Skip to content

In a Fine Arts class… (Class 10 Maths Basic 2025, Q36)

questionbuddy.in/q/cbse-10-maths-2025-430-5-1-q36-fine-arts-class-students-asked

MathsCBSE Class 10 · Basic · Board exam 2025, set 430/5/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2025: question 36

Set in the CBSE Class 10 board exam of 2025, paper 430/5/1, 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

In a Fine Arts class, students were asked to design triangular tiles in geometric pattern.

  1. (i)

    Determine the length OB.

    1 mark
  2. (ii)

    Is DE || CA? Give reason for your answer.

    1 mark
  3. (iii)

    (a) Write all angles of quadrilateral OEBD and show that it is a cyclic quadrilateral.

    2 marks

  4. OR

    (iii)

    (b) Find the perimeter of . (Use )

    2 marks

Figure in the paper: Left: a triangle with a circular design and a star-like black pattern inside a circle. Right: equilateral/isosceles triangle ABC with A at top, B and C at base. A circle with centre O touches sides AB (at E), BC (at D) and AC. OD is perpendicular to BC at D; OE is joined to E; a right-angle mark appears at the point of tangency on AC. A dashed line goes from A through O to D. Angle 30° is marked at B between BC and BO. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/5/1.

Hints

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

Hint 1

OD is a radius drawn to the point of contact, so triangle ODB is right-angled at D.

Hint 2

In an equilateral triangle the incircle touches each side at its mid-point. Think of the mid-point theorem.

Hint 3

A quadrilateral is cyclic if a pair of opposite angles adds up to 180°. For the perimeter, find BD using cos 30° in triangle OBD.

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.

  1. (i) [radius ⟂ tangent], and cm. In right , , so . Hence cm.

    1 mark

  2. (ii) Yes, . is equilateral and , so is also a median. Therefore D is the mid-point of BC.

    ½ mark

  3. (ii) In the same way, the incircle touches AB at its mid-point, so E is the mid-point of AB. By the mid-point theorem in , .

    ½ mark

  4. (iii)(a) [angle of an equilateral triangle]. [tangent ⟂ radius at the point of contact].

    ½ mark

  5. (iii)(a) By the angle sum of a quadrilateral, .

    ½ mark

  6. (iii)(a) Opposite angles: . Also .

    ½ mark

  7. (iii)(a) Since the opposite angles are supplementary, OEBD is a cyclic quadrilateral.

    ½ mark

  8. (iii)(b) In right , . So cm.

    1 mark

  9. (iii)(b) D is the mid-point of BC, so cm. The perimeter of is cm.

    1 mark

Answer: (i) OB = 8 cm; (ii) Yes, D and E are mid-points of BC and AB, so DE ∥ CA (mid-point theorem); (iii)(a) ∠EBD = 60°, ∠OEB = ∠ODB = 90°, ∠EOD = 120°, opposite angles sum to 180°, so cyclic; (iii)(b) 41.52 cm

Where the marks are

  • (i) Uses sin 30° = 4/OB to get OB = 8 cm (1 mark)
  • (ii) States Yes, and that D is the mid-point of BC as AD ⟂ BC in the equilateral triangle (½ mark)
  • (ii) E is the mid-point of AB, hence DE ∥ CA by the mid-point theorem (½ mark)
  • (iii)(a) ∠EBD = 60° and ∠EOD = 120° (½ mark)
  • (iii)(a) ∠OEB = ∠ODB = 90° with reason (½ mark)
  • (iii)(a) Sum of opposite angles = 180° (½ mark)
  • (iii)(a) Concludes OEBD is cyclic (½ mark)
  • (iii)(b) BD = 6.92 cm using cos 30° (1 mark)
  • (iii)(b) BC = 13.84 cm and perimeter = 41.52 cm (1 mark)

Revise the chapter

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 430/5/1 (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.