Question

Find the perimeter of quadrilateral ABCD. Round to the nearest tenth.

Answer 1

Answer:

14.6 units

Step-by-step explanation:

AB = $\sqrt{(1-3)^2+(3-5)^2}$

AB =$\sqrt{8}$

BC = $\sqrt{(3-1)^2+(-1-3)^2}=\sqrt{20}$

CD = $\sqrt{(5-3)^2+(3-(-1))} = \sqrt{20}$

DA = $\sqrt{(5-3)^2+(3-5)^2} = \sqrt{8}$

Primeter = $\sqrt{8} + \sqrt{20} + \sqrt{20} +\sqrt{8} = 14.6$