Correct solutions were received from Thomas and from Isabelle of Lathallan School. Well done to both! Here is Isabelle's solution, with diagrams added by the editor.

I redrew the pentagon inside a 3x3 grid of 9 small squares (all the same size). The across lines were ABCD; EFGH; IJKL; MNOP. The down lines were AEIM; BFJN; CGKO; DHLP. The pentagon was BGOMEB. (This is shown on the diagram below, with the pentagon drawn in blue. Can you see why this is the same as the original pentagon? --Ed.)

Pentagon on grid

As lines EB and BG cut their small squares into halves, the triangle EBG has the same area as each of the nine small squares. The rest of the pentagon is 4 small squares so all of it is the same as 5 small squares. The whole area is 9 small squares so I need to find a square inside, which is just 4 small squares smaller. Then I noticed that if I drew line IO, it would cut the rectangle IKOM in half, making triangle IMO the same as 1 small square. Then I saw that I could draw BIOH as a square which is surrounded by 4 triangles each 1 small square in size. (This is shown in red on the second diagram -- Ed.)
Pentagon on grid with cuts

So this new square is 9 - 4 = 5 small squares in size. Now I could see that if I cut out EBI from the original pentagon, it would fit into BGH, and IMO would fit into OGH. This means that I could turn the pentagon into a square by cutting along BI and along IO and simply rearranging the pieces.

A slightly different approach was taken by Thomas, whose solution is given below. Can you see the similarities and differences between the two ways of solving the problem? (Again, the editor has added diagrams.)

From the midpoint of the left side of the square, draw a line to the midpoint of the diagonal connecting the bottom and right sides (BF). From this point, draw the second line to the upper right corner of the square (FE).
Pentagon with cuts

Why? 3/8 of the square is lost when the triangles are cut away, so, assuming a unit square, the area of the pentagon (and subsequent square) is 5/8; this requires a new square with sides of length

\frac{\sqrt{10}}{4}

The first line makes a triangle (BCF) that, once flipped over and placed with BC along edge AB, makes a right angle and two sides of the desired length. We can check that BF is of the right length using Pythagoras' theorem.

Triangle DEF, formed by the second line, has a longest side also of the desired length. We know the lengths of DE and DF ( $\frac{1}{2}$ and $\frac{\sqrt{2}}{4}$ ) and the angle EDF between them (135 degrees) and so can use the law of cosines $(c^{2} = a^{2} + b^{2} - 2 a b \cos C)$ . This gives

${EF}^{2} = {\frac{1}{2}}^{2} + {\frac{\sqrt{2}}{4}}^{2} - 2 \frac{1}{2} \frac{\sqrt{2}}{4} \frac{- \sqrt{2}}{2} = \frac{5}{8};$

thus

$EF = \sqrt{\frac{5}{8}} = \frac{\sqrt{10}}{4} .$

Thus if we move the second triangle so that DE lies along AD, we see that we have formed a square of side $\frac{\sqrt{10}}{4}$ .

The rearrangement is shown below.

Pentagon solution