SOLUTION: A square having an area of 48 m2 is inscribed in a circle which is inscribed in a regular hexagon. Compute the a. area of the circle, b. area of the regular hexagon and c. perimete

 

The area of the square is 48 m² 
So a side of the square is √48 = √16×3 = 4√3 

AB is half a side of the square, so it's 2√3

OA is also 2√3

So by the Pythagorean theorem hypotenuse OB of right triangle OAB is



OB is the raius of the circle so by that formula for a circle's area:



Angle BDO is half of an interior angle of a regular hexagon. An
interior angle of a regular hexagon is given by:


So angle BDO is 60°, so right triangle BDO is a 30°-60°-90°,
so its hypotenuse OD is twice BD.  So by the Pythagorean theorem








BD is half a side of the hexagon, so one side of the
hgecagon is , so the perimeter of the
hexagon is 6 times that, or 

Edwin