a = 2atan(2)°
b = ½(180-a)
b = ½(180-2atan(2))°
tan(½(180-2atan(2))) = c/(0.5x)
0.5 = c/(0.5x)
c = 0.25x cm.
d = x-0.25x
d = 0.75x cm.
Calculating x.
0.5*x*0.75x = 24
x² = 24/(0.75*0.5)
x² = 64...
Let AB = a
Notice.
AB = CD
a² = 2²+1²+2*2*1cosx
a² = 5+4cosx --- (1).
a² = 3²+4²-2*3*4cosx
a² = 25-24cosx --- (2).
Calculating x.
Equating (1) and (2).
5+4cosx = 25-24cosx
28cosx = 20
cosx = (2...
Diameter of the inscribed circle is;
½(AB)
= ½(8)
= 4 units.
a = ½(4)
a = 2 units.
a is the radius of the inscribed circle.
Calculating red length.
Notice, the red length is the side of the inscr...
Let the side length of the regular hexagon be 1 unit.
a = ½(180-108)
108° = single interior angle of the regular pentagon.
180° = sum of angles on a straight line.
a = ½(72)
a = 36°
tan36 = b/0.5...
Let the side length of the regular pentagon be 1 unit.
a² = 1²+1²
a = √(2) units.
b = ½(a)
b = ½√(2) units.
c = ½(360-108-108)
c = ½(360-216)
c = ½(144)
c = 72°
d = 180-108-45
d = 72-45
d = 27°...
Let the side length of the ascribed regular pentagon be 1 unit.
Let a be the side length of the inscribed regular hexagon.
b = 180-108-(108-120)
b = 72-60
b = 12°
Where;
108° is the singl...
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...
