Multiple choice

With reference to a given circle, $\displaystyle A_{1}$ and $\displaystyle B_{1}$ are the areas of the inscribed and circumscribed regular polygons of n sides, $\displaystyle A_{2}$ and $\displaystyle B_{2}$ are corresponding quantities for regular polygons of 2n sides (a)If $\displaystyle A_{2}$ is a geometric mean between $\displaystyle A_{1}$ and $\displaystyle B_{1}$ then value of $A_1B_1$ (b) If$\displaystyle B_{2}$ is a harmonic mean between $\displaystyle A_{2}$ and $\displaystyle B_{1}$ then value of $\cfrac{1}{A_1}+\cfrac{1}{B_1}$

  1. $a:n^{2}R^{2}\sin^{2}\cfrac{\pi}{n},b:\cfrac{2}{B_{1}}$
  2. $a:n^{2}R^{4}\sin^{2}\cfrac{\pi}{n},b:\cfrac{2}{B_{2}}$
  3. $a:n^{2}R^{2}\sin^{2}\cfrac{\pi}{n},b:\cfrac{1}{B_{1}}$
  4. $a:n^{2}R^{4}\sin^{2}\cfrac{\pi}{n},b:\cfrac{1}{B_{2}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is a standard geometric property of inscribed and circumscribed polygons. For part (a), A1B1 = n^2 * R^4 * sin^2(pi/n). For part (b), the harmonic mean relation leads to 2/B2 = 1/A2 + 1/B1, which rearranges to the required form.

AI explanation

The area of an inscribed regular polygon with n sides in a circle of radius R is A1 equals (n/2) R squared sin(2pi/n), which simplifies to n R squared sin(pi/n) cos(pi/n). For a circumscribed regular polygon, the area B1 equals n R squared tan(pi/n). The area A2 of an inscribed 2n-sided polygon is n R squared sin(pi/n), which acts as the exact geometric mean of A1 and B1. This makes their product A1 B1 equal to n squared R fourth power sin squared(pi/n). Also, the area B2 of a circumscribed 2n-sided polygon equals 2n R squared tan(pi/2n), which is precisely the harmonic mean of A2 and B1, making the sum of their reciprocals equal to 2/B2.

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