EH=EB=3, AE=4, so AH=5.

Note that △AEH is all equal to △CEB(BE=EH, two 90 degrees, ∠EAH=∠ECB(△HEA is similar to △HDC)).

So BC=5, CE=4, HC= 1, so everything can come out.

The area is AB*EC/2=7*4/2= 14.