Let's solve the two problems from the provided geometry exam.

Problems 1 and 2: Classify Quadrilateral $ABCD$

Given the vertices of quadrilateral $ABCD$ are:

• $A(-5, 3)$
• $B(3, 2)$
• $C(2, -6)$
• $D(-6, -5)$

To classify the quadrilateral, we need to find the lengths of the sides and the diagonals.

1. Calculate the lengths of the sides using the distance formula:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

• $AB$: $AB = \sqrt{(3 - (-5))^2 + (2 - 3)^2} = \sqrt{8^2 + (-1)^2} = \sqrt{64 + 1} = \sqrt{65}$

• $BC$: $BC = \sqrt{(2 - 3)^2 + (-6 - 2)^2} = \sqrt{(-1)^2 + (-8)^2} = \sqrt{1 + 64} = \sqrt{65}$

• $CD$: $CD = \sqrt{(2 - (-6))^2 + (-6 - (-5))^2} = \sqrt{8^2 + (-1)^2} = \sqrt{64 + 1} = \sqrt{65}$

• $DA$: $DA = \sqrt{(-6 - (-5))^2 + (-5 - 3)^2} = \sqrt{(-1)^2 + (-8)^2} = \sqrt{1 + 64} = \sqrt{65}$

2. Calculate the lengths of the diagonals:

• $AC$: $AC = \sqrt{(2 - (-5))^2 + (-6 - 3)^2} = \sqrt{7^2 + (-9)^2} = \sqrt{49 + 81} = \sqrt{130}$

• $BD$: $BD = \sqrt{(3 - (-6))^2 + (2 - (-5))^2} = \sqrt{9^2 + 7^2} = \sqrt{81 + 49} = \sqrt{130}$

Since all four sides are equal and the diagonals are equal, $ABCD$ is a square.

Problems 3 and 4: Angle Measures in the Regular Hexagon

Given $ABCDEF$ is a regular hexagon. Regular hexagons have equal sides and angles. Each internal angle in a regular hexagon is $120^\circ$.

To find all the angle measures in the provided diagram, let's label the unlabeled points:

• Let the intersection of lines $DE$ and $AB$ be point $P$.
• Let the intersection of lines $AF$ and $BC$ be point $Q$.

The central angles of a regular hexagon are $60^\circ$ each, dividing the circle into six equal parts.

Angle measures:

1. Central angles: $\angle AOB = \angle BOC = \angle COD = \angle DOE = \angle EOF = \angle FOA = 60^\circ$

2. Interior angles: Each interior angle of the hexagon is $120^\circ$.

3. Angles at points $A$ and $B$: Since $AB$ and $BC$ are extended lines, and $DE$ and $AF$ intersect at those points:

   • $\angle APB = \angle BQC = 180^\circ - 120^\circ = 60^\circ$

4. Angle $\angle PAQ$: Point $P$ lies on line $DE$, so:

   • $\angle PAQ = 180^\circ - 60^\circ = 120^\circ$

Therefore, the key angles are:

• $\angle APB = 60^\circ$
• $\angle PAQ = 120^\circ$
• $\angle PAF = 60^\circ$

Do you need any further details or have any questions?

Suggested Questions for Further Practice:

1. What is the formula to find the distance between two points in the coordinate plane?
2. How do you determine the type of quadrilateral given its vertices?
3. What are the properties of a regular hexagon?
4. How do you find the measure of an angle formed by extending sides of a polygon?
5. Can you explain how to find the length of a diagonal in a polygon?

Tip:

Always double-check your calculations, especially when dealing with square roots and distances, to ensure accuracy in geometry problems.