Mathematics Question and Solution
Shaded region/area is;
2(sector area with radius 2cm and angle 60°) - (equilateral triangle area with sides 2cm each side) + (square area with sides 2cm each side) - (square area with sides 4cm each side - circle area with radius 2cm)/4 - (2(sector area with radius 2cm and angle 60°) - (equilateral triangle area with sides 2cm each side)).
Therefore;
2(4π)/6 - (4sin60)/2 + (4) - (16 - 4π)/4 - (2(4π)/6 - (4sin60)/2)
= (4π)/3 - √(3) + 4 - (4 - π) - ((4π)/3 - √(3))
= (4π)/3 - √(3) + 4 - 4 + π - ((4π)/3 - √(3))
= (4π)/3 - √(3) + π - (4π)/3 + √(3)
= π cm²
Therefore shaded region/area is;
π cm².
Or
Shaded region/area is;
¼(area circle with radius 2cm inscribed square ABCD)
=¼(2*2π)
=¼(4π)
= π cm²
Therefore shaded region/area is;
π cm².
