MathsCBSE Class 10 · Basic · Board exam 2026, set 430/3/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2026: question 38

Set in the CBSE Class 10 board exam of 2026, paper 430/3/1, 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

There is a semicircular park in Aman's society. He wishes to plant saplings along the boundary of the park. There is a borewell at the centre O of the park along the diameter AB as shown in the figure below.

Based on the above information, answer the following questions:

(i)
Find the coordinates of point O.
1 mark
(ii)
Find the radius of the semicircular park.
1 mark
(iii)
(a) One sapling is kept at point C(12, y). Find the coordinates of C.
2 marks
OR
(iii)
(b) One sapling is kept at point P along AB so that $P A = \frac{1}{3} P B$ . Find the coordinates of P.
2 marks

Figure in the paper: Semicircular park with diameter AB along a horizontal line; A(2, 3) at left end and B(22, 3) at right end. O is the centre on AB, P is a point on AB where a sapling is planted, and C(12, y) is the lowest point of the semicircular arc. A borewell is drawn at O. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 430/3/1.

Hints

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

Hint 1

The centre of a semicircle is the midpoint of its diameter.

Hint 2

Every point on the arc is one radius away from O.

Hint 3

$P A = \frac{1}{3} P B$ means AP : PB = 1 : 3. Then use the section formula.

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) O is the midpoint of AB: $O = (\frac{2 + 22}{2}, \frac{3 + 3}{2}) = (12, 3)$ .
1 mark
(ii) $A B = (22 - 2)^{2} + (3 - 3)^{2} = 20$ , so the radius $= \frac{20}{2} = 10$ units.
1 mark
(iii)(a) C lies on the boundary, so $O C = 10$ : $(12 - 12)^{2} + (y - 3)^{2} = 10 \Rightarrow (y - 3)^{2} = 100 \Rightarrow y = 13$ or $y = - 7$ .
1 mark
(iii)(a) C is the lowest point of the arc, which lies below AB, so $y = - 7$ and $C = (12, - 7)$ .
1 mark
(iii)(b) $P A = \frac{1}{3} P B \Rightarrow P A : P B = 1 : 3$ , so P divides AB in the ratio $1 : 3$ .
½ mark
(iii)(b) By the section formula, $P = (\frac{1 ( 22 ) + 3 ( 2 )}{1 + 3}, \frac{1 ( 3 ) + 3 ( 3 )}{1 + 3}) = (7, 3)$ .
1½ marks

Answer: (i) O(12, 3); (ii) 10 units; (iii)(a) C(12, −7); (iii)(b) P(7, 3)

Where the marks are

Midpoint O = (12, 3) (1 mark)
Radius = 10 units (1 mark)
(iii)(a) Uses OC = radius to form $(y - 3)^{2} = 100$ (1 mark) and selects C(12, −7) (1 mark)
(iii)(b) Ratio 1 : 3 (½ mark) and P(7, 3) by the section formula (1½ marks)

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 paper2026 board paper, set 430/3/1 (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 2026: question 38

Hints

Step-by-step solution

Revise the chapter

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