Question:
Consider the triangle $ABC$ where $BC = 12$ cm, $DB = 9$ cm, $CD = 6$ cm, and $\angle BCD = \angle BAC$.
What is the ratio of the perimeter of $\triangle ADC$ to that of $\triangle BDC$?
Consider the triangle $ABC$ where $BC = 12$ cm, $DB = 9$ cm, $CD = 6$ cm, and $\angle BCD = \angle BAC$.
What is the ratio of the perimeter of $\triangle ADC$ to that of $\triangle BDC$?
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Use similarity to relate corresponding sides, then sum for perimeters.
Updated On: Mar 23, 2026
- $\frac{7}{9}$
- $\frac{8}{9}$
- $\frac{6}{9}$
- $\frac{5}{9}$
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The Correct Option is B
Solution and Explanation
Given $\angle BCD = \angle BAC$, triangles $ADC$ and $CBD$ are similar by AA similarity (common angle at $C$).
From similarity:
\[
\frac{AD}{DB} = \frac{DC}{BC}
\]
$DB = 9$, $BC = 12$, $DC = 6$:
\[
\frac{AD}{9} = \frac{6}{12} \Rightarrow AD = 4.5
\]
Perimeter of $\triangle ADC$: $AD + DC + AC$. From similarity, $\frac{AC}{BC} = \frac{AD}{DB} = \frac12$, so $AC = 6$. Perimeter ADC = $4.5 + 6 + 6 = 16.5$.
Perimeter of $\triangle BDC$: $BD + DC + BC = 9 + 6 + 12 = 27$.
Ratio = $\frac{16.5}{27} = \frac{8}{9}$. \[ \boxed{\frac{8}{9}} \]
Perimeter of $\triangle BDC$: $BD + DC + BC = 9 + 6 + 12 = 27$.
Ratio = $\frac{16.5}{27} = \frac{8}{9}$. \[ \boxed{\frac{8}{9}} \]
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