# Towering Trapeziums

Can you find the areas of the trapezia in this sequence?

$OGH$ is an isosceles right-angled triangle:

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Lines $AB$, $CD$, $EF$, and $GH$ are parallel.

**Suppose the area of the smallest triangle $OAB$ is one square unit.**

- If lines $OC$ and $AB$ have the same length, calculate the area of trapezium $ABDC$.
- If lines $OE$ and $CD$ also have the same length, calculate the area of trapezium $CDFE$.
- If lines $OG$ and $EF$ also have the same length, calculate the area of trapezium $EFHG$.

**What would be the area of the $n^{th}$ trapezium in the chain?**

Can you explain your results?

Construct the line perpendicular to $CD$ that passes through the midpoint $M$ of $CD$:

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What do you notice about $\triangle OAB$ and $\triangle OCM?$

**T. Hakim from Annie Gale School in Canada defined a new point, $M$, as the midpoint of $CD$.**

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),

$$ABO = ODM$$

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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

**Matt from Swindon Academy in the UK, Amrit from Hymers College in the UK and Pablo from King's School Alicante in Spain considered the sides $OA$ and $OB$ as the base and height of triangle OAB, since the angle between them is $90^\text o$.**

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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 $2^{n-1}$.

**Guruvignesh and Adithya, also from Hymers College, considered the base and height of the triangle as horizontal and vertical.**

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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 $AF = SF^2$.

Why do this problem?

This problem gives students an opportunity to do some geometrical reasoning and apply what they know about Pythagoras' theorem, area of triangles, and sequences.

Possible approach

Show the image:

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"This diagram was created by starting with a right angled isosceles triangle $OAB$ with area 1 unit."

"$AB$, $CD$, $EF$, and $GH$ are all parallel."

"$OC$ and $AB$ have the same length."

"$OE$ and $CD$ have the same length."

"$OG$ and $EF$ have the same length."

Invite students to construct the diagram for themselves, and then pose the following questions:

"Can you find the area of the trapezium $ABDC?$"

"What about trapezium $CDFE?$"

"What about trapezium $EFHG?$"

Give students some time to explore the problem, at first on their own and then with a partner. As they are working, circulate and listen out for useful strategies to discuss in a mini-plenary. If no useful strategies emerge, share with students the advice offered in Getting Started.

Any students who work out the areas quickly can consider the following: "If the pattern continued, what would be the area of the $n^{th}$ trapezium in the chain?"

Finish by discussing what students found and what strategies they used to work out the relevant areas.

Key questions

Can you find an expression for the area of an isosceles right-angled triangle with two sides of length $a$?

Can you find an expression for the area of an isosceles right-angled triangle with hypotenuse $b$?

Possible extension

Another challenge involving areas and Pythagoras' Theorem is Compare Areas.

### Possible support

Some students may benefit from working on some of the Pythagoras' Theorem Short Problems before looking at this problem.