MathsCBSE Class 10 · Basic · Board exam 2024, set 430/1/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2024: question 36

Set in the CBSE Class 10 board exam of 2024, paper 430/1/1, 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-basedQuadratic Equations

To keep the lawn green and cool, Sadhna uses water sprinklers which rotate in circular shape and cover a particular area.

The diagram below shows the circular areas covered by two sprinklers:

Two circles touch externally. The sum of their areas is $130 π$ sq m and the distance between their centres is 14 m.

Based on above information, answer the following questions:

(i)
Obtain a quadratic equation involving R and r from above.
1 mark
(ii)
Write a quadratic equation involving only r.
1 mark
(iii)
(a) Find the radius r and the corresponding area irrigated.
2 marks
OR
(iii)
(b) Find the radius R and the corresponding area irrigated.
2 marks

Figure in the paper: Two circles touching externally: larger circle with radius R and centre marked with a dot joined by a radius; smaller circle with radius r and centre marked with a dot joined by a radius. Caption (R > r). A photo of lawn sprinklers spraying water is also shown. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 430/1/1.

Hints

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

Hint 1

Write the total area of the two circles and cancel π.

Hint 2

When circles touch externally, the distance between their centres equals R + r. Use this to remove R.

Hint 3

Factorise the quadratic, then use R > r to decide which root is which radius.

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 sum of the areas is $π R^{2} + π r^{2} = 130 π$ , so $R^{2} + r^{2} = 130$ .
1 mark
(ii) The circles touch externally, so the distance between the centres is $R + r = 14$ , giving $R = 14 - r$ . Substituting: $(14 - r)^{2} + r^{2} = 130 \Rightarrow 2 r^{2} - 28 r + 66 = 0 \Rightarrow r^{2} - 14 r + 33 = 0$ .
1 mark
(iii)(a) Factorising: $r^{2} - 14 r + 33 = 0 \Rightarrow (r - 11) (r - 3) = 0 \Rightarrow r = 3$ or $11$ . If $r = 11$ , then $R = 3$ , which contradicts $R > r$ . Hence $r = 3$ m.
1 mark
(iii)(a) Area irrigated by the smaller sprinkler $= π r^{2} = 9 π m^{2}$ .
1 mark
(iii)(b) Put $r = 14 - R$ in $R^{2} + r^{2} = 130$ . This gives $R^{2} - 14 R + 33 = 0 \Rightarrow (R - 11) (R - 3) = 0$ . Since $R > r$ , $R = 11$ m and not 3 m.
1 mark
(iii)(b) Area irrigated by the larger sprinkler $= π R^{2} = 121 π m^{2}$ .
1 mark

Answer: (i) $R^{2} + r^{2} = 130$ ; (ii) $r^{2} - 14 r + 33 = 0$ ; (iii)(a) r = 3 m, area = 9π m²; (iii)(b) R = 11 m, area = 121π m²

Where the marks are

Writes $R^{2} + r^{2} = 130$ (1 mark)
Uses $R + r = 14$ to get $r^{2} - 14 r + 33 = 0$ (1 mark)
(iii)(a) Solves to get r = 3 m, rejecting 11 because r < R (1 mark)
(iii)(a) Area $= 9 π$ m² (1 mark)
(iii)(b) Solves to get R = 11 m, rejecting 3 because R > r (1 mark)
(iii)(b) Area $= 121 π$ m² (1 mark)

Revise the chapter

Chapter: Quadratic Equations (notes, formulas, where marks are lost and every past question)
Topic: Word Problems on Quadratic Equations

Practise this chapterBoard-style questions from Quadratic Equations, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 430/1/1 (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 2024: question 36

Hints

Step-by-step solution

Revise the chapter

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