Mazana Job Shared The Questions.
Number 1.
a = 180-30-45
a = 105°
It implies;
(4/sin105) = (b/sin30)
b = 2.0705523608 units.
c = 45+90
c = 135°
Therefore, the required length,...
Let the side length of the regular octagon be a.
b = ⅛*180(8-2)
b = 135°
b is the single interior angle of the regular octagon.
2c² = a²
c = √(a²/2)
c = ½(√(2)a) units.
c = 0.7071067812a...
r2 will be;
Let it be x.
Therefore;
(4+x)²=(8-x)²+(4-x)²
16+8x+x²=64-16x+x²+16-8x+x²
x²-32x+64=0
(x-16)²=-64+256
x = 16±8√(3)
x ≠ 16+8√(3)
x = 16-8√(3)
Therefore;
x = r2 = 8(2-√...
r2 will be;
Let it be x.
Therefore;
(4+x)²=(8-x)²+(4-x)²
16+8x+x²=64-16x+x²+16-8x+x²
x²-32x+64=0
(x-16)²=-64+256
x = 16±8√(3)
x ≠ 16+8√(3)
x = 16-8√(3)
Therefore;
x = r2 = 8(2-√(3)) unit.
The radius, r if the small circle will be;
(5+r)²=2(5-r)²
25+10r+r²=50-20r+2r²
Therefore
r²-30r+25=0
(r-15)²=-25+225
r = 15±√(200)
r = 15±10√(2)
It implies;
r ≠ 15+10√(2)
r = 15...
Sir Mike Ambrose is the author of the question.
Notice!
AB = BC
Calculating AB.
AB = (20sin60)/sin104
AB = BC = 17.85075169007 cm.
Calculating BD
BD = (20sin16)/sin104
BD = 5.6815...
Let the side length of the inscribed green square be a.
b = ½(10-a) units.
c = b+a
c = ½(10-a+2a)
c = ½(10+a) units.
Therefore, a² (area green inscribed triangle) is;
10² = a²+(½(10+a...
Let OC be a.
Let the radius of the inscribed blue half circle be b.
Calculating area inscribed blue half circle.
20² = 2c²
c = √(200)
c = 10√(2) units.
d = ½(c)
d = 5√(2) units.
d i...
Let the side length of the two congruent inscribed squares be 1 unit each.
a² = 1+2²
a = √(5) units.
a is the radius of the half circle.
tanb = (½)
b = atan(½)°
(√(5)/sin(atan(½))) = (...
Let alpha be a.
Let AB be b.
tana = 2/b --- (1).
tana = b/(2+6)
tana = b/8 --- (2).
Equating (1) and (2).
2/b = b/8
b² = 16
b = 4 units.
Again, b is AB.
Therefore, the requir...
