question.
Area red is;
Area triangle with two side 4 cm and 8 cm, and angle 110°.
= 0.5*4*8sin110
= 15.0350819326 cm²
Area blue is;
Area triangle wit...
a² = 2²+2²-2*2*2cos120
a = √(8+½(8))
a = √(12)
a = 2√(3) units.
a is the side length of each of the 4 inscribed small regular hexagon.
sin30 = b/a
b = asin30
b = ½*2√(3)
b = √(3) units....
Let a be the congruent side lengths each.
b = ½(2)
b = 1 unit.
c²+1² = a²
c = √(a²-1) units.
d = ½(180-45)
d = ½(135)
d = 67.5°
tan67.5 = e/1
e = 2.4142135624 units.
e = (1+√(2))...
a = ¼(60)
a= 15 units.
b² = 40²+15²
b = 5√(73) units.
b = 42.7200187266 units.
c = ½(b)
c = 2.5√(73) units.
c = 21.3600093633 units.
c is each of the congruent lengths.
d = 60-15
d...
a = 180-54-30-18
a = 180-102
a = 78°
a is angle ABC.
b = 78+54
b = 132°
b is angle BAD.
c = 180-b-18
c = 180-150
c = 30°
c is angle ADB.
Let AD = 1 unit.
(1/sin18) = (d/sin30)...
Calculating h.
tan60 = h/a
a = ⅓(√(3)h) units.
tan30 = h/b
b = √(3)h units.
It implies;
a + b = 100
⅓(√(3)h)+√(3)h = 100
√(3)h+3√(3)h = 300
4√(3)h = 300
√(3)h = 75
h = ⅓(75√(3))...
Let a be the radius of the half circle.
a-x = 4
x = a-4 --- (1).
b = (a-2x) units.
It implies;
a² = 4²+(a-2x)²
a² = 16+a²-4ax+4x²
4ax = 16+4x² --- (2).
Substituting (1) in (2) to...
Radius of the quarter circle is (2+3) = 5 units.
Let a be the side length of the inscribed square.
b²+3² = a²
b = √(a²-9) units.
c = 3+b
c = (3+√(a²-9)) units.
Calculating a.
(3+√(...
Sir Mike Ambrose is the author of the question.
Let the ascribed quarter circle radius be 4 units.
Therefore;
Area blue is;
Area triangle with two side 1 unit and 3 units, and angle 2asin(...
