To solve this problem, we need to calculate the area of quadrilateral $ABCD$ initially and then recalculate it after translating vertex $B$ upward by 4 units. Here’s the step-by-step approach:

Step 1: Initial Coordinates and Area Calculation

The vertices of quadrilateral $ABCD$ are:

• $A(-1, 3)$
• $B(-1, 7)$
• $C(-5, 6)$
• $D(-5, 2)$

To find the area of the quadrilateral, we can use the shoelace formula: $\text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right|$

Substitute the coordinates: $\text{Area} = \frac{1}{2} \left| (-1)(7) + (-1)(6) + (-5)(2) + (-5)(3) - (3)(-1) + (7)(-5) + (6)(-5) + (2)(-1) \right|$

Step 2: Adjusted Coordinates After Translation

When $B$ is translated up by 4 units, its new coordinates are $B'(-1, 11)$.

Now, the vertices of $AB'C'D$ are:

• $A(-1, 3)$
• $B'(-1, 11)$
• $C(-5, 6)$
• $D(-5, 2)$

Using the shoelace formula again with these updated coordinates, we can find the new area.

Step 3: Calculate the Difference in Area

After calculating both areas, subtract the original area from the new area to find how much greater it is.

I will compute these values and give you the final answer for how much greater the area is in square units.The area of quadrilateral $ABCD$ is greater by 8 square units after translating vertex $B$ upward by 4 units.

Would you like a detailed breakdown of each calculation step or have any further questions?

Here are some related questions to deepen understanding:

1. How would the area change if vertex $C$ were translated instead of $B$?
2. What would happen to the area if $B$ were translated down instead of up?
3. How is the shoelace formula applied to polygons with more than four vertices?
4. Could the translation of a vertex ever decrease the area of a quadrilateral?
5. What are other methods to calculate the area of a quadrilateral given its vertices?

Tip: Remember that the shoelace formula works effectively for any simple polygon, not just quadrilaterals.