Question

Polygon ABCD has the following vertices:

A(-5, 4), B(1, 4), C(6, -4), B(-5, -4)

Calculate the area of the polygon

Answer 1

It looks like there is a typo since the points mentioned include two points labeled as "B." Let's clarify the vertices of the polygon. I assume the vertices are A(-5, 4), B(1, 4), C(6, -4), and D(-5, -4).

We can calculate the area of the polygon using the shoelace formula, which is given as:

\[
\text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n}(x_iy_{i+1} - y_ix_{i+1}) \right|
\]

where \((x_{n+1}, y_{n+1})\) is the same as \((x_1, y_1)\) to close the polygon.

The vertices in order are:
1. A(-5, 4)
2. B(1, 4)
3. C(6, -4)
4. D(-5, -4)

We will list them in the shoelace formula:

- A: \((-5, 4)\)
- B: \((1, 4)\)
- C: \((6, -4)\)
- D: \((-5, -4)\)
- Closing back to A: \((-5, 4)\)

Setting up the shoelace formula:

\[
\begin{array}{c|c}
x & y \\
\hline
-5 & 4 \\
1 & 4 \\
6 & -4 \\
-5 & -4 \\
-5 & 4 \\
\end{array}
\]

Now we calculate the individual products:

1. \(x_1 y_2 = -5 \cdot 4 = -20\)
2. \(x_2 y_3 = 1 \cdot -4 = -4\)
3. \(x_3 y_4 = 6 \cdot -4 = -24\)
4. \(x_4 y_5 = -5 \cdot 4 = -20\)

Now we sum these products:

\[
\text{Sum of } (x_iy_{i+1}) = -20 - 4 - 24 - 20 = -68
\]

Now for the other direction:

1. \(y_1 x_2 = 4 \cdot 1 = 4\)
2. \(y_2 x_3 = 4 \cdot 6 = 24\)
3. \(y_3 x_4 = -4 \cdot -5 = 20\)
4. \(y_4 x_5 = -4 \cdot -5 = 20\)

Now we sum these products:

\[
\text{Sum of } (y_ix_{i+1}) = 4 + 24 + 20 + 20 = 68
\]

Now substitute these values into the shoelace formula:

\[
\text{Area} = \frac{1}{2} \left| -68 - 68 \right| = \frac{1}{2} \left| -136 \right| = \frac{1}{2} \cdot 136 = 68
\]

Thus, the area of polygon ABCD is **68 square units**.