Let the side length of the regular Pentagon be 1 unit.
Therefore Area Pentagon is;
(5*1)÷(4tan(180/5))
= 5/(4tan36) space unit.
= 1.72047740059 space unit.
Area star is;
5(area triangle with heig...
At x = 0, y = 2.
At y = 16, x = 1.
Gradient of the curve at x = 1 or at the tangent is;
dy/dx = 3In2*2^(3x+1)
For x = 1
dy/dx = 48In2
Therefore coordinate Q is (x¹, 0).
Calculating x¹.
(0-16)/(x¹...
Let the side length of the be 2 unit.
Therefore;
BC = 2 unit.
It implies;
Area orange is;
½(1*2) = 1 square unit.
Area green is;
½(1*2) + ½(2/√(5))(4/√(5))
= 1 + (4/5)
= (9/5)...
Let the side length of the square be 1 unit.
Therefore;
Area square is;
1² = 1 square unit.
It implies;
Area R is;
Area square - Area equilateral triangle with side 1 unit - 2(area sector with r...
Notice.
The ascribed triangle is isosceles with angles 70°, 70° and 40°.
Let the two equal lengths be 1 unit.
(1/sin70) = (a/sin40)
a = (sin40/sin70)
a = 0.6840402867 units.
sin40 = b/a...
a² = 11*11+10*10-2*11*10cos50
a = 8.92113926968 cm.
(8.92113926968/sin50) = (10/sinb)
b = 59.16920318624°
c = 180-50-59.16920318624
c = 70.83079681376°
d = 0.5a
d = 4.46056963484 cm.
e² = 4.460...
Let AB = 1 unit.
Notice;
AB = BE
Therefore;
Length AE is;
√(2-2cos135)
= √(2+√(2)) unit
Length CG is;
√(2-2cos45)
= √(2-√(2)) unit.
Notice;
BD = BF = √(2)
Therefore;
Length DF is;
√(4-4cos4...
Let BC be 1 unit.
Therefore;
AB = 2 unit.
Radius of the circle, r is;
r = √(5) unit.
Therefore;
Area R is;
Area sector with radius √(5) unit and angle (2asin(√(2)/(2√(5))))° - 2(area triangle...
Let the side length of the regular pentagon be 1 unit.
tan36 = a/0.5
a = 0.363271264 unit.
b² = 0.363271264²+0.5²
b = 0.61803398875 unit.
tan72 = c/0.5
c = 1.53884176859 units.
Area Green is;
0...
At point P;
x = (π/6)~30°
y = √(3)-2
Gradient at point P is;
dy/dx = 2/(1+2sinxcosx)
At x = (π/6)~30°
dy/dx = 8-4√(3)
It implies;
yQ exactly as a single fraction in terms of π is;
The coordin...
