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.

Image

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.

Image

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.