The corresponding sides of ΔABC and ΔPQR are in the ratio 3 : 5. AD ⊥ BC and PS ⊥ QR as shown in the following figures : (i) Prove that ΔADC ~ ΔPSR (ii) If AD = 4 cm, find the length of PS. (iii) Using (ii) find ar(ΔABC) : ar(ΔPQR)

MathsCBSE Class 10 · Standard · Board exam 2025, set 30/4/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2025: question 33(A)

Set in the CBSE Class 10 board exam of 2025, paper 30/4/1, section D, question 33(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(A) · 5 marks · Long answerTriangles

The corresponding sides of $Δ$ ABC and $Δ$ PQR are in the ratio $3 : 5$ . $A D ⊥ B C$ and $P S ⊥ QR$ as shown in the following figures :

(i)
Prove that $Δ$ ADC $\sim$ $Δ$ PSR
0 marks
(ii)
If AD = 4 cm, find the length of PS.
0 marks
(iii)
Using (ii) find ar( $Δ$ ABC) : ar( $Δ$ PQR)
0 marks

Figure in the paper: Triangle ABC with altitude AD perpendicular to BC (D on BC, right angle marked at D). Triangle PQR with altitude PS perpendicular to QR (S on QR, right angle marked at S). See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 30/4/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Sides in the same ratio means the whole triangles are similar, so their angles match.

Hint 2

For (iii), write each area as ½ × base × altitude and use the ratios 3/5 for both base and altitude.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Given $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{A C}{P R} = \frac{3}{5}$ , so $Δ A B C \sim Δ P QR$ (SSS similarity), hence $∠ C = ∠ R$ (corresponding angles).
½ mark
(i) In $Δ A D C$ and $Δ P S R$ :
$∠ A D C = ∠ P S R = 9 0^{\circ}$ [AD ⟂ BC, PS ⟂ QR]
$∠ C = ∠ R$ [proved above]
1 mark
$∴ Δ A D C \sim Δ P S R$ [AA similarity]
½ mark
(ii) Corresponding sides of similar triangles are proportional: $\frac{A D}{P S} = \frac{A C}{P R} = \frac{3}{5}$
½ mark
$\frac{4}{P S} = \frac{3}{5}$
½ mark
$P S = \frac{20}{3}$ cm
½ mark
(iii) $\frac{a r ( Δ A B C )}{a r ( Δ P QR )} = \frac{\frac{1}{2} \times B C \times A D}{\frac{1}{2} \times QR \times P S} = \frac{B C}{QR} \times \frac{A D}{P S}$
1 mark
$= \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$ , so $a r (Δ A B C) : a r (Δ P QR) = 9 : 25$
½ mark

Answer: (i) Proved by AA; (ii) PS = 20/3 cm; (iii) 9 : 25

Where the marks are

$Δ A B C \sim Δ P QR$ so $∠ C = ∠ R$ (½ mark)
Two pairs of equal angles in $Δ A D C$ , $Δ P S R$ (1 mark)
Concludes $Δ A D C \sim Δ P S R$ (½ mark)
Writes $\frac{A D}{P S} = \frac{A C}{P R} = \frac{3}{5}$ (½ mark)
Substitutes AD = 4 (½ mark)
PS = 20/3 cm (½ mark)
Area ratio as ratio of $\frac{1}{2} \times$ base × height (1 mark)
Ratio 9 : 25 (½ mark)

Revise the chapter

Chapter: Triangles (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Triangles, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 30/4/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2025: question 33(A)

Hints

Step-by-step solution

Revise the chapter

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