Areas Related to Circles Class 10 Previous Year Questions (Chapter 11)

1. 1

   Step 1: Area of the circular sheet

   1 mark

   Area of the circular sheet = πr² = 22/7 × 70 × 70 = 15400 cm².

   Why? The area of a circle is π times the square of its radius. This gives the whole sheet before the quadrant is cut.

2. 2

   Step 2: Area of the remaining sheet

   1 mark

   Area of the remaining sheet = 15400 − (90/360 × 15400) = 15400 − 3850 = 11550 cm².

   Why? A quadrant is a sector of angle 90°, so its area is one-fourth of the circle. Subtracting the quadrant from the full circle leaves the remaining sheet.

Marks: 1 + 1 = 2 · Split as in the CBSE marking scheme, 2025 · Basic 430/1/3 Q21 (the marks split is taken from the 2025 Basic paper, where this question also appeared)

1. 1

   Step 1: Draw the figure and note the sector fraction

   ½ mark

   Draw a circle with centre O and radius 10 cm. Draw chord AB so that ∠AOB is a right angle. Shade the region between chord AB and the minor arc AB. Then θ/360 = 90/360 = 0.25.

   Why? A segment is the region enclosed by a chord and the corresponding arc. The angle at the centre fixes what fraction of the circle the sector occupies.

2. 2

   Step 2: Write the sector area formula

   ½ mark

   Area of sector OAPB = θ/360 × πr² = 90/360 × 3.14 × 10².

   Why? A sector is the region enclosed by two radii and the corresponding arc. Its area is the same fraction of the whole circular region as its angle is of 360°.

3. 3

   Step 3: Find the area of the sector

   ½ mark

   Area of sector OAPB = 78.5 cm².

   Why? The sector is one quarter of the circular region, so this value is one quarter of πr².

4. 4

   Step 4: Find the area of triangle OAB

   ½ mark

   In ΔOAB, OA and OB are radii of length 10 cm and the angle between them is a right angle, so area of ΔOAB = 1/2 × 10 × 10 = 50 cm².

   Why? The two radii are perpendicular, so one acts as the base and the other as the height of the triangle.

5. 5

   Step 5: Subtract the triangle from the sector

   ½ mark

   Area of minor segment = area of sector OAPB − area of ΔOAB = 78.5 − 50 = 28.5 cm².

   Why? The minor segment is what remains of the minor sector after triangle OAB is removed.

6. 6

   Step 6: State the area of the minor segment

   ½ mark

   Therefore, the area of the corresponding minor segment is 28.5 cm².

   Why? This is the required area of the region bounded by the chord and the minor arc.

Marks: ½ + ½ + ½ + ½ + ½ + ½ = 3 · Split as in the CBSE marking scheme, 2025 · Standard 30/3/1 Q29

1. 1

   Step 1: Find the area swept by one wiper

   1 mark (suggested)

   Each blade sweeps a sector of radius 21 cm and angle 120°. Area of one sector = 120/360 × 22/7 × 21² = 462 cm².

   Why? A wiper blade sweeping through an angle traces out a sector of a circle, and the area of a sector is θ/360 × πr².

2. 2

   Step 2: Find the total area for two wipers

   1 mark (suggested)

   The wipers do not overlap, so total area cleaned = 2 × 462 = 924 cm².

   Why? Non-overlapping regions have areas that add, and the two wipers sweep equal sectors.

3. 3

   Step 3: State the result

   1 mark (suggested)

   The area cleaned at each sweep of the blades is 924 cm².

   Why? The total area for one sweep is the sum of the two non-overlapping swept areas.

Marks: 1 + 1 + 1 = 3 · Suggested split (CBSE’s usual pattern for 3-mark questions)

1. 1

   Step 1: Draw the sector and set up the perimeter

   2 marks

   Let the length of the arc be l. Draw sector OAPB with centre O and radius 6.5 cm. Perimeter of the sector = 2 × 6.5 + l = 31, so l = 31 − 13 = 18 cm.

   Why? The perimeter of a sector is the sum of its two radii and its arc, so subtracting the two radii from the given perimeter gives the arc length.

2. 2

   Step 2: Find the area of the sector

   2 marks

   Area of the sector = (l × r)/2 = (18 × 6.5)/2 = 58.5 cm².

   Why? By the NCERT formulas, arc length l = (θ/360) × 2πr, so (θ/360) × πr² equals l × r divided by 2. Substituting the arc length and the radius gives the area.

3. 3

   Step 3: State the area

   1 mark

   Hence, the area of the sector is 58.5 cm².

   Why? The final statement gives the value asked for, with the unit of area.

Marks: 2 + 2 + 1 = 5 · Split as in the CBSE marking scheme, 2024 · Basic 430(B) Q32

1. 1

   Step 1: Draw the figure and find the area of the circle

   1 mark (suggested)

   Area of the circle = πr² = 3.14 × 10² = 314 cm².

   Why? The area of a circular region is πr²; the whole-circle area is the base for finding any sector.

2. 2

   Step 2: Find the area of the minor sector

   1 mark (suggested)

   Area of the minor sector = (90/360) × 314 = 78.5 cm².

   Why? A sector's area is the same fraction of the circle as its angle is of 360°.

3. 3

   Step 3: Find the area of the major sector

   1 mark (suggested)

   Area of the major sector = πr² − area of the minor sector = 314 − 78.5 = 235.5 cm². Therefore, the area of the corresponding major sector is 235.5 cm².

   Why? The minor and major sectors formed by the same arc together make the whole circle, so subtracting the minor sector leaves the major sector.

Marks: 1 + 1 + 1 = 3 · Suggested split (CBSE’s usual pattern for 3-mark questions)

1. 1

   Step 1: Draw the sector and find its area

   1 mark

   Draw a circle with centre O and radius 21 cm; mark points A and B on the circle so that the angle AOB is 60°, and shade the sector OAPB. Area of the sector = 60/360 × 22/7 × 21 × 21 = 231 cm².

   Why? A sector is the part of the circular region enclosed by two radii and the corresponding arc. Its area is the fraction θ/360 of the area of the whole circle, πr².

2. 2

   Step 2: Find the length of the arc

   1 mark

   Length of the arc = 60/360 × 2 × 22/7 × 21 = 22 cm.

   Why? The arc belonging to a sector of angle θ is the same fraction θ/360 of the whole circumference, 2πr.

3. 3

   Step 3: State the required answers

   1 mark

   Area of the sector = 231 cm² and length of the arc = 22 cm.

   Why? The question asks for both the area and the arc length, so the final sentence states both results with their correct units.

Marks: 1 + 1 + 1 = 3 · Split as in the CBSE marking scheme, 2023 · Standard 30/6/1 Q31