The sides of a triangular path measure 7 m, 9 m, and 11 m. Find the perimeter of a similar triangle in which the shortest side has a length of 21 meters.

Answer 1

Let the ratio of the lengths of corresponding sides in the two triangles be $k$. We know that $\frac{7k}{21} = \frac{7}{9}$, so $k = \frac{3}{2}$. Therefore, the perimeter of the larger triangle is $21 + 18 + 33 = \boxed{72}$.