Question

What is the perimeter of a polygon with vertices at

(−1, 3) , ​ (−1, 6) ​, (2, 10) , ​ (5, 6) ​​, and ​​ ​ (5, 3) ​?

Enter your answer in the box. Do not round any side lengths.​

Answer 1

(-6,6) to (2,10) and (2,10) to (5,6) = square root of 3 squared plus 4 squared = 5 3 plus 5 plus 5 plus 3 plus 6 = 22

Answer 2

we know that

the perimeter of a polygon is the sum of the length sides

in this problem we have five vertices

so

the polygon has five sides

Let

$A(-1,3)\\B(-1,6)\\C(2,10)\\D(5,6)\\E(5,3)$

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 1

Find the distance AB

$A(-1,3)\\B(-1,6)$

substitute the values in the formula

$d=\sqrt{(6-3)^{2}+(-1+1)^{2}}$

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

$dAB=3\ units$

Step 2

Find the distance BC

$B(-1,6)\\C(2,10)$

substitute the values in the formula

$d=\sqrt{(10-6)^{2}+(2+1)^{2}}$

$d=\sqrt{(4)^{2}+(3)^{2}}$

$dBC=5\ units$

Step 3

Find the distance CD

$C(2,10)\\D(5,6)$

substitute the values in the formula

$d=\sqrt{(6-10)^{2}+(5-2)^{2}}$

$d=\sqrt{(-4)^{2}+(3)^{2}}$

$dCD=5\ units$

Step 4

Find the distance DE

$D(5,6)\\E(5,3)$

substitute the values in the formula

$d=\sqrt{(3-6)^{2}+(5-5)^{2}}$

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

$dDE=3\ units$

Step 5

Find the distance AE

$A(-1,3)\\E(5,3)$

substitute the values in the formula

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

$d=\sqrt{(0)^{2}+(6)^{2}}$

$dAE=6\ units$

Step 6

Find the perimeter

the perimeter is equal to

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

substitute the values

$P=3+5+5+3+6=22\ units$

therefore

the answer is

the perimeter of the polygon is $22\ units$