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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.

Ask by Murray Pollard. in the United States
Mar 12,2025

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Answer

The polygon is a trapezoid with a perimeter of \(16 + 2\sqrt{65}\) and an area of 56.

Solution

We first plot the points: - \( J(12,-4) \) - \( U(0,-4) \) - \( S(4,3) \) - \( T(8,3) \) Notice that \( J \) and \( U \) lie on the line \( y=-4 \) (the bottom) and \( S \) and \( T \) lie on the line \( y=3 \) (the top). Arranging the points in order along the convex hull gives the quadrilateral with vertices \[ U(0,-4),\ J(12,-4),\ T(8,3),\ S(4,3). \] This is a trapezoid with bases \( \overline{UJ} \) and \( \overline{ST} \). **Step 1. Identify the base lengths.** - The bottom base from \( U(0,-4) \) to \( J(12,-4) \) has length \[ b_1 = |12-0| = 12. \] - The top base from \( S(4,3) \) to \( T(8,3) \) has length \[ b_2 = |8-4| = 4. \] **Step 2. Find the lengths of the legs.** Calculate the distance between \( U(0,-4) \) and \( S(4,3) \): \[ \text{Leg}_1 = \sqrt{(4-0)^2 + (3-(-4))^2} = \sqrt{4^2 + 7^2} = \sqrt{16+49} = \sqrt{65}. \] Similarly, calculate the distance between \( J(12,-4) \) and \( T(8,3) \): \[ \text{Leg}_2 = \sqrt{(8-12)^2 + (3-(-4))^2} = \sqrt{(-4)^2 + 7^2} = \sqrt{16+49} = \sqrt{65}. \] **Step 3. Compute the perimeter.** The perimeter \( P \) is the sum of all side lengths: \[ P = b_1 + b_2 + \text{Leg}_1 + \text{Leg}_2 = 12 + 4 + \sqrt{65} + \sqrt{65} = 16 + 2\sqrt{65}. \] **Step 4. Calculate the area.** The area \( A \) of a trapezoid is given by: \[ A = \frac{1}{2} \left(b_1 + b_2\right) \times h, \] where \( h \) is the height. Here, the height is the vertical distance between the bases: \[ h = 3 - (-4) = 7. \] Thus, the area is: \[ A = \frac{1}{2} (12 + 4) \times 7 = \frac{1}{2} (16) \times 7 = 8 \times 7 = 56. \] **Summary** - The polygon is a trapezoid. - Perimeter: \(16 + 2\sqrt{65}\) - Area: \(56\)

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Beyond the Answer

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 \( 4 \).

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