Question

the coordinates of the vertices of a rectangle in order are (5,2), (-1,4), (-2,1) and (4,-1). what is the perimeter of the rectangle to the nearest tenth of a unit​

Answer 1

Answer:

19 units

Step-by-step explanation:

Plot points  A(5,2), B(-1,4), C(-2,1) and D(4,-1) on the coordinate plane. Find lengths of all sides as distances between two points:

$AB=\sqrt{(-1-5)^2+(4-2)^2}=\sqrt{6^2+2^2}=\sqrt{36+4}=\sqrt{40}=2\sqrt{10}\ units\\ \\BC=\sqrt{(-2-(-1))^2+(1-4)^2}=\sqrt{1^2+3^2}=\sqrt{1+9}=\sqrt{10}\ units\\ \\CD=\sqrt{(4-(-2))^2+(-1-1)^2}=\sqrt{6^2+2^2}=\sqrt{36+4}=\sqrt{40}=2\sqrt{10}\ units\\ \\DA=\sqrt{(5-4)^2+(2-(-1))^2}=\sqrt{1^2+3^2}=\sqrt{1+9}=\sqrt{10}\ units$

The perimeter of the rectangle ABCD is

$AB+BC+CD+DA=2\sqrt{10}+\sqrt{10}+2\sqrt{10}+\sqrt{10}=6\sqrt{10}\approx 18.97\ units\approx 19\ units$