Let a be the side length of the square.
tanb = a/0.5a
b = atan(2)°
c² = a²+(0.5a)²
c = 1.1180339887a m.
Observing similar plane shape (right-angled) side length ratios.
1.1180339887a...
Let a be the radius of the ascribed circle.
a² = (0.5*6)²+b²
b = √(a²-9) units.
c = a-b
c = (a-√(a²-9)) units.
d = ½(c)
d = ½(a-√(a²-9)) units.
d is the radius of the small inscribed c...
Sir Mike Ambrose is the author of the question.
Area Green is;
Area triangle two lengths 16/3 units and ⅓√(181) units, and angle 138.0127875°, the angle they form + Area triangle two lengths...
Purple area will be;
(⅓) of (1/6) the hexagon area.
= (1/18) of 1 square unit.
= (1/18) square unit.
a² = 3+1-2√(3)cos120
a = 2.394170171 units.
a is AC.
(2.394170171/sin120) = (1/sinb)
b = 21.206023113°
sinc = √(2)/2.394170171
c = 36.2060231125°
c is angle BAC.
Therefore, the requir...
Area Blue will be;
Area sector with radius 3 unit and angle 60° - Area equilateral triangle with two lengths 2 unit each and angle 60°, the angle they form - 2(Area sector with radius 1 unit and...
AD is;
√(8²+3²-2*8*3cos60)
= √(73-24)
= √(49)
= 7 unit.
Therefore radius of area S1 is;
3r+7r+8r=12√(3)
r = (12√(3))/18
r = ⅔(√(3)) unit.
Area S2 = π(⅔(√(3)))²
= ⅓(4π) square unit...
Let the side of the ascribed regular hexagon be 2 units.
Therefore;
Area Blue is;
Area triangle with height 1.5 units and base ½(√(3)) units.
= ½*1.5*½(√(3))
= ⅛(3√(3)) square units.
Area Red is...
Calculating the area of the inscribed circle.
a = (9+r) cm.
b = (9-r) cm.
It implies;
(9+r)² = c²+(9-r)²
81+18r+r²-(81-18r+r²) = c²
c² = 36r
c = 6√(r) cm.
d = (6√(r)-9) cm.
e =...
Let Phi be a.
let Beta be b.
Notice.
b = 2a
And b = 45°
It implies;
2a = 45
a = 22.5°
Let CF = c.
tan22.5 = c/d
d = c/tan22.5 units.
d is CE.
Therefore;
c+d = 12
c+(...
