Question

find the perimeter of the trapezoid with these vertices. (2, 2), (-5, 2), (-2, -5), (-5, -5) give an exact answer (not a decimal approximation). simplify your answer as much as possible.

Explanation:

Step1: Recall the distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Calculate the lengths of the sides

Let $A=(2,2)$, $B = (-5,2)$, $C=(-2,-5)$, $D=(-5,-5)$.
For side $AB$: $x_1 = 2,y_1 = 2,x_2=-5,y_2 = 2$. Then $d_{AB}=\sqrt{(-5 - 2)^2+(2 - 2)^2}=\sqrt{(-7)^2+0^2}=7$.
For side $BC$: $x_1=-5,y_1 = 2,x_2=-2,y_2=-5$. Then $d_{BC}=\sqrt{(-2+5)^2+(-5 - 2)^2}=\sqrt{3^2+(-7)^2}=\sqrt{9 + 49}=\sqrt{58}$.
For side $CD$: $x_1=-2,y_1=-5,x_2=-5,y_2=-5$. Then $d_{CD}=\sqrt{(-5 + 2)^2+(-5+5)^2}=\sqrt{(-3)^2+0^2}=3$.
For side $DA$: $x_1=-5,y_1=-5,x_2=2,y_2 = 2$. Then $d_{DA}=\sqrt{(2 + 5)^2+(2 + 5)^2}=\sqrt{7^2+7^2}=\sqrt{49+49}=\sqrt{98}=7\sqrt{2}$.

Step3: Calculate the perimeter

The perimeter $P$ of the trapezoid is $P=d_{AB}+d_{BC}+d_{CD}+d_{DA}=7+\sqrt{58}+3+7\sqrt{2}=10+\sqrt{58}+7\sqrt{2}$.

Answer:

$10+\sqrt{58}+7\sqrt{2}$