(5-r)² = r² + (3+r)²
Where r is the radius of the blue inscribed circle.
25-10r+r²=r²+9+6r+r²
25-10r=9+6r+r²
9+6r+r²+10r-25=0
r²+16r-16=0
(r+8)²-16-64=0
(r+8)² = 80
r = -8±4...
Sir Mike Ambrose is the author of the question.
Side length of the square is;
Let it be x.
x = √((16*34)-144)
x = √(400)
x = 20 cm.
Therefore;
Area Orange exactly in decimal cm² is;...
Let the radius of the blue inscribed circle be r.
Calculating r.
(5+r)² = (5-r)² + (5+√(25-10r))²
It implies;
20r = 25+10√(25-10r)+25-10r
30r-50 = 10√(25-10r)
(3r-5)² = 25-10r...
Let the side length if the regular hexagon be 1 unit.
Therefore the side length of the square, y is;
½+1+½
y = 2 units.
It implies;
The yellow shaded angle, x is;
(180 - 2atan(2-√...
Let the side of the regular pentagon be 1 unit.
a = 108-90
a = 18°
b = 180-108-18
b = 54°
(1/sin54) = (c/sin108)
c = 1.17557050458 units.
(1/sin54) = (d/sin18)
d = 0.38196601125 un...
Notice;
The diameter of the semicircle is 8 units.
Therefore radius is 4 units.
tan30 = a/2
a = ⅔(√(3)) units.
b = 3a
b = 2√(3) units.
c² = 2²+(⅔(√(3)))²
c = (4/√(3)) units.
d...
Let the inscribed regular pentagon side be 1 unit.
Area small/inscribed pentagon is;
0.5*5(1/(2tan(180/5)))
= 1.72047740059 square units.
Calculating Shaded Area.
a² = 2-2cos(60+108)...
Let the side of the regular pentagon be 1 unit.
a = 360-2(108)-90
a = 54°
tan72 = b/0.5
b = 1.53884176859 units.
tan54 = 1.53884176859/c
c = 1.11803398875 units.
d = c-0.5
d = 0.618...
Let the single side length of the two congruent inscribed regular pentagon be 1 unit.
Area Green is;
0.5*5(1/(2tan(180/5)))
= 1.72047740059 square units.
Calculating Area Shaded.
a = ⅕...
Let the side of the regular decagon which is equal the side of the regular octagon be 1 unit.
Calculating Area Gold.
a = ⅛(180*6)
a = 135°
b = ½(360-2(135))
b = 45°
sin45 = c/1
c = ½...
