MathsCBSE Class 10 · Basic · Board exam 2023, set 430/2/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2023: question 36

Set in the CBSE Class 10 board exam of 2023, paper 430/2/1, 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

Case Study 1. Flower beds look beautiful growing in gardens. One such circular park of radius 'r' m, has two segments with flowers. One segment which subtends an angle of $9 0^{\circ}$ at the centre is full of red roses, while the other segment with central angle $6 0^{\circ}$ is full of yellow coloured flowers. [See figure]

It is given that the combined area of the two segments (of flowers) is $256 \frac{2}{3}$ sq m.

Based on the above, answer the following questions:

(i)
Write an equation representing the total area of the two segments in terms of 'r'.
1 mark
(ii)
Find the value of 'r'.
1 mark
(iii)
(a) Find the area of the segment with red roses.
2 marks
OR
(iii)
(b) Find the area of the segment with yellow flowers.
2 marks

Figure in the paper: A circle with centre O. A horizontal chord near the top bounds a small segment shaded and labelled 'Yellow flowers'. A horizontal chord near the bottom bounds a segment shaded and labelled 'Red roses'. The red segment subtends $9 0^{\circ}$ at the centre and the yellow segment subtends $6 0^{\circ}$ at the centre. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 430/2/1.

Hints

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

Hint 1

A segment's area is the sector's area minus the area of the triangle formed by the two radii and the chord.

Hint 2

For 90° the triangle is right-angled (area $\frac{1}{2} r^{2}$ ); for 60° it is equilateral (area $\frac{3}{4} r^{2}$ ).

Hint 3

Add both segment expressions, take $r^{2}$ common, and equate to $\frac{770}{3}$ .

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) Area of a segment of central angle $θ$ = area of sector − area of triangle $= \frac{θ}{36 0 ^{\circ}} π r^{2} - \frac{1}{2} r^{2} sin θ$ .
Red segment ( $9 0^{\circ}$ ): $\frac{1}{4} π r^{2} - \frac{1}{2} r^{2}$ . Yellow segment ( $6 0^{\circ}$ , triangle is equilateral): $\frac{1}{6} π r^{2} - \frac{3}{4} r^{2}$ .
Equation: $\frac{1}{4} π r^{2} - \frac{1}{2} r^{2} + \frac{1}{6} π r^{2} - \frac{3}{4} r^{2} = 256 \frac{2}{3} = \frac{770}{3}$
1 mark
(ii) $(\frac{5 π}{12} - \frac{1}{2} - \frac{3}{4}) r^{2} = \frac{770}{3}$ . With $π = \frac{22}{7}$ , $3 = 1.732$ : $(1.3095 - 0.5 - 0.4330) r^{2} = 256.67 \Rightarrow 0.3765 r^{2} = 256.67 \Rightarrow r^{2} \approx 681.7$ , so $r \approx 26.1$ m.
1 mark
(iii)(a) Area of red segment $= (\frac{π}{4} - \frac{1}{2}) r^{2} = (\frac{11}{14} - \frac{1}{2}) r^{2} = \frac{2}{7} r^{2}$
1 mark
$= \frac{2}{7} \times 681.7 \approx 194.8$ sq m (≈ 194.63 sq m if $r = 26.1$ is used).
1 mark
(iii)(b) Area of yellow segment $= (\frac{π}{6} - \frac{3}{4}) r^{2} = (0.5238 - 0.4330) r^{2} = 0.0908 r^{2}$
1 mark
$= 0.0908 \times 681.7 \approx 61.9$ sq m (check: $194.8 + 61.9 \approx 256.7$ sq m).
1 mark

Answer: (i) $\frac{1}{4} π r^{2} - \frac{1}{2} r^{2} + \frac{1}{6} π r^{2} - \frac{3}{4} r^{2} = 256 \frac{2}{3}$ ; (ii) r ≈ 26.1 m; (iii)(a) ≈ 194.8 sq m; (iii)(b) ≈ 61.9 sq m

Where the marks are

Writes the sum of the two segment areas (sector − triangle) equated to $256 \frac{2}{3}$ (1 mark)
Simplifies and finds r ≈ 26.1 m (1 mark)
(iii)(a) Correct expression $\frac{1}{4} π r^{2} - \frac{1}{2} r^{2}$ and value ≈ 194.6–194.8 sq m (2 marks)
(iii)(b) Correct expression $\frac{1}{6} π r^{2} - \frac{3}{4} r^{2}$ and value ≈ 62 sq m (2 marks)

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 paper2023 board paper, set 430/2/1 (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 2023: question 36

Hints

Step-by-step solution

Revise the chapter

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