The sum of the number of diagonals and sides of a convex polygon is 55. Find the sum of the interior angles of that polygon.

1. Problem Description

The sum of the number of diagonals and sides of a convex polygon is

5

5. Find the sum of the interior angles of that polygon.

2. Solution Steps

Let $n$ be the number of sides of the convex polygon.

The number of diagonals in a polygon with $n$ sides is given by the formula:

$D = \frac{n(n-3)}{2}$

The problem states that the sum of the number of diagonals and the number of sides is

5

5. Therefore, we have:

$\frac{n(n-3)}{2} + n = 55$

Multiplying both sides by 2, we get:

$n(n-3) + 2n = 110$

$n^2 - 3n + 2n = 110$

$n^2 - n = 110$

$n^2 - n - 110 = 0$

We need to solve this quadratic equation for $n$. We look for two numbers that multiply to -110 and add to -

1. These numbers are -11 and

1

0. $n^2 - 11n + 10n - 110 = 0$

$n(n-11) + 10(n-11) = 0$

$(n-11)(n+10) = 0$

Therefore, $n = 11$ or $n = -10$. Since the number of sides of a polygon must be positive, we have $n = 11$.

Now, we need to find the sum of the interior angles of a polygon with 11 sides. The formula for the sum of the interior angles of a polygon with $n$ sides is:

$S = (n-2) \times 180^\circ$

Substituting $n = 11$, we get:

$S = (11-2) \times 180^\circ$

$S = 9 \times 180^\circ$

$S = 1620^\circ$

3. Final Answer

The sum of the interior angles of the polygon is $1620^\circ$.