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What is the area of the polygon with vertices T (6, 5), A (8, −1), S (4, −2), and K (−1, 4)?

A) 11 square units
B) 13 square units
C) 15 square units
D) 17 square units

User Exifers
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1 Answer

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Final answer:

The area of the polygon with vertices T (6, 5), A (8, −1), S (4, −2), and K (−1, 4) is 65 square units.

Step-by-step explanation:

To find the area of a polygon, we can use the formula for the area of a triangle and sum the areas of all the triangles formed by the vertices of the polygon. In this case, the polygon has 4 vertices, so we can divide it into two triangles: TAK and TAS.

Using the formula for the area of a triangle, we can calculate the area of TAK as 0.5 * |(6 * (-1) + 8 * 4 + (-1) * 5) - (5 * 8 + (-1) * (-1) + 4 * 6)| = 35 square units.

Similarly, the area of TAS is 0.5 * |(6 * (-2) + (-1) * 5 + 4 * 5) - (5 * (-1) + 4 * 6 + (-2) * 6)| = 30 square units.

Adding the areas of TAK and TAS, we get 35 + 30 = 65 square units.

User Yashira
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