Q1. [1] § 6.3 Similarity of Triangles § 6.4 Criteria for Similarity of Triangles § 6.5 Summary
In the given figure $\triangle ABC$ is shown. DE is parallel to BC. If $AD = 5$ cm, $DB = 2.5$ cm and $BC = 12$ cm, then DE is equal to
- (A) $10$ cm
- (B) $6$ cm
- (C) $8$ cm
- (D) $7.5$ cm
Previously asked in CBSE board exam
2024 30/2/1 Q6
Generated by claude-sonnet-4-6 · 2026-06-15 10:22 · grounding rag
Model Answer
(C) 8 cm
Since DE ∥ BC, by AA similarity, △ADE ~ △ABC.
$$\frac{AD}{AB} = \frac{DE}{BC}$$
AB = AD + DB = 5 + 2.5 = 7.5 cm
$$\frac{DE}{12} = \frac{5}{7.5} = \frac{2}{3}$$
$$DE = 12 \times \frac{2}{3} = 8 \text{ cm}$$
Source: Chapter 6, Section 6.3 (Basic Proportionality Theorem / Similarity of Triangles)
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Explanation
- Since DE ∥ BC, triangles ADE and ABC are similar by AA criterion (∠A common, ∠ADE = ∠ABC as corresponding angles).
- The key ratio is AD/AB (not AD/DB). Students often mistakenly use AD/DB = DE/BC — that is a common error.
- Always find the full side AB = AD + DB first, then apply the ratio of corresponding sides.
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cbse.gov.in.