MathsCBSE Class 10 · Standard · Board exam 2026, set 30/3/1 · 3 marks · Short answer

CBSE Class 10 Mathematics Standard 2026: question 27

Set in the CBSE Class 10 board exam of 2026, paper 30/3/1, section C, question 27, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q27 · 3 marks · Short answerAreas Related to Circles

In the given figure, chord AB subtends an angle of $12 0^{\circ}$ at the centre of the circle with radius 7 cm. Find (i) perimeter of major sector OACB, and (ii) area of the shaded segment, if area of $Δ$ OAB = $21.2$ cm $^{2}$ .

Figure in the paper: Circle with centre O and radius 7 cm. Chord AB with central angle AOB = 120°. Arc ACB (major arc) passes through C at the top. The minor segment between chord AB and minor arc is shaded. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 30/3/1.

Hints

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

Hint 1

The major sector's angle is 360° − 120°. Its perimeter is the arc plus two radii.

Hint 2

Shaded segment = minor sector − triangle OAB, and the triangle's area is given.

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 angle of the major sector is $36 0^{\circ} - 12 0^{\circ} = 24 0^{\circ}$ .
Perimeter of major sector OACB = length of major arc ACB + $2 r$
$= \frac{240}{360} \times 2 \times \frac{22}{7} \times 7 + 2 \times 7 = \frac{88}{3} + 14$
1 mark
(i) $= \frac{130}{3}$ cm $\approx 43.3$ cm.
½ mark
(ii) Area of shaded segment = area of minor sector OAB − area of $△ O A B$
$= \frac{120}{360} \times \frac{22}{7} \times 7 \times 7 - 21.2 = \frac{154}{3} - 21.2 \approx 51.33 - 21.2$
1 mark
(ii) $\approx 30.1$ cm².
½ mark

Answer: (i) $\frac{130}{3}$ cm ≈ 43.3 cm; (ii) ≈ 30.1 cm²

Where the marks are

Major arc length with angle 240° plus two radii, correctly substituted (1 mark)
Perimeter = 130/3 cm ≈ 43.3 cm (½ mark)
Segment area = sector area (120°) − 21.2, correctly substituted (1 mark)
Segment area ≈ 30.1 cm² (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 paper2026 board paper, set 30/3/1 (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 2026: question 27

Hints

Step-by-step solution

Revise the chapter

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