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

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Question 927882: 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. perimeter of the regular hexagon.
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
 

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:

A=pi%2Ar%5E2=pi%2A%282sqrt%286%29%29=2pi%2Asqrt%286%29

Angle BDO is half of an interior angle of a regular hexagon. An
interior angle of a regular hexagon is given by:
%28%28n-2%29%2A%22180%B0%22%29%2Fn=%28%286-2%29%2A%22180%B0%22%29%2F6=%22120%B0%22

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
OB%5E2%2BBD%5E2=OD%5E2
24%2BBD%5E2=%282%2ABD%29%5E2
24%2BBD%5E2=4%2ABD%5E2
24=3%2ABD%5E2
8=BD%5E2
sqrt%288%29=BD
2sqrt%282%29=BD

BD is half a side of the hexagon, so one side of the
hgecagon is 4sqrt%282%29, so the perimeter of the
hexagon is 6 times that, or 24sqrt%282%29

Edwin