Question

Find the perimeter of the triangle with the vertices at (2, 2), (-2, 2), and (-2, -3). I keep getting 18 but my online homework website says it is wrong. Find the point (0,b) on the y-axis that is equidistant from the points (5,5) and (4,−3)?

Answer 1

The perimeter is `9+sqrt41`, or approximately 15.4.

Explanation:

The perimeter of the triangle is equal to the sum of the lengths of the 3 sides.

The 3 sides are the segments from:
`(2,2) to (-2,2)`
`(-2,2) to (-2,-3)`
`(-2, -3) to (2, 2)`

The side from (2, 2) to (-2, 2) is going to be a horizontal line from -2 to 2 on the x-axis. In other words, this side has a length of 4 `(2-(-2))`.

The side from (-2, 2) to (-2, -3) is going to be a vertical line (because both x-coordinates are the same) from -3 to 2 on the y-axis. In other words, this side has a length of 5 `(2-(-3))`.

The side from (-2, -3) to (2, 2) is not horizontal or vertical. We're going to have to use the pythagorean theorem for this side. The x-value changes by 4 (-2 to 2), and the y-value changes by 5 (-3 to 2). So, this side's length is equal to `sqrt(4^2+5^2)`

`sqrt(4^2+5^2) = sqrt(16+25) = sqrt41` (this is a prime number so this is as simple as you can make it)

Therefore, the perimeter of the triangle is 4 + 5 + `sqrt41`, or `9 + sqrt41`