Calculating Area Orange.
Let the side length of the inscribed regular triangle be a.
It implies;
2a = 15
a = 7.5 units.
Area sector with radius 15 units and angle 60° - Area regular tr...
Calculating inscribed shaded area green.
a = ⅛*180(8-2)
a = ¼*180*3
a = 135°
a is the single interior angle of the regular octagon.
2b² = 4²
b² = 8
b = 2√(2) units.
c = 2b
c = 4√(2)u...
Let the side length of the square be a.
b²+(0.5a)² = 5²
b = √(25-¼(a²)) units.
c = a-b
c = (a-√(25-¼(a²))) units.
It implies, calculate a.
8² = (0.5a)²+(a-√(25-¼(a²)))²
64 = ¼(a²)+a²...
a = ½(360-2*50)
a = ½(260)
a = 130°
b² = 50²+50²-2*50*50cos130
b = 90.6307787037 units.
sin(0.5*50) = r/c
c = 2.3662015832r units.
sin(0.5*50) = R/d
d = 2.3662015832R units.
It im...
a = (6-b) cm.
b is the radius of the two congruent inscribed yellow half circles.
6 cm is the radius of the ascribed half circle and also, the diameter of the inscribed circle.
c = ½(6)
c =...
x²+(x²/(x+1)²) = 3
(x+1)²*x²+x² = 3(x+1)²
x²(x²+2x+1)+x² = 3(x²+2x+1)
x⁴+2x³+2x² = 3x²+6x+3
x⁴+2x³-x²-6x-3 = 0
It implies;
x = ½(1-√(5)) = -0.6180339887
or
x = ½(1+√(5)) = 1.6180339887...
Let the centre of the circle be N
Therefore;
N coordinate is;
N(2, 4).
Point P is;
P(-2, (17/2))
Point Q is;
Q(x, 0)
Gradient of length NP is;
(4-(17/2))/(2+2)
= -9/8...
πa² = 73π
a = √(73) units.
a is the radius of the circle.
b = 2a
b = 2√(73) units.
b is the diameter of the circle.
c²+7² = (2√(73))²
c² = 292-49
c² = 243
c = √(243)
c = 9√(3) units....
Calculating green area.
c² = a²+b²
c = √(4²+6²)
c = √(52)
c = 2√(13) units.
c = 7.2111025509 units.
d = 90-atan(4/6)
d = atan(3/2)°
e = 180-2d
e = (180-2ata(3/2))°
sin(0.5(180-2a...
Calculating AB.
Notice.
1 unit is the side length of each of the composite plane shape.
a = ½*180(6-2)
a = 120°
a is the single interior angle of the regular hexagon.
b = 180(12-2)/12...
