When Max is 8m from a lamp post which is 6m high his shadow is 2m long. When he is 3m from the lamp post, what is the length of his shadow?

Question

When Max is 8 m from a lamp post which is 6 m high his shadow is 2 m long. When Max is 3 m from the lamp post, what is the length of his shadow?

Answer 1

When he is $8m$ from the lamp post, the end of the shadow is $10m$ from the lamp post meaning we have two legs of a triangle with $6$ and $10$ respectively.

By proportion, then, Max is X-high where $$\frac{X}{2}=\frac{6}{10}\implies X=\frac{12}{10}=1.2\space m=3.927'\quad \text{not very tall}$$

If we now move him closer ($3\space m$) from the post, we can work out the proportions the other way where here, the shadow will be $Y$. $$\frac{X}{Y}=\frac{6}{3+Y}\implies 3X+XY=6Y\implies Y = -\frac{3 X}{X - 6}\land X\ne6$$

$$Y = -\frac{3(1.2)}{(1.2 - 6)}=-\frac{3.6}{-4.8}=0.75\space m$$

Answer 2

Let the top of the lamp post be at $A = (0,6)$, and Max be at $B = (8,0)$, with his head being at height $h$ or $H = (8, h)$.

The gradient of line $AH$ is $\frac{h-6}{8-0} = \frac{h-6}{8}$. When Max moves to the point $B' = (3,0)$, with his head at $H' = (3,h)$, the gradient of line $AH'$ is $\frac{h-6}{3}$.

Now $\text{gradient} = \frac{\text{change in } y}{\text{change in } x}$. Since Max's height does not change, the change in $y$ is the same for both lines. Therefore, for both lines, gradient $\times$ change in $x$ are equal, and the change in $x$ is the length of Max's shadow ($s$):

$$\frac{h-6}{8} \cdot 2 = \frac{h-6}{3} \cdot s\Rightarrow \frac{1}{8} \cdot 2=\frac{1}{3}s \Rightarrow s = \boxed{\frac{3}{4}}.$$

and here is a diagram (look at triangles $H'B'E$ and $HBD$):

Note that we don't need to know Max's height.

Answer 3

Alternately, from the gradient of $AH'$ being $\frac{h-6}{3}$, we can work out $h$ using similarity: $\text{small $\Delta$ ratio} = \text{big $\Delta$ ratio} \Rightarrow \frac{h}{2} = \frac{6}{10} \Rightarrow h = 1.2$, so the gradient of $AH'$ is $\frac{1.2-6}{3} = -1.6$.

Then the light ray cast when Max is $3$ m from the lamppost has the equation $y = -1.6x + 6$ ($y$-intercept must be $6$). When the light hits the ground at $y = 0, -6 = -1.6x \Rightarrow x = \frac{15}{4}$, so his shadow is $\frac{15}{4} - 3 = \frac{3}{4}$ m long.