a²+2² = 8²
a² = 64-4
a = √(60)
a= 2√(15) units.
a is BD
b = ½(a)
b = √(15) units.
c = ½(AB)
c = 4 units.
c is the radius of the half circle.
cosd = √(15)/4
d = acos(¼√(15))°
d is a...
Let the two equal lengths be a units each.
a² = 2b²
b² = a²/2
b = ½√(2)a unit.
It implies;
14+½√(2)a* ½√(2)a = 2a*2a
14+½(a)² = 4a²
14 = 4a²-½(a)²
14 = ½(7)a²
2 = ½(a)²
a² = 4
a =...
a² = 4²+2²
a = √(20)
a = 2√(5) units.
a is the diameter of the circle.
b = ½(a)
b = √(5) units.
b is the radius of the circle.
tanc = 4/2
c = atan(2)°
d = 180-c
d = (180-atan(2))°...
Let the small inscribed regular triangle side length be x.
It implies;
Calculating x.
2² = x²+(3x)²-2*x*3xcos60
4 = 10x²-3x²
4 = 7x²
x = ⅐(2√(7)) units.
Therefore;
3x is;
= 3*⅐(2...
tan60 = √(3)/a
a = 1 unit.
b = 8-a
b = 7 units.
tanc = √(3)/7
c = atan(√(3)/7)°
d = 2c
d = 2atan(√(3)/7)°
e = 180-60-d
e = (120-2atan(√(3)/7))°
f = ½(e)
f = (60-atan(√(3)/7))°...
a² = 2²+b²
a = √(4+b²)
a is AB.
b is the height of the yellow inscribed triangle.
0.5*b*c = 6
bc = 12
c = 12/b units.
c is the base of the inscribed yellow triangle.
d² = b²+(12/b)²
d²...
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...
