1. **Problem Statement:** We have a pentagon ABCDE with vertices \( A(-3, -1), B(-3, 5), C(1, 8), D(5, 5), E(5, -1) \). We need to find its area and perimeter, then dilate it by a factor of 2 from the origin to get A'B'C'D'E', find the new vertices, and compute the new area and perimeter. 2. **Calculate the perimeter of ABCDE:** Calculate the distance between consecutive points using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). - \( AB = \sqrt{(-3 + 3)^2 + (5 + 1)^2} = \sqrt{0 + 36} = 6 \) - \( BC = \sqrt{(1 + 3)^2 + (8 - 5)^2} = \sqrt{16 + 9} = 5 \) - \( CD = \sqrt{(5 - 1)^2 + (5 - 8)^2} = \sqrt{16 + 9} = 5 \) - \( DE = \sqrt{(5 - 5)^2 + (-1 - 5)^2} = \sqrt{0 + 36} = 6 \) - \( EA = \sqrt{(-3 - 5)^2 + (-1 + 1)^2} = \sqrt{64 + 0} = 8 \) Sum: \( 6 + 5 + 5 + 6 + 8 = 30 \) 3. **Calculate the area of ABCDE using the shoelace formula:** \[ \text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n} (x_i y_{i+1} - y_i x_{i+1}) \right| \] where \( (x_{n+1}, y_{n+1}) = (x_1, y_1) \). Calculate: \[ \begin{aligned} S_1 &= (-3)(5) + (-3)(8) + 1(5) + 5(-1) + 5(-1) = -15 - 24 + 5 - 5 - 5 = -44 \\ S_2 &= (-1)(-3) + 5(1) + 8(5) + 5(5) + (-1)(-3) = 3 + 5 + 40 + 25 + 3 = 76 \end{aligned} \] Area: \[ \frac{1}{2} |S_2 - S_1| = \frac{1}{2} |76 - (-44)| = \frac{1}{2} \times 120 = 60 \] 4. **Dilate each vertex by a factor of 2 from the origin:** Multiply each coordinate by 2: \[ A' = (-6, -2), B' = (-6, 10), C' = (2, 16), D' = (10, 10), E' = (10, -2) \] 5. **Calculate the perimeter of A'B'C'D'E':** Distances: - \( A'B' = \sqrt{(-6 + 6)^2 + (10 + 2)^2} = \sqrt{0 + 144} = 12 \) - \( B'C' = \sqrt{(2 + 6)^2 + (16 - 10)^2} = \sqrt{64 + 36} = 10 \) - \( C'D' = \sqrt{(10 - 2)^2 + (10 - 16)^2} = \sqrt{64 + 36} = 10 \) - \( D'E' = \sqrt{(10 - 10)^2 + (-2 - 10)^2} = \sqrt{0 + 144} = 12 \) - \( E'A' = \sqrt{(-6 - 10)^2 + (-2 + 2)^2} = \sqrt{256 + 0} = 16 \) Sum: \( 12 + 10 + 10 + 12 + 16 = 60 \) 6. **Calculate the area of A'B'C'D'E' using the shoelace formula:** \[ \begin{aligned} S_1' &= (-6)(10) + (-6)(16) + 2(10) + 10(-2) + 10(-2) = -60 - 96 + 20 - 20 - 20 = -176 \\ S_2' &= (-2)(-6) + 10(2) + 16(10) + 10(10) + (-2)(-6) = 12 + 20 + 160 + 100 + 12 = 304 \end{aligned} \] Area: \[ \frac{1}{2} |S_2' - S_1'| = \frac{1}{2} |304 - (-176)| = \frac{1}{2} \times 480 = 240 \] 7. **Summary:** - Original area = 60 - Original perimeter = 30 - Dilated vertices: \( A'(-6, -2), B'(-6, 10), C'(2, 16), D'(10, 10), E'(10, -2) \) - Dilated area = 240 - Dilated perimeter = 60 Note: The perimeter doubles approximately (from 30 to 60), and the area scales by the square of the dilation factor (from 60 to 240).