How to find the area of a quardrilateral with vertices 3,7), (-3,3), (1,-3), and (7,1)?!?

quadrilateral
vertex coordinates
(3,7), (-3,3), (1,-3), and (7,1)
Area: 52

A(-3;3) , B(3;7) , C(7;1) , D(1;-3) . Prove that ABCD is a rectangle as follows gradient(AB) = gradient(CD)=2/3 , therefore AB is parallel to CD , gradient(AD)=gradient(BC) =-3/2 therefore AD is parallel to BC hence ABCD is a parallelogram ,also gradient(AB)*gradient(AD)=-1 therefore ABCD is now a rectangle . area=AB*AD= (2sqrt13)*(2sqrt13)=52

Let A=(3,7) B=(-3,3) C=(1,-3) D=(7,1)

Sketch the coordinates in the xy-plane.

Now you need to calculate the distance AB by using sqrt(x1-x2)^2+(y1-y2)^2

sqrt [(3--3)^2+(7-3)^2]

AB=2sqrt 13 or 2 root 13

Now you need to find the height from C to B, note B(-3,3), the y-coordinate already tell you the height.

Area of quardrilateral= base X height

Area of quardrilateral= 2sqrt 13 X 3

Area of quardrilateral= 6sqrt 13 or 6 root 13 unit^2

Note don't forget the unit. Hope it helps.