In a convex polygon, the sum of the number of diagonals and the number of sides is 55. What is the sum of the interior angles of that polygon?

1. Problem Description

In a convex polygon, the sum of the number of diagonals and the number of sides is

5

5. What is 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 number of sides is $n$.

The sum of the number of diagonals and the number of sides is

5

5. Therefore,

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

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 can factor this quadratic equation as follows:

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

Since the number of sides must be a positive integer, we have $n = 11$.

The sum of the interior angles of a polygon with $n$ sides is given by the formula:

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

For $n=11$, the sum of the interior angles is:

$S = (11-2) \times 180^{\circ} = 9 \times 180^{\circ} = 1620^{\circ}$

3. Final Answer

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