Question

Find the perimeter of quadrilateral PQRS with the vertices P(2,4), Q(2,3), R(-2,-2), and S(-2,3).

Answer 1

Answer:

$P=16.53\ units$

Step-by-step explanation:

we know that

The perimeter of quadrilateral PQRS is equal to the sum of its four length sides

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

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

we have

the vertices P(2,4), Q(2,3), R(-2,-2), and S(-2,3)

step 1

Find the distance PQ

P(2,4), Q(2,3)

substitute in the formula

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

$d=\sqrt{(-1)^{2}+(0)^{2}}$

$d=\sqrt{1}$

$dPQ=1\ unit$

step 2

Find the distance QR

Q(2,3), R(-2,-2)

substitute in the formula

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

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

$dQR=\sqrt{41}\ units$

step 3

Find the distance RS

R(-2,-2), and S(-2,3)

substitute in the formula

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

$d=\sqrt{(5)^{2}+(0)^{2}}$

$dRS=5\ units$

step 4

Find the distance PS

P(2,4), S(-2,3)

substitute in the formula

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

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

$dPS=\sqrt{17}\ units$

step 5

Find the perimeter

$P=PQ+QR+RS+PS$

substitute the values

$P=1+\sqrt{41}+5+\sqrt{17}$

$P=6+\sqrt{41}+\sqrt{17}$

$P=16.53\ units$