find the perimeter of the polygon

Question

Find the perimeter of the polygon

with vertices A (-5,2) B (1,5) C (3,1) D (-3,-2)

Answer 1

the distance formula is √(x₂-x₁)²+(y₂-y₁)²

A(-5,2) and B(1,5)

√(1-(-5))²+(5-2)²=√36+9=√45=3√5

B(1,5) and C(3,1)

√(3-1)²+(1-5)²=√4+16=√20=2√5

when I did the problem I found the opposite sides to be equal so you get the same two distances again; try it for yourself

 

Answer 2

The shape formed by these vertices would be a quadrilateral. 

Normally one could measure the unit difference between the x and y coordinates, respectively. But since this quadrilateral is at an angle, that method doesn't apply here. So we would use the distance formula. 

The formula is as follows: the square root of [ (x₁+x₂)² + (y₁+y₂)² ] x1  and x2 would be the x-coordinates of the first line (say, AB) and y1 and y2 would be the y coordinates of that line. 

So we use AB as our example

and plug those values A(-5, 2) and B(1,5) into the formula

we get square root of [(-5+1)² + (2+5)²]

and now solve for the square root of (-4)² + (7)²
which would be the square root of 16+49. That would be the length of side AB.

Do that for all other sides (BC, CD, AD), then add up your results, and you get the perimeter of the quadrilateral.