# Snapped Palm Tree

A palm tree has snapped in a storm. What is the height of the piece that is still standing?

A palm tree that was 16 metres tall has snapped in a storm, so that the top of the palm tree has landed 8 metres from the base.

What is the height of the piece of the trunk that is still standing (vertically)?

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*This problem is adapted from the World Mathematics Championships*

**Labelling the two parts of the trunk**

Suppose the vertical part of the trunk is $a$ and the slanted part is $b$, where $a$ and $b$ are in metres, as shown below.

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Then $a+b=16$, and $a$, $b$ and 8 are the sides of a right-angled triangle with hypotenuse $b$, so $a^2+8^2=b^2$.

This gives us the simultaneous equations $a+b=16$ and $a^2+64=b^2$.

__Solving by substitution of $b$__

Since we want $a$, it is sensible to make $b$ the subject of one of the equations and then substitute that into the other one.

$a+b=16$, so $b=16-a$.

Substituting that into the other equation gives $$\begin{align}a^2+64&=\left(16-a\right)^2\\

a^2+64&=256-32a+a^2\\

64&=256-32a\\

32a&=256-64\\

32a&=192\\

a&=6\end{align}$$

So the height of the part of the trunk that is still standing is 6 metres.

__Solving by substitution of $b^2$__

Since we want $a$, it is sensible to use substitution to remove $b$ from the equations. As $b^2$ is already the subject of the second equation, we could also make $b^2$ the subject of the first equation, and then equate them.

$$\begin{align}a+b&=16\\

b&=16-a\\

b^2&=\left(16-a\right)^2\\

b^2&=256-32a+a^2\end{align}$$

But from the second equation, $b^2=a^2+64$, so

$$\begin{align}a^2+64&=256-32a+a^2\\

64&=256-32a\\

32a&=256-64\\

32a&=192\\

a&=6\end{align}$$

So the height of the part of the trunk that is still standing is 6 metres.

**Labelling the parts of the trunk in terms of the height**

The total length of the trunk of the palm tree was 16 metres, so if the height of the part that is still standing is $h$, where $h$ is in metres, then the slanted section must be 16$-h$. This is labelled in the diagram below.

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The three lengths labelled in the diagram form a right-angled triangle with hypotenuse $16-h$, so $h^2+8^2=\left(16-h\right)^2$ by Pythagoras.

Solving for $h$,

$$\begin{align}h^2+8^2&=\left(16-h\right)^2\\

h^2+64&=256-32h+h^2\\

64&=256-32h\\

32h&=256-64\\

32h&=192\\

h&=6\end{align}$$

So the height of the part of the trunk that is still standing is 6 metres.