To the nearest tenth, what is the perimeter of the triangle with vertices at (−2, 3), (3, 6), and (2, −2)?
1 answer:
9514 1404 393
Answer:
20.3
Step-by-step explanation:
The distance formula can be used to find the side lengths.
d = √((x2 -x1)^2 +(y2 -y1)^2)
For the first two points, ...
d = √((3 -(-2))^2 +(6 -3)^2) = √(5^2 +3^2) = √34 ≈ 5.83
For the next two points, ...
d = √((2 -3)^2 +(-2-6)^2) = √(1 +64) = √65 ≈ 8.06
For the last and first points, ...
d = √((-2-2)^2 +(3-(-2)^2) = √(16 +25) = √41 ≈ 6.40
Then the sum of the side lengths is ...
5.83 +8.06 +6.40 = 20.29 ≈ 20.3
The perimeter of the triangle is about 20.3 units.
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