QuestionAugust 26, 2025

In Exercises 4 and 5, find the area of the polygon with the given vertices. T(0,-2),U(3,5),V(-3,5) A(-3,3),B(-3,-1),C(4,-1),D(4,3)

In Exercises 4 and 5, find the area of the polygon with the given vertices. T(0,-2),U(3,5),V(-3,5) A(-3,3),B(-3,-1),C(4,-1),D(4,3)
In Exercises 4 and 5, find the area of the polygon with the given vertices.
T(0,-2),U(3,5),V(-3,5)
A(-3,3),B(-3,-1),C(4,-1),D(4,3)

Solution
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Answer

Triangle TUV: 0 ### Rectangle ABCD: 28 Explanation 1. Calculate the area of triangle TUV Use the formula for the area of a triangle with vertices (x_1, y_1), (x_2, y_2), (x_3, y_3): **A = \frac{1}{2} | x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) |**. For T(0,-2), U(3,5), V(-3,5): A = \frac{1}{2} | 0(5-5) + 3(5+2) + (-3)(-2-5) | = \frac{1}{2} | 0 + 21 - 21 | = \frac{1}{2} \times 0 = 0 2. Calculate the area of rectangle ABCD Use the formula for the area of a rectangle: **A = \text{length} \times \text{width}**. Length from A(-3,3) to D(4,3) is |4 - (-3)| = 7. Width from B(-3,-1) to A(-3,3) is |3 - (-1)| = 4. Area = 7 \times 4 = 28.

Explanation

1. Calculate the area of triangle TUV<br /> Use the formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$: **$A = \frac{1}{2} | x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) |$**.<br /> For $T(0,-2), U(3,5), V(-3,5)$: <br /> $A = \frac{1}{2} | 0(5-5) + 3(5+2) + (-3)(-2-5) | = \frac{1}{2} | 0 + 21 - 21 | = \frac{1}{2} \times 0 = 0$<br /><br />2. Calculate the area of rectangle ABCD<br /> Use the formula for the area of a rectangle: **$A = \text{length} \times \text{width}$**.<br /> Length from $A(-3,3)$ to $D(4,3)$ is $|4 - (-3)| = 7$.<br /> Width from $B(-3,-1)$ to $A(-3,3)$ is $|3 - (-1)| = 4$.<br /> Area $= 7 \times 4 = 28$.
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