Question

If $\triangle ABC \sim \triangle DEF$ and $AB = 4$ cm, $DE = 6$ cm, $EF = 9$ cm, and $FD = 12$ cm, find the perimeter of $\triangle ABC$.

Answer 1

Step 1: Understand the problem:
We are given that \( \triangle ABC \sim \triangle DEF \), which means that the two triangles are similar. The corresponding sides of similar triangles are proportional. We are also given the following side lengths:
- \( AB = 4 \, \text{cm} \)
- \( DE = 6 \, \text{cm} \)
- \( EF = 9 \, \text{cm} \)
- \( FD = 12 \, \text{cm} \)
We need to find the perimeter of \( \triangle ABC \).

Step 2: Use the property of similar triangles:
Since the triangles are similar, the corresponding sides are proportional. Therefore, we have the following ratios of corresponding sides:
\[ \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} \] Substitute the known values for \( AB \), \( DE \), \( EF \), and \( FD \):
\[ \frac{4}{6} = \frac{BC}{9} = \frac{CA}{12} \] Simplify the ratios:
\[ \frac{2}{3} = \frac{BC}{9} = \frac{CA}{12} \]

Step 3: Solve for \( BC \) and \( CA \):
We now solve for the missing sides of \( \triangle ABC \) using the proportion relationships.
For \( BC \):
\[ \frac{BC}{9} = \frac{2}{3} \] Cross-multiply to solve for \( BC \):
\[ BC = 9 \times \frac{2}{3} = 6 \, \text{cm} \]
For \( CA \):
\[ \frac{CA}{12} = \frac{2}{3} \] Cross-multiply to solve for \( CA \):
\[ CA = 12 \times \frac{2}{3} = 8 \, \text{cm} \]

Step 4: Find the perimeter of \( \triangle ABC \):
The perimeter of \( \triangle ABC \) is the sum of the lengths of its sides:
\[ \text{Perimeter of } \triangle ABC = AB + BC + CA = 4 + 6 + 8 = 18 \, \text{cm} \]

Conclusion:
The perimeter of \( \triangle ABC \) is \( 18 \, \text{cm} \).