This solution is from David Lowe, age 16, from Trinity School, Carlisle, Cumbria.

A square is folded so that the corner E coincides with the midpoint of an opposite edge as shown in the diagram. The length of the edge of the square is 1 unit and the problem is to find the lengths and areas of the three triangles.

David uses trigonometry but alternatively, once you have found the dimensions of triangle EGF you can find all the other lengths and areas using the fact that the triangles EGF, CED and CAB are similar 3:4:5 triangles.

The table shows the solutions:

This is David's method:

The edge, $\mathrm{EF}=1/2$. Taking $\mathrm{FG}=a$ then $\mathrm{GE}=1-a$

As it has a corner of the paper as part of it, we know $\mathrm{EGF}$ is a right angled triangle, and so ${\mathrm{GE}}^2={\mathrm{EF}}^2+{\mathrm{FG}}^2$, which gives

${\displaystyle (1-a){}^2=1/4+a^2}$

so

${\displaystyle a=3/8.}$

Now that we have the length of $a$ we have the lengths of all the sides of this triangle: $\mathrm{EF}=1/2$ , $\mathrm{FG}=3/8$ and $\mathrm{GE}=5/8.$

The area of triangle $\mathrm{EFG}=1/2$ base $\times$ height $=3/32$.

Using this information and trigonometry, we can work out the angle $\mathrm{GEF}$: $\angle \mathrm{GEF}=\tan {}^{-1}(0.375/0.5)=\tan {}^{-1}0.75=36.9^{\circ }.$

taking the angle to the nearest tenth of a degree. Knowing this, we can work out the angle $\mathrm{DEC}=53.1^{\circ }$ and use it with the length $\mathrm{DE}$ to work out the length $\mathrm{DC}=2/3.$ The last part needed of this triangle is $\mathrm{CE}$ and, using Pythagoras theorem, we get $\mathrm{CE}=5/6.$

So $\mathrm{DE}=\frac{1}{2}$, $\mathrm{DC}=\frac{2}{3}$, $\mathrm{CE}=\frac{5}{6}$ and the area of triangle $\mathrm{DEC}=1/2\times 2/3\times 1/2=\frac{1}{6}.$

The last triangle needed is $\mathrm{ABC}$. We know the length $\mathrm{BC}=1-\mathrm{CE}=1/6.$ We can also work out the angle $\mathrm{BCA}$ which is equal to angle $\mathrm{DCE}$ and to angle $\mathrm{FEG}.$ We can now use trigonometry to work out the length $\mathrm{AB}=1/6\tan \mathrm{BCA}=1/8.$

And so once again, using Pythagoras, we can work out the length of line $\mathrm{AC}$ which is $5/24.$

And so we have the final set of lengths: $\mathrm{AB}=1/8$, $\mathrm{BC}=1/6$ and $\mathrm{AC}=5/24.$

The area of triangle $\mathrm{ABC}=1/2\times 1/8\times 1/6=1/96$.

This means the total area of all the paper with a single thickness = $26/96=.270833..$

This can be backed up by working out the area of the trapezium $\mathrm{ABEG}$, subtracting the area of triangle $\mathrm{ABC}$ and then multiplying the result by 2 to give the area of the original square that is now double thickness.

Area of trapezium = $1/2[1/8+5/8]\times 1=3/8.$ Double thickness area = $3/8-1/96=35/96.$

The total area is $2\times 35/96+26/96=1$ and so all areas are worked out and recorded, and the total area of single thickness paper and double thickness areas are recorded.