Question 29803
see the following example and try

SEE THE TRAPEZOID ....ABCD below
AREA OF TRAPEZOID IS GIVEN BY 
AREA=(1/2)*ALTITUDE*(SUM OF PARALLEL SIDES)=(1/2)*DP*(AB+CD)
WHERE ALTITUDE IS THE PERPENDICULAR DISTANCE BETWEEN PARALLEL SIDES =DP
PARALLEL SIDES ARE AB AND CD.
WE HAVE AB=40
BC=AD=20
ANGLE DAP=63
DAP IS A RIGHT ANGLED TRIANGLE...DP/DA=SIN(63)
DP=DA*SIN(63)=20*0.89=17.8
SO ALTITUDE=17.8
AP/DA=COS(63)
AP=DA*COS(63)=20*0.4545=9.1
SO CD =AB-2*AP SINCE TRAPEZOID IS ISOCELLES..
CD=40-2*9.1=21.8
AREA=(1/2)17.8(40+21.8)=550 SQUARE UNITS.
{{{drawing( 800, 800,-10,50,-10,50,red( line( 0, 0, 40,0 ) ),locate( 0, 0, A ),red( line( 40, 0, 30.9,17.8 ) ),locate( 40, 0, B ),red( line( 30.9, 17.8, 9.1,17.8 ) ),locate( 30.9, 17.8, C ),red( line( 9.1, 17.8, 0,0 ) ),locate( 9.1, 17.8, D ),green( line( 9.1, 17.8, 9.1,0 ) ),locate( 9.1, 0, P ) ))}}}