MathsCBSE Class 10 · Standard · Board exam 2025, set 30/4/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2025: question 36

Set in the CBSE Class 10 board exam of 2025, paper 30/4/1, 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-basedAreas Related to Circles

The Olympic symbol comprising five interlocking rings represents the union of the five continents of the world and the meeting of athletes from all over the world at the Olympic games. In order to spread awareness about Olympic games, students of Class-X took part in various activities organised by the school. One such group of students made 5 circular rings in the school lawn with the help of ropes. Each circular ring required 44 m of rope.

Also, in the shaded regions as shown in the figure, students made rangoli showcasing various sports and games. It is given that $Δ$ OAB is an equilateral triangle and all unshaded regions are congruent.

Based on above information, answer the following questions :

(i)
Find the radius of each circular ring.
1 mark
(ii)
What is the measure of $∠$ AOB ?
1 mark
(iii)(a)
Find the area of shaded region $R_{1}$ .
2 marks
OR
(iii)(b)
Find the length of rope around the unshaded regions.
2 marks

Figure in the paper: Left: five interlocking rings (Olympic symbol) in two rows, three on top and two below. Right: a rangoli of five overlapping shaded circles with a central equilateral triangle OAB formed by the centres O, A, B; the unshaded congruent regions lie between the circles near the triangle. Region R1 is marked in the upper-left circle, with radius OR1 and points A, B, O labelled. 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

The rope needed for a ring is its circumference.

Hint 2

What are the angles of an equilateral triangle? A $6 0^{\circ}$ angle at a centre cuts off a segment and an arc.

Hint 3

Segment area = sector area − triangle area; arc length $= \frac{θ}{360} \times 2 π r$ . Count how many arcs border the unshaded regions.

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) The rope used for one ring is its circumference: $2 π r = 44$ , so $2 \times \frac{22}{7} \times r = 44$ .
½ mark
(i) Therefore $r = \frac{44 \times 7}{44} = 7$ m.
½ mark
(ii) $Δ O A B$ is equilateral, so each of its angles is $6 0^{\circ}$ . Hence $∠ A O B = 6 0^{\circ}$ .
1 mark
(iii)(a) Region $R_{1}$ is the circle with centre O minus two congruent segments, each cut off by a chord subtending $6 0^{\circ}$ at O.
Area of one segment $= \frac{60}{360} \times \frac{22}{7} \times 7^{2} - \frac{3}{4} \times 7^{2} = \frac{77}{3} - \frac{49 3}{4}$
Area of $R_{1} = \frac{22}{7} \times 7 \times 7 - 2 (\frac{77}{3} - \frac{49 3}{4})$
1 mark
(iii)(a) $= 154 - \frac{154}{3} + \frac{49 3}{2} = (\frac{308}{3} + \frac{49 3}{2})$ m² $\approx 145.05$ m² (using $3 = 1.73$ )
1 mark
(iii)(b) There are 4 congruent unshaded regions. Each is bounded by two equal arcs, each subtending $6 0^{\circ}$ at a centre, so the rope needed is $8 \times$ (length of one arc).
½ mark
(iii)(b) Length of one arc $= \frac{60}{360} \times 2 \times \frac{22}{7} \times 7 = \frac{22}{3}$ m, so the total length $= 8 \times \frac{22}{3}$ .
1 mark
(iii)(b) $= \frac{176}{3}$ m $\approx 58.67$ m
½ mark

Answer: (i) 7 m; (ii) 60°; (iii)(a) (308/3 + 49√3/2) m² ≈ 145.05 m²; (iii)(b) 176/3 m ≈ 58.67 m

Where the marks are

Writes $2 π r = 44$ (½ mark)
Gets $r = 7$ m (½ mark)
States $∠ A O B = 6 0^{\circ}$ because the triangle is equilateral (1 mark)
(iii)(a) Writes area = circle − 2 segments with correct substitution (1 mark)
(iii)(a) Area $= \frac{308}{3} + \frac{49 3}{2} \approx 145.05$ m² (1 mark)
(iii)(b) Recognises that the rope length is 8 arcs of $6 0^{\circ}$ (½ mark)
(iii)(b) Correct arc-length substitution (1 mark)
(iii)(b) Length $= \frac{176}{3} \approx 58.67$ m (½ mark)

Revise the chapter

Chapter: Areas Related to Circles (notes, formulas, where marks are lost and every past question)
Topic: Area of a Sector of a Circle

Practise this chapterBoard-style questions from Areas Related to Circles, 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.

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 Standard 2025: question 36

Hints

Step-by-step solution

Revise the chapter

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