MathsCBSE Class 10 · Basic · Board exam 2023, set 430/6/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2023: question 38

Set in the CBSE Class 10 board exam of 2023, paper 430/6/1, 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-basedCircles

People of a circular village Dharamkot want to construct a road nearest to it. The road cannot pass through the village. But the people want the road at a shortest distance from the centre of the village. Suppose the road starts from A which is outside the circular village (as shown in the figure) and touch the boundary of the circular village at B such that AB = 20 m. Also the distance of the point A from the centre O of the village is 25 m.

Based on the above information, answer the following questions:

(i)
If B is the mid-point of AC, then find the distance AC.
1 mark
(ii)
Find the shortest distance of the road from the centre of the village.
1 mark
(iii)
Find the circumference of the village.
2 marks
OR
(iii)
Find the area of the village.
2 marks

Figure in the paper: Left: a circular village with a road drawn as a horizontal line above and below the circle (labelled Road). Right: a circle with centre O; the horizontal tangent line passes through points C (left), B (top of circle, point of tangency) and A (right), with CBA a straight line. AB = 20 m. Segment OA is drawn from centre O to A, labelled 25 m. Segment OB is a vertical radius from O up to B, perpendicular to the tangent line at B. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 430/6/1.

Hints

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

Hint 1

If B is the mid-point of AC, how does BC compare with AB?

Hint 2

The road touches the circle at B. What angle does the radius OB make with the road?

Hint 3

Find OB using Pythagoras in triangle OBA. OB is the radius you need for (iii).

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) B is the mid-point of AC, so $B C = A B = 20$ m.
$A C = A B + B C = 20 + 20 = 40$ m
1 mark
(ii) The shortest distance from O to the road is OB, and $O B ⊥ A B$ [tangent ⟂ radius at the point of contact].
In right $△ O B A$ : $O B = O A^{2} - A B^{2} = 2 5^{2} - 2 0^{2} = 625 - 400 = 225 = 15$ m
1 mark
(iii)(a) The radius is $r = O B = 15$ m.
Circumference $= 2 π r = 2 π (15) = 30 π$ m
1 mark
$= 30 \times \frac{22}{7} = \frac{660}{7}$ m $\approx 94.29$ m
1 mark
(iii)(b) Area $= π r^{2} = π (15)^{2} = 225 π m^{2}$
1 mark
$= 225 \times \frac{22}{7} = \frac{4950}{7} m^{2} \approx 707.14 m^{2}$
1 mark

Answer: (i) 40 m; (ii) 15 m; (iii)(a) 30π m = 660/7 m; (iii)(b) 225π m² = 4950/7 m²

Where the marks are

AC = 40 m (1 mark)
Uses tangent ⟂ radius and Pythagoras to get OB = 15 m (1 mark)
(iii)(a) Circumference = $2 π (15)$ (1 mark) = 30π or 660/7 m (1 mark)
(iii)(b) Area = $π (15)^{2}$ (1 mark) = 225π or 4950/7 m² (1 mark)

Revise the chapter

Chapter: Circles (notes, formulas, where marks are lost and every past question)
Topic: Tangent Perpendicular to Radius Theorem

Practise this chapterBoard-style questions from Circles, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 430/6/1 (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 2023: question 38

Hints

Step-by-step solution

Revise the chapter

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