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CBSE Class 10 Maths Chapter 11 Areas Related to Circles Model Questions - 3 Marks - Part 2

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SECTION C — Short Answer Type Questions [3 Marks Each]

• 3 Marks Q26. The inner circumference of a circular track is $352\text{ m}$ and the track is $7\text{ m}$ wide everywhere. Find the cost of putting up a fence along the outer circumference at $\text{₹}2$ per meter.
  Solution:
  Inner circumference $2\pi r = 352\text{ m} \implies r = \frac{352 \times 7}{2 \times 22} = 56\text{ m}$.
  Width of the track $w = 7\text{ m}$.
  Outer radius $R = r + w = 56 + 7 = 63\text{ m}$.
  Outer circumference = $2\pi R = 2 \times \frac{22}{7} \times 63 = 396\text{ m}$.
  Cost of fencing outer circumference = $396 \times 2 = \text{₹}792$.
  Answer: $\text{₹}792$.
• 3 Marks Q27. A park is in the form of a circle of radius $14\text{ m}$. A path $7\text{ m}$ wide runs inside the park. Find the area of the path and the cost of gravelling it at $\text{₹}5$ per $\text{m}^2$.
  Solution:
  Outer radius $R = 14\text{ m}$, width of path $w = 7\text{ m}$.
  Inner radius $r = R - w = 14 - 7 = 7\text{ m}$.
  Area of the path = $\pi(R^2 - r^2) = \frac{22}{7}(14^2 - 7^2) = \frac{22}{7}(196 - 49) = \frac{22}{7} \times 147 = 462\text{ m}^2$.
  Cost of gravelling = $462 \times 5 = \text{₹}2310$.
  Answer: Area = $462\text{ m}^2$; Cost = $\text{₹}2310$.
• 3 Marks Q28. The difference between the outer and inner circumferences of a circular ring is $44\text{ cm}$. If the outer radius is $21\text{ cm}$, find the inner radius and the area of the ring (annulus).
  Solution:
  Outer radius $R = 21\text{ cm}$.
  Given difference in circumferences: $2\pi R - 2\pi r = 44 \implies 2\pi(21 - r) = 44$.
  $2 \times \frac{22}{7} \times (21 - r) = 44 \implies 21 - r = 7 \implies r = 14\text{ cm}$.
  Area of the ring = $\pi(R^2 - r^2) = \frac{22}{7}(21^2 - 14^2) = \frac{22}{7}(441 - 196) = \frac{22}{7} \times 245 = 770\text{ cm}^2$.
  Answer: Inner radius = $14\text{ cm}$; Area = $770\text{ cm}^2$.
• 3 Marks Q29. A bicycle wheel makes $5000$ revolutions in moving $11\text{ km}$. Find the diameter of the wheel and calculate the area swept by a spoke of half the length in $100$ revolutions.
  Solution:
  Distance traveled = $11\text{ km} = 1,100,000\text{ cm}$.
  Circumference = $\frac{1,100,000}{5000} = 220\text{ cm}$.
  Diameter $d = \frac{220}{\pi} = \frac{220 \times 7}{22} = 70\text{ cm}$ (radius $r = 35\text{ cm}$).
  Spoke length = half of radius = $17.5\text{ cm}$.
  Area swept in 1 revolution = $\pi (17.5)^2 = \frac{22}{7} \times 306.25 = 962.5\text{ cm}^2$.
  Area swept in $100$ revolutions = $100 \times 962.5 = 96,250\text{ cm}^2$.
  Answer: Diameter = $70\text{ cm}$; Area swept = $96,250\text{ cm}^2$.
• 3 Marks Q30. The short and long hands of a clock are $4\text{ cm}$ and $6\text{ cm}$ long respectively. Find the sum of the areas swept by their tips in $24\text{ hours}$.
  Solution:
  Short hand completes $2$ rotations in $24\text{ hours}$ ($r_1 = 4\text{ cm}$).
  Long hand completes $24$ rotations in $24\text{ hours}$ ($r_2 = 6\text{ cm}$).
  Area swept by short hand = $2 \times (\pi \times 4^2) = 32\pi\text{ cm}^2$.
  Area swept by long hand = $24 \times (\pi \times 6^2) = 864\pi\text{ cm}^2$.
  Total sum of areas = $32\pi + 864\pi = 896\pi = 896 \times 3.14 = 2813.44\text{ cm}^2$.
  Answer: $2813.44\text{ cm}^2$.
• 3 Marks Q31. If the perimeter and the area of a circle are numerically equal, find its radius. Using this radius, find the perimeter and area of a corresponding square inscribed in that circle.
  Solution:
  Perimeter of circle = Area of circle $\implies 2\pi r = \pi r^2 \implies r = 2\text{ units}$.
  Diameter of circle = $4\text{ units}$.
  For a square inscribed in the circle, the diagonal equals the diameter ($4\text{ units}$).
  Side of square $s = \frac{4}{\sqrt{2}} = 2\sqrt{2}\text{ units}$.
  Perimeter of square = $4s = 4 \times 2\sqrt{2} = 8\sqrt{2}\text{ units}$.
  Area of square = $s^2 = (2\sqrt{2})^2 = 8\text{ square units}$.
  Answer: Radius = $2$; Square Perimeter = $8\sqrt{2}$; Square Area = $8$.
• 3 Marks Q32. The area of a sector is $\frac{5}{18}$ of the area of the circle. Find the central angle of the sector and the length of its arc if the circle's radius is $21\text{ cm}$.
  Solution:
  Central angle $\theta = \frac{5}{18} \times 360^\circ = 100^\circ$.
  Radius $r = 21\text{ cm}$.
  Arc length = $\frac{100^\circ}{360^\circ} \times 2\pi(21) = \frac{5}{18} \times 2 \times \frac{22}{7} \times 21 = \frac{110}{3}\text{ cm} \approx 36.67\text{ cm}$.
  Answer: Central angle = $100^\circ$; Arc length = $36.67\text{ cm}$.
• 3 Marks Q33. If the radius of a circle is increased by $20\%$, find the percentage increase in its area and its circumference.
  Solution:
  New radius $r' = 1.20r$.
  Circumference ($C = 2\pi r$) scales linearly, so percentage increase in circumference = $20\%$.
  Area ($A = \pi r^2$) scales quadratically: $A' = \pi (1.20r)^2 = 1.44\pi r^2$.
  Percentage increase in area = $\frac{1.44A - A}{A} \times 100 = 44\%$.
  Answer: Area increase = $44\%$; Circumference increase = $20\%$.
• 3 Marks Q34. A wire when bent in the form of a square encloses an area of $484\text{ cm}^2$. If the same wire is bent into the form of a circle, find the area of the circle and compare it with the square's area.
  Solution:
  Area of square = $484\text{ cm}^2 \implies \text{side } s = 22\text{ cm}$.
  Perimeter of wire = $4 \times 22 = 88\text{ cm}$.
  Circumference of circle = $88\text{ cm} \implies 2\pi r = 88 \implies r = 14\text{ cm}$.
  Area of circle = $\pi r^2 = \frac{22}{7} \times 14^2 = 616\text{ cm}^2$.
  Comparison: Circle area ($616\text{ cm}^2$) is greater than square area ($484\text{ cm}^2$).
  Answer: Circle area = $616\text{ cm}^2$.
• 3 Marks Q35. Two circles touch internally. The sum of their areas is $116\pi\text{ cm}^2$ and the distance between their centers is $6\text{ cm}$. Find the radii of the two circles.
  Solution:
  Let radii be $R$ and $r$ ($R > r$).
  Circles touch internally $\implies R - r = 6 \implies R = r + 6$.
  Sum of areas: $\pi R^2 + \pi r^2 = 116\pi \implies R^2 + r^2 = 116$.
  $(r + 6)^2 + r^2 = 116 \implies 2r^2 + 12r + 36 = 116 \implies 2r^2 + 12r - 80 = 0$.
  $r^2 + 6r - 40 = 0 \implies (r + 10)(r - 4) = 0 \implies r = 4\text{ cm}$ (since radius > 0).
  $R = 4 + 6 = 10\text{ cm}$.
  Answer: Radii are $10\text{ cm}$ and $4\text{ cm}$.
• 3 Marks Q36. Find the area of a regular hexagon circumscribing a circle of radius $6\text{ cm}$.
  Solution:
  For a regular hexagon circumscribing a circle of inradius $r = 6\text{ cm}$, the side length $a = \frac{2r}{\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}\text{ cm}$.
  Area of the regular hexagon = $6 \times \left(\frac{\sqrt{3}}{4} a^2\right) = 6 \times \frac{\sqrt{3}}{4} (48) = 72\sqrt{3} \approx 72 \times 1.732 = 124.7\text{ cm}^2$.
  Answer: $124.7\text{ cm}^2$ (or $72\sqrt{3}\text{ cm}^2$).
• 3 Marks Q37. A pendulum of length $21\text{ cm}$ oscillates, describing an arc of length $33\text{ cm}$. Find the angle described by the pendulum at the center in degrees and the area of the sector formed.
  Solution:
  Radius $r = 21\text{ cm}$, arc length $s = 33\text{ cm}$.
  $\theta = \frac{s}{2\pi r} \times 360^\circ = \frac{33}{2 \times \frac{22}{7} \times 21} \times 360^\circ = \frac{33}{132} \times 360^\circ = 90^\circ$.
  Area of sector = $\frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 21^2 = \frac{1386}{4} = 346.5\text{ cm}^2$.
  Answer: Angle = $90^\circ$; Sector area = $346.5\text{ cm}^2$.
• 3 Marks Q38. A field is in the form of a circle of radius $28\text{ m}$. Find the cost of fencing the field at $\text{₹}15$ per meter and ploughing it at $\text{₹}2$ per $\text{m}^2$.
  Solution:
  Radius $r = 28\text{ m}$.
  Circumference = $2 \times \frac{22}{7} \times 28 = 176\text{ m}$.
  Cost of fencing = $176 \times 15 = \text{₹}2640$.
  Area = $\frac{22}{7} \times 28^2 = 2464\text{ m}^2$.
  Cost of ploughing = $2464 \times 2 = \text{₹}4928$.
  Answer: Fencing cost = $\text{₹}2640$; Ploughing cost = $\text{₹}4928$.
• 3 Marks Q39. The wheel of a car of diameter $80\text{ cm}$ makes complete revolutions at $66\text{ km/h}$. Find the number of revolutions each wheel makes in $10\text{ minutes}$ and the angular distance covered.
  Solution:
  Number of revolutions in $10\text{ minutes} = 4375$ (calculated similar to Q10).
  Angular distance covered = $4375 \times 360^\circ = 1,575,000^\circ$ (or $8750\pi\text{ radians}$).
  Answer: $4375$ revolutions; $1,575,000^\circ$ angular distance.
• 3 Marks Q40. A circular pond is of diameter $17.5\text{ m}$. It is surrounded by a path of width $3.5\text{ m}$ running outward. Find the area of the path and the cost of paving it at $\text{₹}12$ per $\text{m}^2$.
  Solution:
  Inner radius $r = \frac{17.5}{2} = 8.75\text{ m}$.
  Outer radius $R = 8.75 + 3.5 = 12.25\text{ m}$.
  Area of the path = $\pi(R^2 - r^2) = \frac{22}{7}(12.25^2 - 8.75^2) = \frac{22}{7}(150.0625 - 76.5625) = \frac{22}{7} \times 73.5 = 231\text{ m}^2$.
  Cost of paving = $231 \times 12 = \text{₹}2772$.
  Answer: Path area = $231\text{ m}^2$; Cost = $\text{₹}2772$.
• 3 Marks Q41. Find the area of the shaded design in a square $ABCD$ of side $14\text{ cm}$ where four semicircles are drawn with sides as diameters.
  Solution:
  Area of square = $14 \times 14 = 196\text{ cm}^2$.
  Radius of each semicircle = $\frac{14}{2} = 7\text{ cm}$.
  Area of four semicircles = $4 \times \left(\frac{1}{2} \pi r^2\right) = 2 \times \frac{22}{7} \times 49 = 308\text{ cm}^2$.
  Shaded region area = Area of 4 semicircles - Area of square = $308 - 196 = 112\text{ cm}^2$.
  Answer: $112\text{ cm}^2$.
• 3 Marks Q42. A square water tank has a side length of $10\text{ m}$. Four cows are tied at the four corners of the tank with ropes of length $3.5\text{ m}$ each. Find the total area that can be grazed by the cows and the ungrazed area of the tank field.
  Solution:
  Total grazed area (4 quadrants of radius $3.5\text{ m}$) = $\pi r^2 = \frac{22}{7} \times (3.5)^2 = 38.5\text{ m}^2$.
  Total area of the tank = $10 \times 10 = 100\text{ m}^2$.
  Ungrazed area = $100 - 38.5 = 61.5\text{ m}^2$.
  Answer: Grazed area = $38.5\text{ m}^2$; Ungrazed area = $61.5\text{ m}^2$.
• 3 Marks Q43. A chord of a circle of radius $28\text{ cm}$ subtends an angle of $60^\circ$ at the center. Find the area of the minor segment and the length of the chord.
  Solution:
  Since $\theta = 60^\circ$ and radii are equal, the triangle is equilateral, so length of the chord = $28\text{ cm}$.
  Area of sector = $\frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 28^2 = \frac{1}{6} \times 2464 = 410.67\text{ cm}^2$.
  Area of equilateral triangle = $\frac{\sqrt{3}}{4} \times 28^2 = 196\sqrt{3} \approx 196 \times 1.732 = 339.47\text{ cm}^2$.
  Area of minor segment = $410.67 - 339.47 = 71.2\text{ cm}^2$.
  Answer: Chord length = $28\text{ cm}$; Segment area = $71.2\text{ cm}^2$.
• 3 Marks Q44. A track is $7\text{ m}$ wide and the inner circumference is $440\text{ m}$. Find the area of the track and the cost of leveling it at $\text{₹}3$ per $\text{m}^2$.
  Solution:
  Inner circumference $2\pi r = 440\text{ m} \implies r = 70\text{ m}$.
  Outer radius $R = 70 + 7 = 77\text{ m}$.
  Area of the track = $\pi(R^2 - r^2) = \frac{22}{7}(77^2 - 70^2) = \frac{22}{7}(5929 - 4900) = \frac{22}{7} \times 1029 = 3234\text{ m}^2$.
  Cost of leveling = $3234 \times 3 = \text{₹}9702$.
  Answer: Area = $3234\text{ m}^2$; Cost = $\text{₹}9702$.
• 3 Marks Q45. In a circle of radius $14\text{ cm}$, an arc subtends an angle of $45^\circ$ at the center. Find: (i) length of the arc, (ii) area of the sector, and (iii) area of the segment formed by the corresponding chord.
  Solution:
  (i) Arc length = $\frac{45^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 14 = \frac{1}{8} \times 88 = 11\text{ cm}$.
  (ii) Sector area = $\frac{45^\circ}{360^\circ} \times \frac{22}{7} \times 14^2 = \frac{1}{8} \times 616 = 77\text{ cm}^2$.
  (iii) Triangle area = $\frac{1}{2} r^2 \sin(45^\circ) = \frac{1}{2} (196) \times 0.7071 = 69.3\text{ cm}^2$.
  Segment area = $77 - 69.3 = 7.7\text{ cm}^2$.
  Answer: (i) $11\text{ cm}$; (ii) $77\text{ cm}^2$; (iii) $7.7\text{ cm}^2$.
• 3 Marks Q46. A copper wire bent in the shape of an equilateral triangle encloses an area of $121\sqrt{3}\text{ cm}^2$. If the same wire is bent into a circle, find the area of the circle.
  Solution:
  Area of triangle = $\frac{\sqrt{3}}{4} a^2 = 121\sqrt{3} \implies a^2 = 484 \implies a = 22\text{ cm}$.
  Perimeter of wire = $3 \times 22 = 66\text{ cm}$.
  Circumference of circle = $66\text{ cm} \implies 2\pi r = 66 \implies r = \frac{66 \times 7}{2 \times 22} = 10.5\text{ cm}$.
  Area of circle = $\pi r^2 = \frac{22}{7} \times (10.5)^2 = 346.5\text{ cm}^2$.
  Answer: $346.5\text{ cm}^2$.
• 3 Marks Q47. Two circles touch externally at point $C$. Their radii are $4\text{ cm}$ and $9\text{ cm}$. Find the area of the region enclosed between the common tangents and the circles.
  Solution:
  Radii $R = 9\text{ cm}$, $r = 4\text{ cm}$. Distance between centers $d = R + r = 13\text{ cm}$.
  The region is bounded by the external tangent and the circular arcs. Using the trapezoid method formed by the centers and tangents, the area equals the area of the right trapezoid minus the area of the two sector quadrants: $\text{Area} = 2\sqrt{Rr} \times \frac{R+r}{2} - \left(\frac{\pi R^2}{4} + \frac{\pi r^2}{4}\right) \approx 78 - (\approx 48.7) = 29.3\text{ cm}^2$ (approx).
  Answer: $\approx 29.3\text{ cm}^2$.
• 3 Marks Q48. A circular field has a perimeter of $352\text{ m}$. A road $7\text{ m}$ wide runs around it from the outside. Find the cost of gravelling the road at $\text{₹}4$ per $\text{m}^2$.
  Solution:
  Inner circumference $2\pi r = 352\text{ m} \implies r = 56\text{ m}$.
  Outer radius $R = 56 + 7 = 63\text{ m}$.
  Area of the road = $\pi(R^2 - r^2) = \frac{22}{7}(63^2 - 56^2) = \frac{22}{7}(3969 - 3136) = \frac{22}{7} \times 833 = 2618\text{ m}^2$.
  Cost of gravelling = $2618 \times 4 = \text{₹}10,472$.
  Answer: $\text{₹}10,472$.
• 3 Marks Q49. The radii of the inner and outer boundaries of a circular running track are $70\text{ m}$ and $77\text{ m}$ respectively. Find the area of the track and calculate the extra distance run by an athlete on the outer boundary compared to the inner boundary in one full lap.
  Solution:
  Area of the track = $\pi(77^2 - 70^2) = \frac{22}{7}(5929 - 4900) = \frac{22}{7} \times 1029 = 3234\text{ m}^2$.
  Extra distance = Outer circumference - Inner circumference = $2\pi(77) - 2\pi(70) = 2\pi(7) = 2 \times \frac{22}{7} \times 7 = 44\text{ m}$.
  Answer: Area = $3234\text{ m}^2$; Extra distance = $44\text{ m}$.
• 3 Marks Q50. An equilateral triangle has side $12\text{ cm}$. With each vertex as center, circles are drawn with radius equal to half the side length ($6\text{ cm}$). Find the area of the triangle outside the three circular sectors. ($\text{Use }\sqrt{3} = 1.732$).
  Solution:
  Area of equilateral triangle = $\frac{\sqrt{3}}{4} \times 12^2 = 36\sqrt{3} = 36 \times 1.732 = 62.352\text{ cm}^2$.
  Combined area of three sectors at the vertices ($60^\circ$ each) = $\frac{180^\circ}{360^\circ} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 6^2 = \frac{396}{7} \approx 56.57\text{ cm}^2$.
  Area outside the sectors = $62.352 - 56.57 = 5.782\text{ cm}^2$.
  Answer: $5.782\text{ cm}^2$.

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