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Exercise 6.4 · Q5

Q.The number of diagonals of a polygon is twice the number of its sides. Find the number of sides of the polygon.

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Express the diagonal count of an nn-sided polygon in terms of nn, set it equal to twice the number of sides, and solve the resulting equation.

[!FORMULA] A polygon with nn vertices has nC2^{n}C_{2} total line segments joining pairs of vertices; of these, nn are the sides themselves, so diagonals =nC2−n=n(n−1)2−n=n(n−3)2={}^{n}C_{2}-n=\dfrac{n(n-1)}{2}-n=\dfrac{n(n-3)}{2}.

  1. Let the polygon have nn sides. Number of diagonals =n(n−3)2=\dfrac{n(n-3)}{2}.
  2. Given: diagonals =2×(number of sides)=2n=2\times(\text{number of sides})=2n.
  3. Equation: n(n−3)2=2n\dfrac{n(n-3)}{2}=2n.
  4. Multiply both sides by 2: n(n−3)=4nn(n-3)=4n. …

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