Question

The coordinates of the vertices of a polygon are (-2, 1). (-3, 3), (-1, 5), (2, 4), and (2, 1). What is the perimeter of the polygon to the nearest tenth of a unit.
Do not label your answer. Answer with a number only.

Answer 1

Answer:

$15.2\ units$

Step-by-step explanation:

step 1

Plot the vertices of the polygon to better understand the problem

we have

$A(-2, 1). B(-3, 3), C(-1, 5), D(2, 4),E(2, 1)$

using a graphing tool

The polygon is a pentagon (the number of sides is 5)

see the attached figure

The perimeter is equal to

$P=AB+BC+CD+DE+AE$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 2

Find the distance AB

$A(-2, 1). B(-3, 3)$

substitute in the formula

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

$d=\sqrt{(2)^{2}+(-1)^{2}}$

$d_A_B=\sqrt{5}=2.24\ units$

step 3

Find the distance BC

$B(-3, 3), C(-1, 5)$

substitute in the formula

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

$d=\sqrt{(2)^{2}+(2)^{2}}$

$d_B_C=\sqrt{8}=2.83\ units$

step 4

Find the distance CD

$C(-1, 5), D(2, 4)$

substitute in the formula

$d=\sqrt{(4-5)^{2}+(2+1)^{2}}$

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

$d_C_D=\sqrt{10}=3.16\ units$

step 5

Find the distance DE

$D(2, 4),E(2, 1)$

substitute in the formula

$d=\sqrt{(1-4)^{2}+(2-2)^{2}}$

$d=\sqrt{(-3)^{2}+(0)^{2}}$

$d_D_E=\sqrt{9}\ units$

$d_D_E=3\ units$

step 6

Find the distance AE

$A(-2, 1).E(2, 1)$

substitute in the formula

$d=\sqrt{(1-1)^{2}+(2+2)^{2}}$

$d=\sqrt{(0)^{2}+(4)^{2}}$

$d_A_E=\sqrt{16}\ units$

$d_A_E=4\ units$

step 7

Find the perimeter

$P=AB+BC+CD+DE+AE$

substitute the values

$P=2.24+2.83+3.16+3+4=15.23\ units$

Round to the nearest tenth of a unit

$P=15.2\ units$