Question

The image of a tree on the film of a camera is of length $35mm$, the distance from the lens to the film is $42mm$ and the distance from the lens to the tree is $6m$. How tall is the portion of the tree being photographed?

Answer 1

Let $AB$ and $EF$ be the heights of the portion of the tree and its image on the film respectively.
Let the point $C$ denote the lens.
Let $CG$ and $CH$ be altitudes of
$△ACB$ and $△FEC$.
Clearly, $AB\parallel FE$.
In $△ACB$ and $△FEC$,
$\angle BAC=\angle FEC$
$\angle ECF=\angle ACB$ (vertically opposite angles)
$△ACB\sim △ECF$ ($AA$ criterion)
Thus, $\frac{AB}{EF}=\frac{CG}{CH}$
$\Rightarrow AB=\frac{CG}{CH}\times EF=\frac{6\times 0.035}{0.042}=5$.
Hence, the height of the tree photographed is $5m$.