A quadrilateral has $n=4$ sides. Therefore, the sum of the interior angles of a quadrilateral must be \begin{align*} 180^\circ\cdot(n-2) &=180^\circ \cdot (4-2) \\ &= 180^\circ \cdot 2 \\ &= 360^{\circ}. \end{align*}

Let's check this for the above three options: \begin{align} 64^{\circ} + 73^{\circ} + 98^{\circ} +124^{\circ} = 359^{\circ}\quad \color{red}\mathsf{X}\\ & \\ 73^{\circ} + 112^{\circ} + 93^{\circ} + 82^{\circ} = 360^{\circ}\quad \color{green}\checkmark\\ & \\ 80^{\circ} + 140^{\circ} + 49^{\circ} + 84^{\circ} = 353^{\circ}\quad \color{red}\mathsf{X} \end{align}

Therefore, only option II could be the angles of a quadrilateral.

A quadrilateral has $n=4$ sides. Therefore, the sum of the interior angles of a quadrilateral must be \begin{align*} 180^\circ\cdot(n-2) &=180^\circ \cdot (4-2) \\ &= 180^\circ \cdot 2 \\ &= 360^{\circ}. \end{align*}

Let's check this for the above three options: \begin{align} 85^{\circ} + 115^{\circ} + 79^{\circ} +90^{\circ} = 369^{\circ}\quad \color{red}\mathsf{X}\\ & \\ 59^{\circ} + 161^{\circ} + 90^{\circ} + 40^{\circ} = 350^{\circ}\quad \color{red}\mathsf{X}\\ & \\ 80^{\circ} + 141^{\circ} + 39^{\circ} + 100^{\circ} = 360^{\circ}\quad \color{green}\checkmark \end{align}

Therefore, only option III could be the angles of a quadrilateral.

A quadrilateral has $n=4$ sides. Therefore, the sum of the interior angles of a quadrilateral must be \begin{align*} 180^\circ\cdot(n-2) &=180^\circ \cdot (4-2) \\[3pt] &= 180^\circ \cdot 2 \\[3pt] &= 360^{\circ}. \end{align*}

Adding up the internal angles and solving for $m \angle C,$ we get

\begin{align} m \angle A + m \angle B + m \angle C + m \angle D & = 360^\circ\\[3pt] 70^\circ + 75^\circ + m \angle C +110^\circ&= 360^\circ \\[3pt] m \angle C +255^\circ&= 360^\circ \\[3pt] m \angle C &= 105^\circ. \end{align}

A quadrilateral has $n=4$ sides. Therefore, the sum of the interior angles of a quadrilateral must be \begin{align*} 180^\circ\cdot(n-2) &=180^\circ \cdot (4-2) \\[3pt] &= 180^\circ \cdot 2 \\[3pt] &= 360^{\circ} \end{align*}

Adding up the internal angles and solving for $x,$ we get

\begin{align} m \angle A + m \angle B + m \angle C + m \angle D & = 360^\circ\\[5pt] x + 90^{\circ} + 3x+6x &= 360^\circ \\[5pt] 10x +90^{\circ} & = 360^\circ \\[5pt] 10x &= 270^\circ \\[5pt] x &= 27^\circ \end{align}

Therefore, \begin{align} m\angle C & = 3x \\[3pt] & = 3(27^{\circ}) \\[3pt] & = 81^\circ . \end{align}

Remember that the sum of the interior angles of a polygon with $n$ sides is $180^\circ \cdot (n-2).$

To find the number of sides in our polygon, we start by setting up the following equation:

\[ 180^\circ \cdot (n-2) = 900^\circ \]

Now, we solve for $n,$ as follows:

\begin{align} 180^\circ \cdot (n-2) & = 900^\circ\\[5pt] n - 2 & = \dfrac{900^\circ}{180^\circ}\\[5pt] n - 2 & = 5\\[5pt] n & = 7 \end{align}

Therefore, the polygon has $n = 7$ sides (so it is a heptagon).

Remember that the sum of the interior angles of a polygon with $n$ sides is $180^\circ \cdot (n-2).$

To find the number of sides in our polygon, we start by setting up the following equation:

\[ 180^\circ \cdot (n-2) = 4\,140^\circ \]

Now, we solve for $n,$ as follows:

\begin{align} 180^\circ \cdot (n-2) & = 4\,140^\circ\\[5pt] n - 2 & = \dfrac{4\,140^\circ}{180^\circ}\\[5pt] n - 2 & = 23\\[5pt] n & = 25 \end{align}

Therefore, the polygon has $n = 25$ sides.