Identify the polygon that has vertices \( J(12,-4), U(0,-4) \), \( S(4,3) \), and \( T(8,3) \), and then find the perimeter and area of the polygon.
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The polygon identified by the vertices \( J(12,-4) \), \( U(0,-4) \), \( S(4,3) \), and \( T(8,3) \) is a quadrilateral. To find the perimeter, we calculate the lengths of each side: 1. \( JU \): \[ JU = \sqrt{(12 - 0)^2 + (-4 - (-4))^2} = \sqrt{12^2} = 12 \] 2. \( US \): \[ US = \sqrt{(4 - 0)^2 + (3 - (-4))^2} = \sqrt{4^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65} \] 3. \( ST \): \[ ST = \sqrt{(8 - 4)^2 + (3 - 3)^2} = \sqrt{4^2} = 4 \] 4. \( TJ \): \[ TJ = \sqrt{(12 - 8)^2 + (-4 - 3)^2} = \sqrt{4^2 + (-7)^2} = \sqrt{16 + 49} = \sqrt{65} \] Adding these lengths together for the perimeter: \[ \text{Perimeter} = JU + US + ST + TJ = 12 + \sqrt{65} + 4 + \sqrt{65} = 16 + 2\sqrt{65} \] Next, to find the area, we can use the shoelace formula: \[ \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right| \] Substituting the vertices \( J(12,-4) \), \( U(0,-4) \), \( S(4,3) \), \( T(8,3) \): \[ = \frac{1}{2} \left| 12 \cdot (-4) + 0 \cdot 3 + 4 \cdot 3 + 8 \cdot (-4) - (-4 \cdot 0 + -4 \cdot 4 + 3 \cdot 8 + 3 \cdot 12) \right| \] \[ = \frac{1}{2} \left| -48 + 0 + 12 - 32 + 0 + 16 + 24 + 36 \right| \] \[ = \frac{1}{2} \left| -68 + 76 \right| = \frac{1}{2} \left| 8 \right| = 4 \] So, the perimeter of the polygon is \( 16 + 2\sqrt{65} \) and the area is \( 4 \).