Class 10 Maths Chapter 10 Circles Practice Paper 1

PRACTICE PAPER 01 (2025-26) CHAPTER 11: AREAS RELATED TO CIRCLES SUBJECT: MATHEMATICS MAX. MARKS: 40 CLASS: X DURATION: 1½ hrs

General Instructions:

1. All questions are compulsory.

2. This question paper contains 20 questions divided into five Sections A, B, C, D and E.

3. Section A comprises of 10 MCQs of 1 mark each. Section B comprises of 4 questions of 2 marks each.

Section C comprises of 3 questions of 3 marks each. Section D comprises of 1 question of 5 marks and

Section E comprises of 2 Case Study Based Questions of 4 marks each.

4. There is no overall choice.

5. Use of Calculators is not permitted. SECTION – A Questions 1 to 10 carry 1 mark each.

1. The circumference of a circle is 44 cm. What is its radius?
(a) 7 cm
(b) 14 cm
(c) 21 cm
(d) 22 cm

2. If the radius of a circle is tripled, by what factor does the area increase?
(a) 3
(b) 6
(c) 9
(d) 27

3. A wheel has a diameter of 70 cm. How many complete revolutions will it take to cover 1.1 km?
(a) 400
(b) 500
(c) 600
(d) 700

4. The area of a sector of angle θ° in a circle of radius r is:
(a) θ/180 × πr²
(b) θ/360 × πr²
(c) θ/90 × πr²
(d) θ/720 × πr²

5. If the perimeter of a semicircle is 36 cm, then its radius is: [Use π = 22/7]
(a) 3.5 cm
(b) 7 cm
(c) 10.5 cm
(d) 14 cm

6. The length of the minute hand of a clock is 21 cm. What is the area swept by it in 10 minutes?
(a) 115.5 cm²
(b) 231 cm²
(c) 462 cm²
(d) 693 cm²

7. A circular park of radius 20 m has a path of width 5 m running around it on the outside. The area of the path is:
(a) 225π m²
(b) 275π m²
(c) 325π m²
(d) 400π m²

8. If the area of a circle is 154 cm², then its circumference is: [Use π = 22/7]
(a) 22 cm
(b) 44 cm
(c) 66 cm
(d) 88 cm In the following questions 9 and 10, a statement of assertion
(a) is followed by a statement of reason (R). Mark the correct choice as:
(a) Both assertion
(a) and reason (R) are true and reason (R) is the correct explanation of assertion
(a) .
(b) Both assertion
(a) and reason (R) are true but reason (R) is not the correct explanation of assertion
(a) .
(c) Assertion
(a) is true but reason (R) is false.

(d) Assertion
(a) is false but reason (R) is true.

9. Assertion
(a) : The area of a circle of radius 7 cm is equal to the total area of two circles of radii 3.5 cm and 3.5√2 cm. Reason (R): If two circles have radii r₁ and r₂, then the sum of their areas equals πr₁² + πr₂².

10. Assertion
(a) : If a square of side 14 cm is inscribed in a circle, then the area of the circle is 308 cm². Reason (R): When a square is inscribed in a circle, the diagonal of the square equals the diameter of the circle. SECTION – B Questions 11 to 14 carry 2 marks each.

11. A horse is tied to a pole with a 28 m long rope. Find the area over which the horse can graze. [Use π = 22/7]

12. Find the area of a quadrant of a circle whose circumference is 88 cm. [Use π = 22/7]

13. The wheels of a locomotive have diameter 2.1 m. What distance will the locomotive travel when its wheels make 500 revolutions? [Use π = 22/7]

14. A sector of 56° is cut off from a circle of radius 18 cm. Find the length of the arc and the area of the sector. [Use π = 22/7] SECTION – C Questions 15 to 17 carry 3 marks each.

15. A copper wire when bent in the form of a square encloses an area of 484 cm². If the same wire is bent in the form of a circle, find the area enclosed by it. [Use π = 22/7]

16. Four equal circles are described about the four corners of a square so that each circle touches two other circles externally. If each side of the square is 14 cm, find the area enclosed between the circumferences of the circles. [Use π = 22/7]

17. A circular pond is 21 m in diameter. It is surrounded by a 3.5 m wide path. Find the cost of gravelling the path at ₹25 per m². [Use π = 22/7] SECTION – D Question 18 carries 5 marks.

18. A park is in the form of a rectangle 120 m × 100 m. At the centre of the park, there is a circular lawn. The area of the park excluding the lawn is 11304 m². Find the radius of the circular lawn. Further, find the cost of fencing the circular lawn at the rate of ₹50 per metre. [Use π = 3.14] SECTION – E (Case Study Based Questions) Questions 19 to 20 carry 4 marks each.

19. Flower Garden Design: A landscape designer is creating a circular flower garden with a radius of 14 m. The garden is divided into four equal sectors, each planted with a different type of flower. A circular fountain with a radius of 3.5 m is placed at the center of the garden. Based on the above information, answer the following questions:
(a) What is the area of one sector of the garden (excluding the fountain)? (1 mark)
(b) If a walking path of width 1.4 m is created around the outer boundary of the garden, find the area of the path.

(2 marks) OR What is the perimeter of one sector of the garden (including the arc and two radii)? (2 marks)
(c) Find the cost of planting flowers in one sector (excluding fountain area) at the rate of ₹80 per m². (1 mark)

20. Pizza Serving Calculation: A pizza restaurant serves two sizes of pizzas: Regular pizza: diameter 21 cm Large pizza: diameter 35 cm The restaurant cuts each pizza into 8 equal slices. A customer wants to compare the sizes to make a better choice. Based on the above information, answer the following questions:
(a) Find the area of one slice of the regular pizza. (1 mark)
(b) How many regular pizza slices would equal the area of one large pizza slice? (2 marks)
(c) If a regular pizza costs ₹250 and a large pizza costs ₹400, which is more economical per square cm? (1 mark) DETAILED ANSWER KEY - PRACTICE PAPER 01

SECTION A - ANSWERS

1. Answer:
(a) 7 cm Given: Circumference = 44 cm We know, Circumference = 2πr 44 = 2 × (22/7) × r 44 = (44/7) × r r = 44 × 7/44 = 7 cm Answer:
(a) 7 cm

2. Answer:
(c) 9 Original area = πr² New radius = 3r New area = π(3r)² = π × 9r² = 9πr² Factor of increase = 9πr²/πr² = 9 Answer:
(c) 9

3. Answer:
(b) 500 Diameter = 70 cm, so radius = 35 cm Distance in one revolution = Circumference = 2πr = 2 × (22/7) × 35 = 220 cm = 2.2 m Total distance = 1.1 km = 1100 m Number of revolutions = 1100/2.2 = 500 Answer:
(b) 500

4. Answer:
(b) θ/360 × πr² Area of full circle = πr² (corresponds to 360°) Area of sector with angle θ° = (θ/360) × πr² Answer:
(b) θ/360 × πr²

5. Answer:
(b) 7 cm Perimeter of semicircle = πr + 2r = r(π + 2) 36 = r(22/7 + 2) 36 = r(22/7 + 14/7) 36 = r(36/7) r = 36 × 7/36 = 7 cm Answer:
(b) 7 cm

6. Answer:
(b) 231 cm² Length of minute hand = radius = 21 cm In 60 minutes, minute hand completes 360° In 10 minutes, angle covered = (10/60) × 360° = 60° Area swept = (θ/360) × πr² = (60/360) × (22/7) × 21² = (1/6) × (22/7) × 441 = (22 × 441)/(7 × 6) = 9702/42 = 231 cm² Answer:
(b) 231 cm²

7. Answer:
(a) 225π m² Inner radius = 20 m Outer radius = 20 + 5 = 25 m Area of path = π(R² - r²) = π(25² - 20²) = π(625 - 400) = 225π m² Answer:
(a) 225π m²

8. Answer:
(b) 44 cm Area = 154 cm² πr² = 154 (22/7) × r² = 154 r² = 154 × 7/22 = 49 r = 7 cm Circumference = 2πr = 2 × (22/7) × 7 = 44 cm Answer:
(b) 44 cm

9. Answer:
(a) Area of circle with radius 7 cm = π(7)² = 49π Area of circle with radius 3.5 cm = π(3.5)² = 12.25π Area of circle with radius 3.5√2 cm = π(3.5√2)² = π × 12.25 × 2 = 24.5π Sum = 12.25π + 24.5π = 36.75π ≠ 49π So Assertion is false. Reason is true as it correctly states the formula. Answer:
(d) Assertion
(a) is false but reason (R) is true

10. Answer:
(a) Side of square = 14 cm Diagonal = 14√2 cm Diameter of circle = diagonal = 14√2 cm Radius = 7√2 cm Area of circle = π(7√2)² = π × 49 × 2 = 98π = 98 × (22/7) = 308 cm² Both assertion and reason are true, and reason correctly explains assertion. Answer:
(a)

SECTION B - ANSWERS

11. Solution: The horse can graze in a circular area with radius = 28 m Area = πr² = (22/7) × 28² = (22/7) × 784 = 22 × 112 = 2464 m² The horse can graze over an area of 2464 m²

12. Solution: Circumference = 88 cm 2πr = 88 2 × (22/7) × r = 88 r = (88 × 7)/(2 × 22) = 14 cm Area of quadrant = (1/4) × πr² = (1/4) × (22/7) × 14² = (1/4) × (22/7) × 196 = (22 × 196)/(7 × 4) = 4312/28 = 154 cm² Area of quadrant = 154 cm²

13. Solution: Diameter = 2.1 m, so radius = 1.05 m Distance in one revolution = Circumference = 2πr = 2 × (22/7) × 1.05 = 2 × (22/7) × (105/100) = (2 × 22 × 105)/(7 × 100) = 4620/700 = 6.6 m Distance in 500 revolutions = 500 × 6.6 = 3300 m = 3.3 km The locomotive will travel 3.3 km or 3300 m

14. Solution: Radius = 18 cm, Angle θ = 56° Length of arc = (θ/360) × 2πr = (56/360) × 2 × (22/7) × 18 = (56 × 2 × 22 × 18)/(360 × 7) = 44352/2520 = 17.6 cm Area of sector = (θ/360) × πr² = (56/360) × (22/7) × 18² = (56/360) × (22/7) × 324 = (56 × 22 × 324)/(360 × 7) = 399168/2520 = 158.4 cm² Length of arc = 17.6 cm, Area of sector = 158.4 cm²

SECTION C - ANSWERS

15. Solution: Area of square = 484 cm² Side of square = √484 = 22 cm Perimeter of square = 4 × 22 = 88 cm This becomes the circumference of circle: 2πr = 88 2 × (22/7) × r = 88 r = (88 × 7)/(2 × 22) = 14 cm Area of circle = πr² = (22/7) × 14² = (22/7) × 196 = 22 × 28 = 616 cm² Area enclosed by circle = 616 cm²

16. Solution: Side of square = 14 cm Area of square = 14² = 196 cm² Each circle has radius = 14/2 = 7 cm Each circle at corner contributes a quadrant inside the square Total area of 4 quadrants = 4 × (1/4) × πr² = πr² = (22/7) × 7² = (22/7) × 49 = 22 × 7 = 154 cm² Area enclosed between circumferences = Area of square - Area of quadrants = 196 - 154 = 42 cm² Area enclosed between circles = 42 cm²

17. Solution: Diameter of pond = 21 m, so radius = 10.5 m Width of path = 3.5 m Outer radius = 10.5 + 3.5 = 14 m Area of path = π(R² - r²) = (22/7) × (14² - 10.5²) = (22/7) × (196 - 110.25) = (22/7) × 85.75 = (22 × 85.75)/7 = 1886.5/7 = 269.5 m² Cost of gravelling = 269.5 × 25 = ₹6737.50 Cost of gravelling the path = ₹6737.50

SECTION D - ANSWER

18. Solution: Dimensions of park = 120 m × 100 m Area of park = 120 × 100 = 12000 m² Area of park excluding lawn = 11304 m² Area of circular lawn = 12000 - 11304 = 696 m² πr² = 696 3.14 × r² = 696 r² = 696/3.14 r² = 221.66 r ≈ 14.89 m Circumference of lawn = 2πr = 2 × 3.14 × 14.89 = 93.51 m Cost of fencing = 93.51 × 50 = ₹4675.50 Radius of circular lawn ≈ 14.89 m Cost of fencing = ₹4675.50

SECTION E - ANSWERS

19. Solution:
(a) Radius of garden = 14 m Radius of fountain = 3.5 m Area of one sector of garden = (1/4) × π × 14² = (1/4) × (22/7) × 196 = 154 m² Area of fountain in one sector = (1/4) × π × 3.5² = (1/4) × (22/7) × 12.25 = 9.625 m² Area of one sector excluding fountain = 154 - 9.625 = 144.375 m²
(a) Area = 144.375 m²
(b) Outer radius = 14 + 1.4 = 15.4 m Area of path = π(R² - r²) = (22/7) × (15.4² - 14²) = (22/7) × (237.16 - 196) = (22/7) × 41.16 = 129.36 m²
(b) Area of path = 129.36 m² OR Arc length of one sector = (1/4) × 2πr = (1/4) × 2 × (22/7) × 14 = 22 m Perimeter = Arc length + 2 radii = 22 + 14 + 14 = 50 m OR: Perimeter = 50 m
(c) Cost = 144.375 × 80 = ₹11,550
(c) Cost = ₹11,550

20. Solution:
(a) Regular pizza: diameter = 21 cm, radius = 10.5 cm Total area = π × 10.5² = (22/7) × 110.25 = 346.5 cm² Area of one slice = 346.5/8 = 43.3125 cm²
(a) Area of one regular slice = 43.3125 cm²
(b) Large pizza: diameter = 35 cm, radius = 17.5 cm Total area = π × 17.5² = (22/7) × 306.25 = 962.5 cm² Area of one large slice = 962.5/8 = 120.3125 cm² Number of regular slices = 120.3125/43.3125 ≈ 2.78 ≈ 3 slices
(b) About 3 regular slices equal one large slice
(c) Regular pizza: Cost per cm² = 250/346.5 = ₹0.72 per cm² Large pizza: Cost per cm² = 400/962.5 = ₹0.42 per cm²
(c) Large pizza is more economical at ₹0.42 per cm²