This time the area computation is not going to help at all as all the pieces have an area of 4 squares. We can still avoid to actually solve the puzzle by doing some consideration: let's start with a bit of paint ...

This is the problem once the squares to be filled have been painted like on a chessboard.

Looking at the puzzle we can see that we've to fill an equal number of dark squares and light squares. Looking at the pieces we can see that all of them but one have an interesting property: no matter how you rotate them or flip them over they'll cover an equal number of dark and light squares.

No matter how you rotate or flip a piece if a square of type "A" is dark then the others of type "A" are dark and the "B" type ones are light. All pieces excluding the T-shaped one have an equal number of "A" and "B" type squares.

The only piece that hasn't this property is the T-shaped one; no matter how you rotate or flip it over it will always cover three dark and one light squares or three light and one dark squares. Because it's the only piece with this property it must be the extra one as there is no way to re-equilibrate the light and dark square count using the other pieces once you place this piece on the puzzle.