Area green is;
(Area triangle with height 4 units and base 8√(3) units - Area sector with radius 8 units and angle 30°) + (Area sector with radius 8 units and angle 60° - Area triangle with heig...
Let x³ be a.
Therefore;
a²-a-2 = 0
Resolving the above quadratic equation via completing the square approach.
(a-(1/2))² = 2+(1/2)²
(a-(1/2))² = (9/4)
a-(1/2) = ±(3/2)
a = ½±(3/2)
It implies;
a =...
a = ¼(7*7)π-½(7*7)
a = ¼(49π)-½(49)
a = ¼(49π-98) square units.
b = 2a
b = 2*¼(49π-98)
b = ½(49π-98) square units.
It implies;
Area of red region is;
Half area square with side 14 units - b
= ½...
Let the radius of the small yellow inscribed circle be a.
Let the radius of the big green inscribed circle be b.
Notice!
PQ = a+b
Calculating radius of the ascribed semi circle.
c² = 12²+16²
c =...
Let the side of the three inscribed congruent regular hexagon be 1 unit.
It implies, the side length of the square is;
a = 6 units.
b² = 2-2cos120
b = √(3) units.
c = 1+(½)
c = ½(3) units.
c = 1...
Radius of the semi circle r is;
r = ½(10)
r = 5 cm.
a = 1+5
a = 6 cm.
b = 5-1
b = 4 cm.
a² = b²+c²
c = √(6²-4²)
c = √(20)
c = 2√(5) cm.
cosd = 4/6
d = acos(2/3)
d = 48.189685104...
Calculating Area triangle ABC.
Let a be the radius of the inscribed circle.
b² = 2a²
b = √(2)a units.
b is EG.
c² = a²+(2+a)²
c² = a²+4+4a+a²
c = √(2a²+4a+4) units.
c is CE.
Theref...
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Notice!
The area of the regular pentagon and the square are the same.
Let it be 4 square units.
Therefore the side length of the square a, is;
a = √(4)
a = 2 units.
Calculating b, the side length...
Notice.
The polygon is regular hexagon.
The three circles are congruent.
The inscribed triangle is equilateral.
Let the side length of the regular hexagon be a.
Let the a the congruent circles' rad...
