You had to understand the pattern of the triangles and the relationship between triangles $ABO$ and $ODM$.

This relation states that because the hypotenuse of $ODM$ is the same as $ABO$, and because it is an isosceles right triangle (so the other sides are the same),

$$ABO = ODM$$

With this information, I was able to figure out the rest of the triangle, and even the $n^\text{th}$ trapezium with these steps!

1) last full smaller-looking triangle = half of the next triangle (in the first case, $OAB=\frac12OCD$)

2) multiply half-triangle by two to create full triangle (therefore $2\times$ size of the previous triangle) (in the first case, $OCD$ has area $2\times$ area of $OAB$ which is $2\times1=2$)

3) subtract larger from smaller to get trapezium (in the first case, $2-1=1$)

After solving 4 trapeziums, I noticed a pattern, which was doubling the last trapezium's size, which made it a lot easier to solve.

next trapezium = $2\times$ last trapeizum

Amrit said:

Let $OA=OB=x$, then

$\frac12x^2=1$

Therefore $x^2=2\Rightarrow x=\sqrt2$

Let $AB=OC=y$, then by the Pythagorean theorem,

$2x^2=y^2$

Plugging in the value of $x$, we have $2\times2=y^2\Rightarrow y=2$

The area of triangle $OCD$ is $\frac12y^2=2$. Therefore the area of the trapezium is $2-1=1$.

Pablo and Matt continued using the same methods to find the areas of the next trapeziums and generalise. This is Pablo's work:

TRAPEZIUM 2

$OC = OD = 2$

$CD^2 = 2^2 + 2^2$

$CD^2 = 8$

$CD =OE= 2\sqrt2$

Area of $OEF = 2\sqrt2 \times2\sqrt2\times \frac12= 4$

Area of trapezium $CDFE = $Area$ OEF - $Area$ OCD = 4 - 2 = 2 $ units$^2$

TRAPEZIUM 3

$OE = OF = 2\sqrt2$

$EF^2 = OE^2+ OF^2 = (2\sqrt2)^2+(2\sqrt2)^2$

$EF^2 = 8 + 8$

$EF = 4$

$OG = 4$

Area $OGH = 4\times4\times\frac12 = 8$ units$^2$

Area of trapezium $EFHG = $Area$ OGH - $Area$ OEF = 8 - 4 = 4 $ units$^2$

NTH TRAPEZIUM

For the first trapezium, $ABDC$, we had to subtract the area of the triangle $OAB$ from triangle $OCD$. If we call $OA$ the 'first' length, $OC$ is the 'second' length.

$OA = \sqrt{2^1}$ [first]

$OC = 2 = \sqrt{2^2}$ [second]

this continues for $OE$ and $OG$

So the area of the trapezium was $\frac12 OC^2 - \frac12 OA^2$

If instead of $OA$ and $OC$ we had the $n^\text{th}$ and the $(n+1)^\text{th}$ lengths, we can

say that:

the area of the $n^\text{th}$ trapezium

$= \frac12\left(\sqrt{2^{n+1}}^2\right) - \frac12\left(\sqrt{2^n}^2\right)$

$= \frac12 \times 2^{n+1} - \frac12\times 2^n$

$= 2^n - 2^{n-1}$

$= 2^n\left(1 - 2^{-1}\right)$

$= 2^n\left(1 - \frac12\right)$

$= 2^n\left(\frac12\right)$

$= 2^n\left(2^{-1}\right)$

$= 2^{n-1}$

Amrit used similarity to find the area of the $n^\text{th}$ trapezium:

It is clear to see that all the trapeziums are similar.

Say $OA$ was length $y_0$ and $OC$ was length $y_1$, then $y_1=\sqrt2y_0$

If $CD=OE$ was $y_2$, then $y_2=\sqrt2y_1$

If we call $y_n$ the length of the $n^\text{th}$ parallel line, then $y_{n+1}=\sqrt2 y_n$

So $\dfrac{y_{n+1}}{y_n}=\sqrt2$

This is the scale factor between the sides of adjacent trapezia, so the area factor between them is $2$. We now need to find the area of the first trapezia to find a formula for the others.

The area of the first trapezium was $1$, so the area of the second trapezium is $2$, and the area of the third trapezium is $4$, and the area of te fourth trapezium is $8$, so

the area of the $n^\text{th}$ trapezium is $2^{n-1}$.

Firstly, we must use the formula for the area of a triangle: $\frac12b\times h=1\Rightarrow b\times h=2$

Then, they observed that splitting triangle OAB in half along the height $h$ creates two more right-angled triangles. The new triangles are also isosceles, so $\frac12b=h$.

So $b=AB=2$ and $h=1$ since the area is $1$ square unit.

Given that $AB=OC$, $OC= 2$ and similarly $OD=2$ since $OCD$ is a similar triangle to that of triangle $OGH$. Using the relative sizes of a 45 degree right angled triangle: $1:1: \sqrt2$ (this comes from $1^2+1^2=(\sqrt2)^2$),

$CD:OC=\sqrt2:1$

So $CD=2\sqrt2$

Then, they observed that splitting triangle OAB in half along the height $h$ creates two more right-angled triangles. The new triangles are also isosceles, so $\frac12b=h$.

So $b=AB=2$ and $h=1$ since the area is $1$ square unit.

Given that $AB=OC$, $OC= 2$ and similarly $OD=2$ since $OCD$ is a similar triangle to that of triangle $OGH$. Using the relative sizes of a 45 degree right angled triangle: $1:1: \sqrt2$ (this comes from $1^2+1^2=(\sqrt2)^2$),

$CD:OC=\sqrt2:1$

So $CD=2\sqrt2$

Because we now know that the height of each right angled isosceles triangle in this shape is half the base, the height of triangle $OCD$ is $2^\frac12$.

Thus, the height of the trapezium is $2^{\frac12} – 1$.

$$\begin{split}A&=\frac{a+b}2h\\

&=\frac{AB+CD}2\left(2^{\frac12} – 1\right)\\

&=\frac{2\sqrt2+2}2\left(2^{\frac12} – 1\right)\\

&=\left(\sqrt2+1\right)\left(\sqrt2-1\right)\\

&=1\end{split}$$ So the area is $1$ square unit.

For the next trapezium,

$OE = 2\sqrt2$

Using the relative sizes ratio for a right angled isosceles triangle:

$EF = 4$

Therefore the height of triangle $OEF$ is $2$

But the height of the trapezium is $2 – \sqrt2$

Area of trapezium $CDFE:$

$$\begin{split}A &= \frac12 (a + b) h\\

&= \frac12 (EF + CE) (2 – \sqrt2)\\

&= \frac12 (4 + 2\sqrt2) (2 – \sqrt2)\\

&=(2+\sqrt2)(2-\sqrt2)

&= 2\end{split}$$

Adithya and Guruvignesh both created a table to show how the sequence continues:

Term | 1 | 2 | 3 | 4 | 5 |

Area | 1 | 2 | 4 | 8 | 16 |

And they spotted that the $n^\text {th}$ term is equal to $2^{n-1}$.

The reason for this is because each trapezium is a similar shape to the one before. So, if we increase by scale factor $\sqrt2$ (due to the relative sizes of a right angled isosceles triangle), the area factor will increase by 2. Because $AF = SF^2$.

Creating and manipulating expressions and formulae. 2D shapes and their properties. Making and proving conjectures. Regular polygons and circles. Area - triangles, quadrilaterals, compound shapes. Networks/Graph Theory. Inequalities. Mathematical reasoning & proof. Quadratic equations. Generalising.