Question 1147593
<pre>{{{drawing(400,15600/97,-5,92,-5,34,
locate(0,0,A), locate(87.6,0,B),locate(27,33,D), locate(60,33,C),
green(line(27.78633482,0,27.78633482,28.77359201),
line(59.78633482,0,59.78633482,28.77359201)),
locate(27.7,0,E),locate(59.7,0,F),locate(25,28,134^o),
locate(9,17,40),
red(arc(27.78633482,27.78633482,15,-15,222,365)),
locate(43,33,32),
line(0,0,87.57266964,0),line(0,0,27.78633482,28.77359201),
line(27.78633482,28.77359201,27.78633482+32,28.77359201),
line(87.57266964,0,27.78633482+32,28.77359201) )}}}

AB∥DC tells us that ABCD is a trapezoid (trapezium in the UK).
AD = BC tells us it is an isosceles trapezoid. 
m∠ADC = 134°, ∠ADC and ∠A are supplementary, so ∠A = 180°-134°=46°

{{{DE/AD=sin(""<A)}}}

{{{DE/AD=sin(46^o)}}}

{{{DE=AD*sin(46^o)}}}

{{{DE=40*0.7193398003}}}

{{{DE=28.77359201}}}

{{{AE/AD=cos(""<A)}}}

{{{AE/AD=cos(46^o)}}}

{{{AE=AD*cos(46^o)}}}

{{{AE=40*0.6946583705}}}

{{{AE=27.78633482}}}

{{{matrix(1,4,
Area, of, triangle, ADE = expr(1/2)*AE*DE = 399.7563308)}}}

Area of triangle BCF is also 399.7563308

Area of rectangle DEFC = EF∙DE = 32∙28.77359201 = 920.7549443

Adding the two triangles' and the rectangle's areas:

399.7563308 + 399.7563308 + 920.7549443 = 1720.267606

Edwin</pre>