This is a problem of similar triangles. Remember that, in geometry, two triangles, ΔABC and ΔA′B′C′, are similar if and only if corresponding angles have the same measure. This statement implies that these two triangles are similar if and only if the lengths of corresponding sides are proportional. This is expressed as ΔABC ~ ΔA′B′C′.
Then, applying the previous statement, we have:
[tex] \frac{x}{5} = \frac{18-x}{25}[/tex]
Isolating x:
[tex]25x=90-5x[/tex]
[tex]30x=90[/tex]
[tex]x=3[/tex]
