Question

What is the perimeter of this? (100 points)

Answer 1

19.2 units

Explanation:

To find the perimeter (
`\sf P`) of the quadrilateral ABCD formed by points A(-3,5), B(2,6), C(0,2), and D(-5,1) using the distance formula, we'll calculate the distances between consecutive pairs of these points and then sum them up.

The distance formula is given by:

`\sf d = √((x_2 - x_1)^2 + (y_2 - y_1)^2)`

Let's calculate each distance:

Distance between A and B (
`\sf AB`):

`\sf AB = √((2 - (-3))^2 + (6 - 5)^2) \\\\ = √(5^2 + 1^2) = √(25 + 1) \\\\= √(26) \\\\ \approx 5.1`

Distance between B and C (
`\sf BC`):

`\sf BC = √((0 - 2)^2 + (2 - 6)^2) \\\\ = √((-2)^2 + (-4)^2) \\\\= √(4 + 16) \\\\= √(20) \\\\= 2√(5)\\\\ \approx 4.5`

3. Distance between C and D (
`\sf CD`):

`\sf CD = √(((-5) - 0)^2 + (1 - 2)^2) \\\\ = √((-5)^2 + (-1)^2) \\\\ = √(25 + 1) \\\\ = √(26) \approx 5.1`

4. Distance between D and A (
`\sf DA`):

`\sf DA = √(((-3) - (-5))^2 + (5 - 1)^2) \\\\ = √(2^2 + 4^2)\\\\ = √(4 + 16)\\\\ = √(20) \\\\= 2√(5) \\\\ \approx 4.5`

Now, find the perimeter (
`\sf P`) by summing up these distances:

`\sf P = AB + BC + CD + DA`

`\sf P = 5.1 + 4.5 + 5.1 + 4.5`

`\sf P = 19.2`

Therefore, the perimeter of quadrilateral ABCD is approximately 19.2 units, rounded to the nearest tenth.