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°
Let the single side length of the regular pentagon be 4 unit.
Therefore angle x degree will be;
Sinx = (2/4)
Sinx = ½
Therefore;
x = sin–¹(½)
x = 30°
Let R be radius of the quarter circle.
Let r be radius of the inscribed circle.
R will be;
√(3²+4²)
R=√(25)
R= 5 units.
Therefore r will be;
(5-r)²=(3+r)²+r²
25-10r+r²=9+6r+r²+r...
Kindly move the question right/left twice to review the complete solution.
Thank you amazingly.
Sir Mike Ambrose is the author of the question.
Exact Area R in its simplest from will be;
Area semi circle of radius 2 units - Area right-angled triangle of height 2√(2+√(2)) units and width 2√...
