Find the Number of Sides in a Regular Polygon Inscribed in a Circle of Radius r

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To determine the number of sides in a regular polygon inscribed in a circle of radius r with an area of 2r^2√2, the polygon is divided into n equal triangles. Each triangle's area can be calculated using the formula 1/2 * a * b * sin(C), where a and b are the lengths of two sides, and C is the included angle. By equating the total area of the n triangles to the given area of the polygon, a relationship can be established to solve for n. The discussion emphasizes the importance of understanding the geometric properties of inscribed polygons and the use of trigonometric formulas in calculating areas. Ultimately, the solution leads to finding the specific value of n that satisfies the area condition.
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Hi guy's I am havin trouble with this problem... can anyone help me out...A regular polygon of n sides is inscribed in a circle of radius r. if the area of the polygon is 2r^2 root 2 how many sides does it have
 
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Take the polygon has n sides. Now draw radius to each vertex of the polygon. we form n triangles whose area are equal. now use the formula that the area of a triangle ABC is given by 1/2absinC.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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