Q1. [1] § 6.1 Introduction § 6.3 Similarity of Triangles § 6.4 Criteria for Similarity of Triangles
In the given figure, in $\triangle ABC$, $DE \parallel BC$. If $AD = 2 \cdot 4$ cm, $DB = 4$ cm and $AE = 2$ cm, then the length of $AC$ is :
- A $\dfrac{10}{3}$ cm
- B $\dfrac{3}{10}$ cm
- C $\dfrac{16}{3}$ cm
- D $1 \cdot 2$ cm
Previously asked in CBSE board exam
2024 30/5/1 Q17
Generated by claude-sonnet-4-6 · 2026-06-15 10:23 · grounding rag
Model Answer
Option A: $\dfrac{10}{3}$ cm
By Basic Proportionality Theorem (DE ∥ BC):
$$\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{2.4}{4} = \frac{2}{EC} \implies EC = \frac{2 \times 4}{2.4} = \frac{10}{3} \text{ cm}$$
$$AC = AE + EC = 2 + \frac{10}{3} = \frac{16}{3} \text{ cm}$$
Wait — $AC = \dfrac{16}{3}$ cm → Option C: $\dfrac{16}{3}$ cm
Source: Chapter 6, Section 6.3 (Theorem 6.1 — Basic Proportionality Theorem)
Explanation
- Apply BPT: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ → $\dfrac{2.4}{4} = \dfrac{2}{EC}$ → $EC = \dfrac{10}{3}$ cm.
- Then $AC = AE + EC = 2 + \dfrac{10}{3} = \dfrac{16}{3}$ cm.
- Option A ($\frac{10}{3}$) is just EC, a common trap. The question asks for AC, so always add AE. Correct answer is C.
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