Area of Trapezoid Does Not Equal Sum of Constituent Triangles Trying to calculate the area of this figure. It's two right triangles put together in the shape of a trapezoid. However, the sum of the areas of triangles doesn't equal the area of the trapezoid.
Sum of Area of Triangles: 
$$((4*3) * 1/2) + ((12 * 5) * 1/2) = 36;$$
Area of Trapezoid: 
$$(1/2 * (3 + 13))(4) = 32;$$
 A: Are you sure that's a trapezoid? The top and bottom look parallel in the picture, but is the picture a proof? Or is your calculation perhaps a proof that they aren't parallel?
A: 
LET us consider the following angles
We see that in $triangle ABD...90+b+c=180$
We see that in $triangle CBD... 90+a+d=180$
then we can say $a+d=b+c$
IF THIS WAS TRAPEZIUM THEN $a$=$c$
WHICH IS ONLY POSSIBLE IF $d$=$b$
so if  $d$=$b$ we could have said it was a trapezium
but if  $d$ is not equal to n$b$ we cannot say it is a trapezium
IN THE AFORESAID CASE IT IS NOT A TRAPEZIUM  which you can prove using trigonometry
$cos(CDB)=\frac {5}{13}$, but $cos(ABD)=\frac {3}{5}$. Therefore those two lines are not parallel ie $a$ not equal to $c$
A: $AB$ is not parallel to $DC$, because $cos(CDB)=\frac {5}{13}$, but $cos(ABD)=\frac {3}{5}$. Therefore those two lines are not parallel and the figure is not a trapezoid.
A: As you observed, the two triangles are Pythagorean:  $\triangle ABD$ is $3$-$4$-$5$, and $\triangle BCD$ is $5$-$12$-$13$.
Now, suppose $ABCD$ is a trapezoid with $AB \parallel DC$ as the drawn figure seems to suggest.  Then if we were to draw the altitude from $DC$ to $B$, this would have length $AD = 4$.  If we call the foot of this altitude $E$, then $DE = AB = 5$ and $CE = CD - DE = 13 - 5 = 8$.  It would then follow that $\triangle BCE$ is right, with legs of length $4$ and $8$, and hypotenuse $12$.  But $4$-$8$-$12$ is not a Pythagorean triple since $4^2 + 8^2 = 80 \ne 144 = 12^2$.
An accurately drawn figure is shown below:

