To find the perimeter of triangle ABC, we need to find the distance between each pair of vertices using the distance formula:
AB = √[(6-(-2))^2 + (2-2)^2] = √[64 + 0] = √64 = 8
BC = √[(0-6)^2 + (8-2)^2] = √[36 + 36] = √72 ≈ 8.49
AC = √[(-2-0)^2 + (2-8)^2] = √[4 + 36] = √40 ≈ 6.32
Perimeter = AB + BC + AC
Perimeter ≈ 8 + 8.49 + 6.32 ≈ 22.81
To find the area of triangle ABC, we can use the formula for the area of a triangle given its vertices:
Area = 0.5 * |x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|
Area = 0.5 * |-2(2-8) + 6(8-2) + 0(2-2)|
Area = 0.5 * |-12 + 48 + 0|
Area = 0.5 * 36
Area = 18
Therefore, the perimeter of triangle ABC is approximately 22.81 units and the area of triangle ABC is 18 square units.
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The vertices of triangle ABC are A(-2, 2), B(6, 2), and C(0, 8).
What is the perimeter and area of triangle ABC?
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