MathsCBSE Class 10 · Basic · Board exam 2025, set 430/6/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2025: question 36

Set in the CBSE Class 10 board exam of 2025, paper 430/6/1, section E, question 36, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q36 · 4 marks · Case-basedCompetency-basedTriangles

A triangular window of a building is shown above. Its diagram represents a $△$ ABC with $∠$ A = $9 0^{\circ}$ and AB = AC. Points P and R trisect AB and PQ $∥$ RS $∥$ AC.

Based on the above, answer the following questions:

(i)
Show that $△$ BPQ $\sim$ $△$ BAC.
1 mark
(ii)
Prove that $P Q = \frac{1}{3} A C$ .
1 mark
(iii)(a)
If AB = 3 m, find length BQ and BS. Verify that $B Q = \frac{1}{2} B S$ .
2 marks
OR
(iii)(b)
Prove that $B R^{2} + R S^{2} = \frac{4}{9} B C^{2}$ .
2 marks

Figure in the paper: Right-angled isosceles triangle ABC with right angle at A (bottom left), AC vertical, AB horizontal. Points P and R lie on AB with A, R, P, B in order, and S and Q lie on hypotenuse CB. Segments RS and PQ are drawn vertically, parallel to AC, with small arrows marked. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/6/1.

Hints

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

Hint 1

PQ ∥ AC gives equal corresponding angles; ∠B is shared.

Hint 2

Trisecting AB means BP = AB/3 and BR = 2AB/3; use ratios of similar triangles.

Hint 3

For lengths along the hypotenuse, use Pythagoras in the small right triangles at P and R.

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.

(i) In $△ B P Q$ and $△ B A C$ : $∠ B P Q = ∠ B A C = 9 0^{\circ}$ [corresponding angles, PQ ∥ AC], and $∠ B$ is common. Hence $△ B P Q \sim △ B A C$ [AA similarity criterion].
1 mark
(ii) Since P and R trisect AB (order A, R, P, B), $B P = \frac{1}{3} A B$ . From $△ B P Q \sim △ B A C$ : $\frac{P Q}{A C} = \frac{B P}{B A} = \frac{1}{3}$
½ mark
$\Rightarrow P Q = \frac{1}{3} A C$ . Hence proved.
½ mark
(iii)(a) $A B = A C = 3$ m, so $B P = 1$ m. From $\frac{B P}{B A} = \frac{P Q}{A C}$ and $B A = A C$ , $P Q = B P = 1$ m. In right $△ B P Q$ ( $∠ P = 9 0^{\circ}$ ): $B Q = 1^{2} + 1^{2} = 2$ m.
1 mark
(iii)(a) Similarly $△ B R S \sim △ B A C$ with $B R = 2$ m, so $R S = B R = 2$ m and $B S = 2^{2} + 2^{2} = 22$ m.
½ mark
(iii)(a) $\frac{1}{2} B S = \frac{1}{2} \times 22 = 2 = B Q$ . Hence verified.
½ mark
(iii)(b) $△ B R S \sim △ B A C$ [RS ∥ AC, AA], and $B R = \frac{2}{3} A B$ , so $R S = \frac{2}{3} A C$ . Hence $B R^{2} + R S^{2} = (\frac{2}{3} A B)^{2} + (\frac{2}{3} A C)^{2}$
1 mark
(iii)(b) $= \frac{4}{9} (A B^{2} + A C^{2})$
½ mark
(iii)(b) $= \frac{4}{9} B C^{2}$ [Pythagoras theorem in $△ A B C$ , $∠ A = 9 0^{\circ}$ ]. Hence proved.
½ mark

Answer: (i) AA similarity; (ii) $P Q = \frac{1}{3} A C$ ; (iii)(a) $B Q = 2$ m, $B S = 22$ m, so $B Q = \frac{1}{2} B S$ ; (iii)(b) $B R^{2} + R S^{2} = \frac{4}{9} B C^{2}$

Where the marks are

(i) Shows $△ B P Q \sim △ B A C$ by AA (1 mark)
(ii) Writes $\frac{P Q}{A C} = \frac{B P}{B A} = \frac{1}{3}$ (½ mark) and concludes $P Q = \frac{1}{3} A C$ (½ mark)
(iii)(a) Finds PQ = 1 m and $B Q = 2$ m (1 mark)
(iii)(a) Finds $B S = 22$ m (½ mark) and verifies $B Q = \frac{1}{2} B S$ (½ mark)
(iii)(b) Writes $B R = \frac{2}{3} A B$ , $R S = \frac{2}{3} A C$ (½ + ½ mark)
(iii)(b) Gets $\frac{4}{9} (A B^{2} + A C^{2})$ (½ mark) and uses Pythagoras to get $\frac{4}{9} B C^{2}$ (½ mark)

Revise the chapter

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

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 430/6/1 (Basic): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Basic 2025: question 36

Hints

Step-by-step solution

Revise the chapter

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