What is the quickest way to solve such a question below? I have tried different rotations to fit them as a square, but couldn't make it work.
Four of the five jigsaw pieces shown below fit together to make a square. Which one of these diagrams does not fit?
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The figures have $4, 5, 6, 7, 8$ squares respectively. See which of the four numbers added together gives a perfect square number.
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As all others have pointed out, you can just count the number of boxes to figure out that the square will have to be a 5x5 square and that B is the piece that will not be used. But if you are still interested in how that square is made:
Put D to the left of C to form the base of the square. Rotate E 180 degrees and put on top of this. The A piece will now fit the remaining space.
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Each of the choices has $4$, $5$, $6$, $7$ and $8$ small squares respectively. Since we need to include four of these figures and exclude one, I would add up these five numbers and look at each of the possibilities you get from subtracting one of the numbers from that sum. Then we just see which result is a square.
The sum of the numbers is $30$. The five possibilities you get from subtracting the choices is $$26 \quad 25 \quad 24 \quad 23 \quad 22\,.$$ The only square among these numbers is $25$, so if there is a way to arrange four of the figures into a square, that arrangement must exclude the figure with $5$ little squares, figure $B$. Then see Bram28's answer for the actual construction of the square (so we know it really can be done).
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Four of the five jigsaw pieces shown below fit together to make a square. Which one of these diagrams does not fit?
Having figured that the square must be $5\times 5$ (i.e. $30-5$) and the block $B$ (i.e. $=5$) is the odd one, how to construct it step-by-step?
The block $E$ must be in the corner, otherwise other blocks can not squeeze through the remaining sidewalk:
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Only block $A$ can fit the bottom left corner:
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The block $D$ must be rotated $180^\circ$ and put in one corner:
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And finally, the block $C$ must also be rotated by $180^\circ$ and placed in the suitable spot.
Observe that the least number of little squares in the final figure would be $4+5+6+7=22$ and the most number of little squares equals $5+6+7+8=26$. Since the final figure is a square, it obviously has $25$ little squares. This means the second figure isn't used.