Q1. [1] § 6.3 Similarity of Triangles
In the adjoining figure, PQ $\parallel$ XY $\parallel$ BC, AP = 2 cm, PX = 1.5 cm and BX = 4 cm. If QY = 0.75 cm, then AQ + CY =
- A 6 cm
- B 4.5 cm
- C 3 cm
- D 5.25 cm
Previously asked in CBSE board exam
2025 30/6/1 Q6
Generated by claude-sonnet-4-6 · 2026-06-15 10:26 · grounding rag
Model Answer
Answer: (D) 5.25 cm
By BPT (Thales' Theorem), since PQ ∥ XY ∥ BC:
$$\frac{AP}{PX} = \frac{AQ}{QY} \Rightarrow \frac{2}{1.5} = \frac{AQ}{0.75} \Rightarrow AQ = 1 \text{ cm}$$
For CY, using $\frac{AX}{XB} = \frac{AY}{YC}$: AX = AP + PX = 3.5 cm, XB = 4 cm.
$$\frac{AY}{YC} = \frac{3.5}{4} \Rightarrow AY = AQ + QY = 1.75 \text{ cm} \Rightarrow CY = \frac{1.75 \times 4}{3.5} = 2 \text{ cm}$$
Wait — let me recheck: $\frac{AX}{XB}=\frac{AY}{YC} \Rightarrow \frac{3.5}{4}=\frac{1.75}{YC} \Rightarrow YC = \frac{1.75\times4}{3.5}=2$ cm
$$AQ + CY = 1 + 2 = \boxed{3 \text{ cm}}$$
Correct answer: (C) 3 cm
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Explanation
- Apply BPT to triangle ABC with PQ ∥ BC: $\dfrac{AP}{PX} = \dfrac{AQ}{QY}$ gives AQ = 1 cm.
- Apply BPT again with XY ∥ BC: $\dfrac{AX}{XB} = \dfrac{AY}{YC}$, where AX = 2 + 1.5 = 3.5 cm, AY = AQ + QY = 1.75 cm, XB = 4 cm → CY = 2 cm.
- AQ + CY = 1 + 2 = 3 cm. Examiners expect clear ratio setup using BPT (Theorem 6.1) at each step.
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