Area Black will be;
Area triangle BCD - area sector of radius 4m and angle 45° + area sector of radius 4m and angle 45° - quarter circle of radius 2m - 2(right-angled triangle of height and base...
Let MN=NP=PD=MD=y.
Let AR=x.
Therefore;
AD=AB=BC=CD=2+x.
RD=2.
RM=2-y.
And;
RC=2+2x.
There deriving an equation.
Considering similar triangle RCD and RNM.
It implies;
2-y=2
...
Radius R of the big circle is 50 unit.
(R-10)²=R(R-18)
R²-20R+100=R²-18R
2R=100
R=50 units.
Radius r of the small inscribed unshaded circle is 41 unit.
2r=2R-18, and R=50
2r=100-18
2r=82...
Let the radius of the ascribed quarter circle be 2x.
Therefore the radius of the inscribed yellow half circle is x.
Calculating x.
¼(π(2x)²)-½(πx²) = 17
πx²-½(πx²) = 17
2πx²-πx² = 34...
a = 180-30-45
a = 105°
(1/sin105) = (b/sin30)
b = 0.51763809021 units.
c = √(2)-b
c = 0.89657547217 units.
(0.89657547217/sin105) = (d/sin30)
d = 0.46410161514 units.
It implies, ar...
a = (r+R)
b = (R-r)
It implies;
8²+b² = a²
R²+2Rr+r² = 64+R²-2Rr+r²
4Rr = 64
Rr = 16
R = 16/r --- (1).
It implies;
a = r+(16/r)
a = (16+r²)/r
b = (16/r)-r
b = (16-r²)/r
The...
Each of the triangles is equilateral with single side length;
(√(24)÷3^(1/4)) units.
Total yellow area will be;
(√(24)÷3^(1/4))²x(3/4)x0.5sin60 + (√(24)÷3^(1/4))²x(2/3)x(0.5)²sin60 + (√(24...
Diameter of circle Z will be;
½(Diameter of the ascribed circle - length of the inscribed square)
= ½(√(16²+16²) - 16)
= ½(16√(2)-16)
= (8√(2)-8) cm
The radius of circle Z will be;...
The required angle is;
180° - (11.25°+33.75°)
= 180° - 45°
= 135°
