Questions: In Exercises 15-18, use the diagram to find the perimeter and the area of the polygon. 15. triangle CDE 16. triangle ABF

Solution Steps

The vertices of the polygon are given as:

• A(-5, 4)
• B(0, 3)
• C(4, -1)
• D(4, -5)
• E(2, -3)
• F(-2, -1)

To find the perimeter, calculate the distance between each pair of consecutive vertices using the distance formula: \[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

1. Distance AB: \[ \sqrt{(0 - (-5))^2 + (3 - 4)^2} = \sqrt{5^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \]

2. Distance BC: \[ \sqrt{(4 - 0)^2 + (-1 - 3)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} \]

3. Distance CD: \[ \sqrt{(4 - 4)^2 + (-5 - (-1))^2} = \sqrt{0 + (-4)^2} = \sqrt{16} = 4 \]

4. Distance DE: \[ \sqrt{(2 - 4)^2 + (-3 - (-5))^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} \]

5. Distance EF: \[ \sqrt{(-2 - 2)^2 + (-1 - (-3))^2} = \sqrt{(-4)^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} \]

6. Distance FA: \[ \sqrt{(-5 - (-2))^2 + (4 - (-1))^2} = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} \]

Sum of distances: \[ \sqrt{26} + \sqrt{32} + 4 + \sqrt{8} + \sqrt{20} + \sqrt{34} \]

Use the Shoelace formula for the area of a polygon: \[ \text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n-1} (x_i y_{i+1} - y_i x_{i+1}) + (x_n y_1 - y_n x_1) \right| \]

Substitute the coordinates: \[ \text{Area} = \frac{1}{2} \left| (-5 \cdot 3 + 0 \cdot (-1) + 4 \cdot (-5) + 4 \cdot (-3) + 2 \cdot (-1) + (-2 \cdot 4)) - (4 \cdot 0 + 3 \cdot 4 + (-1) \cdot 4 + (-5) \cdot 2 + (-3) \cdot (-2) + (-1) \cdot (-5)) \right| \]

\[ = \frac{1}{2} \left| (-15 + 0 - 20 - 12 - 2 - 8) - (0 + 12 - 4 - 10 + 6 + 5) \right| \]

\[ = \frac{1}{2} \left| (-57) - (9) \right| \]

\[ = \frac{1}{2} \left| -66 \right| \]

\[ = \frac{1}{2} \times 66 \]

\[ = 33 \]

Final Answer