Let the square side be 2 units.
tan30 = a/2
a = ⅓(2√(3)) unit.
sin60 = 2/b
b = ⅓(4√(3)) units.
c = 2-a
c = ⅓(6-2√(3)) units.
It implies;
Area Shaded is;
2(½*(⅓(6-2√(3)))²sin60)...
Sir Mike Ambrose is the author of the question.
Let the side of the inscribed square be 2 units.
Therefore;
Area S is;
Area triangle with two side 1.03528 units and 0.53589838486 units, a...
cos15 = 12/a
a = 12.42331416492 cm.
Where a is the side length of the inscribed regular triangle.
Radius, r of the inscribed circle is;
tan30 = r/(0.5* 12.42331416492)
r = 3.58630188867...
a = 90+45
a = 135°
b² = 6²+(4√(2))²-2*6*4√(2)cos135
b = 10.7703296143 units.
b = 2√(29) units.
b is BC.
(10.7703296143/sin135) = (6/sinc)
c = 23.1985905136°
c is angle ACD.
tan23.198...
Let the diameter of the blue semi circle be a.
Calculating a.
It implies;
(½*a*6)+(½*10*16) = 140
6a+160 = 2*140
6a = 280-160
6a = 120
a = 20 cm.
Again, a is the diameter of the blue...
Let x be the side of the small inscribed square.
Calculating x.
Notice;
The side length of the ascribed square is 14 cm.
A small triangle with height ½ cm on the same vertical length w...
Sir Mike Ambrose is the author of the question.
Let the small inscribed circle's radius be x.
x = 6 cm.
Let the big inscribed circle's radius be y.
y = 8 cm.
Let the ascribed semi circle's...
a = ½(10)
a = 5 units.
a is the radius of the circle.
b = (5-a) units.
c = (5-2a) units.
Calculating a.
5² = (5-2a)²+(5-a)²
25 = 25-20a+4a²+25-10a+a²
5a²-30a+25 = 0
a²-6a+5 = 0
a²...
a = 4+6
a = 20 cm.
a is the radius of the quarter circle.
tanb = 10/4
b = atan(5/2)°
c = b+60
c = (60+atan(5/2))°
c = 128.1985905136°
(10/sin128.1985905136) = (4/sind)
d = 18.3215084...
Calculating x.
(2√(3)/sin100) = (a/sin60)
a = 3.0462798357 units.
b = 180-(180-100-60)
b = 160°
Therefore, the required length, x is;
x² = 2(3.0462798357)²-2(3.0462798357)²cos160
x...
