Let the side length of the inscribed regular pentagon be 1 unit.
a = ½(120)
a = 60°
b = 180-108
b = 72°
c = 180-60-72
c = 48°
It implies;
(1/sin60) = (d/sin48)
d = 0.8581097301 u...
Let CE be 1 unit.
Observing Sum of Interior Angles of a Triangle.
a = 180-64-10
a = 106°
a is angle CED.
Observing Sine Rule.
(1/sin10) = (b/sin106)
b = 5.5356854811 units.
b is DE....
Let the two equal lengths be a
a²+x² = 10²
a = √(100-x²) units.
b = ½(8+10)
b = 9 units.
c²+(9-8)² = x²
c = √(x²-1) units.
Observing similar plane shape (right-angled) side length r...
Radius of the two congruent inscribed white circles a, is;
a = ½*4√(3)
a = 2√(3) cm.
b = 4√(3) cm.
b is the radius of the ascribed circle.
Area yellow is;
2(2(area sector with radius...
Let a be the radius of the big inscribed circle.
Let b be the radius of the bigger inscribed circle.
Therefore;
b = 2a cm.
c = 2a+a
c = 3a cm.
c is the radius of the ascribed circle....
Equation of the curve is;
y = ⅔(8x-x²)
Calculating Area E.
(Area under the curve at x = (3/2) and x = 4) - Area trapezium with parallel sides (13/2) units and (10⅔) units, and height (5/3)...
Calculating angle x.
Notice!
x (angle BCD) will sum up the interior reflex angle BCD of the plane shape ABCD to get 360°
Plane shape ABCD is a quadrilateral, therefore the sum of its inte...
Calculating h, height of the triangle.
Notice!
a+b = 28
a = (28-b) units.
It implies;
b² = 5²+h²
h² = b²-25 --- (1).
a² = 9²+h²
And a = (28-b) units.
(28-b)² = 81+h²
h² = 784-5...
2(20x)+2(30x)+2x² = (x+30)(x+20)
Where x is the radius of the inscribed circle.
It implies;
40x+60x+2x² = x²+20x+30x+600
100x+2x² = x²+50x+600
x²+50x-600 = 0
Calculating x via completing th...
Let a be the side length of regular hexagon and the regular triangle (equilateral triangle).
Calculating a.
0.5a²sin60 = √(3)
¼*√(3)a² = √(3)
¼(a²) = 1
a² = 4
a = 2 units.
Again, a is th...
