Let the radius of the two equal inscribed semi circles be 1 unit each.
2² = a²+1²
a² = 3
a = √(3) units.
b = 1+1+a
b = (2+√(3)) units.
b is the height of the ascribed right-angled triangl...
Sir Mike Ambrose is the author of the question.
Let the side length of the regular ascribed pentagon be 1 unit.
a = ⅕*180(5-2)
a = 108°
a is the single interior angle of the regular pentagon....
a = ½(10)
a = 5 cm.
a is the radius of the ascribed semi circle.
b = (5-c) cm.
c is the radius of the inscribed blue circle.
It implies;
(5-c)² = c²+d²
25-10c+c² = c²+d²
d² = 25-10c
d = √(25-10c)...
Notice!
A regular triangle (equilateral triangle) with side length 2 units is inscribed the quarter circle due to the apex vertices of the two equilateral triangles on point the circumference of t...
Area R is;
Area sector with radius 8 cm and angle 20.16632290743° - Area triangle with height 8 cm and base (4.2sin 20.16632290743) cm
= ( 20.16632290743 *64π÷360) - (½*8*4.2sin 20.166322907...
Let BC = 1 unit.
(1/sin36) = (a/sin72)
a = 1.6180339887 units.
a is BD = CD.
b = 180-12-30
b = 138°
b is angle BAD.
(1.6180339887/sin138) = (c/sin30)
c = 1.2090569265 units.
c is AB....
Area coloured is;
Area sector with radius 3 units and angle 120° - Area triangle with height 3 units and base 3sin120 units + Area triangle with height 3 units and base 3sin60
= (120*9π÷360) - (½*...
Let the congruent two inscribed semi circles radius be r.
a = (3-2r) units.
b = 2r units.
It implies, calculating r.
2²+a² = b²
2²+(3-2r)² = (2r)²
4+9-12r+4r² = 4r²
12r = 13
r = (13...
a = 36+18
a = 54°
Let one of the green triangle area side be b.
cos54 = c/b
c = bcos54 units.
c = 0.5877852523b units.
sin54 = d/b
d = bsin54 units.
d = 0.8090169944b units.
e = c+...
sin36 = c/b
c = bsin36 units.
c = 0.5877852523b units.
Calculating b.
b² = (0.5877852523b)²+(0.5(b+1))²
0.6545084972b² = ¼(b²+2b+1)
1.6180339888b²-2b-1 = 0
It implies;
b = 1.61803 unit...
