Close Menu
Maths

Areas Related to Circles Class 10 Maths Worksheet with Answers Chapter 11

Updated:9 Mins Read

We are near the end of the book and are solving the 1th chapter here, Areas Related to Circles. We suggest you try solving only the questions below after you have thoroughly completed the entire chapter. The questions in this worksheet from the NCERT book are arranged into four levels: Basic, Standard, Advance, and HOTS for tricky, logic-based questions.

The primary topics covered in this chapter –

  • Arc length formula
  • Sector area formula
  • Segment area concept
  • Circle-based word problems
  • Mixed units conversion

Class 10 Maths Worksheet – Chapter 11: Areas Related to Circles

Use π = 22/7.
Use correct units (cm², m², mm²). Keep √ form where possible and round only when asked.

Basic

  1. Write the meaning (1 line each):

    • Sector
    • Segment
    • Arc
  2. Find the area of a sector of a circle with radius 6 cm and angle 60°.
  3. Find the area of a quadrant of a circle whose circumference is 22 cm.
  4. Find the area of a circle of radius 14 cm.
  5. Find the length of an arc of a circle of radius 7 cm making an angle 90° at the centre.
  6. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre.
    Find the length of the arc.
  7. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre.
    Find the area of the sector formed by the arc.

Standard

  1. The length of the minute hand of a clock is 14 cm.
    Find the area swept by the minute hand in 5 minutes.
  2. A chord of a circle of radius 10 cm subtends a right angle (90°) at the centre.
    Find the area of the corresponding:

    • (i) minor segment
    • (ii) major sector

    (Use π = 3.14)

  3. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre.
    Find the areas of the corresponding minor and major segments.
    (Use π = 3.14 and √3 = 1.73)
  4. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre.
    Find the area of the corresponding segment of the circle.
    (Use π = 3.14 and √3 = 1.73)
  5. A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by a 5 m rope.
    Find the area in which the horse can graze.
    (Use π = 3.14)
  6. In the same field question above, find the increase in grazing area if the rope were 10 m instead of 5 m.
    (Use π = 3.14)
  7. An umbrella has 8 ribs equally spaced. Assume the umbrella is a flat circle of radius 45 cm.
    Find the area between two consecutive ribs.

Advance

  1. A brooch is made with silver wire in the form of a circle of diameter 35 mm.
    The wire is also used in making 5 diameters which divide the circle into 10 equal sectors.
    Find:

    • (i) Total length of silver wire required
    • (ii) Area of each sector
  2. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°.
    Find the total area cleaned at each sweep (both wipers together).
  3. To warn ships for underwater rocks, a lighthouse spreads a red-coloured light over a sector of angle 80° to a distance of 16.5 km.
    Find the area of the sea over which the ships are warned.
    (Use π = 3.14)
  4. Find the length of an arc of radius 7 cm making an angle of 270° at the centre.
  5. Find the area of the major sector of a circle of radius 4 cm if the minor sector angle is 30°.
    (Use π = 3.14)
  6. Find the area of a segment of a circle with radius 21 cm and central angle 120°.
    (Use π = 22/7 and √3 = 1.73)
  7. Write the formulas (with units) for:

    • Length of arc of angle θ
    • Area of sector of angle θ
    • Area of segment

HOTS

  1. Tick the correct option:
    Area of a sector of angle p (in degrees) of a circle with radius R is:

    • (A) (p/180) × πR²
    • (B) (p/180) × 2πR
    • (C) (p/360) × πR²
    • (D) (p/720) × 2πR²
  2. If the area of a circle is 200 cm² and the area of its minor sector is 50 cm²,
    find the area of the corresponding major sector.
  3. A sector has radius 14 cm and arc length 22 cm.
    Find the angle of the sector (in degrees).
  4. A circle has radius 10 cm. A sector of angle 72° is cut out.
    Without calculating decimals, write the area of the remaining part as a fraction of the circle’s area.
  5. True/False (1–2 lines reason):
    “If two sectors have the same angle, the one with larger radius always has larger area.”
  6. A chord subtends 60° at the centre. Which is bigger: the minor segment or the minor sector?
    Answer with a reason (no numbers needed).
  7. A student finds the area of a segment using only “area of sector” and forgets to subtract the triangle area.
    What kind of error will happen: overestimate or underestimate? Explain in 2 lines.

Answer Key

Basic – Answers

  1. Ans:

    • Sector: Region between two radii and the arc.
    • Segment: Region between a chord and its arc.
    • Arc: Curved part of the circle boundary.

    Hint: Sector uses radii; segment uses chord.

  2. Ans: Area = (60/360) × π × 6² = 6π = 132/7 cm².
    Hint: Area of sector = (θ/360) × πr².
  3. Ans: r = 3.5 cm, quadrant area = 77/8 cm².
    Hint: 2πr = 22 → find r, then take 1/4 of πr².
  4. Ans: Area = π × 14² = 616 cm².
    Hint: Use π = 22/7.
  5. Ans: Arc length = (90/360) × 2π × 7 = 11 cm.
    Hint: Arc length = (θ/360) × 2πr.
  6. Ans: Arc length = (60/360) × 2π × 21 = 7π = 22 cm.
    Hint: (1/6) of circumference.
  7. Ans: Sector area = (60/360) × π × 21² = 231 cm².
    Hint: (1/6) of circle area.

Standard – Answers

  1. Ans: Area swept = (5/60) × π × 14² = (1/12) × π × 196 = 154/3 cm² (≈ 51.33 cm²).
    Hint: 5 minutes is 1/12 of a full rotation.
  2. Ans:

    • Sector (90°) area = 25π = 78.5 cm²
    • Δ area = (1/2) × 10 × 10 = 50 cm²
    • Minor segment = 78.5 − 50 = 28.5 cm²
    • Major sector = 3.14 × 10² − 78.5 = 314 − 78.5 = 235.5 cm²

    Hint: Segment = sector − triangle.

  3. Ans:

    • Sector area = (60/360) × 3.14 × 15² = 117.75 cm²
    • Triangle area = (1/2) × 15 × 15 × sin60° = 56.25√3 ≈ 97.31 cm²
    • Minor segment ≈ 117.75 − 97.31 = 20.44 cm²
    • Major segment ≈ 3.14 × 15² − 20.44 = 706.5 − 20.44 = 686.06 cm²

    Hint: sin60° = √3/2.

  4. Ans:

    • Sector area = (120/360) × 3.14 × 12² = 150.72 cm²
    • Triangle area = (1/2) × 12 × 12 × sin120° = 36√3 ≈ 62.28 cm²
    • Segment area ≈ 150.72 − 62.28 = 88.44 cm²

    Hint: sin120° = sin60°.

  5. Ans: Grazing area = quadrant of r = 5:
    (1/4) × 3.14 × 5² = 19.625 m².
    Hint: Corner peg gives 90° sector.
  6. Ans: Increase = quadrant(r=10) − quadrant(r=5)
    = 78.5 − 19.625 = 58.875 m².
    Hint: Both are quadrants.
  7. Ans: Angle between ribs = 360°/8 = 45°
    Area = (45/360) × π × 45² = (1/8) × π × 2025
    = 795.5 cm² (approx).
    Hint: Each gap is a 45° sector.

Advance – Answers

  1. Ans:

    • Circumference = πd = (22/7)×35 = 110 mm
    • 5 diameters = 5×35 = 175 mm
    • Total wire length = 110 + 175 = 285 mm
    • Circle area = πr² = (22/7)×(17.5²) = 962.5 mm²
    • Each sector area = 962.5/10 = 96.25 mm²

    Hint: Wire includes outer circle + 5 diameters.

  2. Ans: One wiper area = (115/360) × π × 25²
    = (115/360) × π × 625 ≈ 627.48 cm²
    Total (2 wipers) ≈ 1254.96 cm² (≈ 1255 cm²).
    Hint: Two equal sectors, no overlap.
  3. Ans: Area = (80/360) × 3.14 × 16.5²
    = (2/9) × 3.14 × 272.25 ≈ 189.97 km² (≈ 190.0 km²).
    Hint: Sector area with r in km gives km².
  4. Ans: Arc length = (270/360) × 2π × 7 = (3/4)×14π = 10.5π = 33 cm.
    Hint: 270° is 3/4 of the circle.
  5. Ans: Major angle = 360° − 30° = 330°
    Major sector area = (330/360) × 3.14 × 4²
    = (11/12) × 50.24 ≈ 46.05 cm².
    Hint: Major = full circle − minor sector.
  6. Ans: Sector area = (120/360) × (22/7) × 21² = (1/3) × (22/7) × 441 = 462 cm²
    Triangle area = (1/2)×21×21×sin120° = 220.5×(√3/2) = 110.25√3 ≈ 110.25×1.73 = 190.73 cm²
    Segment area ≈ 462 − 190.73 = 271.27 cm².
    Hint: Segment = sector − triangle.
  7. Ans:

    • Arc length = (θ/360) × 2πr
    • Sector area = (θ/360) × πr²
    • Segment area = sector area − triangle area

    Hint: Always keep θ in degrees here.

HOTS – Answers

  1. Ans: (C) (p/360) × πR²
    Hint: Sector is fraction of 360°.
  2. Ans: Major sector area = 200 − 50 = 150 cm².
    Hint: Major + minor = full circle.
  3. Ans: Arc length = (θ/360)×2πr = 22
    So θ = (22×360)/(2π×14) = (22×360)/(28π)
    Using π=22/7 ⇒ θ = 360/4 = 90°.
    Hint: Solve for θ first, then simplify.
  4. Ans: Remaining fraction = 1 − (72/360) = 1 − 1/5 = 4/5.
    Hint: Area scales with angle.
  5. Ans: True.
    Hint: Same θ ⇒ area ∝ r², bigger r gives bigger area.
  6. Ans: Minor sector is bigger.
    Hint: Minor segment = sector − triangle, so it must be smaller than the sector.
  7. Ans: Overestimate.
    Hint: Segment is sector minus triangle; forgetting subtraction makes the answer too large.

Worksheet for Other chapters

Amit, a BE in Mechanical Engineering, is a math enthusiast dedicated to making math fun and accessible for kids in classes 1 to 10. With a knack for simplifying complex concepts, Amit offers easy-to-understand solutions, fostering a love for math in young minds across India.

Related Posts

Leave A Reply