Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Determine the value of n if the number of triangles that can be formed using the vertices of a regular polygon of n+1 sides is 21 more than the number of triangles formed

Determine the value of n if the number of triangles that can be formed using the vertices of a regular polygon of n+1 sides is 21 more than the number of triangles formed using n sides.

 

Option: 1

9


Option: 2

6


Option: 3

5


Option: 4

7


Answers (1)

best_answer

Let's assume the number of triangles formed using the vertices of a regular polygon with \mathrm{n+1} sides is \mathrm{\mathrm{T} 1}, and the number of triangles formed using the vertices of a regular polygon with n sides is \mathrm{\mathrm{T} 2}.

We are given that T1 is 21 more than T2. Mathematically, this can be represented as:

\mathrm{ \mathrm{T} 1=\mathrm{T} 2+21 }

Using the formula for the number of triangles formed using the vertices of a regular polygon, we have:

\mathrm{ \begin{aligned} & T 1=(n+1)(n)(n-1) / 6 \\\\ & T 2=n(n-1)(n-2) / 6 \end{aligned} }

Substituting these values into the equation \mathrm{\mathrm{T} 1=\mathrm{T} 2+21}, we get:

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

Multiplying both sides of the equation by 6 to eliminate the denominator, we have:

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

Expanding the terms, we get:

\mathrm{ n^3-n=n^3-3 n^2+2 n+126 }

Simplifying the equation, we have:

\mathrm{ 0=-3 n^2+2 n+126 }

Rearranging the equation, we get:

\mathrm{ 3 n^2-2 n-126=0 }

Now, we can solve this quadratic equation to find the value of n. Using factoring or the quadratic formula, we find that the roots of the equation are \mathrm{n=-6 \, \, and\, \, n=7.33} (approximately).

Since the number of sides of a regular polygon cannot be negative, the only valid solution is \mathrm{n=7}. Therefore, the value of n is 7 .

Posted by

Rishabh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question