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Step-by-Step Solution
Step 1: Identify the expression for the number of triangles
The number of triangles, $T_n$, that can be formed by joining the vertices of an $n$-sided regular polygon is given by the binomial coefficient:
$T_n = \binom{n}{3}$
Step 2: Express $T_{n+1}$ in terms of $n$
Similarly, for an $(n+1)$-sided polygon, the number of triangles is:
$T_{n+1} = \binom{n+1}{3}$
Step 3: Set up the given condition
According to the problem, the difference $T_{n+1} - T_n$ is given as 10:
$\binom{n+1}{3} - \binom{n}{3} = 10$
Step 4: Simplify the binomial coefficients
Recall that $\binom{n}{3} = \frac{n(n-1)(n-2)}{6}$ and $\binom{n+1}{3} = \frac{(n+1)n(n-1)}{6}$. Substituting these forms into the equation, we get:
$\frac{(n+1)\,n\,(n-1)}{6} - \frac{n(n-1)(n-2)}{6} = 10$
Step 5: Factor and simplify
Combine the terms on the left-hand side:
$ \frac{1}{6}\bigl[(n+1)n(n-1) - n(n-1)(n-2)\bigr] = 10 $
Simplify inside the brackets:
$(n+1)n(n-1) = n(n-1)(n+1)$
$n(n-1)(n-2)$ is subtracted.
You can factor out $n(n-1)$:
$ n(n-1)\bigl[(n+1) - (n-2)\bigr] = n(n-1)\bigl(n+1 - n + 2\bigr) = n(n-1)(3).$
Thus, we have:
$ \frac{n(n-1)\cdot 3}{6} = 10 $
Which reduces to:
$ \frac{3n(n-1)}{6} = 10 $
$ n(n-1) = 20 $
Step 6: Solve for $n$
We have the quadratic equation:
$n^2 - n - 20 = 0$
Factorizing this, we get:
$(n - 5)(n + 4) = 0$
Thus, $n = 5$ or $n = -4$. Since $n$ represents the number of sides of a polygon, it must be positive. Therefore, $n = 5$.
Final Answer
Hence, the value of $n$ is $\boxed{5}$.
