CBSE Class 10 Mathematics Standard 2025: question 32(a)
Set in the CBSE Class 10 board exam of 2025, paper 30/6/1, section D, question 32(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.
Q32(a) · 5 marks · Long answerCompetency-basedCircles
There is a circular park of diameter 65 m as shown in the following figure, where AB is a diameter. An entry gate is to be constructed at a point P on the boundary of the park such that distance of P from A is 35 m more than the distance of P from B. Find distance of point P from A and B respectively.
Figure in the paper: A circle with centre O. AB is a horizontal diameter with A on the left and B on the right. P is a point on the upper arc, joined to A and B forming triangle APB with a right angle at P. 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
What is the angle at P, since AB is a diameter?
Hint 2
Use Pythagoras in triangle APB with PB = and PA = , AB = 65.
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.
- ½ mark
Let m; then m.
- 1 mark
[angle in a semicircle], so by Pythagoras theorem:
- 2 marks
- ½ mark
- ½ mark
(rejecting as distance cannot be negative)
- ½ mark
m and m.
Answer: Distance from A = 60 m, from B = 25 m
Where the marks are
- Takes , (½ mark)
- Applies Pythagoras in right (1 mark)
- Forms (2 marks)
- Factorises (½ mark)
- (½ mark)
- Distances 60 m and 25 m (½ mark)
Revise the chapter
- Chapter: Circles (notes, formulas, where marks are lost and every past question)
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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.