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2 years ago

7

To the nearest tenth, what is the perimeter of the triangle with vertices at (−2, 3), (3, 6), and (2, −2)?

1 answer:

katovenus [111]2 years ago

7 0

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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1 year ago

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Bad White [126]

Answer:

Option (1)

Step-by-step explanation:

From the picture attached,

Triangle CAB is a right triangle.

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Similarly, m∠ACD = 90°

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Since m∠4 = 35°,

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m∠3 = 90° - 35°

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In the triangle ABC,

m∠ACB + m∠CBA + m∠BAC = 180° [Property of a triangle]

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