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The vertices of AABC are A(2,8), B(16,2), and (6,2). The perimeter of AABC is __ units, and its area is __ square units.
a) Perimeter: 46.15; Area: 49.5
b) Perimeter: 39.94; Area: 42.78
c) Perimeter: 45.68; Area: 48.21
d) Perimeter: 42.12; Area: 44.75

1 Answer

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Final answer:

The perimeter of the triangle is approximately 32.44 units and the area is 12 square units.

Step-by-step explanation:

To find the perimeter of a triangle, we need to calculate the sum of the lengths of its sides. In this case, the vertices of the triangle A(2,8), B(16,2), and C(6,2) are given. We can use the distance formula to find the lengths of each side. Let's calculate:

Side AB: sqrt((16-2)^2 + (2-8)^2) = sqrt(196 + 36) = sqrt(232) ≈ 15.23

Side BC: sqrt((6-16)^2 + (2-2)^2) = sqrt(100 + 0) = sqrt(100) = 10

Side CA: sqrt((2-6)^2 + (8-2)^2) = sqrt(16 + 36) = sqrt(52) ≈ 7.21

Now, we can find the perimeter: Perimeter = AB + BC + CA ≈ 15.23 + 10 + 7.21 ≈ 32.44 units.

To find the area of a triangle, we can use the formula: Area = 0.5 * base * height. The base and height can be found by taking the difference in x-coordinates and y-coordinates respectively:

Base = |2-6| = 4 units

Height = |8-2| = 6 units

Area = 0.5 * 4 * 6 = 12 square units.

User Alap Anerao
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