Question

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

Answer 1

Answer:

72.14 units

Step-by-step explanation:

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

Apply the distance formula as;

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Finding the lengths of the segments

For AB

$AB=\sqrt{(5-8)^2+(-18-13)^2} =\sqrt{(-3^2)+(-31^2)} =\sqrt{970} =31.14$

For BC

$BC=\sqrt{(-6-5)^2+(-6--18)^2} =\sqrt{(-11^2)+12^2} =\sqrt{265} =16.28$

For CD

$CD=\sqrt{(-2--6)^2+(-5--6)^2} =\sqrt{4^2+1^2} =\sqrt{16+1} =\sqrt{17} =4.12$

For DA

$DA=\sqrt{(8--2)^2+(13--5)^2} =\sqrt{10^2+18^2} =\sqrt{424} =20.6$

Finding the perimeter will be;

31.14+16.28+4.12+20.6=72.14 units