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.
12.

Explanation:

Step1: Recall distance formula

The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For vertical or horizontal lines, if $x_1=x_2$, $d=\vert y_2 - y_1\vert$; if $y_1 = y_2$, $d=\vert x_2 - x_1\vert$.

Step2: Solve for problem 7

• $GH$: Since $x_G=x_H = 2$, $GH=\vert4-(-3)\vert=7$.
• $HJ$: Since $y_H=y_J=-3$, $HJ=\vert2 - (-2)\vert = 4$.
• $JK$: Since $x_J=x_K=-2$, $JK=\vert4-(-3)\vert=7$.
• $KG$: Since $y_K=y_G = 4$, $KG=\vert2-(-2)\vert = 4$.
• Perimeter $P=7 + 4+7 + 4=22$.

Step3: Solve for problem 8

• $QR$: Since $y_Q=y_R = 2$, $QR=\vert-3 - 1\vert=4$.
• $RS$: Since $x_R=x_S = 1$, $RS=\vert2-(-2)\vert = 4$.
• $ST$: Since $y_S=y_T=-2$, $ST=\vert1-(-3)\vert = 4$.
• $TQ$: Since $x_T=x_Q=-3$, $TQ=\vert2-(-2)\vert = 4$.
• Perimeter $P=4 + 4+4 + 4=16$.

Step4: Solve for problem 9

• $UV$: Since $y_U=y_V = 4$, $UV=\vert-2 - 3\vert=5$.
• $VW$: Since $x_V=x_W = 3$, $VW=\vert4-(-4)\vert = 8$.
• $WU$: $WU=\sqrt{(-2 - 3)^2+(4-(-4))^2}=\sqrt{(-5)^2+8^2}=\sqrt{25 + 64}=\sqrt{89}$.
• Perimeter $P=5 + 8+\sqrt{89}=13+\sqrt{89}$.

Step5: Solve for problem 10

• $XY=\sqrt{(3-(-1))^2+(0 - 3)^2}=\sqrt{4^2+(-3)^2}=\sqrt{16 + 9}=5$.
• $YZ=\sqrt{(-1 - 3)^2+(-2 - 0)^2}=\sqrt{(-4)^2+(-2)^2}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}$.
• $ZX$: Since $x_Z=x_X=-1$, $ZX=\vert3-(-2)\vert=5$.
• Perimeter $P=5+2\sqrt{5}+5=10 + 2\sqrt{5}$.

Step6: Solve for problem 11

• $LP=\sqrt{(1-(-1))^2+(4-(-2))^2}=\sqrt{2^2+6^2}=\sqrt{4 + 36}=\sqrt{40}=2\sqrt{10}$.
• $PM=\sqrt{(4-(-1))^2+(0-(-2))^2}=\sqrt{5^2+2^2}=\sqrt{25 + 4}=\sqrt{29}$.
• $MN=\sqrt{(4 - 2)^2+(0-0)^2}=2$.
• $NL=\sqrt{(1 - 2)^2+(4-0)^2}=\sqrt{(-1)^2+4^2}=\sqrt{1 + 16}=\sqrt{17}$.
• Perimeter $P=2\sqrt{10}+\sqrt{29}+2+\sqrt{17}$.

Step7: Solve for problem 12

• $FA$: Since $y_F=y_A = 4$, $FA=\vert-2-0\vert=2$.
• $AB=\sqrt{(2 - 0)^2+(0 - 4)^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$.
• $BC$: Since $x_B=x_C = 2$, $BC=\vert0-(-2)\vert=2$.
• $CD$: Since $y_C=y_D=-2$, $CD=\vert2-0\vert=2$.
• $DE$: Since $x_D=x_E = 0$, $DE=\vert-2 - 2\vert=4$.
• $EF$: Since $y_E=y_F = 2$, $EF=\vert0-(-2)\vert=2$.
• Perimeter $P=2+2\sqrt{5}+2+2+4+2=12 + 2\sqrt{5}$.

Answer:

Problem 7: 22
Problem 8: 16
Problem 9: $13+\sqrt{89}$
Problem 10: $10 + 2\sqrt{5}$
Problem 11: $2\sqrt{10}+\sqrt{29}+2+\sqrt{17}$
Problem 12: $12 + 2\sqrt{5}$