CBSE Class 10 Areas related to circles Sure Shot Questions

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PRACTICE QUESTIONS
CLASS X: CHAPTER - 12
AREAS RELATED TO CIRCLES

1. If the perimeter of a semicircular protractor is 36 cm, find its diameter.

Answer:
1. Perimeter of a semicircle = πr + 2r = r(π + 2).
2. So r × \( \frac{36}{7} \) = 36, because π + 2 = \( \frac{22}{7} \) + 2 = \( \frac{36}{7} \).
3. r = 36 × \( \frac{7}{36} \) = 7 cm.
4. Diameter = 2r = 14 cm.

Teacher's Note:
a) Perimeter of a semicircle includes the straight diameter, so add 2r to the arc πr.
b) π = 22/7 is used because no value of π is given in the question.

2. A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.

Answer:
1. Distance covered in one revolution = circumference = 11 km ÷ 5000 = 11000 m ÷ 5000 = 2.2 m = 220 cm.
2. 2πr = 220, so 2 × \( \frac{22}{7} \) × r = 220.
3. r = 220 × \( \frac{7}{44} \) = 35 cm.
4. Diameter = 2 × 35 = 70 cm.

Teacher's Note:
a) In one revolution a wheel covers a distance equal to its circumference.
b) Convert all lengths to the same unit before dividing.

3. The diameter of the wheels of a bus is 140 cm. How many revolutions per minute must a wheel make in order to move at a speed of 66 km per hour?

Answer:
1. Radius = 70 cm, so circumference = 2 × \( \frac{22}{7} \) × 70 = 440 cm.
2. Speed = 66 km/h = 66 × 1000 ÷ 60 m/min = 1100 m/min = 110000 cm/min.
3. Revolutions per minute = 110000 ÷ 440 = 250.

Teacher's Note:
a) Revolutions = total distance ÷ circumference.
b) Convert km per hour to cm per minute before dividing.
c) π = 22/7 is used because no value of π is given in the question.

4. Two circles touch externally. The sum of their areas is 130 sq. cm and the distance between their centres is 14 cm. Find the radii of the circles.

Answer:
1. Let the radii be r₁ and r₂. For circles touching externally, r₁ + r₂ = 14.
2. The sum of the areas is 130π, so r₁² + r₂² = 130.
3. (r₁ + r₂)² = r₁² + r₂² + 2r₁r₂, so 196 = 130 + 2r₁r₂ and r₁r₂ = 33.
4. r₁ and r₂ are the roots of t² - 14t + 33 = 0, that is (t - 11)(t - 3) = 0.
5. The radii are 11 cm and 3 cm.

Teacher's Note:
a) The printed sum of areas 130 sq. cm cannot be correct: without π the radii would not be real. The sum of the areas must be 130π sq. cm, and the answer is worked on that basis.
b) For external contact the distance between the centres is the sum of the radii.

5. Two circles touch internally. The sum of their areas is 116 sq. cm and the distance between their centres is 6 cm. Find the radii of the circles.

Answer:
1. Let the radii be r₁ and r₂ with r₁ > r₂. For circles touching internally, r₁ - r₂ = 6.
2. The sum of the areas is 116π, so r₁² + r₂² = 116.
3. (r₁ - r₂)² = r₁² + r₂² - 2r₁r₂, so 36 = 116 - 2r₁r₂ and r₁r₂ = 40.
4. (r₁ + r₂)² = 116 + 80 = 196, so r₁ + r₂ = 14.
5. Solving r₁ + r₂ = 14 and r₁ - r₂ = 6 gives r₁ = 10 cm and r₂ = 4 cm.

Teacher's Note:
a) The printed sum of areas 116 sq. cm is taken as 116π sq. cm; otherwise no pair of radii exists.
b) For internal contact the distance between the centres is the difference of the radii.

6. A paper is in the form of a rectangle ABCD in which AB = 18 cm and BC = 14 cm. A semicircular portion with BC as diameter is cut off. Find the area of the remaining paper.

Answer:
1. Area of the rectangle = 18 × 14 = 252 cm².
2. The semicircle has diameter BC = 14 cm, so its radius = 7 cm.
3. Area of the semicircle = \( \frac{1}{2} \) × \( \frac{22}{7} \) × 7 × 7 = 77 cm².
4. Remaining area = 252 - 77 = 175 cm².

Teacher's Note:
a) The diameter of the semicircle is the side BC, so the radius is half of 14 cm.
b) Subtract the cut-off semicircle from the rectangle.

7. A square ABCD is inscribed in a circle of radius r. Find the area of the square.

Answer:
1. The diagonal of the inscribed square is a diameter of the circle, so the diagonal = 2r.
2. Area of a square = \( \frac{1}{2} \) × (diagonal)².
3. Area = \( \frac{1}{2} \) × (2r)² = 2r².

Teacher's Note:
a) The diagonal of a square inscribed in a circle equals the diameter.
b) Area of a square in terms of its diagonal d is d²/2.

8. Find the area of a right-angled triangle, if the radius of its circumcircle is 2.5cm and the altitude drawn to the hypotenuse is 2cm long.

Answer:
1. The circumcentre of a right-angled triangle is the midpoint of the hypotenuse, so the hypotenuse = 2 × 2.5 = 5 cm.
2. Area = \( \frac{1}{2} \) × hypotenuse × altitude = \( \frac{1}{2} \) × 5 × 2 = 5 cm².

Teacher's Note:
a) The hypotenuse is the diameter of the circumcircle of a right-angled triangle.
b) Any side can be taken as the base: here the hypotenuse with its altitude.

9. A steel wire, bent in the form of a square, encloses an area of 121 sq. cm. The same wire is bent in the form of a circle. Find the area of the circle.

Answer:
1. Side of the square = √121 = 11 cm, so the length of the wire = 4 × 11 = 44 cm.
2. Circumference of the circle = 44 cm, so 2 × \( \frac{22}{7} \) × r = 44 and r = 7 cm.
3. Area of the circle = \( \frac{22}{7} \) × 7 × 7 = 154 cm².

Teacher's Note:
a) The length of the wire stays the same: the perimeter of the square equals the circumference of the circle.
b) π = 22/7 is used because no value of π is given in the question.

10. A wire is looped in the form of a circle of radius 28 cm. It is rebent into a square form. Determine the length of the side of the square.

Answer:
1. Length of the wire = circumference = 2 × \( \frac{22}{7} \) × 28 = 176 cm.
2. Perimeter of the square = 176 cm, so 4 × side = 176.
3. Side of the square = 44 cm.

Teacher's Note:
a) The wire length is unchanged when it is rebent.
b) π = 22/7 is used because no value of π is given in the question.

11. A circular park, 42 m diameter, has a path 3.5 m wide running round it on the outside. Find the cost of gravelling the path at Rs. 4 per m².

Answer:
1. Radius of the park r = 21 m. Radius with the path R = 21 + 3.5 = 24.5 m.
2. Area of the path = π(R² - r²) = \( \frac{22}{7} \) × (24.5² - 21²) = \( \frac{22}{7} \) × (24.5 + 21)(24.5 - 21).
3. = \( \frac{22}{7} \) × 45.5 × 3.5 = 500.5 m².
4. Cost = 500.5 × 4 = Rs. 2002.

Teacher's Note:
a) Area of a path round a circle = π(R² - r²) = π(R + r)(R - r).
b) Multiply the area by the rate to get the cost.

12. A road, which is 7m wide, surrounds a circular park whose circumference is 352m. Find the area of the road.

Answer:
1. 2πr = 352, so r = 352 × \( \frac{7}{44} \) = 56 m.
2. Outer radius R = 56 + 7 = 63 m.
3. Area of the road = π(R² - r²) = \( \frac{22}{7} \) × (63 + 56)(63 - 56) = \( \frac{22}{7} \) × 119 × 7 = 2618 m².

Teacher's Note:
a) First find the radius of the park from its circumference.
b) The road is a circular ring, so use π(R² - r²).

13. A racetrack is in the form of a ring whose inner and outer circumference are 437 m and 503 m respectively. Find the width of the track and also it area.

Answer:
1. 2πr = 437 and 2πR = 503, so 2π(R - r) = 66.
2. Width = R - r = 66 ÷ (2 × \( \frac{22}{7} \)) = 66 × \( \frac{7}{44} \) = 10.5 m.
3. Area = π(R² - r²) = π(R + r)(R - r). Since 2π(R + r) = 437 + 503 = 940, π(R + r) = 470.
4. Area = 470 × 10.5 = 4935 m².

Teacher's Note:
a) The width of a ring track is R - r, found from the difference of the two circumferences.
b) Use π(R + r) = half of the sum of the circumferences to avoid finding R and r separately.

14. From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)

Answer:
1. Area of the remaining sheet = π(4² - 3²) = 3.14 × (16 - 9).
2. = 3.14 × 7 = 21.98 cm².

Teacher's Note:
a) The remaining sheet is the difference of the two circular areas.
b) Use π = 3.14 as stated.

15. Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs Rs 15. (Take π = 3.14)

Answer:
1. Length of the lace = circumference = πd = 3.14 × 1.5 = 4.71 m.
2. Cost = 4.71 × 15 = Rs. 70.65.

Teacher's Note:
a) The lace goes along the edge, so the circumference is needed.
b) Cost = length × rate.

16. A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14)

Answer:
1. Area of the square = 6 × 6 = 36 cm².
2. Area of the circle = 3.14 × 2 × 2 = 12.56 cm².
3. Left over area = 36 - 12.56 = 23.44 cm².

Teacher's Note:
a) Left over area = area of the square - area of the circle.
b) Check that the circle (diameter 4 cm) fits inside the square of side 6 cm.

17. The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take π = 3.14)

Answer:
1. 2πr = 31.4, so r = 31.4 ÷ (2 × 3.14) = 5 cm.
2. Area = πr² = 3.14 × 5 × 5 = 78.5 cm².

Teacher's Note:
a) Find the radius first, then the area.
b) Use π = 3.14 throughout.

18. The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of this garden

Answer:
1. The total length is 20 m and the width is 7 m, so each semicircle has radius 3.5 m.
2. Length of the rectangular part = 20 - 7 = 13 m.
3. Area = rectangle + two semicircles = 13 × 7 + \( \frac{22}{7} \) × 3.5 × 3.5 = 91 + 38.5 = 129.5 m².
4. Perimeter = 2 × 13 + 2 × \( \frac{22}{7} \) × 3.5 = 26 + 22 = 48 m.

Teacher's Note:
a) Two semicircular ends make one full circle of diameter 7 m.
b) The perimeter has only the two straight sides and the two semicircular arcs.
c) π = 22/7 is used because no value of π is given in the question.

19. From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1cm are removed. (as shown in the right sided adjoining figure). Find the area of the remaining sheet.

Answer:
1. Area of the card sheet = \( \frac{22}{7} \) × 14 × 14 = 616 cm².
2. Area of the two small circles = 2 × \( \frac{22}{7} \) × 3.5 × 3.5 = 77 cm².
3. Area of the rectangle = 3 × 1 = 3 cm².
4. Remaining area = 616 - 77 - 3 = 536 cm².

Teacher's Note:
a) Subtract the areas of all the removed parts from the area of the big circle.
b) π = 22/7 is used because no value of π is given in the question.

20. A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)

Answer:
1. Radius of the flower bed r = 33 m. Radius with the path R = 33 + 4 = 37 m.
2. Area of the path = π(R² - r²) = 3.14 × (37 + 33)(37 - 33) = 3.14 × 70 × 4.
3. = 3.14 × 280 = 879.2 m².

Teacher's Note:
a) The path is a ring, so its area is the difference of two circular areas.
b) Always convert the diameter into a radius first.

21. Find the circumference of the inner and the outer circles, shown in the right sided adjoining figure? (Take π = 3.14)

Answer:
1. From the figure, the inner circle has radius 10 m and the outer circle has radius 19 m.
2. Inner circumference = 2 × 3.14 × 10 = 62.8 m.
3. Outer circumference = 2 × 3.14 × 19 = 119.32 m.

Teacher's Note:
a) Both radii are measured from the common centre.
b) Use C = 2πr for each circle separately.

22. Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square?

Answer:
1. Circumference = 44 cm, so 2 × \( \frac{22}{7} \) × r = 44 and r = 7 cm.
2. Area of the circle = \( \frac{22}{7} \) × 7 × 7 = 154 cm².
3. Perimeter of the square = 44 cm, so side = 11 cm and its area = 121 cm².
4. 154 > 121, so the circle encloses more area.

Teacher's Note:
a) For a fixed perimeter the circle encloses the greatest area.
b) π = 22/7 is used because no value of π is given in the question.

23. A circular flower garden has an area of 314 m². A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? (Take π = 3.14)

Answer:
1. πr² = 314, so r² = 314 ÷ 3.14 = 100 and r = 10 m.
2. The sprinkler covers a radius of 12 m, which is more than 10 m.
3. Yes, the sprinkler will water the entire garden.

Teacher's Note:
a) Compare the radius of the garden with the reach of the sprinkler.
b) Since the sprinkler is at the centre, covering 10 m in every direction is enough.

24. How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = \( \frac{22}{7} \))

Answer:
1. Circumference = 2 × \( \frac{22}{7} \) × 28 = 176 cm.
2. Distance = 352 m = 35200 cm.
3. Number of rotations = 35200 ÷ 176 = 200.

Teacher's Note:
a) Number of rotations = total distance ÷ circumference.
b) Change metres into centimetres before dividing.

25. Three horses are tethered with 7 m long ropes at the three corners of a triangular field having sides 20m, 34 m and 42 m. Find the area of the plot which can be grazed by the horses. Also, find the area of the plot, which remains ungrazed.

Answer:
1. The three sectors have the same radius 7 m, and the angles at the corners add up to 180°.
2. Grazed area = \( \frac{180}{360} \) × \( \frac{22}{7} \) × 7 × 7 = 77 m².
3. Semi-perimeter s = (20 + 34 + 42) ÷ 2 = 48 m.
4. Area of the field = √(48 × 28 × 14 × 6) = √112896 = 336 m².
5. Ungrazed area = 336 - 77 = 259 m².

Teacher's Note:
a) The angle sum of a triangle is 180°, so the three sectors together make a semicircle.
b) Use Heron's formula for the area of the field.

26. Find the area of a ΔCAB with ∠ACB = 120° & CA = CB = 18 cm.

Answer:
1. Area = \( \frac{1}{2} \) × CA × CB × sin 120° = \( \frac{1}{2} \) × 18 × 18 × \( \frac{\sqrt{3}}{2} \).
2. = 81√3 cm² ≈ 140.30 cm².
3. Alternatively, the altitude from C bisects ∠C: height = 18 cos 60° = 9 cm and half base = 18 sin 60° = 9√3 cm, so area = \( \frac{1}{2} \) × 18√3 × 9 = 81√3 cm².

Teacher's Note:
a) The area of a triangle with two sides and the included angle is \( \frac{1}{2} \)ab sin C.
b) This triangle is needed in the next two questions to find the area of the segment.

27. Find the area of sector of angle 120° and radius 18 cm.

Answer:
1. Area of the sector = \( \frac{\theta}{360} \) × πr² = \( \frac{120}{360} \) × \( \frac{22}{7} \) × 18 × 18.
2. = \( \frac{2376}{7} \) cm² ≈ 339.43 cm².

Teacher's Note:
a) Use sector area = \( \frac{\theta}{360} \)πr² with θ in degrees.
b) π = 22/7 is used because no value of π is given in the question.

28. Find the area of the segment AOB of angle 120° and radius 18 cm.

Answer:
1. Area of the sector = \( \frac{2376}{7} \) cm² ≈ 339.43 cm².
2. Area of the triangle OAB = \( \frac{1}{2} \) × 18 × 18 × sin 120° = 81√3 ≈ 140.30 cm².
3. Area of the segment = sector - triangle = 339.43 - 140.30 = 199.13 cm².

Teacher's Note:
a) Segment area = sector area - triangle area for a minor segment.
b) Questions 26, 27 and 28 together build this answer step by step.

29. The minute hand of a circular clock is 15 cm long. Find the area of the face of the clock and how far does the tip of the minute hand move in 35 minutes? (Take π = 3.14)

Answer:
1. Area of the clock face = πr² = 3.14 × 15 × 15 = 706.5 cm².
2. In 60 minutes the tip covers the circumference 2 × 3.14 × 15 = 94.2 cm.
3. In 35 minutes the tip moves \( \frac{35}{60} \) × 94.2 = 54.95 cm.

Teacher's Note:
a) The minute hand is the radius of the clock face.
b) The tip travels an arc: 35 minutes is \( \frac{35}{60} \) of a full circle.

30. Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is Rs 15/m². (Take π = 3.14)

Answer:
1. Radius = 0.8 m, so area = 3.14 × 0.8 × 0.8 = 2.0096 m².
2. Cost = 2.0096 × 15 = Rs. 30.14 (approximately).

Teacher's Note:
a) Use the radius, not the diameter, in πr².
b) Round the money to two decimal places.

31. A chord of a circle of radius 14 cm makes a right angle at the centre. Find the areas of the minor and the major segments of the circle.

Answer:
1. Area of the minor sector = \( \frac{90}{360} \) × \( \frac{22}{7} \) × 14 × 14 = 154 cm².
2. Area of the right triangle at the centre = \( \frac{1}{2} \) × 14 × 14 = 98 cm².
3. Minor segment = 154 - 98 = 56 cm².
4. Area of the circle = \( \frac{22}{7} \) × 14 × 14 = 616 cm², so major segment = 616 - 56 = 560 cm².

Teacher's Note:
a) The triangle formed by the radii and the chord has a right angle at the centre.
b) Major segment = area of the circle - area of the minor segment.

32. A square tank has area of 1600 m². There are four semicircular plots around it. Find the cost of turfing the plots at Rs. 1.25 per m². Take π = 3.14.

Answer:
1. Side of the tank = √1600 = 40 m, so each semicircle has diameter 40 m and radius 20 m.
2. Area of four semicircles = 2 full circles = 2 × 3.14 × 20 × 20 = 2512 m².
3. Cost = 2512 × 1.25 = Rs. 3140.

Teacher's Note:
a) Four semicircles make two complete circles of the same radius.
b) The semicircles are built on the sides of the square, so each side is a diameter.

33. A lawn is rectangular in the middle and it has semicircular portions along the shorter sides of the rectangle. The rectangular portion measures 50m by 35m. Find the area of the lawn.

Answer:
1. The shorter side is 35 m, so each semicircle has diameter 35 m and radius 17.5 m.
2. Area of the rectangle = 50 × 35 = 1750 m².
3. Area of the two semicircles = \( \frac{22}{7} \) × 17.5 × 17.5 = 962.5 m².
4. Area of the lawn = 1750 + 962.5 = 2712.5 m².

Teacher's Note:
a) The two semicircles together are one circle of radius 17.5 m.
b) π = 22/7 is used because no value of π is given in the question.

34. A rope by which a cow is tethered is increased from 16 m to 23 m. How much additional ground does it have now to graze?

Answer:
1. The cow grazes a circle: additional area = π(23² - 16²).
2. = \( \frac{22}{7} \) × (23 + 16)(23 - 16) = \( \frac{22}{7} \) × 39 × 7 = 858 m².

Teacher's Note:
a) The grazing region is a circle whose radius is the length of the rope.
b) π = 22/7 is used because no value of π is given in the question.

35. The perimeter of a certain sector of a circle of radius 6.5 cm is 31 cm. Find the area of the sector.

Answer:
1. Perimeter of a sector = 2r + arc length, so 31 = 2 × 6.5 + l.
2. Arc length l = 31 - 13 = 18 cm.
3. Area of the sector = \( \frac{1}{2} \) × l × r = \( \frac{1}{2} \) × 18 × 6.5 = 58.5 cm².

Teacher's Note:
a) Perimeter of a sector includes two radii and the arc.
b) Area of a sector = \( \frac{1}{2} \) × arc length × radius.

36. The area of the sector of a circle of radius 10.5 cm is 69.3 cm². Find the central angle of the sector.

Answer:
1. Area of the sector = \( \frac{\theta}{360} \) × \( \frac{22}{7} \) × 10.5 × 10.5 = \( \frac{\theta}{360} \) × 346.5.
2. 69.3 = \( \frac{\theta}{360} \) × 346.5, so θ = 69.3 × 360 ÷ 346.5 = 72°.

Teacher's Note:
a) Rearrange the sector formula to find θ.
b) Check: \( \frac{72}{360} \) = \( \frac{1}{5} \) of 346.5 is 69.3.

37. A sector of 56° cut out from a circle, contains 17.6 cm². Find the radius of the circle.

Answer:
1. \( \frac{56}{360} \) × \( \frac{22}{7} \) × r² = 17.6.
2. r² = 17.6 × 360 × 7 ÷ (56 × 22) = 44352 ÷ 1232 = 36.
3. r = 6 cm.

Teacher's Note:
a) Write the sector formula and solve for r².
b) Take the positive square root since a radius cannot be negative.

38. The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of distances travelled be their tips in 2 days. Take π = 3.14.

Answer:
1. In 2 days the short (hour) hand makes 2 × 2 = 4 revolutions and the long (minute) hand makes 2 × 24 = 48 revolutions.
2. Distance by the short hand = 4 × 2 × 3.14 × 4 = 100.48 cm.
3. Distance by the long hand = 48 × 2 × 3.14 × 6 = 1808.64 cm.
4. Sum = 100.48 + 1808.64 = 1909.12 cm.

Teacher's Note:
a) The hour hand makes 2 revolutions a day and the minute hand makes 24 a day.
b) The tip of each hand travels one circumference per revolution.

39. Find the lengths of the arcs cut off from a circle of radius 12 cm by a chord 12 cm long. Also find the area of the minor segment. Take √3 = 1.73 and π = 3.14.

Answer:
1. The chord equals the radius, so the triangle formed with the centre is equilateral and the central angle is 60°.
2. Minor arc = \( \frac{60}{360} \) × 2 × 3.14 × 12 = 12.56 cm.
3. Major arc = 2 × 3.14 × 12 - 12.56 = 75.36 - 12.56 = 62.8 cm.
4. Area of the minor sector = \( \frac{60}{360} \) × 3.14 × 144 = 75.36 cm².
5. Area of the equilateral triangle = \( \frac{\sqrt{3}}{4} \) × 144 = 36 × 1.73 = 62.28 cm².
6. Minor segment = 75.36 - 62.28 = 13.08 cm².

Teacher's Note:
a) A chord equal to the radius subtends 60° at the centre.
b) Major arc = circumference - minor arc.

40. The perimeter of a sector of a circle of radius 5.6 cm is 27.2 cm. Find the area of the sector.

Answer:
1. 27.2 = 2 × 5.6 + l, so the arc length l = 27.2 - 11.2 = 16 cm.
2. Area of the sector = \( \frac{1}{2} \) × l × r = \( \frac{1}{2} \) × 16 × 5.6 = 44.8 cm².

Teacher's Note:
a) The arc length is found from the perimeter by subtracting the two radii.
b) Area = \( \frac{1}{2} \)lr avoids needing π.

41. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the following: (i) Area of minor sector (ii) Area of major sector (iii) Area of major segment (iv) Area of minor segment. (Use π = 3.14)

Answer:
1. (i) Minor sector = \( \frac{90}{360} \) × 3.14 × 10 × 10 = 78.5 cm².
2. (ii) Area of the circle = 314 cm², so major sector = 314 - 78.5 = 235.5 cm².
3. (iii) Triangle at the centre = \( \frac{1}{2} \) × 10 × 10 = 50 cm². Major segment = 235.5 + 50 = 285.5 cm².
4. (iv) Minor segment = 78.5 - 50 = 28.5 cm².

Teacher's Note:
a) The minor and major sectors add up to the full circle.
b) The triangle is added to the major sector and subtracted from the minor sector to get the segments.

42. In a circle of radius 10.5 cm, the minor arc is one-fifth of the major arc. Find the area of the sector corresponding to the major arc.

Answer:
1. Minor arc : major arc = 1 : 5, so the minor arc is \( \frac{1}{6} \) of the circumference and the major arc is \( \frac{5}{6} \).
2. Angle of the major sector = \( \frac{5}{6} \) × 360° = 300°.
3. Area = \( \frac{300}{360} \) × \( \frac{22}{7} \) × 10.5 × 10.5 = \( \frac{5}{6} \) × 346.5 = 288.75 cm².

Teacher's Note:
a) Arcs are proportional to their central angles.
b) Divide the circumference in the ratio 1 : 5 to get the fractions \( \frac{1}{6} \) and \( \frac{5}{6} \).

43. It is proposed to add two circular ends, to a square lawn whose side measures 58 cm, the centre of each circle being the point of intersection of the diagonals of the square. Find the area of the whole lawn.

Answer:
1. Let the side be a = 58 cm. The circle through the vertices has radius r with r² = \( \frac{a^{2}}{2} \) = \( \frac{3364}{2} \) = 1682 cm².
2. Each circular end is a segment of this circle cut by a side of the square; the side subtends 90° at the centre.
3. Area of one segment = \( \frac{90}{360} \) × \( \frac{22}{7} \) × 1682 - \( \frac{1}{2} \) × 1682 = 1682 × \( \frac{2}{7} \) = \( \frac{3364}{7} \) ≈ 480.57 cm².
4. Area of two segments = \( \frac{6728}{7} \) ≈ 961.14 cm².
5. Area of the square = 58 × 58 = 3364 cm².
6. Area of the whole lawn = 3364 + \( \frac{6728}{7} \) = \( \frac{30276}{7} \) ≈ 4325.14 cm².

Teacher's Note:
a) The two circular ends are segments on opposite sides of the square, each cut off by a side that subtends 90° at the centre.
b) Area of the whole lawn = square + two segments. Take π = 22/7 as no value is given.

44. It is proposed to add two circular ends, to a square lawn whose side measures 50 m, the centre of each circle being the point of intersection of the diagonals of the square. Find the area of the whole lawn. Take π = 3.14

Answer:
1. Side a = 50 m, so r² = \( \frac{2500}{2} \) = 1250 m².
2. Area of one segment = \( \frac{90}{360} \) × 3.14 × 1250 - \( \frac{1}{2} \) × 1250 = 981.25 - 625 = 356.25 m².
3. Area of two segments = 712.5 m².
4. Area of the square = 50 × 50 = 2500 m².
5. Area of the whole lawn = 2500 + 712.5 = 3212.5 m².

Teacher's Note:
a) Same method as the previous question with a different side and π = 3.14.
b) Radius squared of the circle through the vertices is half of the side squared.

45. In an equilateral triangle of side 12 cm, a circle is inscribed touching its sides. Find the area of the portion of the triangle not included in the circle. Take √3 = 1.73 and π = 3.14.

Answer:
1. Area of the triangle = \( \frac{\sqrt{3}}{4} \) × 12² = 36√3 = 36 × 1.73 = 62.28 cm².
2. Inradius r = \( \frac{12}{2\sqrt{3}} \) = 2√3 cm, so r² = 12.
3. Area of the circle = 3.14 × 12 = 37.68 cm².
4. Required area = 62.28 - 37.68 = 24.60 cm².

Teacher's Note:
a) Inradius of an equilateral triangle of side a is \( \frac{a}{2\sqrt{3}} \).
b) Required area = triangle - inscribed circle.

46. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find (i) length of the arc (ii) area of sector formed by the arc (iii) area of segment formed by the corresponding chord of the arc.

Answer:
1. (i) Arc length = \( \frac{60}{360} \) × 2 × \( \frac{22}{7} \) × 21 = 22 cm.
2. (ii) Area of the sector = \( \frac{60}{360} \) × \( \frac{22}{7} \) × 21 × 21 = 231 cm².
3. (iii) Triangle with the centre is equilateral of side 21 cm: area = \( \frac{\sqrt{3}}{4} \) × 441 = 190.96 cm².
4. Segment = 231 - 190.96 = 40.04 cm².

Teacher's Note:
a) The two radii and the chord form an equilateral triangle when the angle is 60°.
b) Segment area = sector area - triangle area.

47. If three circles of radius r each, are drawn such that each touches the other two, the find the area included between them. Take π = 3.14 and √3 = 1.73.

Answer:
1. The centres form an equilateral triangle of side 2r.
2. Area of the triangle = \( \frac{\sqrt{3}}{4} \) × (2r)² = √3 r².
3. The three sectors inside the triangle have angle 60° each, total area = \( \frac{180}{360} \) × πr² = \( \frac{1}{2} \)πr².
4. Area between the circles = √3 r² - \( \frac{1}{2} \) × 3.14 r² = (1.73 - 1.57) r² = 0.16 r².

Teacher's Note:
a) The required region is the triangle of centres minus three 60° sectors.
b) Three 60° sectors together make a semicircle.

48. If four circles of radius r each, are drawn such that each touches the other two, the find the area included between them. Take π = 3.14.

Answer:
1. The centres form a square of side 2r, with area 4r².
2. The four sectors inside the square have angle 90° each, total area = πr² = 3.14 r².
3. Area between the circles = 4r² - 3.14r² = 0.86 r².

Teacher's Note:
a) Each circle touches the two neighbouring circles, so the centres form a square of side 2r.
b) Four quadrants make one full circle.

49. The length of an arc subtending an angle of 72° at the centre is 44 cm. Find the area of the circle.

Answer:
1. 72° is \( \frac{1}{5} \) of 360°, so the circumference = 5 × 44 = 220 cm.
2. 2 × \( \frac{22}{7} \) × r = 220, so r = 35 cm.
3. Area = \( \frac{22}{7} \) × 35 × 35 = 3850 cm².

Teacher's Note:
a) Arc length is proportional to the central angle.
b) Find the radius from the circumference, then the area.

50. A park is in the form of rectangle 120 m by 100 m. At the centre of the park, there is a circular lawn. The area of the park excluding the lawn is 11384 sq. m. Find the radius of the circular lawn.

Answer:
1. Area of the park = 120 × 100 = 12000 m².
2. Area of the lawn = 12000 - 11384 = 616 m².
3. πr² = 616, so r² = 616 × \( \frac{7}{22} \) = 196 and r = 14 m.

Teacher's Note:
a) Area of the lawn = area of the park - area excluding the lawn.
b) π = 22/7 is used because no value of π is given in the question.

51. Find the area of shaded portion in the below figure

Answer:
1. From the figure, ABO is a quadrant of a circle with centre O and radius OA = 3.5 cm, and D is a point on OB with OD = 2 cm.
2. Area of the quadrant = \( \frac{1}{4} \) × \( \frac{22}{7} \) × 3.5 × 3.5 = 9.625 cm².
3. Area of the right triangle AOD = \( \frac{1}{2} \) × OA × OD = \( \frac{1}{2} \) × 3.5 × 2 = 3.5 cm².
4. Shaded area = 9.625 - 3.5 = 6.125 cm².

Teacher's Note:
a) The shaded part is the quadrant with the triangle AOD removed.
b) The values OA = 3.5 cm and OD = 2 cm are read from the figure.

52. Find the area of shaded portion in the above right-sided figure

Answer:
1. From the figure, OAB(C) is a square of side OA = 10 cm inscribed in the quadrant OPQ with its vertex B on the arc.
2. Radius of the quadrant = OB = diagonal of the square = 10√2 cm, so r² = 200.
3. Area of the quadrant = \( \frac{1}{4} \) × 3.14 × 200 = 157 cm².
4. Area of the square = 10 × 10 = 100 cm².
5. Shaded area = 157 - 100 = 57 cm².

Teacher's Note:
a) The radius of the quadrant is the diagonal of the square.
b) Shaded area = quadrant - square. Take π = 3.14, as in the similar question 83.

53. Find the area of shaded portion in the below figure

Answer:
1. The figure is the same as in question 92: PS is a diameter of a circle of radius 6 cm and PQ = QR = RS = 4 cm.
2. Radius of the semicircle on PS = 6 cm, on QS = 4 cm and on PQ = 2 cm.
3. Shaded area = semicircle on PS - semicircle on QS + semicircle on PQ.
4. = \( \frac{1}{2} \)π(6² - 4² + 2²) = \( \frac{1}{2} \)π × 24 = 12π.
5. With π = 22/7, area = \( \frac{264}{7} \) ≈ 37.71 cm².

Teacher's Note:
a) The figure carries no numbers; the dimensions are taken from question 92, which shows the same figure.
b) The shaded part is the big semicircle with the semicircle on QS removed and the small semicircle on PQ added.

54. Find the area of shaded portion in the above right-sided figure

Answer:
1. From the figure, the square has side 100 cm and four circles of radius 14 cm have their centres at the corners of the square.
2. A quarter of each circle lies inside the square, so the four quadrants make one circle of radius 14 cm.
3. Area of the circle = \( \frac{22}{7} \) × 14 × 14 = 616 cm².
4. Shaded area = 100 × 100 - 616 = 9384 cm².

Teacher's Note:
a) The four corner quadrants add up to one full circle.
b) The figure gives 14 cm as the radius and 100 cm as the side of the square.

55. An athletic track, 14 m wide, consists of two straight sections 120 m long joining semicircular ends whose inner radius is 35 m. Calculate the area of the track.

Answer:
1. Outer radius of the semicircular ends = 35 + 14 = 49 m.
2. Area of the two straight parts = 2 × 120 × 14 = 3360 m².
3. Area of the two semicircular rings = π(49² - 35²) = \( \frac{22}{7} \) × (49 + 35)(49 - 35) = \( \frac{22}{7} \) × 84 × 14 = 3696 m².
4. Area of the track = 3360 + 3696 = 7056 m².

Teacher's Note:
a) The two semicircular ends together form one circular ring.
b) Add the area of the two straight rectangles.

56. The cost of fencing a circular field at the rate of Rs 24 per metre is Rs 5280. The field is to be ploughed at the rate of Rs 0.50 per m². Find the cost of ploughing the field.

Answer:
1. Length of the fence = 5280 ÷ 24 = 220 m = circumference.
2. 2 × \( \frac{22}{7} \) × r = 220, so r = 35 m.
3. Area of the field = \( \frac{22}{7} \) × 35 × 35 = 3850 m².
4. Cost of ploughing = 3850 × 0.50 = Rs. 1925.

Teacher's Note:
a) Fencing is along the circumference; ploughing is over the area.
b) Find the circumference from the cost, then the radius, then the area.

57. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Answer:
1. πR² = π × 8² + π × 6², so R² = 64 + 36 = 100.
2. R = 10 cm.

Teacher's Note:
a) Cancel π from both sides.
b) The radii 6, 8 and 10 form a Pythagorean triple.

58. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Answer:
1. 2πR = 2π × 19 + 2π × 9, so R = 19 + 9 = 28 cm.

Teacher's Note:
a) Circumference is proportional to the radius, so the radii simply add.
b) Compare this with question 57, where the squares of the radii add.

59. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Answer:
1. Speed = 66 km/h = 1100 m/min, so distance in 10 minutes = 11000 m = 1100000 cm.
2. Circumference of a wheel = π × 80 = \( \frac{22}{7} \) × 80 = \( \frac{1760}{7} \) cm.
3. Number of revolutions = 1100000 ÷ \( \frac{1760}{7} \) = 1100000 × \( \frac{7}{1760} \) = 4375.

Teacher's Note:
a) Revolutions = distance travelled ÷ circumference.
b) The result is a whole number, so the number of complete revolutions is 4375.

60. Find the area of the segment AYB shown in Fig., if radius of the circle is 21 cm and ∠AOB = 120°. (Use π = 22/7).

Answer:
1. Area of the sector OAYB = \( \frac{120}{360} \) × \( \frac{22}{7} \) × 21 × 21 = 462 cm².
2. Area of △OAB = \( \frac{1}{2} \) × 21 × 21 × sin 120° = \( \frac{441\sqrt{3}}{4} \) ≈ 190.96 cm².
3. Area of the segment = 462 - 190.96 = 271.04 cm².

Teacher's Note:
a) Segment area = sector area - area of the triangle formed by the two radii and the chord.
b) Area of the triangle uses \( \frac{1}{2} \)r² sin θ.

61. Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector (Use π = 3.14).

Answer:
1. Area of the sector = \( \frac{30}{360} \) × 3.14 × 4 × 4 = \( \frac{50.24}{12} \) ≈ 4.19 cm².
2. Area of the circle = 3.14 × 16 = 50.24 cm².
3. Area of the major sector = 50.24 - 4.19 = 46.05 cm².

Teacher's Note:
a) The major sector has angle 360° - 30° = 330°.
b) Major sector = area of the circle - minor sector.

62. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Answer:
1. Area = \( \frac{60}{360} \) × \( \frac{22}{7} \) × 6 × 6 = \( \frac{132}{7} \) ≈ 18.86 cm².

Teacher's Note:
a) Use sector area = \( \frac{\theta}{360} \)πr².
b) π = 22/7 is used because no value of π is given in the question.

63. Find the area of a quadrant of a circle whose circumference is 22 cm.

Answer:
1. 2πr = 22, so r = 22 × \( \frac{7}{44} \) = 3.5 cm.
2. Area of the quadrant = \( \frac{1}{4} \) × \( \frac{22}{7} \) × 3.5 × 3.5 = 9.625 cm².

Teacher's Note:
a) Find the radius from the circumference first.
b) A quadrant is one-fourth of the circle.

64. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Answer:
1. In 60 minutes the hand turns 360°, so in 5 minutes it turns 30°.
2. Area swept = \( \frac{30}{360} \) × \( \frac{22}{7} \) × 14 × 14 = \( \frac{154}{3} \) ≈ 51.33 cm².

Teacher's Note:
a) The area swept is a sector with the minute hand as radius.
b) 5 minutes is \( \frac{1}{12} \) of the full circle.

65. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use π = 3.14)

Answer:
1. (i) Minor sector = \( \frac{1}{4} \) × 3.14 × 100 = 78.5 cm². Triangle = \( \frac{1}{2} \) × 10 × 10 = 50 cm². Minor segment = 78.5 - 50 = 28.5 cm².
2. (ii) Major sector = 314 - 78.5 = 235.5 cm².

Teacher's Note:
a) Minor segment = minor sector - triangle.
b) Major sector = circle - minor sector.

66. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord

Answer:
1. (i) Arc length = \( \frac{60}{360} \) × 2 × \( \frac{22}{7} \) × 21 = 22 cm.
2. (ii) Area of the sector = \( \frac{60}{360} \) × \( \frac{22}{7} \) × 21 × 21 = 231 cm².
3. (iii) The triangle is equilateral with side 21 cm: area = \( \frac{\sqrt{3}}{4} \) × 441 ≈ 190.96 cm². Segment = 231 - 190.96 = 40.04 cm².

Teacher's Note:
a) This is the same data as question 46.
b) A 60° central angle makes an equilateral triangle with the chord.

67. 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 of the circle. (Use π = 3.14 and √3 = 1.73)

Answer:
1. Area of the minor sector = \( \frac{60}{360} \) × 3.14 × 225 = 117.75 cm².
2. Area of the equilateral triangle = \( \frac{\sqrt{3}}{4} \) × 225 = 1.73 × 56.25 = 97.3125 cm².
3. Minor segment = 117.75 - 97.3125 = 20.4375 ≈ 20.44 cm².
4. Area of the circle = 3.14 × 225 = 706.5 cm², so major segment = 706.5 - 20.4375 = 686.0625 ≈ 686.06 cm².

Teacher's Note:
a) Minor and major segments add up to the whole circle.
b) Keep the working in decimals and round only at the end.

68. 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)

Answer:
1. Area of the sector = \( \frac{120}{360} \) × 3.14 × 144 = 150.72 cm².
2. Area of △OAB = \( \frac{1}{2} \) × 12 × 12 × sin 120° = 36√3 = 36 × 1.73 = 62.28 cm².
3. Segment = 150.72 - 62.28 = 88.44 cm².

Teacher's Note:
a) sin 120° = \( \frac{\sqrt{3}}{2} \); the height of the triangle is r cos 60° = 6 cm and half the base is 6√3 cm.
b) Segment = sector - triangle.

69. In Fig, two circular flower beds have been shown on two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds.

Answer:
1. Area of the square lawn = 56 × 56 = 3136 m².
2. The radius of each bed is OA, with OA² = \( \frac{56^{2}}{2} \) = 1568 m².
3. Each flower bed is a segment on one side of the square. The angle at O is 90°.
4. Area of one segment = \( \frac{1}{4} \) × \( \frac{22}{7} \) × 1568 - \( \frac{1}{2} \) × 1568 = 1568 × \( \frac{2}{7} \) = 448 m².
5. Area of the two beds = 896 m².
6. Sum of the areas = 3136 + 896 = 4032 m².

Teacher's Note:
a) The diagonals of a square are equal and meet at right angles, so each side subtends 90° at O.
b) Sum of areas = lawn + two segments.

70. Find the area of the shaded region in the below Fig., where ABCD is a square of side 14 cm.

Answer:
1. Each circle touches the sides of the square and its neighbours, so its diameter = 14 ÷ 2 = 7 cm and radius = 3.5 cm.
2. Area of the four circles = 4 × \( \frac{22}{7} \) × 3.5 × 3.5 = 154 cm².
3. Area of the square = 14 × 14 = 196 cm².
4. Shaded area = 196 - 154 = 42 cm².

Teacher's Note:
a) Two circles fit across the side of the square, so each diameter is half the side.
b) Shaded area = square - four circles.

71. The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded region. . (Use π = 3.14 and √3 = 1.73205)

Answer:
1. Area = \( \frac{\sqrt{3}}{4} \) a² = 17320.5, so a² = 17320.5 × 4 ÷ 1.73205 = 40000 and a = 200 cm.
2. Radius of each circle = 100 cm.
3. The three sectors inside the triangle are 60° each, total area = \( \frac{1}{2} \) × 3.14 × 100² = 15700 cm².
4. Shaded area = 17320.5 - 15700 = 1620.5 cm².

Teacher's Note:
a) The shaded region is the triangle minus three 60° sectors.
b) Three 60° sectors together make a semicircle.

72. An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Answer:
1. The 8 ribs divide the circle into 8 equal sectors, each with angle 360° ÷ 8 = 45°.
2. Area between two consecutive ribs = \( \frac{45}{360} \) × \( \frac{22}{7} \) × 45 × 45 = \( \frac{1}{8} \) × \( \frac{44550}{7} \) = \( \frac{22275}{28} \) ≈ 795.54 cm².

Teacher's Note:
a) Equal spacing means equal sectors: divide the circle area by 8.
b) π = 22/7 is used because no value of π is given in the question.

73. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find (i) the area of that part of the field in which the horse can graze. (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. . (Use π = 3.14)

Answer:
1. (i) The horse grazes a quadrant of radius 5 m: area = \( \frac{1}{4} \) × 3.14 × 25 = 19.625 m².
2. (ii) With a 10 m rope the quadrant has area \( \frac{1}{4} \) × 3.14 × 100 = 78.5 m².
3. Increase = 78.5 - 19.625 = 58.875 m².

Teacher's Note:
a) At a corner of a square field the horse can graze only a quadrant (90°).
b) Both rope lengths are less than 15 m, so the quadrant stays inside the field.

74. In Fig., ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Answer:
1. Each circle touches the two neighbouring circles, so 2r = 14 and r = 7 cm.
2. The four quadrants inside the square make one circle: area = \( \frac{22}{7} \) × 7 × 7 = 154 cm².
3. Shaded area = 196 - 154 = 42 cm².

Teacher's Note:
a) Adjacent centres are 14 cm apart, so each radius is 7 cm.
b) Four quadrants make a full circle.

75. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in above sided Fig. Find the area of the remaining portion of the square.

Answer:
1. Area of the square = 4 × 4 = 16 cm².
2. Four quadrants of radius 1 cm make one circle: area = \( \frac{22}{7} \) × 1 × 1 = \( \frac{22}{7} \) cm².
3. Circle of diameter 2 cm has radius 1 cm: area = \( \frac{22}{7} \) cm².
4. Remaining area = 16 - \( \frac{44}{7} \) = \( \frac{68}{7} \) ≈ 9.71 cm².

Teacher's Note:
a) The four corner quadrants add up to one circle of radius 1 cm.
b) π = 22/7 is used because no value of π is given in the question.

76. Find the area of the shaded design in the below Fig., where ABCD is a square of side 10 cm and semicircles are drawn with each side of the square as diameter. (Use π = 3.14)

Answer:
1. Radius of each semicircle = 5 cm.
2. Area of the four semicircles = 4 × \( \frac{1}{2} \) × 3.14 × 25 = 157 cm².
3. The four semicircles cover the square once and the four petals twice, so petals = 157 - 100 = 57 cm².
4. Area of the shaded design = 57 cm².

Teacher's Note:
a) The shaded design is made of the overlaps of adjacent semicircles.
b) Sum of the four semicircles - square = area of the overlaps.

77. In the above sided Fig., AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Answer:
1. Radius of the big circle = 7 cm; radius of the small circle = \( \frac{7}{2} \) = 3.5 cm.
2. Area of the small circle = \( \frac{22}{7} \) × 3.5 × 3.5 = 38.5 cm².
3. The shaded region on the right is the two segments cut by chords BC and CA. Area of the semicircle ACB = \( \frac{1}{2} \) × \( \frac{22}{7} \) × 49 = 77 cm².
4. Area of △ABC = \( \frac{1}{2} \) × AB × OC = \( \frac{1}{2} \) × 14 × 7 = 49 cm².
5. Area of the two segments = 77 - 49 = 28 cm².
6. Shaded area = 38.5 + 28 = 66.5 cm².

Teacher's Note:
a) The shaded region is the small circle together with the two segments on chords BC and CA.
b) OC is perpendicular to AB, so OC is the height of △ABC.

78. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. Find the area of the design (shaded region).

Answer:
1. Area of the circle = \( \frac{22}{7} \) × 32 × 32 = \( \frac{22528}{7} \) ≈ 3218.29 cm².
2. For an equilateral triangle inscribed in a circle of radius R, side = R√3 = 32√3 cm.
3. Area of △ABC = \( \frac{\sqrt{3}}{4} \) × (32√3)² = 768√3 = 768 × 1.73 = 1328.64 cm².
4. Area of the design = 3218.29 - 1328.64 = 1889.65 cm².

Teacher's Note:
a) The design is the circle minus the inscribed equilateral triangle.
b) Take π = 22/7 and √3 = 1.73, as no values are given.

79. Find the area of the shaded region in above sided Fig., if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Answer:
1. Area of the square = 14 × 14 = 196 cm².
2. Each semicircle has diameter 14 cm, radius 7 cm; the two semicircles together make one circle: area = \( \frac{22}{7} \) × 7 × 7 = 154 cm².
3. Shaded area = 196 - 154 = 42 cm².

Teacher's Note:
a) The semicircles touch at P, the centre of the square.
b) Shaded area = square - two semicircles.

80. 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)

Answer:
1. Area of the sector = \( \frac{80}{360} \) × 3.14 × 16.5 × 16.5.
2. = \( \frac{2}{9} \) × 854.865 = 189.97 km².

Teacher's Note:
a) The light covers a sector with radius 16.5 km and angle 80°.
b) The area is in square kilometres.

81. In Fig., ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Answer:
1. BC² = 14² + 14² = 392, so the radius of the semicircle has r² = 98.
2. Area of the semicircle on BC = \( \frac{1}{2} \) × \( \frac{22}{7} \) × 98 = 154 cm².
3. Area of the quadrant = \( \frac{1}{4} \) × \( \frac{22}{7} \) × 14 × 14 = 154 cm².
4. Area of △ABC = \( \frac{1}{2} \) × 14 × 14 = 98 cm².
5. Segment between chord BC and the quadrant arc = 154 - 98 = 56 cm².
6. Shaded area = 154 - 56 = 98 cm².

Teacher's Note:
a) The shaded crescent equals the area of △ABC.
b) Semicircle - segment of the quadrant = shaded region.

82. Calculate the area of the designed region in the above sided Fig. common between the two quadrants of circles of radius 8 cm each.

Answer:
1. Area of the two quadrants = 2 × \( \frac{1}{4} \) × \( \frac{22}{7} \) × 8 × 8 = \( \frac{704}{7} \) ≈ 100.57 cm².
2. Area of the square = 8 × 8 = 64 cm².
3. Common region = sum of the two quadrants - square = 100.57 - 64 = 36.57 cm².

Teacher's Note:
a) The two quadrants together cover the square and overlap in the common region, which is counted twice.
b) Common area = quadrant + quadrant - square.

83. In Fig., a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

Answer:
1. OB is the diagonal of the square, so OB² = 20² + 20² = 800.
2. Area of the quadrant = \( \frac{1}{4} \) × 3.14 × 800 = 628 cm².
3. Area of the square = 20 × 20 = 400 cm².
4. Shaded area = 628 - 400 = 228 cm².

Teacher's Note:
a) The radius of the quadrant is the diagonal of the square.
b) Shaded area = quadrant - square.

84. On a square handkerchief, nine circular designs each of radius 7 cm are made (see in the above sided Fig.). Find the area of the remaining portion of the handkerchief.

Answer:
1. Three circles fit along each side, so the side of the handkerchief = 3 × 14 = 42 cm.
2. Area of the handkerchief = 42 × 42 = 1764 cm².
3. Area of nine circles = 9 × \( \frac{22}{7} \) × 7 × 7 = 1386 cm².
4. Remaining area = 1764 - 1386 = 378 cm².

Teacher's Note:
a) Each circle has diameter 14 cm; three of them span one side.
b) Remaining area = square - nine circles.

85. In the given figure, ΔABC is right angled at A. Semicircles are drawn on AB, AC and BC as diameters. It is given that AB = 3cm and AC = 4cm. Find the area of the shaded region.

Answer:
1. BC = √(3² + 4²) = 5 cm.
2. Area of the semicircle on AB = \( \frac{1}{2} \)π × \( \frac{3}{2} \)² and on AC = \( \frac{1}{2} \)π × 2²; on BC = \( \frac{1}{2} \)π × \( \frac{5}{2} \)².
3. Shaded area = semicircle on AB + semicircle on AC + △ABC - semicircle on BC.
4. The semicircles satisfy \( \frac{1}{2} \)π × \( \frac{9}{4} \) + \( \frac{1}{2} \)π × 4 - \( \frac{1}{2} \)π × \( \frac{25}{4} \) = 0.
5. So shaded area = area of △ABC = \( \frac{1}{2} \) × 3 × 4 = 6 cm².

Teacher's Note:
a) This is the lunes of Hippocrates: the area of the two lunes equals the area of the right triangle.
b) The three semicircle areas satisfy the Pythagoras relation, so they cancel out.

86. Find the area of the shaded region in the below figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Answer:
1. QR passes through the centre O, so QR is a diameter and ∠QPR = 90°.
2. QR² = 24² + 7² = 625, so QR = 25 cm and the radius = 12.5 cm.
3. Area of the semicircle = \( \frac{1}{2} \) × \( \frac{22}{7} \) × 12.5 × 12.5 = \( \frac{1}{2} \) × 491.07 = 245.54 cm².
4. Area of △PQR = \( \frac{1}{2} \) × 24 × 7 = 84 cm².
5. Shaded area = 245.54 - 84 = 161.54 cm².

Teacher's Note:
a) An angle in a semicircle is a right angle.
b) The shaded part is the semicircle minus the triangle.

87. Find the areas of the shaded region in the above right sided figure.

Answer:
1. From the figure, the upper semicircle has radius 14 cm and each of the two lower semicircles has diameter 14 cm, that is radius 7 cm.
2. Area of the upper semicircle = \( \frac{1}{2} \) × \( \frac{22}{7} \) × 14 × 14 = 308 cm².
3. Area of the two lower semicircles = \( \frac{22}{7} \) × 7 × 7 = 154 cm².
4. Shaded area = 308 + 154 = 462 cm².

Teacher's Note:
a) The two small semicircles together make one circle of radius 7 cm.
b) The three parts are added because the whole figure is shaded.

88. In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides. Find the area of the remaining portion of the triangle. Take √3 = 1.732

Answer:
1. Area of the triangle = \( \frac{\sqrt{3}}{4} \) × 24² = 144√3 = 144 × 1.732 = 249.408 cm².
2. Inradius r = \( \frac{24}{2\sqrt{3}} \) = 4√3 cm, so r² = 48.
3. Area of the circle = 3.14 × 48 = 150.72 cm².
4. Remaining area = 249.408 - 150.72 = 98.688 ≈ 98.69 cm².

Teacher's Note:
a) Only √3 is given; π = 3.14 is used as in question 45.
b) Remaining portion = triangle - inscribed circle.

89. Find to the three places of decimals the radius of the circle whose area is the sum of the areas of two triangles whose sides are 35, 53, 66 and 33, 56, 65 measured in cms. (Take π = \( \frac{22}{7} \))

Answer:
1. First triangle: s = (35 + 53 + 66) ÷ 2 = 77, area = √(77 × 42 × 24 × 11) = √853776 = 924 cm².
2. Second triangle: s = (33 + 56 + 65) ÷ 2 = 77, area = √(77 × 44 × 21 × 12) = √853776 = 924 cm².
3. Sum of the areas = 1848 cm².
4. πr² = 1848, so r² = 1848 × \( \frac{7}{22} \) = 588.
5. r = √588 = 14√3 = 24.249 cm.

Teacher's Note:
a) Use Heron's formula for both triangles.
b) Give the final radius correct to three decimal places.

90. A square park has each side of 100m. At each corner of the park, there is a slower bed in the form of a quadrant of radius 14 m. Find the area of the remaining part of the park. (Take π = \( \frac{22}{7} \))

Answer:
1. Area of the park = 100 × 100 = 10000 m².
2. Four quadrants of radius 14 m make one circle: area = \( \frac{22}{7} \) × 14 × 14 = 616 m².
3. Remaining area = 10000 - 616 = 9384 m².

Teacher's Note:
a) The four corner quadrants make one full circle.
b) Remaining area = park - flower beds.

91. Find the area of the shaded region in below figure, where radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Answer:
1. Area of the whole ring = π(14² - 7²) = \( \frac{22}{7} \) × 147 = 462 cm².
2. The unshaded part is the portion of the ring inside the 40° angle: area = \( \frac{40}{360} \) × 462 = \( \frac{154}{3} \) cm².
3. Shaded area = 462 - \( \frac{154}{3} \) = \( \frac{1232}{3} \) ≈ 410.67 cm².

Teacher's Note:
a) The shaded part is the ring with the 40° wedge removed, so it covers 320° of the ring.
b) Shaded area = \( \frac{320}{360} \) × 462.

92. PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal. Semicircles are drawn on PQ and QS as diameters as shown in above right sided figure. Find the perimeter and area of the shaded region.

Answer:
1. PS = 12 cm, so PQ = QR = RS = 4 cm and QS = 8 cm. The radii are 6 cm (on PS), 4 cm (on QS) and 2 cm (on PQ).
2. Perimeter = arc on PS + arc on QS + arc on PQ = π × 6 + π × 4 + π × 2 = 12π.
3. With π = 22/7, perimeter = \( \frac{264}{7} \) ≈ 37.71 cm.
4. Area = \( \frac{1}{2} \)π(6² - 4² + 2²) = 12π = \( \frac{264}{7} \) ≈ 37.71 cm².

Teacher's Note:
a) The boundary of the shaded region is made of three semicircular arcs.
b) Area = big semicircle - semicircle on QS + semicircle on PQ.

93. An athletic track 14 m wide consists of two straight sections 120 m long joining semicirculars ends whose inner radius is 35m. Calculate the area of the shaded region.

Answer:
1. Outer radius of the ends = 35 + 14 = 49 m.
2. Area of the two straight parts = 2 × 120 × 14 = 3360 m².
3. Area of the two semicircular rings = \( \frac{22}{7} \) × (49² - 35²) = \( \frac{22}{7} \) × 84 × 14 = 3696 m².
4. Area of the track = 3360 + 3696 = 7056 m².

Teacher's Note:
a) This is the same data as question 55.
b) The ends make one complete ring.

94. The above right-sided figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find : (i) the distance around the track along its inner edge (ii) the area of the track.

Answer:
1. The inner semicircles have diameter 60 m, so the inner radius = 30 m and the outer radius = 40 m.
2. (i) Distance along the inner edge = 2 × 106 + 2 × \( \frac{22}{7} \) × 30 = 212 + \( \frac{1320}{7} \) = 400.57 m (approximately).
3. (ii) Area of the two straight parts = 2 × 106 × 10 = 2120 m².
4. Area of the two semicircular rings = \( \frac{22}{7} \) × (40² - 30²) = \( \frac{22}{7} \) × 700 = 2200 m².
5. Area of the track = 2120 + 2200 = 4320 m².

Teacher's Note:
a) The inner edge has two straight parts and two semicircles forming one circle of radius 30 m.
b) The two semicircular ends together form one complete ring.

95. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see below Figure). If ∠AOB = 30°, find the area of the shaded region.

Answer:
1. Area of the shaded region = \( \frac{30}{360} \) × π(21² - 7²).
2. = \( \frac{1}{12} \) × \( \frac{22}{7} \) × 392 = \( \frac{1}{12} \) × 1232 = \( \frac{308}{3} \) ≈ 102.67 cm².

Teacher's Note:
a) The shaded part is the difference of two sectors with the same angle 30°.
b) Use the same fraction \( \frac{30}{360} \) for both sectors.

96. In the above right sided Figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB, (ii) shaded region.

Answer:
1. (i) Area of the quadrant = \( \frac{1}{4} \) × \( \frac{22}{7} \) × 3.5 × 3.5 = 9.625 cm².
2. (ii) Area of the right triangle BOD = \( \frac{1}{2} \) × OB × OD = \( \frac{1}{2} \) × 3.5 × 2 = 3.5 cm².
3. Shaded area = 9.625 - 3.5 = 6.125 cm².

Teacher's Note:
a) The shaded region is the quadrant minus triangle BOD.
b) OB is a radius, so OB = 3.5 cm.

97. A path of 4m width runs round a semicircular grassy plot whose circumference is \( 163 \frac{3}{7} \) m. Find (i) the area of the path (ii) the cost of gravelling the path at the rate of Rs. 1.50 per sq. m (iii) the cost of turfing the plot at the rate of 45 paise per sq. m.

Answer:
1. The curved boundary of the plot is \( 163 \frac{3}{7} \) = \( \frac{1144}{7} \) m, so π r = \( \frac{1144}{7} \) and r = \( \frac{1144}{7} \) × \( \frac{7}{22} \) = 52 m.
2. Outer radius of the path R = 52 + 4 = 56 m.
3. (i) Area of the path = \( \frac{1}{2} \) × \( \frac{22}{7} \) × (56² - 52²) = \( \frac{11}{7} \) × 432 = \( \frac{4752}{7} \) ≈ 678.86 m².
4. (ii) Cost of gravelling = \( \frac{4752}{7} \) × 1.50 = Rs. 1018.29 (approximately).
5. (iii) Area of the plot = \( \frac{1}{2} \) × \( \frac{22}{7} \) × 52 × 52 = \( \frac{29744}{7} \) ≈ 4249.14 m². Cost of turfing = 4249.14 × 0.45 = Rs. 1912.11 (approximately).

Teacher's Note:
a) The given length 163 3/7 m is the curved edge of the semicircle, which gives a whole-number radius of 52 m.
b) The path is a semicircular ring: \( \frac{1}{2} \)π(R² - r²).
c) 45 paise = Rs. 0.45.

98. Find the area of the shaded region in below figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

Answer:
1. Area of the equilateral triangle OAB = \( \frac{\sqrt{3}}{4} \) × 12² = 36√3 = 36 × 1.73 = 62.28 cm².
2. The circle with centre O and radius 6 cm lies partly inside the triangle (a 60° sector). The shaded region is the triangle plus the remaining 300° of the circle.
3. Area of the 300° sector = \( \frac{300}{360} \) × \( \frac{22}{7} \) × 6 × 6 = \( \frac{660}{7} \) ≈ 94.29 cm².
4. Shaded area = 62.28 + 94.29 = 156.57 cm².

Teacher's Note:
a) The triangle and the major sector do not overlap, so their areas are added.
b) Take √3 = 1.73 and π = 22/7, as no values are given.

99. A round table cover has six equal designs as shown in the above right-sided figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs 0.35 per cm². (Use √3 = 1.7)

Answer:
1. The hexagon inside the circle is regular, so it is made of six equilateral triangles of side 28 cm.
2. Area of the hexagon = 6 × \( \frac{\sqrt{3}}{4} \) × 28² = 6 × 1.7 × 196 = 1999.2 cm².
3. Area of the circle = \( \frac{22}{7} \) × 28 × 28 = 2464 cm².
4. Area of the six designs = 2464 - 1999.2 = 464.8 cm².
5. Cost = 464.8 × 0.35 = Rs. 162.68.

Teacher's Note:
a) The six designs are the six equal segments between the hexagon and the circle.
b) Side of a regular hexagon inscribed in a circle = radius.

100. Find the area of the shaded region in the below figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Answer:
1. Area of the shaded region = \( \frac{40}{360} \) × π(14² - 7²).
2. = \( \frac{1}{9} \) × \( \frac{22}{7} \) × 147 = \( \frac{1}{9} \) × 462 = \( \frac{154}{3} \) ≈ 51.33 cm².

Teacher's Note:
a) Here only the 40° part of the ring is shaded, unlike question 91.
b) Ring sector = \( \frac{\theta}{360} \)π(R² - r²).

101. Find the area of the shaded region in the above right-sided figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Answer:
1. QR passes through the centre O, so QR is a diameter and ∠QPR = 90°.
2. QR² = 24² + 7² = 625, so QR = 25 cm and the radius = 12.5 cm.
3. Area of the semicircle = \( \frac{1}{2} \) × \( \frac{22}{7} \) × 12.5 × 12.5 ≈ 245.54 cm².
4. Area of △PQR = \( \frac{1}{2} \) × 24 × 7 = 84 cm².
5. Shaded area = 245.54 - 84 = 161.54 cm².

Teacher's Note:
a) This is the same data as question 86.
b) Angle in a semicircle is 90°.

102. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in below figure. Find : (i) the total length of the silver wire required. (ii) the area of each sector of the brooch.

Answer:
1. Radius = 17.5 mm.
2. (i) Circumference = \( \frac{22}{7} \) × 35 = 110 mm. Five diameters = 5 × 35 = 175 mm. Total wire = 110 + 175 = 285 mm.
3. (ii) Area of the circle = \( \frac{22}{7} \) × 17.5 × 17.5 = 962.5 mm². Area of each sector = 962.5 ÷ 10 = 96.25 mm².

Teacher's Note:
a) The wire forms the rim and the five diameters.
b) Ten equal sectors share the circle equally.

103. The area of a sector is one-twelfth that of the complete circle. Find the angle of the sectort.

Answer:
1. Area of the sector = \( \frac{1}{12} \) of the area of the circle.
2. So \( \frac{\theta}{360} \) = \( \frac{1}{12} \) and θ = 360° ÷ 12 = 30°.

Teacher's Note:
a) Sector area is proportional to its angle.
b) One-twelfth of 360° is 30°.

104. Find the area of the circle in which a square of area 64 sq. cm is inscribed . (use π = 3.14)

Answer:
1. Side of the square = √64 = 8 cm, so its diagonal = 8√2 cm.
2. The diagonal is the diameter, so r = 4√2 cm and r² = 32.
3. Area of the circle = 3.14 × 32 = 100.48 cm².

Teacher's Note:
a) The diagonal of the inscribed square equals the diameter of the circle.
b) r² = \( \frac{d^{2}}{4} \) = \( \frac{128}{4} \) = 32.

105. In the below figure, ABC is right angled triangle at A. Find the area of the shaded region, if AB = 6cm and BC = 10cm..

Answer:
1. AC = √(10² - 6²) = 8 cm, so the area of the triangle = \( \frac{1}{2} \) × 6 × 8 = 24 cm².
2. Inradius r = \( \frac{AB + AC - BC}{2} \) = \( \frac{6 + 8 - 10}{2} \) = 2 cm.
3. Area of the circle = \( \frac{22}{7} \) × 2 × 2 = \( \frac{88}{7} \) ≈ 12.57 cm².
4. Shaded area = 24 - \( \frac{88}{7} \) = \( \frac{80}{7} \) ≈ 11.43 cm².

Teacher's Note:
a) For a right triangle with the right angle at A, the inradius is (AB + AC - BC) ÷ 2.
b) Shaded area = triangle - inscribed circle.

106. In the above right-sided figure, ABC is an equilateral triangle inscribed in a circle of radius 4 cm with centre O. Find the area of the shaded region.

Answer:
1. The three sides of the triangle cut off three equal segments, and the shaded region is one of them.
2. Side of the triangle = 4√3 cm, so area of △ABC = \( \frac{\sqrt{3}}{4} \) × 48 = 12√3 ≈ 20.78 cm².
3. Area of the circle = \( \frac{22}{7} \) × 4 × 4 = \( \frac{352}{7} \) ≈ 50.29 cm².
4. Total area of the three segments = 50.29 - 20.78 = 29.51 cm².
5. Shaded area = 29.51 ÷ 3 ≈ 9.84 cm².

Teacher's Note:
a) By symmetry the three segments have equal area.
b) Each segment = \( \frac{1}{3} \) × (circle - triangle).

107. The diameter of a coin is 1 cm see the below figure. If four such coins be placed on a table so that the rim of each touches that of the other two, find the area of the shaded region. (use π = 3.1416)

Answer:
1. Radius of a coin = 0.5 cm. The centres form a square of side 1 cm.
2. Area of the square = 1 cm².
3. The four quadrants inside the square make one circle: area = 3.1416 × 0.5 × 0.5 = 0.7854 cm².
4. Shaded area = 1 - 0.7854 = 0.2146 cm².

Teacher's Note:
a) The shaded region is the square of centres minus four quadrants.
b) Four quadrants make one full coin.

108. In the above right-sided figure, ABCD is a rectangle, having AB = 14 cm and BC = 20 cm. Two sectors of 180° have been cut off. Calculate (i) area of the shaded region (ii) length of the boundary of the shaded region.

Answer:
1. Each sector of 180° is a semicircle on a 14 cm side, so r = 7 cm.
2. (i) Area of the rectangle = 14 × 20 = 280 cm². Area of the two semicircles = \( \frac{22}{7} \) × 7 × 7 = 154 cm². Shaded area = 280 - 154 = 126 cm².
3. (ii) Boundary = two straight sides of 20 cm + two semicircular arcs = 2 × 20 + 2 × \( \frac{22}{7} \) × 7 = 40 + 44 = 84 cm.

Teacher's Note:
a) The two semicircles on opposite sides make one circle of radius 7 cm.
b) The boundary of the shaded region has two straight edges and two arcs.

109. Find the area of the shaded region given in below Figue

Answer:
1. From the figure the flower in the middle has overall width 14 - 3 - 3 = 8 cm.
2. It is a square of side 4 cm with a semicircle of diameter 4 cm on each side, so each semicircle has radius 2 cm.
3. Area of the flower = 4 × 4 + 4 × \( \frac{1}{2} \) × \( \frac{22}{7} \) × 2 × 2 = 16 + \( \frac{176}{7} \) = \( \frac{288}{7} \) ≈ 41.14 cm².
4. Area of the square sheet = 14 × 14 = 196 cm².
5. Shaded area = 196 - \( \frac{288}{7} \) = \( \frac{1084}{7} \) ≈ 154.86 cm².

Teacher's Note:
a) The four semicircles together make one circle of radius 2 cm.
b) The central square has side 4 cm, because the flower is 8 cm wide in all.

110. Find the number of revolutions made by a circular wheel of area 1.54 m² in rolling a distance of 176 m.

Answer:
1. πr² = 1.54, so r² = 1.54 × \( \frac{7}{22} \) = 0.49 and r = 0.7 m.
2. Circumference = 2 × \( \frac{22}{7} \) × 0.7 = 4.4 m.
3. Number of revolutions = 176 ÷ 4.4 = 40.

Teacher's Note:
a) Find the radius from the area, then the circumference.
b) Revolutions = distance ÷ circumference.

111. Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of 90° at the centre.

Answer:
1. The chord is the hypotenuse of the right triangle at the centre: 2r² = 25, so r² = 12.5.
2. Minor segment = \( \frac{1}{4} \)πr² - \( \frac{1}{2} \)r².
3. Major segment = πr² - minor segment.
4. Difference = major - minor = πr² - 2 × (minor) = \( \frac{1}{2} \)πr² + r² = r²(\( \frac{\pi}{2} \) + 1).
5. = 12.5 × (\( \frac{11}{7} \) + 1) = 12.5 × \( \frac{18}{7} \) = \( \frac{225}{7} \) ≈ 32.14 cm².

Teacher's Note:
a) The right angle at the centre makes the triangle isosceles right-angled.
b) Difference of the two segments = (area of circle) - 2 × (minor segment).
c) π = 22/7 is used because no value of π is given in the question.

112. Find the difference of the areas of a sector of angle 120° and its corresponding major sector of a circle of radius 21 cm.

Answer:
1. The major sector has angle 360° - 120° = 240°.
2. Difference = \( \frac{120}{360} \) × \( \frac{22}{7} \) × 21 × 21 = \( \frac{1}{3} \) × 1386 = 462 cm².

Teacher's Note:
a) The corresponding major sector has the remaining angle 240°.
b) Difference of sectors = \( \frac{240^{\circ} - 120^{\circ}}{360} \) of the circle area.

113. The central angles of two sectors of circles of radii 7 cm and 21 cm are respectively 120° and 40°. Find the areas of the two sectors as well as the lengths of the corresponding arcs.

Answer:
1. First sector (r = 7 cm, θ = 120°): area = \( \frac{120}{360} \) × \( \frac{22}{7} \) × 49 = \( \frac{154}{3} \) ≈ 51.33 cm²; arc = \( \frac{120}{360} \) × 2 × \( \frac{22}{7} \) × 7 = \( \frac{44}{3} \) ≈ 14.67 cm.
2. Second sector (r = 21 cm, θ = 40°): area = \( \frac{40}{360} \) × \( \frac{22}{7} \) × 441 = 154 cm²; arc = \( \frac{40}{360} \) × 2 × \( \frac{22}{7} \) × 21 = \( \frac{44}{3} \) ≈ 14.67 cm.

Teacher's Note:
a) The two arcs have the same length even though the areas differ.
b) Arc = \( \frac{\theta}{360} \) × 2πr.

114. The length of the minute hand of a clock is 5 cm. Find the area swept by the minute hand during the time period 6:05 a m and 6:40 a m.

Answer:
1. The time from 6:05 to 6:40 is 35 minutes.
2. 35 minutes correspond to \( \frac{35}{60} \) of a full turn, an angle of 210°.
3. Area swept = \( \frac{35}{60} \) × \( \frac{22}{7} \) × 5 × 5 = \( \frac{7}{12} \) × \( \frac{550}{7} \) = \( \frac{275}{6} \) ≈ 45.83 cm².

Teacher's Note:
a) The hand sweeps a sector with the minute hand as radius.
b) The minute hand turns 6° every minute, so 35 minutes make 210°.

115. All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if area of the circle is 1256 cm². (Use π = 3.14).

Answer:
1. A rhombus whose vertices lie on a circle is cyclic, so it is a square, and its diagonals are diameters.
2. 3.14 r² = 1256, so r² = 400 and r = 20 cm.
3. Each diagonal = 40 cm, so area = \( \frac{1}{2} \) × 40 × 40 = 800 cm².

Teacher's Note:
a) Opposite angles of a cyclic rhombus are supplementary and equal, so each is 90°.
b) Area of a rhombus = \( \frac{1}{2} \) × product of the diagonals.

116. An archery target has three regions formed by three concentric circles as shown in the below figure. If the diameters of the concentric circles are in the ratio 1: 2:3, then find the ratio of the areas of three regions.

Answer:
1. Let the radii be r, 2r and 3r.
2. Innermost region = πr².
3. Middle region = π(2r)² - πr² = 3πr².
4. Outer region = π(3r)² - π(2r)² = 5πr².
5. Ratio of the areas = 1 : 3 : 5.

Teacher's Note:
a) The ratio of the diameters is the same as the ratio of the radii.
b) Areas of rings are found by subtracting the inner circle from the outer circle.

117. Area of a sector of central angle 200° of a circle is 770 cm². Find the length of the corresponding arc of this sector.

Answer:
1. \( \frac{200}{360} \) × \( \frac{22}{7} \) × r² = 770, so r² = 770 × \( \frac{360}{200} \) × \( \frac{7}{22} \) = 441 and r = 21 cm.
2. Arc length = \( \frac{200}{360} \) × 2 × \( \frac{22}{7} \) × 21 = \( \frac{220}{3} \) ≈ 73.33 cm.

Teacher's Note:
a) Find the radius from the area first.
b) Then use arc length = \( \frac{\theta}{360} \) × 2πr.

118. Three circles each of radius 3.5 cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.

Answer:
1. The centres form an equilateral triangle of side 7 cm, with area \( \frac{\sqrt{3}}{4} \) × 49 = 12.25√3 = 12.25 × 1.732 = 21.217 cm².
2. The three 60° sectors inside it have total area \( \frac{1}{2} \) × \( \frac{22}{7} \) × 3.5 × 3.5 = 19.25 cm².
3. Area enclosed = 21.217 - 19.25 = 1.967 ≈ 1.97 cm².

Teacher's Note:
a) The area between the circles = triangle of centres - three 60° sectors.
b) Take √3 = 1.732 and π = 22/7 as no values are given.

119. Find the area of the sector of a circle of radius 5 cm, if the corresponding arc length is 3.5 cm.

Answer:
1. Area of the sector = \( \frac{1}{2} \) × l × r = \( \frac{1}{2} \) × 3.5 × 5 = 8.75 cm².

Teacher's Note:
a) Use the formula \( \frac{1}{2} \)lr for a sector with arc length l.
b) No value of π is needed.

120. Four circular cardboard pieces of radii 7 cm are placed on a paper in such a way that each piece touches other two pieces. Find the area of the portion enclosed between these pieces.

Answer:
1. The centres form a square of side 14 cm, with area 196 cm².
2. The four quadrants inside the square make one circle: area = \( \frac{22}{7} \) × 7 × 7 = 154 cm².
3. Area enclosed = 196 - 154 = 42 cm².

Teacher's Note:
a) The centres of four touching circles form a square of side 2r.
b) Enclosed area = square - four quadrants.

121. On a square cardboard sheet of area 784 cm², four congruent circular plates of maximum size are placed such that each circular plate touches the other two plates and each side of the square sheet is tangent to two circular plates. Find the area of the square sheet not covered by the circular plates.

Answer:
1. Side of the sheet = √784 = 28 cm.
2. Two plates fit along each side, so the diameter of each plate = 14 cm and the radius = 7 cm.
3. Area of the four plates = 4 × \( \frac{22}{7} \) × 7 × 7 = 616 cm².
4. Area not covered = 784 - 616 = 168 cm².

Teacher's Note:
a) The side of the square equals two diameters, giving r = 7 cm.
b) Uncovered area = square - four circles.

122. Floor of a room is of dimensions 5 m × 4 m and it is covered with circular tiles of diameters 50 cm each as shown in the below figure. Find the area of floor which is not covered by tiles.

Answer:
1. Number of tiles along the length = 500 ÷ 50 = 10; along the breadth = 400 ÷ 50 = 8. Total tiles = 80.
2. Radius of a tile = 25 cm = 0.25 m, so area of a tile = \( \frac{22}{7} \) × 0.25 × 0.25 = \( \frac{11}{56} \) m².
3. Area covered by 80 tiles = 80 × \( \frac{11}{56} \) = \( \frac{110}{7} \) ≈ 15.71 m².
4. Area of the floor = 5 × 4 = 20 m².
5. Area not covered = 20 - \( \frac{110}{7} \) = \( \frac{30}{7} \) ≈ 4.29 m².

Teacher's Note:
a) Count the tiles from the dimensions of the floor.
b) Convert all lengths to metres before finding areas.

123. With the vertices A, B and C of a triangle ABC as centres, arcs are drawn with radii 5 cm each as shown in below figure. If AB = 14 cm, BC = 48 cm and CA = 50 cm, then find the area of the shaded region. (Use π = 3.14).

Answer:
1. 14² + 48² = 196 + 2304 = 2500 = 50², so △ABC is right-angled at B.
2. Area of △ABC = \( \frac{1}{2} \) × 14 × 48 = 336 cm².
3. The three sectors have the same radius 5 cm and their angles add up to 180°: area = \( \frac{1}{2} \) × 3.14 × 25 = 39.25 cm².
4. Shaded area = 336 - 39.25 = 296.75 cm².

Teacher's Note:
a) Use the converse of the Pythagoras theorem to find the right angle.
b) The sum of the angles of a triangle is 180°, so the three sectors make a semicircle.

124. Find the area of the shaded region in the above right sided figure, where arcs drawn with centres A, B, C and D intersect in pairs at mid-points P, Q, R and S of the sides AB, BC, CD and DA, respectively of a square ABCD (Use π = 3.14).

Answer:
1. The square has side 12 cm, so each arc has radius 6 cm.
2. Area of the square = 12 × 12 = 144 cm².
3. The four quadrants make one circle: area = 3.14 × 6 × 6 = 113.04 cm².
4. Shaded area = 144 - 113.04 = 30.96 cm².

Teacher's Note:
a) The side 12 cm is read from the figure.
b) Each arc passes through the mid-points of two sides, so the radius is half the side.

125. Find the area of the shaded field shown in the below figure.

Answer:
1. The field is a rectangle 8 m by 4 m together with a semicircle at the bottom left.
2. The total height is 6 m and the rectangle is 4 m high, so the semicircle has radius 2 m (diameter 4 m).
3. Area of the rectangle = 8 × 4 = 32 m².
4. Area of the semicircle = \( \frac{1}{2} \) × \( \frac{22}{7} \) × 2 × 2 = \( \frac{44}{7} \) ≈ 6.29 m².
5. Area of the field = 32 + \( \frac{44}{7} \) = \( \frac{268}{7} \) ≈ 38.29 m².

Teacher's Note:
a) The semicircle fits below the left part of the rectangle and its diameter is half the width 8 m.
b) Add the two areas.

126. A calf is tied with a rope of length 6 m at the corner of a square grassy lawn of side 20 m. If the length of the rope is increased by 5.5m, find the increase in area of the grassy lawn in which the calf can graze.

Answer:
1. Original rope r = 6 m; new rope R = 6 + 5.5 = 11.5 m.
2. At a corner the calf grazes a quadrant, so the increase = \( \frac{1}{4} \)π(R² - r²).
3. = \( \frac{1}{4} \) × \( \frac{22}{7} \) × (11.5² - 6²) = \( \frac{1}{4} \) × \( \frac{22}{7} \) × 96.25 = 75.625 m².

Teacher's Note:
a) The calf can graze only a quarter circle in the corner of the lawn.
b) Both rope lengths are less than the side 20 m, so no part is cut off.

127. In the above right sided figure, ABCD is a trapezium with AB || DC, AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centres A, B, C and D have been drawn, then find the area of the shaded region of the figure.

Answer:
1. Area of the trapezium = \( \frac{1}{2} \) × (18 + 32) × 14 = 350 cm².
2. The sum of the angles of a trapezium is 360°, so the four sectors together make a full circle of radius 7 cm: area = \( \frac{22}{7} \) × 7 × 7 = 154 cm².
3. Shaded area = 350 - 154 = 196 cm².

Teacher's Note:
a) The four sectors of equal radius have total angle 360°.
b) Shaded area = trapezium - four sectors.

128. A circular pond is 17.5 m is of diameter. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of Rs 25 per m²

Answer:
1. Radius of the pond r = 8.75 m. Outer radius R = 8.75 + 2 = 10.75 m.
2. Area of the path = π(R + r)(R - r) = \( \frac{22}{7} \) × 19.5 × 2 = \( \frac{858}{7} \) ≈ 122.57 m².
3. Cost = \( \frac{858}{7} \) × 25 = Rs. 3064.29 (approximately).

Teacher's Note:
a) Convert the diameter to a radius before adding the width of the path.
b) The path is a ring, so use π(R² - r²).

129. A circular park is surrounded by a road 21 m wide. If the radius of the park is 105 m, find the area of the road.

Answer:
1. Outer radius R = 105 + 21 = 126 m.
2. Area of the road = π(R² - r²) = \( \frac{22}{7} \) × (126 + 105)(126 - 105) = \( \frac{22}{7} \) × 231 × 21 = 15246 m².

Teacher's Note:
a) The road is a ring of width 21 m.
b) Use π(R + r)(R - r) to keep the numbers small.

130. In the below figure, arcs are drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm. to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of the shaded region (Use π = 3.14).

Answer:
1. The radius of each arc = 5 cm and each angle of the triangle = 60°.
2. Area of the equilateral triangle = \( \frac{\sqrt{3}}{4} \) × 100 = 25√3 = 25 × 1.732 = 43.30 cm².
3. Area of the three 60° sectors = \( \frac{1}{2} \) × 3.14 × 25 = 39.25 cm².
4. Shaded area = 43.30 - 39.25 = 4.05 cm².

Teacher's Note:
a) Three 60° sectors together make a semicircle.
b) Only π is given, so use √3 = 1.732.

131. In the above right sided figure, arcs have been drawn with radii 14 cm each and with centres P, Q and R. Find the area of the shaded region.

Answer:
1. The three shaded sectors have the same radius 14 cm.
2. The angles at P, Q and R are the angles of a triangle, so they add up to 180°.
3. Shaded area = \( \frac{180}{360} \) × \( \frac{22}{7} \) × 14 × 14 = 308 cm².

Teacher's Note:
a) The sum of the angles of a triangle is 180°, so the three sectors make a semicircle.
b) π = 22/7 is used because no value of π is given in the question.

132. In the below figure, arcs have been drawn of radius 21 cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region.

Answer:
1. The four sectors have the same radius 21 cm.
2. The angles of a quadrilateral add up to 360°, so the four sectors make one full circle.
3. Shaded area = \( \frac{22}{7} \) × 21 × 21 = 1386 cm².

Teacher's Note:
a) The angle sum of a quadrilateral is 360°.
b) π = 22/7 is used because no value of π is given in the question.

133. A piece of wire 20 cm long is bent into the form of an arc of a circle subtending an angle of 60° at its centre. Find the radius of the circle.

Answer:
1. Arc length = \( \frac{60}{360} \) × 2πr = \( \frac{\pi r}{3} \) = 20.
2. r = \( \frac{60}{\pi} \) = 60 × \( \frac{7}{22} \) = \( \frac{210}{11} \) ≈ 19.09 cm.

Teacher's Note:
a) The length of the wire is the length of the arc.
b) π = 22/7 is used because no value of π is given in the question.

134. In the below figure, ABC is a right angled triangle at B, AB = 28 cm and BC = 21 cm. With diameter a semicircle is drawn and with BC as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.

Answer:
1. AC² = 28² + 21² = 1225, so AC = 35 cm. The semicircle is drawn on AC as diameter, with radius 17.5 cm.
2. Area of the semicircle = \( \frac{1}{2} \) × \( \frac{22}{7} \) × 17.5 × 17.5 = 481.25 cm².
3. Area of △ABC = \( \frac{1}{2} \) × 28 × 21 = 294 cm².
4. Area of the quadrant with radius BC = \( \frac{1}{4} \) × \( \frac{22}{7} \) × 21 × 21 = 346.5 cm².
5. Shaded area = semicircle + triangle - quadrant = 481.25 + 294 - 346.5 = 428.75 cm².

Teacher's Note:
a) The printed question leaves out the diameter; the semicircle is on AC, the hypotenuse.
b) The quadrant lies inside the semicircle and the triangle together, so it is subtracted.

135. In the above right-sided figure, O is the centre of a circular arc and AOB is a straight line. Find the perimeter and the area of the shaded region. (use π = 3.142)

Answer:
1. AB is a diameter and ∠ACB = 90°, so AB² = 12² + 16² = 400 and AB = 20 cm, radius = 10 cm.
2. Perimeter of the shaded region = semicircular arc + AC + CB = 3.142 × 10 + 12 + 16 = 31.42 + 28 = 59.42 cm.
3. Area of the semicircle = \( \frac{1}{2} \) × 3.142 × 100 = 157.1 cm².
4. Area of △ACB = \( \frac{1}{2} \) × 12 × 16 = 96 cm².
5. Shaded area = 157.1 - 96 = 61.1 cm².

Teacher's Note:
a) The angle in a semicircle is a right angle.
b) The straight diameter AB is not a part of the shaded boundary; the chords AC and CB are.