A woman discovered a scratch along a straight line on a circular table top of radius 8 cm. She divided the table top into 4 equal quadrants and discovered the scratch passing through the origin inclined at an angle (π)/(4) anticlockwise along the positive direction of x-axis. Find the area of the region enclosed by the x-axis, the scratch and the circular table top in the first quadrant, using integration.

MathsCBSE Class 12 · Board exam 2025, set 65/7/1 · 5 marks · Long answer

CBSE Class 12 Mathematics 2025: question 33

Set in the CBSE Class 12 board exam of 2025, paper 65/7/1, section D, question 33, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33 · 5 marks · Long answerCompetency-basedApplication of Integrals

A woman discovered a scratch along a straight line on a circular table top of radius 8 cm. She divided the table top into 4 equal quadrants and discovered the scratch passing through the origin inclined at an angle $\frac{π}{4}$ anticlockwise along the positive direction of x-axis. Find the area of the region enclosed by the x-axis, the scratch and the circular table top in the first quadrant, using integration.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 65/7/1.

Hints

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

Hint 1

The scratch through the origin at $4 5^{\circ}$ is the line $y = x$ , and the table edge is $x^{2} + y^{2} = 64$ .

Hint 2

Split the area at the intersection point: below the line from 0 to $42$ , then below the circle from $42$ to 8.

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.

Figure: Draw a circle of radius 8 centred at the origin O and the line $y = x$ through O. In the first quadrant they meet at A. Drop the perpendicular from A to the x-axis at $x = 42$ . Shade the region bounded by the x-axis, the line OA and the arc from A to C(8,0).
1 mark
Equation of the circular table top: $x^{2} + y^{2} = 64$ .
½ mark
The scratch passes through the origin at angle $\frac{π}{4}$ , so its slope is $tan \frac{π}{4} = 1$ and its equation is $y = x$ .
½ mark
Solving, $2 x^{2} = 64$ gives $x = 42$ in the first quadrant, so $A (42, 42)$ .
½ mark
Required area $= \int_{0}^{42} x d x + \int_{42}^{8} 64 - x^{2} d x$
1 mark
$= [\frac{x ^{2}}{2}]_{0}^{42} + [\frac{x}{2} 64 - x^{2} + 32 sin^{- 1} \frac{x}{8}]_{42}^{8}$
1 mark
$= 16 + [32 \cdot \frac{π}{2} - (22 \cdot 42 + 32 \cdot \frac{π}{4})] = 16 + 16 π - 16 - 8 π = 8 π cm^{2}$
½ mark

Answer: $8 π$ cm²

Where the marks are

Correct figure with region shaded (1 mark)
Equation of circle $x^{2} + y^{2} = 64$ (½ mark)
Equation of scratch $y = x$ (½ mark)
Point of intersection $x = 42$ (½ mark)
Correct integral set-up (1 mark)
Correct integration (1 mark)
Final area $8 π$ cm² (½ mark)

Revise the chapter

Chapter: Application of Integrals (notes, formulas, where marks are lost and every past question)
Topic: Area of Circle and Ellipse by Integration

Practise this chapterBoard-style questions from Application of Integrals, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 65/7/1: every question, by section.

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CBSE Class 12 Mathematics 2025: question 33

Hints

Step-by-step solution

Revise the chapter

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