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√...
Blue Area is;
Area square of single side 4cm - 2(area circle of radius 1cm) - 8(area square of single side 1cm) + 8(area quarter circle of radius 1cm).
= 16 - 2π - 8 + ¼(8π)
= 8 - 2π + 2π...
Blue Area will be;
Area square of single side 4cm - 3(Area circle of radius 1cm)
It will be;
(4x4) - 3(1²(π))
It is;
16 - 3π
Answer is;
(16-3π) square units.
The radius of the circle with the inscribed square will be;
½(√(4²+4²))
= ½(√(32))
= ½(4√(2))
= 2√(2) cm.
Therefore Blue Area will be;
2(area circle of radius 2cm) - area circle o...
