Solving for the Pentagon's Smallest Interior Angle

The question asks for the measure of the smallest interior angle of a pentagon. The angles are given as expressions: $x, x+10, x+20, x+30,$ and $x+40$. We can find the smallest angle by first solving for $x$.

Polygon Angle Sum Rule

The sum of the interior angles of a polygon with $n$ sides is calculated using the formula:

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

A pentagon has $n=5$ sides. So, the sum of its interior angles is:

$ (5-2) \times 180^{\circ} = 3 \times 180^{\circ} = 540^{\circ} $

Algebraic Equation Setup

The sum of the given angle measures must equal the total sum for a pentagon:

$ x + (x+10) + (x+20) + (x+30) + (x+40) = 540^{\circ} $

Solving for x

Combine like terms:

$ 5x + (10+20+30+40) = 540^{\circ} $

$ 5x + 100 = 540^{\circ} $

Isolate the term with $x$:

$ 5x = 540^{\circ} - 100^{\circ} $

$ 5x = 440^{\circ} $

Solve for $x$:

$ x = \frac{440^{\circ}}{5} $

$ x = 88^{\circ} $

Identifying the Smallest Angle

The angle measures are $x, x+10, x+20, x+30,$ and $x+40$. The smallest angle is the one represented by the smallest value, which is $x$.

Since $x = 88^{\circ}$, the measure of the smallest interior angle of the pentagon is $88^{\circ}$.