Question

Find the perimeter of the quadrilateral that has vertices at (8,13), (5,-18), (-6,-6), and (-2,-5).

Answer 1

72.14 units

Explanation:

Let A(8,13), B(5,-18),C(-6,-6) and D(-2,-5)

Apply the distance formula as;

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

Finding the lengths of the segments

For AB

`AB=√((5-8)^2+(-18-13)^2) =√((-3^2)+(-31^2)) =√(970) =31.14`

For BC

`BC=√((-6-5)^2+(-6--18)^2) =√((-11^2)+12^2) =√(265) =16.28`

For CD

`CD=√((-2--6)^2+(-5--6)^2) =√(4^2+1^2) =√(16+1) =√(17) =4.12`

For DA

`DA=√((8--2)^2+(13--5)^2) =√(10^2+18^2) =√(424) =20.6`

Finding the perimeter will be;

31.14+16.28+4.12+20.6=72.14 units