MathsCBSE Class 10 · Basic · Board exam 2025, set 430/1/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2025: question 38

Set in the CBSE Class 10 board exam of 2025, paper 430/1/1, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedCoordinate Geometry

Case Study – 3. In a society, there is a circular park having two gates. The gates are placed at points A(10, 20) and B(50, 50), as shown in the figure below. Two fountains are installed at points P and Q on AB such that AP = PQ = QB. Based on the above information, answer the following questions :

(i)
Find the coordinates of the centre C.
1 mark
(ii)
Find the radius of the circular park.
1 mark
(iii)(a)
Find the coordinates of the point P.
2 marks
OR
(iii)(b)
Find the distance of the fountain at Q from gate A.
2 marks

Figure in the paper: A circle with a diameter AB drawn as a straight line through the centre C. A is at the lower-left on the circle, B at the upper-right on the circle. Points P and Q lie on AB between A and B, with P closer to A, C between P and Q (near the middle), and Q closer to B. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/1/1.

Hints

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

Hint 1

AB passes through the centre, so it is a diameter and C is its mid-point.

Hint 2

The radius is the distance from C to A.

Hint 3

AP = PQ = QB splits AB into three equal parts. So P divides AB in the ratio 1 : 2, and AQ is two-thirds of AB.

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) AB is a diameter, so the centre C is the mid-point of AB:
$C = (\frac{10 + 50}{2}, \frac{20 + 50}{2}) = (30, 35)$
1 mark
(ii) Radius $= A C = (30 - 10)^{2} + (35 - 20)^{2} = 400 + 225 = 625 = 25$ units.
1 mark
(iii)(a) Since AP = PQ = QB, P divides AB in the ratio $1 : 2$ .
½ mark
(iii)(a) By the section formula:
$P = (\frac{1 \times 50 + 2 \times 10}{1 + 2}, \frac{1 \times 50 + 2 \times 20}{1 + 2})$
1 mark
(iii)(a) $P = (\frac{70}{3}, \frac{90}{3}) = (\frac{70}{3}, 30)$
½ mark
(iii)(b) The diameter is $A B = 2 \times 25 = 50$ units.
½ mark
(iii)(b) Since AP = PQ = QB, we get $A Q = \frac{2}{3} A B = \frac{2}{3} \times 50$ .
1 mark
(iii)(b) $A Q = \frac{100}{3}$ units (about 33.33 units).
Check: Q divides AB in the ratio 2 : 1, so $Q = (\frac{110}{3}, 40)$ . Then $A Q = (\frac{80}{3})^{2} + 2 0^{2} = \frac{100}{3}$ , which agrees.
½ mark

Answer: (i) C(30, 35); (ii) 25 units; (iii)(a) P(70/3, 30); (iii)(b) AQ = 100/3 units

Where the marks are

(i) Uses the mid-point formula to get C(30, 35) (1 mark)
(ii) Uses the distance formula to get radius 25 (1 mark)
(iii)(a) Recognises that P divides AB in the ratio 1 : 2 (½ mark)
(iii)(a) Applies the section formula correctly to both coordinates (½ + ½ marks)
(iii)(a) Gets P(70/3, 30) (½ mark)
(iii)(b) Finds AB = 50 (½ mark)
(iii)(b) Writes AQ = (2/3)AB (1 mark)
(iii)(b) Gets AQ = 100/3 (½ mark)

Revise the chapter

Chapter: Coordinate Geometry (notes, formulas, where marks are lost and every past question)
Topic: Section Formula (Internal Division)

Practise this chapterBoard-style questions from Coordinate Geometry, one at a time, marked against the answer key and marking points.Back to the full paper2025 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 2025: question 38

Hints

Step-by-step solution

Revise the chapter

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