Mathematics Question and Solution

By Ogheneovo Daniel Ephivbotor
5th June, 2025

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.

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