Question bank/JEE Main/Mathematics

Let $A_{1},A_{2},A_{3},\ldots\ldots,A_{14}$be a regular polygon
with 14 side inscribed in a circle of radius R. If
$\left( A_{1}A_{3} \right)^{2} + \left( A_{1}A_{7} \right)^{2} + \left( A_{3}A_{7} \right)^{2} = {KR}^{2}$
then the value of $\frac{K^{2}}{2}$is
ekuk$A_{1},A_{2},A_{3},\ldots\ldots,A_{14}$lecgqHkqt gS ftldh 14
Hkqtk,a] R f= o`r ds vUrxZr
$\left( A_{1}A_{3} \right)^{2} + \left( A_{1}A_{7} \right)^{2} + \left( A_{3}A_{7} \right)^{2} = {KR}^{2}$
rc $\frac{K^{2}}{2}$dk eku

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$\angle A_{1}{OA}_{3} = 2\pi/7$
$\angle A_{3}{OA}_{7} = 4\pi/7$
$\angle A_{1}{OA}_{7} = 6\pi/7$
[IMAGE] Consider ekukfd
$\Rightarrow cos\theta = 1 - \frac{\left( A_{i}A_{j} \right)^{2}}{2R^{2}}$
$\Rightarrow \left( A_{i}A_{j} \right)^{2} = (1 - cos\theta)2R^{2}$
$\because\left( A_{1}A_{3} \right)^{2} + \left( A_{1}A_{7} \right)^{2} + \left( A_{3}A_{7} \right)^{2}$
$= 2R^{2}\left\lbrack \left( 1 - cos\frac{2\pi}{7} \right) + \left( 1 - cos\frac{4\pi}{7} \right) + \left( 1 - cos\frac{6\pi}{7} \right) \right\rbrack = 2R^{2}\left\lbrack 3 + \frac{1}{2} \right\rbrack = 7R^{2}$

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx