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  • If the difference between the number of triangles formed using the vertices of a regular polygon of <img alt="\mathrm{(n+1)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%28na

If the difference between the number of triangles formed using the vertices of a regular polygon of \mathrm{(n+1)} sides and n sides is 21, what is the value of n ?

Option: 1

256


Option: 2

326


Option: 3

126


Option: 4

466


Answers (1)

best_answer

To find the value of n, let's consider the difference between the number of triangles formed using the vertices of a regular polygon of \mathrm{(n+1)} sides and n sides.

The number of triangles that can be formed using the vertices of a regular polygon with n sides is given by the

formula: \mathrm{T 1=(n \times(n-1) \times(n-2)) / 6}.

Similarly, the number of triangles that can be formed using the vertices of a regular polygon with \mathrm{(n+ 1) } sides is given by the formula:

\mathrm{ T 2=((n+1) \times n \times(n-1)) / 6 . }

The difference between the number of triangles formed by these two polygons is

\mathrm{T 2-T 1=((n+1) \times n \times(n-1)) / 6-(n \times(n-1) \times(n-2)) / 6.}

Given that this difference is equal to 21 , we can set up the equation:

\mathrm{ ((n+1) \times n \times(n-1)) / 6-(n \times(n-1) \times(n-2)) / 6=21 }

Multiplying bth sides of the equation by 6 to eliminate the denominator, we get:

\mathrm{ (n+1) \times n \times(n-1)-n \times(n-1) \times(n-2)=126 }

Simplifying the equation, we have:

\mathrm{ \left(n^3-n\right)-\left(n^3-3 n^2+2 n\right)=126 }

Combining like terms, we get:

\mathrm{ -n+2 n=126 }

Simplifying further, we have:

\mathrm{ n=126 }

Therefore, the value of n is 126 .

Posted by

SANGALDEEP SINGH

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