Let the length (base) of the ascribed rectangle be 4 units.
Therefore its width/breadth/height is 2 units.
tana = 4/2
a = atan(2)°
b = atan(½)°
sin(atan(½)) = c/4
c = 3.577708764 unit...
Let the radius of the ascribed quarter circle be 1 unit.
It implies;
R = 1 unit.
2a² = 1²
a² = ½
a = ½√(2) units.
a is the side length of the inscribed square.
b = 2R
b = 2 units.
b...
Sir Mike Ambrose is the author of the question.
The vertices of area purple is;
(6, 12) ((24/5), (36/5)) ((180/29), (168/29)) (12, (21/2))
Area purple is;
½((6, 12) ((24/5), (36/5)) ((180/2...
Notice.
MN = CQ = 6 units.
MQ = 6+2
MQ = 8 units.
a² = 8²+6²
a = √(100)
a = 10 units.
a is CM.
3 - 6
r - 10
Cross Multiply.
6r = 30
r = 5 units.
r is the radius of the insc...
Calculating x.
Let the base of the ascribed triangle be 1 unit.
a = 180-15-45
a = 120°
(1/sin120) = (b/sin15)
b = 0.2988584907 units.
c = 180-15-30
c = 135°
(0.2988584907/sin15) =...
Notice.
Calculating the required angle, alpha.
4 units is the radius of the quarter circle.
It implies;
3² = 4²+4²-2*4*4cosa
9 = 32-32cosa
32cosa = 32-9
a = acos(23/32)
a = 44.04862...
Calculating Area Shaded.
Let r be the radius of the semi circle.
Calculating r.
((6r/√(10))-2)²=(r/2)²+(3r/2)²
72r²-48√(10)r+80=50r²
22r²-48√(10)r+80= 0
11r²-2⁴√(10)r+40= 0
Therefore...
Calculating area of the inscribed red square.
Let a be the side length of the inscribed red square.
2b² = a²
b = √(a²/2)
b = ½√(2)a units.
10² = a²+(½√(2)a)²-2a*½√(2)acos(45+90)
100 = ½...
Let the base of the quadrilateral be 1 unit.
a = 180-58-28-38
a = 180-124
a = 56°
(1/sin56) = (b/sin(58+28))
b = 1.2032796622 units.
(1/sin56) = (c/sin38)
c = 0.7426219217 units.
d...
Let the ascribed quarter circle radius be 1 unit.
2a² = 1²
a² = ½
a = ½√(2) units.
a = 0.7071067812 units.
a is the radius of the blue inscribed quarter circle.
Area blue inscribed quarte...
