Question

in exercises 7 - 12, find the perimeter of the polygon with the given vertices. (see example 2.)

1. g(2, 4), h(2, - 3), j(- 2, - 3), k(- 2, 4)
2. q(- 3, 2), r(1, 2), s(1, - 2), t(- 3, - 2)
3. u(- 2, 4), v(3, 4), w(3, - 4)
4. x(- 1, 3), y(3, 0), z(- 1, - 2)

11.

Explanation:

Step1: Recall distance formula

The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. The perimeter of a polygon is the sum of the lengths of its sides.

Step2: For problem 7

• Side $GH$: $x_1 = 2,y_1 = 4,x_2 = 2,y_2=-3$. Since $x_1=x_2$, $d_{GH}=\vert4-(-3)\vert = 7$.
• Side $HJ$: $x_1 = 2,y_1=-3,x_2=-2,y_2=-3$. Since $y_1 = y_2$, $d_{HJ}=\vert2 - (-2)\vert=4$.
• Side $JK$: $x_1=-2,y_1=-3,x_2=-2,y_2 = 4$. Since $x_1=x_2$, $d_{JK}=\vert4-(-3)\vert = 7$.
• Side $KG$: $x_1=-2,y_1 = 4,x_2 = 2,y_2 = 4$. Since $y_1 = y_2$, $d_{KG}=\vert2-(-2)\vert = 4$.
• Perimeter $P=7 + 4+7 + 4=22$.

Step3: For problem 8

• Side $QR$: $x_1=-3,y_1 = 2,x_2 = 1,y_2 = 2$. Since $y_1 = y_2$, $d_{QR}=\vert1-(-3)\vert = 4$.
• Side $RS$: $x_1 = 1,y_1 = 2,x_2 = 1,y_2=-2$. Since $x_1=x_2$, $d_{RS}=\vert2-(-2)\vert = 4$.
• Side $ST$: $x_1 = 1,y_1=-2,x_2=-3,y_2=-2$. Since $y_1 = y_2$, $d_{ST}=\vert1-(-3)\vert = 4$.
• Side $TQ$: $x_1=-3,y_1=-2,x_2=-3,y_2 = 2$. Since $x_1=x_2$, $d_{TQ}=\vert2-(-2)\vert = 4$.
• Perimeter $P = 4+4+4+4=16$.

Step4: For problem 9

• Side $UV$: $x_1=-2,y_1 = 4,x_2 = 3,y_2 = 4$. Since $y_1 = y_2$, $d_{UV}=\vert3-(-2)\vert = 5$.
• Side $VW$: $x_1 = 3,y_1 = 4,x_2 = 3,y_2=-4$. Since $x_1=x_2$, $d_{VW}=\vert4-(-4)\vert = 8$.
• Side $WU$: $d_{WU}=\sqrt{(-2 - 3)^2+(4+4)^2}=\sqrt{(-5)^2+8^2}=\sqrt{25 + 64}=\sqrt{89}$.
• Perimeter $P=5 + 8+\sqrt{89}=13+\sqrt{89}$.

Step5: For problem 10

• Side $XY$: $d_{XY}=\sqrt{(3 + 1)^2+(0 - 3)^2}=\sqrt{16 + 9}=\sqrt{25}=5$.
• Side $YZ$: $d_{YZ}=\sqrt{(-1 - 3)^2+(-2 - 0)^2}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}$.
• Side $ZX$: $x_1=-1,y_1=-2,x_2=-1,y_2 = 3$. Since $x_1=x_2$, $d_{ZX}=\vert3-(-2)\vert = 5$.
• Perimeter $P=5+2\sqrt{5}+5=10 + 2\sqrt{5}$.

Step6: For problem 11

• Side $LM$: $d_{LM}=\sqrt{(4 - 1)^2+(0 - 4)^2}=\sqrt{9 + 16}=\sqrt{25}=5$.
• Side $MN$: $d_{MN}=\sqrt{(2 - 4)^2+(0 - 0)^2}=2$.
• Side $NP$: $d_{NP}=\sqrt{(-1 - 2)^2+(-2 - 0)^2}=\sqrt{9 + 4}=\sqrt{13}$.
• Side $PL$: $d_{PL}=\sqrt{(1+1)^2+(4 + 2)^2}=\sqrt{4 + 36}=\sqrt{40}=2\sqrt{10}$.
• Perimeter $P=5+2+\sqrt{13}+2\sqrt{10}=7+\sqrt{13}+2\sqrt{10}$.

Answer:

1. 22
2. 16
3. $13+\sqrt{89}$
4. $10 + 2\sqrt{5}$
5. $7+\sqrt{13}+2\sqrt{10}$