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NCERT Class 9 Mathematics Exercise Set 6.1 Solutions: Measuring Space: Perimeter and Area

Step-by-step working for every question of Exercise Set 6.1 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 6.1 Q1

   A circle has perimeter 44 cm (π ≈ 22/7). Find its radius.

   1. The perimeter of a circle is given by the formula 2πr.
      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. We are given that the perimeter is 44 cm and π is 22/7.
   3. Substitute the values into the formula: 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.

   4. Simplify the left side to get (44/7) × r = 44.
      why?

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

   5. Multiply both sides by 7 to get 44r = 308.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   6. Divide both sides by 44 to find r = 7.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   Answer: 7 cm

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2. Ex 6.1 Q2

   Circumference, to 3 significant figures (π ≈ 22/7), of circles of radius (i) 7 cm, (ii) 10 cm, (iii) 12 cm.

   1. (i) The formula for the circumference of a circle is C = 2πr.
      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. (i) Substitute r = 7 cm and π = 22/7 into the formula: C = 2 × (22/7) × 7.
      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. (i) Simplify the expression to get C = 44 cm.
      why?

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

   4. (i) Express the result to 3 significant figures as 44.0 cm.
   5. (ii) Substitute r = 10 cm and π = 22/7 into the formula: C = 2 × (22/7) × 10.
      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.

   6. (ii) Calculate the value: C = 440/7 ≈ 62.857 cm.
   7. (ii) Round the result to 3 significant figures to get 62.9 cm.
   8. (iii) Substitute r = 12 cm and π = 22/7 into the formula: C = 2 × (22/7) × 12.
      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.

   9. (iii) Calculate the value: C = 528/7 ≈ 75.428 cm.
   10. (iii) Round the result to 3 significant figures to get 75.4 cm.

   Answer: (i) 44.0 cm; (ii) 62.9 cm; (iii) 75.4 cm

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3. Ex 6.1 Q3

   Arc length (π ≈ 22/7) for (i) radius 3.5 cm, centre angle 60°; (ii) radius 6.3 m, centre angle 120°.

   1. (i) The formula for arc length is l = (θ/360°) × 2πr.
      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. (i) Substitute r = 3.5 cm, θ = 60°, and π = 22/7 into the formula.
      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. (i) l = (60/360) × 2 × (22/7) × 3.5.
      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) Simplify the fraction 60/360 to 1/6 and calculate 2 × (22/7) × 3.5 = 22.
      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) l = (1/6) × 22 = 22/6 ≈ 3.67 cm.
   6. (ii) Use the same formula l = (θ/360°) × 2πr for the second part.
      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. (ii) Substitute r = 6.3 m, θ = 120°, and π = 22/7 into the formula.
      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.

   8. (ii) l = (120/360) × 2 × (22/7) × 6.3.
      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.

   9. (ii) Simplify the fraction 120/360 to 1/3 and calculate 2 × (22/7) × 6.3 = 44 × 0.9 = 39.6.
      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.

   10. (ii) l = (1/3) × 39.6 = 13.2 m.

   Answer: (i) 3.67 cm; (ii) 13.2 m

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4. Ex 6.1 Q4

   Perimeter of a sector (arc plus both radii) of radius 14 cm and angle 75° (π ≈ 22/7).

   1. The perimeter of a sector is the sum of the arc length and the two radii.
      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. First, calculate the length of the arc using the formula: (θ/360°) × 2πr.
      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. Substitute the given values into the formula: (75/360) × 2 × (22/7) × 14.
      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 expression to find the arc length: 18.33 cm.
      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. Add the lengths of the two radii to the arc length: 18.33 + 14 + 14.
      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. Calculate the total perimeter: 46.33 cm.

   Answer: 18.33 cm, 46.33 cm

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5. Ex 6.1 Q5

   Perimeters of the nine shapes of Fig. 6.14 (π ≈ 22/7): (i) 80 m × 60 m rectangle, semicircles on the 60 m ends; (ii) half-ring, diameters 12 cm and 8 cm; (iii) 10 cm square, semicircle on each side; (iv) 12 cm equilateral triangle, semicircle on each side; (v) plus-shape of five 14 cm squares with end semicircles and corner quarter circles of radius 14 cm; (vi) semicircle on 28 cm with four 7 cm semicircles along the diameter; (vii) right triangle 8 cm, 6 cm with semicircles on all sides; (viii) semicircle on 12 cm with three 4 cm semicircles inside; (ix) semicircle on 20 cm with 10 cm semicircles above the left half and below the right half.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. (i) The boundary is the two straight 80 m sides and two semicircles of diameter 60 m.
   2. (i) The two semicircles make one full circle of diameter 60 m: π × 60 = (22/7) × 60 = 1320/7 ≈ 188.57 m.
   3. (i) Perimeter = 2 × 80 + 188.57 = 160 + 188.57 = 348.57 m.
   4. (ii) Outer semicircle (diameter 12 cm): ½ × (22/7) × 12 = 132/7 cm. Inner semicircle (diameter 8 cm): ½ × (22/7) × 8 = 88/7 cm.
   5. (ii) The two flat ends are each (12 − 8) ÷ 2 = 2 cm long.
   6. (ii) Perimeter = 132/7 + 88/7 + 2 + 2 = 220/7 + 4 = 248/7 ≈ 35.43 cm.
   7. (iii) The square's sides lie inside the shape, so the boundary is four semicircles of diameter 10 cm.
   8. (iii) Each is ½ × (22/7) × 10 = 110/7 cm, so the perimeter is 4 × 110/7 = 440/7 ≈ 62.86 cm.
   9. (iv) The boundary is three semicircles, one on each 12 cm side of the triangle; they meet at its corners.
   10. (iv) Each is ½ × (22/7) × 12 = 132/7 cm, so the perimeter is 3 × 132/7 = 396/7 ≈ 56.57 cm.
   11. (v) The boundary has four semicircles of diameter 14 cm (on the outer sides of the four end squares) and four quarter circles of radius 14 cm (at the corners, each centred at a corner of the middle square).
   12. (v) Each semicircle is ½ × (22/7) × 14 = 22 cm, so the four give 88 cm.
   13. (v) Each quarter circle is ¼ × 2 × (22/7) × 14 = 22 cm, so the four give 88 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.

   14. (v) Perimeter = 88 + 88 = 176 cm.
   15. (vi) The big semicircle on 28 cm is ½ × (22/7) × 28 = 44 cm.
   16. (vi) The 28 cm diameter is replaced by four semicircles on 7 cm parts: 4 × ½ × (22/7) × 7 = 4 × 11 = 44 cm.
   17. (vi) Perimeter = 44 + 44 = 88 cm.
   18. (vii) The hypotenuse is √(8² + 6²) = √100 = 10 cm. The boundary is the three semicircles on 10 cm, 8 cm and 6 cm.
       why?

       In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

   19. (vii) Perimeter = ½ × (22/7) × (10 + 8 + 6) = (11/7) × 24 = 264/7 ≈ 37.71 cm.
   20. (viii) Big semicircle on 12 cm: ½ × (22/7) × 12 = 132/7 cm. Three small semicircles on 4 cm: 3 × ½ × (22/7) × 4 = 132/7 cm.
   21. (viii) Perimeter = 132/7 + 132/7 = 264/7 ≈ 37.71 cm.
   22. (ix) Big semicircle on 20 cm: ½ × (22/7) × 20 = 220/7 cm. The two semicircles on 10 cm (one above the left half, one below the right half): 2 × ½ × (22/7) × 10 = 220/7 cm.
   23. (ix) Perimeter = 220/7 + 220/7 = 440/7 ≈ 62.86 cm.

   Answer: (i) 348.57 m; (ii) 35.43 cm; (iii) 62.86 cm; (iv) 56.57 cm; (v) 176 cm; (vi) 88 cm; (vii) 37.71 cm; (viii) 37.71 cm; (ix) 62.86 cm

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6. Ex 6.1 Q6

   A car tyre has diameter 56 cm (π ≈ 22/7). (i) Distance travelled in one revolution? (ii) Revolutions in 10 km?

   1. (i) The distance travelled in one revolution is the circumference of the tyre.
   2. (i) Circumference = π × diameter = (22/7) × 56 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.

   3. (i) Distance in one revolution = 176 cm.
   4. (ii) Convert the total distance to cm: 10 km = 10 × 1000 m = 10000 m = 1000000 cm.
   5. (ii) Number of revolutions = Total distance ÷ Distance in one revolution.
   6. (ii) Number of revolutions = 1000000 ÷ 176.
   7. (ii) Number of revolutions = 5681.8

   Answer: (i) 176 cm; (ii) 5681.8

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7. Ex 6.1 Q7

   Total perimeter of all the petals (π ≈ 22/7): (i) Fig. 6.15A, four petals in a 14 cm square, arcs centred at the midpoints of the sides; (ii) Fig. 6.15B, six petals in a regular hexagon of side 42 cm, arcs centred at the vertices.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. (i) Each petal runs from a corner of the square to its centre. Its two edges are arcs of radius 7 cm (half of 14 cm), centred at the midpoints of the two sides that meet at that corner.
   2. (i) From such a midpoint, the corner and the centre of the square are both 7 cm away and in perpendicular directions, so each arc is a quarter circle of radius 7 cm.
   3. (i) One quarter arc = ¼ × 2 × (22/7) × 7 = 11 cm. Four petals have 8 arcs: 8 × 11 = 88 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.

   4. (ii) Each petal runs from a vertex of the hexagon to its centre. Its two edges are arcs of radius 42 cm centred at the two neighbouring vertices.
   5. (ii) In a regular hexagon the centre is 42 cm from every vertex, so a neighbouring vertex, the petal's vertex and the centre form an equilateral triangle: each arc turns through 60°.
   6. (ii) One arc = (60/360) × 2 × (22/7) × 42 = (1/6) × 264 = 44 cm.
      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. (ii) Six petals have 12 arcs: 12 × 44 = 528 cm.

   Answer: (i) 88 cm; (ii) 528 cm

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8. Ex 6.1 Q8

   Two circles have perimeters in the ratio 5 : 4. Ratio of their radii?

   1. Let the radii of the two circles be r₁ and r₂.
   2. The perimeter (circumference) of a circle is given by the formula 2πr.
      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. The ratio of their perimeters is given as 5 : 4, so (2πr₁) / (2πr₂) = 5 / 4.
      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.

   4. Cancel the common factor 2π from the numerator and the denominator.
   5. The ratio simplifies to r₁ / r₂ = 5 / 4.
      why?

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

   6. Thus, the ratio of their radii is 5 : 4.

   Answer: 5 : 4

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