MathsCBSE Class 10 · Standard · Board exam 2023, set 30/2/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2023: question 36

Set in the CBSE Class 10 board exam of 2023, paper 30/2/1, 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. In an annual day function of a school, the organizers wanted to give a cash prize along with a memento to their best students. Each memento is made as shown in the figure and its base ABCD is shown from the front side. The rate of silver plating is ₹ $20$ per cm $^{2}$ .

(i)
What is the area of the quadrant ODCO ?
1 mark
(ii)
Find the area of $△ A O B$ .
1 mark
(iii)
(a) What is the total cost of silver plating the shaded part ABCD ?
2 marks
OR
(iii)
(b) What is the length of arc CD ?
2 marks

Figure in the paper: A memento base: a circle with centre O and radius OC = OD = 7 cm, with a right angle at O (angle DOC = 90°). D and C are on the circle. Segment AD = 3 cm extends from D to A, and segment CB = 3 cm extends from C to B, with A and B joined so that ABCD is the base. The shaded region is the part between the arc DC and the segment AB below the arc. Labels: 7 cm along OC and OD, 3 cm along AD and CB. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 30/2/1.

Hints

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

Hint 1

A quadrant is a sector with a central angle of 90°.

Hint 2

OA is made of OD plus DA. What are OA and OB, and what angle do they make at O?

Hint 3

The shaded region is what remains of the triangle AOB once the quadrant is removed 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) The quadrant has $r = 7$ cm and $θ = 9 0^{\circ}$ .
Area $= \frac{θ}{36 0 ^{\circ}} \times π r^{2} = \frac{90}{360} \times \frac{22}{7} \times 7 \times 7$ .
½ mark
$= \frac{77}{2} = 38.5$ cm $^{2}$ .
½ mark
(ii) $O A = O D + D A = 7 + 3 = 10$ cm and $O B = O C + C B = 7 + 3 = 10$ cm, with $∠ A O B = 9 0^{\circ}$ .
ar $(△ A O B) = \frac{1}{2} \times O A \times O B = \frac{1}{2} \times 10 \times 10 = 50$ cm $^{2}$ .
1 mark
(iii)(a) Area of the shaded part ABCD = ar $(△ A O B)$ − area of quadrant ODCO $= 50 - 38.5 = 11.5$ cm $^{2}$ .
Cost $= 11.5 \times 20$ .
1 mark
$= ₹ 230$ .
1 mark
(iii)(b) Length of arc CD $= \frac{θ}{36 0 ^{\circ}} \times 2 π r = \frac{90}{360} \times 2 \times \frac{22}{7} \times 7$ .
1 mark
$= 11$ cm.
1 mark

Answer: (i) 38.5 cm²; (ii) 50 cm²; (iii)(a) ₹230; (iii)(b) 11 cm

Where the marks are

(i) Correct sector-area formula with substitution (½ mark) and answer 38.5 cm² (½ mark)
(ii) Takes OA = OB = 10 cm and gets area 50 cm² (1 mark)
(iii)(a) Shaded area = 50 − 38.5, multiplied by ₹20 (1 mark); cost ₹230 (1 mark)
(iii)(b) Arc-length formula with substitution (1 mark); 11 cm (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 paper2023 board paper, set 30/2/1 (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 2023: question 36

Hints

Step-by-step solution

Revise the chapter

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