CBSE Class 10 Mathematics Standard 2025: question 36
Set in the CBSE Class 10 board exam of 2025, paper 30/6/1, 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
Anurag purchased a farmhouse which is in the form of a semicircle of diameter 70 m. He divides it into three parts by taking a point P on the semicircle in such a way that as shown in the following figure, where O is the centre of semicircle.
In part I, he planted saplings of Mango tree, in part II, he grew tomatoes and in part III, he grew oranges. Based on given information, answer the following questions.
- (i)1 mark
What is the measure of ?
- (ii)1 mark
Find the length of wire needed to fence entire piece of land.
- (iii)2 marks
(a) Find the area of region in which saplings of Mango tree are planted.
OR
(iii)2 marks(b) Find the length of wire needed to fence the region III.
Figure in the paper: A semicircle with diameter AB (A at left, B at right, O centre at the midpoint of the base). P is on the arc, joined to A and to B. Angle PAB is marked . Region I is the triangle PBA side near B (between PB and arc), region II is the lower part between AB, AP and the regions, and region III is the upper-left segment between AP and the arc. Symbols show mango saplings (stars in III), tomatoes (faces in II) and oranges (circles, marked I/II/III as indicated). See it in the official paper.
Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 30/6/1.
Hints
Open one hint at a time and try again before the next.
Hint 1
The angle at the centre is twice the angle at the circumference on the same arc. Use this to find first.
Hint 2
The fence for the whole land runs along the curved boundary and the diameter. Region I is a segment: take the sector minus the triangle, and note that triangle POB is equilateral.
Hint 3
For region III you need chord AP and arc AP. is an angle in a semicircle.
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.
- 1 mark
(i) The angle subtended by arc PB at the centre is twice the angle at the circumference: . Since AOB is a straight line, .
- 1 mark
(ii) Radius m. Fence = semicircular arc + diameter m.
- 1 mark
(iii)(a) Mango region I is the segment cut off by chord PB. Its area = area of sector POB () − area of . is equilateral because m and . So area .
- 1 mark
m m.
- 1 mark
(iii)(b) [angle in a semicircle]. In right , , so m.
- ½ mark
Region III is bounded by chord AP and arc AP, and arc AP subtends at O. Arc AP m.
- ½ mark
Wire needed m m.
Answer: (i) 120°; (ii) 180 m; (iii)(a) (1925/3 − 1225√3/4) m² ≈ 111.2 m²; (iii)(b) (220/3 + 35√3) m ≈ 133.9 m
Where the marks are
- (i) (1 mark)
- (ii) Fence = arc + diameter = 180 m (1 mark)
- (iii)(a) Sets up sector POB minus equilateral triangle POB (1 mark)
- (iii)(a) Area m² ≈ 111.2 m² (1 mark)
- (iii)(b) m (1 mark)
- (iii)(b) Arc AP m set up (½ mark)
- (iii)(b) Total length ≈ 133.9 m (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
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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.