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...
Shaded/green area will be;
Area semi circle of radius 9unit - 2(area sector of radius 9unit and angle 60°) + area equilateral triangle of single side length 9unit - area sector of radius 9unit a...
Notice;
Single interior angle of a regular polygon with 6 sides (regular hexagon) is 120°.
The formula to get it is;
(180°(n-2))/n.
Where n is the side number of the regular polygon....
Let the radius of the bigger circle be 2 unit.
Therefore the radius of the smaller circle will be;
½(2)=1 unit.
Let the radius of the semi circle be r.
It implies;
2² + r² = (1 + r)²...
Blue area will be;
Area square of single side length 4cm - 5(area circle of radius 1cm) + area square of single side 2cm.
= 16 - 5π + 4
= 20 - 5π
= 5(4 - π) cm²
Let x be a side length of the largest possible inscribed square of the scalene triangle with sides 13 cm, 14 cm and 15 cm respectively.
It implies;
x = 14÷((1/tan(53.13))+1+(1/tan(67.38)))...
Calculating x (radius of the small inscribed circle)
28² + x² = (56-x)²
784 + x² = 3136 - 112x + x²
112x = 2351
x = 21 units.
