The first field is a very odd shape; none of the four sides are the same length, but they are all an exact number of metres long. The other odd thing is that two of the opposite corners are exact right angles.
This field is as small as it can be within these constraints - so what is its area? And by implication, how long are the four sides?
When you have worked out the size of the first field, we can consider the second field. It too has four sides, none of which are the same length, but all an exact number of metres long. Also, it too has two opposite corners that are exactly 90 degrees.
And to finish off the peculiarities, its area is precisely twice that of the first field.
So how long are the four sides of the second field?
Each of the two fields consists of two right-angled triangles, which can be resolved using Pythagoras' theory.
To solve the first you need to find the smallest number that is the sum of two pairs of square numbers. This turns out to be 65, which is the sum of 1 (one squared) and 64 (eight squared) and the sum of 16 (four squared) and 49 (seven squared).
So the four sides are 1, 8, 4 and 7 metres long.
Thus the field is made up of two triangles; one with a base of eight and a height of one, and the other with a base of four and a height of seven. Using the half base times height for the area of a triangle we get:
8 / 2 * 1 + 4 / 2 * 7 = 4 + 14 = 18
The field's area is 18 square metres.
In a similar way for the second field, 125 is the sum of 25 (five squared) and 100 (ten squared) and the sum of 121 (eleven squared) and 4 (two squared). And as above, the area is:
10 / 2 * 5 + 2 / 2 * 11 = 25 + 11 = 36
twice that of the first field.
The four sides are 2, 5, 10 and 11 metres.