Find the height of a bar, given the lengths of shadows cast by it and another bar 
What is the height of the red bar?


My try: with respect to the picture, it seems for the green bar $\frac{h}{H}=\frac{2}{3}$. So, I think that ratio is the same for the red bar, and the height of the red bar is 
$$\frac{h}{6+4}=\frac 23\qquad\to\qquad h_{red}=\frac{20}{3}$$
Is this correct?
 A: Basic approach.  The $2$-meter green bar has a shadow of $3$ meters.  It is $6$ meters from the red bar to the wall—how tall would it have to be for its shadow to extend right up to the foot of the wall?  Call that height $h$.
How much taller would a bar have to be for the shadow to go up, vertically, another $4$ meters?  That plus $h$ is your answer.
A: Suppose we had a third bar $2$m high that was at a distance $0.01$m from the back wall. All but $0.01$m of the shadow of that third bar would be on the wall, and the shadow on the wall could not be higher than the bar (because the light is shining at a downward angle), so it must be less than $2$m tall.
So the total length of shadow of the third bar is less than $2.01$m.
Your claim is that a bar's height is $\frac23$ the total length of its shadow on the ground and back wall combined.
So you predict a height of less than 
$\frac23 (2.01)\mathrm m = 1.34\mathrm m.$
That is clearly not the height of the third bar.
Think again about your model. It may help to ask where you should put a notch in the taller bar so that the shadow of the notch is exactly at the corner where the ground shadow meets the back-wall shadow.
Then figure how far is that notch from the top of the bar, and how far from the bottom.
A: Using similar triangles, we have $\dfrac{h_\text{red}-4}6=\dfrac23$, yielding $h_\text{red}=8$.
A: As Zach Boyd mentions, this problem can only be solved if we assume that the light source is sufficiently distant (eg the sun) so that the light rays are parallel. miracle173 mentions that it looks like the light source is far away because the shadows of the two bars are parallel, but of course that could simply be a diagram artifact. 
We also need to assume that the bars and back wall are vertical, and hence perpendicular to the floor (or if they aren't vertical they are at least parallel to each other)
So let's assume that the light rays are parallel, that the floor is horizontal, and that the bars and back wall are vertical. In which case, the lines connecting any object point to its corresponding shadow point are parallel. In particular, the lines from the tops of the bars to their shadow are parallel, so we have two key similar right triangles.
The key triangle of the green bar has a base of 3m and height of 2m. The key triangle of the red bar has a base of 6m and since it's similar to the green key triangle it has a height of 4m since $\frac{3}{2} = \frac{6}{4}$. That height is measured above the top of the shadow of the red bar (as illustrated in the diagram in Doug M's excellent answer). The top of the shadow of the red bar is 4m above floor level, thus the total height of the red bar is 8m.
A: Without doing any real math it's fairly simple to determine that the red bar is 8 meters tall.
Rationale:
The 2m green bar casts a 3m shadow, so it would take a 4m bar to cast the 6m shadow between the red bar and the wall, and thus the lower portion of the red bar is 4m tall. The red bar is parallel to the wall, so the 4m shadow on the wall is cast by the upper 4m of the bar. 4m + 4m = 8m, so the bar is 8 meters tall.
A: 3D histograms are evil according to Edward Tufte. Here, they are used to obfuscate information and make this geometry problem harder than it is. Also, as mentioned by @CandiedOrange and @LamarLatrell, the original drawing isn't to scale.
Here's a 3D render with correct heights:

By playing with perspective and point of view, you can seamlessly merge lengths that appear on distinct axes. It might give you the wrong impression that you could simply add those lengths.

But if you select the correct perspective, the problem becomes much clearer.


A: The vertical bar height is$ \frac23$ length of its uninturrupted shadow length. So $ \dfrac23 \cdot 6 =4.$
Rest of the red bar casts s a direct $1:1$  vertical projection/translation of $4$ length onto the wall. So red height total height of red bar is $4+4=8.$
Else symbolically
$$ h-4 = \frac23 \cdot 6 \, \rightarrow h= 8. $$
A: This image is flawed. At least if those black bars are shadows some one has been playing with the lighting.
If the red bar is meant to be 8m (and I think it is) it should look like this:

Instead it looks like this:

Which I can prove is not 8m by arguing that the 4m mark is halfway up the blue wall. That means the blue wall is 8m and a 8m red bar should be no taller than the blue wall. Yet it is taller, as the yellow box shows.
The red bar given is somewhere between 11-12m. Sure, that isn't what the person who drew this was thinking. But this is a terrible inconsistency to confront students with. This isn't math. This is "guess what I'm thinking".  
Now sure, this assumes the image is drawn in a particular 3D style. But it's one that mathematics students are familiar with. Can you guess which one? 

The essential assumption is that lines that are parallel in 2D are parallel in 3D. 
A: There are three similar right triangles here and the ratios of their catheti (or legs) are all equal:
$$\frac 23=\frac{4}{x}=\frac{h}{6+x} \implies x=6,\quad h=\frac{2(6+x)}{3}=8$$
($x$ is the an imaginary cathetus which goes beyond the blue wall).
A: The floor shadow is twice as long, and anything above that is reflective of the actual height (the bar and the wall are parallel, so their angle with the light source is the same). So, the red bar is twice the green bar $+ \ 4$ meters, which is $2\cdot2+4=8$ meters.
A: Here is a different visualization.

A: Visualising the model is a bit tricky. The eye cannot see the tiny differences. One of the great examples that eyes can be trapped is the missing square puzzle. Therefore, we cannot rely on the visual presentation of the model to calculate the height.
Here, I will try to present the problem mathematically:

We know the length of $BC=6m$ which represents the horizontal shadow, $CD=4m$ which represents the vertical shadow, $BE=4m$, $AE=x$
height of the red bar is $AE + BE = 4m + x$
$\angle{AED}=90^\circ$ and $\angle{ADE}=60^\circ$ and $\angle{ADE}=30^\circ$
So, $AD=2x$ and $ED=x\sqrt{3}=6m \Rightarrow x = \frac{6}{\sqrt{3}}$
$AB = 4m + \frac{6}{\sqrt{3}}m \approx 7.464m$
