Answer: Perimeter \(=22+6\sqrt5+6\sqrt3\); area \(=65+25\sqrt5\).

The horizontal difference between the two parallel sides is

\((11+2\sqrt5)-(4+3\sqrt5)=7-\sqrt5.\)

The height is

\(7+\sqrt5.\)

By Pythagoras' theorem,

\(BC^2=(7-\sqrt5)^2+(7+\sqrt5)^2.\)

Expanding gives

\(BC^2=(54-14\sqrt5)+(54+14\sqrt5)=108.\)

So

\(BC=\sqrt{108}=6\sqrt3.\)

The perimeter is therefore

\((4+3\sqrt5)+(11+2\sqrt5)+(7+\sqrt5)+6\sqrt3.\)

Hence

\(\boxed{22+6\sqrt5+6\sqrt3}.\)

The area of a trapezium is

\(\frac12(\text{sum of parallel sides})(\text{height}).\)

So the area is

\(\frac12\left((4+3\sqrt5)+(11+2\sqrt5)\right)(7+\sqrt5).\)

This is

\(\frac12(15+5\sqrt5)(7+\sqrt5).\)

Expanding,

\(\frac12(105+15\sqrt5+35\sqrt5+25)=65+25\sqrt5.\)

Therefore the area is

\(\boxed{65+25\sqrt5}.\)