Mathematics Question and Solution
Notice.
The inscribed square area is 3 square units.
Therefore the single side length of the inscribed square is √(3) units.
The radius of the ascribed quarter circle will be;
√((√(3))²+(√(3))²)
= √(6) units.
Notice that the green inscribed triangle is a right-angled triangle.
The marked angle will be;
2(90-tan–¹(√(6)/√(3)))
= 2(90-tan–¹(√(2)))
= (180-2tan–(√(2)))
The hypotenuse of the green inscribed right-angle triangle is;
√((√(3))²+(√(6))²) = √(9) = 3 units.
The base of the green inscribed right-angle triangle is;
= 3cos(180-2tan–(√(2))) units.
Therefore;
Area green inscribed right-angled triangle is;
½(3x3cos(180-2tan–(√(2)))sin(180-2tan–(√(2))))
= √(2) square units.
