To solve that question we must look for similar triangles. Lets do a picture.
Take triangle ABC. If we get a line parallel to its base, it divide its sides proportionally. We can write:
[tex]\begin{gathered} \frac{60}{45}=\frac{length\text{ of shadow}}{\text{length of rod}} \\ \end{gathered}[/tex]
Then we get:
[tex]\begin{gathered} \frac{60}{45}=\frac{length\text{ of shadow}}{12} \\ \\ 45\times length\text{ of shadow =60}\times12 \\ \\ \text{length of shadow=}\frac{60\times12}{45} \\ \\ \text{length of shadow=16}cm \end{gathered}[/tex]
The final answer is:
[tex]\text{length of shadow =16}cm[/tex]