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# Trapezium Arch

Sachdave from Doha College in Qatar used exterior angles, marked $\text a$ in the diagram below. Sachdave's method was similar to this:

Tracing around the arch, it is necessary to turn through $\text{a}^\text{o}$ nine times, or ten times to end up facing in the opposite direction from where you started. You would also turn through $180^\text{o}.$

So $\text a = 180\div10 = 18$. Then notice that $\text{a, } x$ and ${x}$ make a straight line, so $\text a + 2x = 180,$ so $2 x = 162,$ so $x = 81.$

Sameer from Hymers College and Kira and Emma from Wycombe High School in the UK considered the angles in a regular icosagon (twenty-sided polygon). Sameer said:

The inner perimeter of this shape forms half of a regular icosagon.

In any regular polygon, the sum of the exterior angles is $360^\text{o}.$ As there are $20$ exterior angles, each one is $360^\text{o}\div20$, which is $18^\text{o}.$

The interior angle is $180^\text o-$ exterior angle, which will be $180^\text o-18^\text o=162^\text o.$ This is shown in the diagram.

The interior angle forms a circle with the smaller angles of the trapeziums. In a circle, all the angles add up to $360^\text o.$

The two angles marked $y^\text o$ intersecting this circle are the same, so $$\begin{align}360&=162+2y\\ \Rightarrow 198&=2y\\ \Rightarrow y&=99\end{align}$$

In a trapezium, the angles add up to $360^\text o,$ so $$\begin{align}2x+2y&=360\\ \Rightarrow x+y&=180\\ \Rightarrow

x+99&=180\\ \Rightarrow x&=81\end{align}$$

Sophie from the Stephen Perse Foundation used triangles:

$$180^\text{o}\div10=18^\text o$$

Considering the trapezium below the triangle,

$$\begin{align}99\times2&=198\\\frac{360-198}{2}&=81\end{align}$$

Sophie could also have used the larger isosceles triangles directly:

Thank you also to Toby from Chairo Christian School in Australia, who submitted a correct solution.

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Age 11 to 14

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Sachdave from Doha College in Qatar used exterior angles, marked $\text a$ in the diagram below. Sachdave's method was similar to this:

Tracing around the arch, it is necessary to turn through $\text{a}^\text{o}$ nine times, or ten times to end up facing in the opposite direction from where you started. You would also turn through $180^\text{o}.$

So $\text a = 180\div10 = 18$. Then notice that $\text{a, } x$ and ${x}$ make a straight line, so $\text a + 2x = 180,$ so $2 x = 162,$ so $x = 81.$

Sameer from Hymers College and Kira and Emma from Wycombe High School in the UK considered the angles in a regular icosagon (twenty-sided polygon). Sameer said:

The inner perimeter of this shape forms half of a regular icosagon.

In any regular polygon, the sum of the exterior angles is $360^\text{o}.$ As there are $20$ exterior angles, each one is $360^\text{o}\div20$, which is $18^\text{o}.$

The interior angle is $180^\text o-$ exterior angle, which will be $180^\text o-18^\text o=162^\text o.$ This is shown in the diagram.

The interior angle forms a circle with the smaller angles of the trapeziums. In a circle, all the angles add up to $360^\text o.$

The two angles marked $y^\text o$ intersecting this circle are the same, so $$\begin{align}360&=162+2y\\ \Rightarrow 198&=2y\\ \Rightarrow y&=99\end{align}$$

In a trapezium, the angles add up to $360^\text o,$ so $$\begin{align}2x+2y&=360\\ \Rightarrow x+y&=180\\ \Rightarrow

x+99&=180\\ \Rightarrow x&=81\end{align}$$

Sophie from the Stephen Perse Foundation used triangles:

$$180^\text{o}\div10=18^\text o$$

Each of the triangles is isosceles.

$\dfrac{180-18}{2}=81$ which is shown on the diagram.

Angles on a straight line add up to $180^\text o$

$180-81=99$ which is shown on the diagram.

$\dfrac{180-18}{2}=81$ which is shown on the diagram.

Angles on a straight line add up to $180^\text o$

$180-81=99$ which is shown on the diagram.

Considering the trapezium below the triangle,

$$\begin{align}99\times2&=198\\\frac{360-198}{2}&=81\end{align}$$

Thank you also to Toby from Chairo Christian School in Australia, who submitted a correct solution.

This problem is taken from the UKMT Mathematical Challenges.

You can find more short problems, arranged by curriculum topic, in our short problems collection.