A water sprinkler is a device used to irrigate agricultural crops, lawns, landscapes, golf courses, and other areas. Water sprinklers can be used for residential, industrial, and agricultural usage. A water sprinkler is set to shoot a stream of water a distance of 21 m and rotate through an angle which is equal to complementary angle of 10°. (i) What is the area of sector in terms of arc length? (ii) What is the area of the watered region (in terms of π)? (iii) (A) If the radius(r) changes to 28m, find the angle θ so that the area of the watered region remains the same. (iii) (B) If the radius(r) is increased from 21m to 28m and the angle remains the same, what is the increase in the area of the watered region?

(i) Area = (l × r)/(2); (ii) 98π m²; (iii)(A) θ = 45°; (iii)(B) ≈ 239.56 m² ((686π)/(9) m²)

MathsCBSE Class 10 · Basic · Sample paper, session 2025-26 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic sample paper 2025-26: question 37

From the official CBSE sample question paper for 2025-26, section E, question 37, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q37 · 4 marks · Case-basedCompetency-basedAreas Related to Circles

A water sprinkler is a device used to irrigate agricultural crops, lawns, landscapes, golf courses, and other areas. Water sprinklers can be used for residential, industrial, and agricultural usage.

A water sprinkler is set to shoot a stream of water a distance of 21 m and rotate through an angle which is equal to complementary angle of $1 0^{\circ}$ .

(i)
What is the area of sector in terms of arc length?
1 mark
(ii)
What is the area of the watered region (in terms of $π$ )?
1 mark
(iii)
(A) If the radius( $r$ ) changes to 28m, find the angle $θ$ so that the area of the watered region remains the same.
2 marks
OR
(iii)
(B) If the radius( $r$ ) is increased from 21m to 28m and the angle remains the same, what is the increase in the area of the watered region?
2 marks

Figure in the paper: A sector of a circle with the sprinkler at the vertex on the right. The arc on the left is labelled $l$ , the bounding radii are straight lines, the central angle is marked $θ$ , and the shaded region is labelled 'Watered region'. The vertex is labelled 'Water sprinkler'. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

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

Hint 1

The complementary angle of $1 0^{\circ}$ is $9 0^{\circ} - 1 0^{\circ}$ ; the shooting distance is the radius.

Hint 2

Arc length is $l = \frac{θ}{360} \times 2 π r$ ; substitute it into the sector area formula to express the area in terms of $l$ .

Hint 3

For (iii), set up the sector area formula with the new radius, then either equate it to the old area or subtract the old area from it.

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) For a sector with arc length $l$ and radius $r$ : area of sector $= \frac{θ}{36 0 ^{\circ}} π r^{2} = \frac{1}{2} \times l \times r$ , i.e. $\frac{l \times r}{2}$ .
1 mark
(ii) The angle of rotation is the complement of $1 0^{\circ}$ , so $θ = 9 0^{\circ} - 1 0^{\circ} = 8 0^{\circ}$ , with $r = 21$ m.
Area watered $= \frac{80}{360} \times π \times 2 1^{2} = \frac{2}{9} \times 441 π = 98 π m^{2}$
1 mark
(iii)(A) For the area to stay the same with $r = 28$ m: $\frac{θ}{360} \times π \times 2 8^{2} = 98 π$
1 mark
(iii)(A) So $θ = \frac{98 \times 360}{784} = 4 5^{\circ}$ .
1 mark
(iii)(B) Increase in area $= \frac{80}{360} π (2 8^{2} - 2 1^{2}) = \frac{2}{9} π (784 - 441) = \frac{2}{9} \times 343 π$
1 mark
(iii)(B) $= \frac{686}{9} \times \frac{22}{7} = \frac{2156}{9} \approx 239.56 m^{2}$ $(i.e. \frac{686 π}{9} m^{2})$
1 mark

Answer: (i) Area = $\frac{l \times r}{2}$ ; (ii) $98 π$ m²; (iii)(A) $θ = 4 5^{\circ}$ ; (iii)(B) ≈ 239.56 m² ( $\frac{686 π}{9}$ m²)

Where the marks are

(i) Area of sector = ½ × arc length × radius (1 mark)
(ii) Uses $θ = 8 0^{\circ}$ to get area $98 π$ m² (1 mark)
(iii)(A) Equates $\frac{θ}{360} π (28)^{2}$ to $98 π$ (1 mark); gets $θ = 4 5^{\circ}$ (1 mark)
(iii)(B) Writes increase as $\frac{80}{360} π (784 - 441)$ (1 mark); gets ≈ 239.56 m² (1 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 paperSample paper 2025-26 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic sample paper 2025-26: question 37

Hints

Step-by-step solution

Revise the chapter

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