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NCERT Class 9 Mathematics Exercise Set 6.3 Solutions: Measuring Space: Perimeter and Area

Step-by-step working for every question of Exercise Set 6.3 in NCERT's Class 9 Mathematics book, chapter “Measuring Space: Perimeter and Area”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Ex 6.3 Q1

   Area of a sector of radius 7 cm and angle 60° (π ≈ 22/7).

   1. The formula for the area of a sector is (θ/360) × πr².
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   2. Substitute the given values: θ = 60°, r = 7 cm, and π = 22/7.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   3. Area = (60/360) × (22/7) × 7².
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   4. Simplify the fraction 60/360 to 1/6.
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   5. Area = (1/6) × 22 × 7.
   6. Area = 154/6.
   7. Area = 77/3.
   8. Area ≈ 25.67 cm².

   Answer: 25.67 cm²

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2. Ex 6.3 Q2

   Area of a quadrant of a circle with circumference 44 cm (π ≈ 22/7).

   1. The circumference of the circle is given as 44 cm.
      why?

      Circumference = 2πr = πd: π is the same ratio (circumference ÷ diameter) for every circle. Take π = 22/7 or 3.14 as the question says.

   2. Using the formula C = 2πr, we have 2 × (22/7) × r = 44.
      why?

      Circumference = 2πr = πd: π is the same ratio (circumference ÷ diameter) for every circle. Take π = 22/7 or 3.14 as the question says.

   3. Solving for r gives r = 44 × 7 ÷ 44 = 7 cm.
   4. The area of a full circle is πr² = (22/7) × 7 × 7 = 154 cm².
      why?

      Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.

   5. A quadrant is one-fourth of a circle, so its area is 154 ÷ 4.
   6. The area of the quadrant is 38.5 cm².
      why?

      Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.

   Answer: 38.5 cm²

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3. Ex 6.3 Q3

   A clock's minute hand is 7 cm long. Area it sweeps in 10 minutes (π ≈ 22/7)?

   1. The minute hand moves 360° in 60 minutes.
   2. In 10 minutes, the angle swept is (10/60) × 360° = 60°.
   3. The area swept is the area of a sector with radius 7 cm and angle 60°.
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   4. Area = (60/360) × (22/7) × 7².
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   5. Area = (1/6) × 22 × 7.
   6. Area = 154/6 cm².
   7. Area ≈ 25.67 cm².

   Answer: 25.67 cm²

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4. Ex 6.3 Q4

   A chord of a circle of radius 10 cm subtends 90° at the centre. Areas of (i) the minor sector (90°), (ii) the major sector (270°), with π ≈ 3.14.

   1. (i) The radius of the circle is r = 10 cm.
   2. (i) The angle subtended by the minor sector is θ = 90°.
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   3. (i) The area of the minor sector is calculated using the formula (θ/360) × πr².
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   4. (i) Substituting the values, Area = (90/360) × 3.14 × (10)².
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   5. (i) Simplifying the fraction and square, Area = (1/4) × 3.14 × 100.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   6. (i) Calculating the product, Area = 0.25 × 314 = 78.5 cm².
   7. (ii) The angle subtended by the major sector is 360° − 90° = 270°.
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   8. (ii) The area of the major sector is calculated using the formula (θ/360) × πr².
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   9. (ii) Substituting the values, Area = (270/360) × 3.14 × (10)².
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   10. (ii) Simplifying the fraction and square, Area = (3/4) × 3.14 × 100.
       why?

       Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   11. (ii) Calculating the product, Area = 0.75 × 314 = 235.5 cm².

   Answer: (i) 78.5 cm²; (ii) 235.5 cm²

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5. Ex 6.3 Q5

   A chord of a circle of radius 15 cm subtends 60° at the centre. Areas of the minor and major segments (π ≈ 3.14, √3 ≈ 1.73).

   1. Let O be the centre, and AB be the chord subtending 60° at O.
   2. In △OAB, OA = OB = 15 cm (radii), so ∠OAB = ∠OBA = (180° − 60°)/2 = 60°.
   3. Thus △OAB is equilateral, and its area is (√3/4) × 15² = (1.73/4) × 225 = 97.3125 cm².
   4. Area of sector OAB = (60/360) × π × 15² = (1/6) × 3.14 × 225 = 117.75 cm².
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   5. Area of minor segment = Area of sector − Area of △OAB = 117.75 − 97.3125 = 20.4375 cm² ≈ 20.44 cm².
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   6. Area of circle = π × 15² = 3.14 × 225 = 706.5 cm².
      why?

      Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.

   7. Area of major segment = Area of circle − Area of minor segment = 706.5 − 20.4375 = 686.0625 cm² ≈ 686.06 cm².
      why?

      Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.

   Answer: Area of minor segment = 20.44 cm²; Area of major segment = 686.06 cm²

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6. Ex 6.3 Q6

   Two non-overlapping car wipers, each blade 28 cm long, each sweeping 120°. Total area cleaned per sweep (π ≈ 22/7)?

   1. The area cleaned by one wiper blade is the area of a sector with radius r = 28 cm and angle θ = 120°.
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   2. Area of one sector = (θ/360) × πr² = (120/360) × (22/7) × 28².
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   3. Simplify the fraction: 120/360 = 1/3.
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   4. Calculate r²: 28² = 784.
   5. Compute the area for one wiper: (1/3) × (22/7) × 784 = (1/3) × 22 × 112 = 2464/3 cm².
   6. Since there are two non-overlapping wipers, multiply the area of one sector by 2.
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   7. Total area cleaned = 2 × (2464/3) = 4928/3 cm².
   8. Convert to decimal: 4928 ÷ 3 ≈ 1642.67 cm².

   Answer: 1642.67 cm²

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7. Ex 6.3 Q7

   A chord of a circle of radius r subtends 60° at the centre. Show the minor segment has area πr²/6 − (√3/4)r², i.e. r²(π/6 − √3/4).

   1. Let O be the centre of the circle and AB be the chord subtending 60° at O.
   2. Since OA = OB = r and ∠AOB = 60°, triangle OAB is equilateral.
   3. The area of equilateral triangle OAB is (√3/4)r².
   4. The area of the sector OAB with angle 60° is (60/360) × πr² = πr²/6.
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   5. The area of the minor segment is the area of the sector minus the area of the triangle.
      why?

      A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.

   6. Area of minor segment = πr²/6 − (√3/4)r².
      why?

      Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.

   7. This can be written as r²(π/6 − √3/4).

   Answer: Area of the minor segment is r²(π/6 − √3/4).

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8. Ex 6.3 Q8

   An equilateral triangle is inscribed in a circle of radius r. Show area of triangle : area of circle = 3√3/(4π) ≈ 0.413.

   1. Let the equilateral triangle be inscribed in a circle of radius r.
   2. The centre of the circle is also the centroid of the equilateral triangle, 2/3 of the way along each median, so the height of the triangle is r + r/2 = 3r/2; from (√3/2)s = 3r/2 the side is s = r√3.
   3. The area of the equilateral triangle is (√3/4) × s².
   4. Substituting s = r√3, the area of the triangle is (√3/4) × (r√3)² = (√3/4) × 3r² = (3√3/4)r².
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   5. The area of the circle is πr².
      why?

      Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.

   6. The ratio of the area of the triangle to the area of the circle is [(3√3/4)r²] / [πr²].
      why?

      Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.

   7. Simplifying the ratio by cancelling r² gives (3√3)/(4π).
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   8. Numerically, 3√3 ≈ 5.196 and 4π ≈ 12.566, so the ratio is approximately 5.196 / 12.566 ≈ 0.413.

   Answer: Area of triangle : Area of circle = 3√3/(4π) ≈ 0.413

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9. Ex 6.3 Q9

   A square is inscribed in a circle of radius r. Show area of square : area of circle = 2/π ≈ 0.637.

   1. Let the radius of the circle be r, so the diameter is 2r.
   2. The diagonal of the inscribed square is equal to the diameter of the circle, which is 2r.
   3. Let the side of the square be a. By the Pythagorean theorem, a² + a² = (2r)².
   4. Simplifying gives 2a² = 4r², so a² = 2r².
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   5. The area of the square is a², which equals 2r².
   6. The area of the circle is πr².
      why?

      Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.

   7. The ratio of the area of the square to the area of the circle is 2r² / πr².
      why?

      Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.

   8. Cancelling r² from the numerator and denominator gives 2/π.
   9. The value 2/π is approximately 0.637.

   Answer: Area of square : Area of circle = 2/π ≈ 0.637

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10. Ex 6.3 Q10

    A regular hexagon is inscribed in a circle of radius r. Show area of hexagon : area of circle = 3√3/(2π) ≈ 0.827, and explain why this is exactly twice the answer of Q8.

    1. A regular hexagon inscribed in a circle of radius r is divided into 6 equilateral triangles by joining the center to each vertex.
    2. The side length of each equilateral triangle is equal to the radius r.
    3. The area of one equilateral triangle with side r is (√3/4)r².
    4. The total area of the hexagon is 6 × (√3/4)r² = (3√3/2)r².
    5. The area of the circle with radius r is πr².
       why?

       Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.

    6. The ratio of the area of the hexagon to the area of the circle is [(3√3/2)r²] / [πr²].
       why?

       Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.

    7. Simplifying the ratio gives 3√3/(2π).
       why?

       Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

    8. Using the values √3 ≈ 1.732 and π ≈ 3.14, the ratio is (3 × 1.732) / (2 × 3.14) = 5.196 / 6.28 ≈ 0.827.
    9. Joining alternate vertices of the hexagon gives the inscribed equilateral triangle of Q8, of area (3√3/4)r²; the hexagon's area (3√3/2)r² is exactly double.
    10. Reason: from the centre, the triangle splits into 3 triangles with two sides r and a 120° angle between them; the hexagon is the triangle plus 3 corner caps, each with two sides r (hexagon sides) and a 120° angle. All 6 pieces are congruent, so the hexagon is twice the triangle, and its ratio is 2 × 3√3/(4π) = 3√3/(2π).

    Answer: The ratio of the area of the regular hexagon to the area of the circle is 3√3/(2π) ≈ 0.827, which is exactly twice the ratio obtained in Q8.

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