​ reposting this

You shouldn't have any trouble with the perimeter.  4 applications of the distance formula and sum.

For the area one way or the other you're going to need the height of the trapezoid.  That's not so easy.

As far as I can tell you're going to need to use the formula for the distance between two parallel lines.

\(\text{the top of the trapezoid has slope}\\ m=\dfrac{5-7}{6-3}=-\dfrac 2 3\\ (y_t-7)=-\dfrac 2 3 (x-3)\\ y_t = -\dfrac 2 3 x+9\)

\(\text{the bottom of the trapezoid has the same slope}\\ (y_b-5)=-\dfrac 2 3 (x+1)\\ y_b=-\dfrac 2 3 x +\dfrac{13}{3}\)

\(\text{The distance between two lines is}\\ d= \dfrac{|b_2-b_1|}{\sqrt{m^2+1}}\)

\(b_1=9,~b_2=\dfrac{13}{3},~m = -\dfrac 2 3\\ d = \dfrac{|9-\frac{13}{3}|}{\sqrt{\dfrac 4 9 + 1}} = \dfrac{\frac{14}{3}}{\sqrt{\dfrac{13}{9}}}=\\ \dfrac{14}{\sqrt{13}} = \dfrac{14\sqrt{13}}{13}\)

\(\text{the length of the top segment is}\\ \ell_t = \sqrt{(6-3)^2+(5-7)^2} = \sqrt{13}\\ \text{the length of the bottom segment is }\\ \ell_b = \sqrt{(-1-5)^2 + (5-1)^2}=\sqrt{52} = 2\sqrt{13}\)

\(\text{Now we can apply the area of a trapezoid formula}\\ A= \dfrac{\ell_t + \ell_b}{2} h = \\ \dfrac{\sqrt{13}+2\sqrt{13}}{2}\times\dfrac{14\sqrt{13}}{13} = \\ \dfrac{3\sqrt{13}}{2}\dfrac{14\sqrt{13}}{13} = 21\)

I kind of rushed through this and there may be an easier method.  I'm sure folks will chime in pointing out my errors.