The corresponding sides of $\triangle ABC$ and $\triangle PQR$ are in the ratio 3 : 5. AD$\perp$BC and PS$\perp$QR as shown in the following figures :
Generated by claude-sonnet-4-6 · 2026-06-15 10:26 · grounding rag
Model Answer
(i) Prove that △ADC ~ △PSR
In △ADC and △PSR:
- ∠ADC = ∠PSR = 90° (AD ⊥ BC and PS ⊥ QR)
- ∠ACD = ∠PRS (corresponding angles of similar triangles △ABC ~ △PQR, since sides are in ratio 3:5)
Therefore, by AA similarity criterion,
$$\triangle ADC \sim \triangle PSR$$
(ii) Find the length of PS
Since △ADC ~ △PSR, their corresponding sides are in the same ratio as △ABC and △PQR, i.e., 3 : 5.
$$\frac{AD}{PS} = \frac{3}{5}$$
$$\frac{4}{PS} = \frac{3}{5}$$
$$PS = \frac{4 \times 5}{3} = \frac{20}{3} \approx 6.67 \text{ cm}$$
(iii) Find ar(△ABC) : ar(△PQR)
The ratio of areas of two similar triangles equals the square of the ratio of their corresponding sides.
$$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)} = \left(\frac{3}{5}\right)^2 = \frac{9}{25}$$
$$\therefore \text{ar}(\triangle ABC) : \text{ar}(\triangle PQR) = \boxed{9 : 25}$$
Source: Chapter 6, Sections 6.3 and 6.4
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Explanation
- Part (i): Use AA similarity — one pair of angles is 90° (given), and the second pair comes from the fact that △ABC ~ △PQR (their corresponding angles are equal, so ∠C = ∠R).
- Part (ii): Since the triangles are similar, the ratio of altitudes equals the ratio of corresponding sides (3:5). Set up a proportion with AD = 4 cm to find PS.
- Part (iii): Key theorem: The ratio of areas of two similar triangles = square of the ratio of their corresponding sides. Since sides are in ratio 3:5, areas are in ratio 9:25. Examiners expect the theorem stated and then applied.