I need to draw an isosceles trapezoid with perpendicular diagonals, but I am not sure where to start. Here is my code for a square with perpendicular diagonals (if it is even helpful).

begintikzpicture[scale=5.5]
 coordinate[label=left:$W$] (W) at (0,0);
 coordinate[label=right:$X$] (X) at (1,0);
 coordinate[label=:$Y$] (Y) at (1, 1);
 coordinate[label=:$Z$] (Z) at (0, 1);
 coordinate[label=:$O$] (O) at (0.5, 0.5);
 draw (W)--(X)--(Y)--(Z)--(W)--(Y)--(Z)--(X);
 draw (O) -- node[sloped] $ (Y);
 draw (O) -- node[sloped] $ (Z);
 draw (O) -- node[sloped] $ (X);
 draw (O) -- node[sloped] $ (W);
endtikzpicture

Would I have to mathematically find the coordinates of one such trapezoid or is there another way to do it? All help is appreciated!

A PSTricks solution just for fun purposes.

documentclass[pstricks,12pt]standalone
usepackagepst-eucl
begindocument
foreach a in 0,10,...,350%
pspicture(-7,-7)(7,7)
 pstGeonode(0,0)O(2;a)A([offset=6]AO)B
 pstRotation[RotAngle=-90]OA[D]
 pstRotation[RotAngle=90]OB[C]
 psline(A)(B)(C)(D)(A)(C)(D)(B)
endpspicture
enddocument

Algorithm

• Define two points O and A.


• Define point B such that OA is perpendicular to OB.


• Define C as the image of rotating point B about O counter-clockwise.


• Define D as the image of rotating point A about O clockwise.


• Draw the lines.

Let ABXY be an isosceles trapezium with perpendicular diagonals AX and BY.

Then, the triangle AOB is isosceles and right at O (ie., the angle at O is right). So, the base angles should have 45 degrees.

Since AB and XY are parallel, to construct the trapezium it is enough to choose the lengths r_1 = OX and r_2 = OA.

If the origin of the coordinate system is O=(0,0) then the vertices can be given in polar coordinates by:

A=(-135:r_2) B=(- 45:r_2) X=( 45:r_1) Y=(135:r_1)

Below, the MWE where the commands

newcommandradioi1cm
newcommandradioii2cm

determines the radius r_1 and r_2.

MWE

documentclass[margin=2mm]standalone
usepackagetikz

begindocument
begintikzpicture[scale=2]
newcommandradioi1cm
newcommandradioii2cm
coordinate[label=below:$O$] (O) at ( 0:0 );
coordinate[label=left:$A$] (A) at (-135:radioii);
coordinate[label=right:$B$] (B) at (- 45:radioii);
coordinate[label=right:$X$] (X) at ( 45:radioi );
coordinate[label=left:$Y$] (Y) at ( 135:radioi );

draw (A) -- (B) -- (X) -- (Y) -- cycle;
draw[dashed] (A) -- (X) (B) -- (Y);
endtikzpicture
enddocument

Conceptually the same as @Sigur's nice answer but with slightly different parametrization. There are two free parameters, which can be taken to be the length of the upper edge and the height. They go into commands of the sort

draw[isosceleles trapezium=of width 2 and height 3 and name my trap];

This is illustrated in

documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture[isosceleles trapezium/.style args=of width #1 and height #2
and name #3insert path=
(45:#1/sqrt(2)) coordinate(#3-TR) -- (-45:sqrt(#2*#2-#1*#1/2)) coordinate(#3-BR) 
-- (-135:sqrt(#2*#2-#1*#1/2)) coordinate(#3-BL) -- (135:#1/sqrt(2)) coordinate(#3-TL) -- cycle]
draw[isosceleles trapezium=of width 2 and height 3 and name my trap];
draw[dashed] (my trap-TL) -- (my trap-BR) (my trap-TR) -- (my trap-BL);
draw[latex-latex] ([yshift=2mm]my trap-TL) -- ([yshift=2mm]my trap-TR)
node[midway,fill=white] $w$;
draw[latex-latex] ([xshift=-2mm]my trap-TL -| my trap-BL) -- 
([xshift=-2mm]my trap-BL) node[midway,fill=white] $h$;
beginscope[xshift=6cm,rotate=30]
 draw[isosceleles trapezium=of width 1 and height 2.5 and name another trap];
 draw[dashed] (another trap-TL) -- (another trap-BR) (another trap-TR) -- (another trap-BL);
endscope
endtikzpicture
enddocument

So the first parameter is w and the second one h. In addition, there is the name which is used to name the corners such that you can draw the diagonals, say. And of course you can rotate the thing and so on.