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# Areas related to circles - Area of sector and segment of a circle (Class 10 maths)

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Areas related to circles - Area of sector and segment of a circle (Class 10 maths).

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### Areas related to circles - Area of sector and segment of a circle (Class 10 maths)

1. 1. Areas Related To Circles Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com
2. 2. Chapter : Areas Related To Circles Website: www.letstute.com Q) 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: a) The area of that part of the field in which the horse can graze b) The increase in the grazing area if the rope were 10 m long instead of 5 m Use [ π = 3.14] Given: Side of a square = 15m Length of the rope = 5 m To Find: a) Area of that part of the field in which the horse can graze. b) Increase in the grazing area if 10 m long rope is used Problems based on Area of sector and segment of a circle C 15m 15m 5 m 5m A B D Q 15m 15m P 10 m Q
3. 3. Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com Solution: a) Let ABCD represent the square shaped grass field of side 15 m. Let D be the corner to which the horse is tethered and let DP (= 5m) be the rope by which it is tied. Area of the part of the field over which the horse can graze = Area of quadrant DPQ Thus, radius r (= 5m) and sector angle θ (= 90⁰) C 15m 15m 5 m 5m A B D Q 15m 15m P 10 m Q
4. 4. Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com = θ x πr2 360 = = 19.625 m2 Hence, the area of the part of the field in which the horse can graze is 19.625 m2      5514.3 360 90 m2      5514.3 4 1 = m2 4 5.78= m2 Area of quadrant DPQ C 15m 15m 5 m 5m A B D Q 15m 15m P 10 m Q
5. 5. Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com b) If r = 10 m, then the grazing area = m2 = 78.5 m2 = m2      101014.3 360 90 = θ x πr2 360      101014.3 4 1 = m2 4 314 C 15m 15m 5 m 5m A B D Q 15m 15m P 10 m Q
6. 6. Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com Increase in grazing area = (78.5 - 19.625) m2 = Hence, the increase in the grazing area is 58.875 m2 58.875m2 C 15m 15m 5 m 5m A B D Q 15m 15m P 10 m Q
7. 7. Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com Q) 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. Given: Number of ribs = 8 Radius = 45 cm To find: Area between two consecutive ribs = ?
8. 8. Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com Solution: Angle made by the two consecutive ribs of the umbrella at the centre = Angle of the full circle Number of ribs = 360⁰ 8 = 45⁰
9. 9. Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com Area between two consecutive ribs = Area of a sector of the circle of radius r (=45cm) and sector angle θ (=450) = θ x πr2 360 = cm2 = cm2      4545 7 22 360 45 = cm2      4545 7 11 4 1 28 22275
10. 10. Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com Hence, the area between two consecutive ribs of the umbrella is 795.53 cm2 = 795.53 cm2
11. 11. Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com Q) The diagram represents the area swept by the wiper of a car with the dimensions given in the figure, calculate the shaded area swept by the wiper. O 300 7 cm 14 cm B D C A Given: OD = 7 cm DC = 14 cm ∠ 𝐂𝐎𝐁 = ∠ 𝐃𝐎𝐀 = 𝛉 = 𝟑𝟎 𝟎 To find: Area swept by the wiper??
12. 12. Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com Solution: Let the radii of the sectors COB and DOA be ‘R’ and ‘r’ respectively Then, R = (7+14) cm = 21 cm and r = 7 cm -------- (1) ∠ 𝐂𝐎𝐁 = ∠ 𝐃𝐎𝐀 = 𝛉 = 𝟑𝟎 𝟎 −−−−−−−− (2) 300 7 cm 14 cm O B D C A
13. 13. Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com 300 O B D C A Area of the shaded = [Area of sector COB – Area of sector DOA] region = θ x πR2 - θ x πr2 360 360 = θ x π (R2 - r2) 360 = 30 x 22 (212 – 72) cm2 360 7 Using (1) and (2), we get, = 1 x 22 (441 - 49) cm2 12 7 = 22 x 392 cm2 84 7 cm 14 cm
14. 14. Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com 300 7 cm 14 cm O B D C A = 11 x 392 cm2 42 Hence, the area swept by the wiper is 102.67 cm2 102.67cm2 = 11 x 56 cm2 6 = 11 x 28 cm2 3 = 308 cm2 3 =
15. 15. Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com Q) In the given figure, O is the center of the concentric circles. Radius of the inner circle is half the radius of the outer circle. Given∠AOC = 1350 and OA = 14 cm, Calculate the area of the shaded region. (Leave your answer in terms of π) Given:∠AOC = 1350 OA = 14 cm To find: Area of the shaded region = ? S R O1350 A D C B P Q 14 cm
16. 16. Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com Solution: Let the radii of the outer circle and inner concentric circles be ‘R’ and ‘r’ respectively. Then, R = 14 cm and r = 7 cm -------- (1) ∠ 𝐀𝐎𝐃 = 𝟏𝟖𝟎 𝟎 − 𝟏𝟑𝟓 𝟎 = 450 ------- (Linear pair) 𝐀𝐥𝐬𝐨 ∠ BOC =∠ AOD = 450 (=θ) ----- V. opp.∠’s ---------- (2) S R O1350 A D C B P Q 14 cm
17. 17. Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com Area of the shaded region = 2[Area of the sector AOD – Area of sector POQ] = 2 𝛉 𝟑𝟔𝟎 × 𝛑𝐑 𝟐 − 𝛉 𝟑𝟔𝟎 × 𝛑𝐫 𝟐 = 2 × 𝟒𝟓 𝟑𝟔𝟎 𝛑 𝟏𝟒 𝟐 − 𝟕 𝟐 𝐜𝐦 𝟐 By using (1) and (2), we get, = 𝟐 × 𝟓 𝟒𝟎 𝛑 𝟏𝟗𝟔 − 𝟒𝟗 𝐜𝐦 𝟐 = 𝟐 𝟖 𝛑 × 𝟏𝟒𝟕 cm2 = 𝟐 × 𝛉 𝟑𝟔𝟎 × 𝛑(𝐑 𝟐 − 𝐫 𝟐 ) S R O1350 A D C B P Q 14 cm
18. 18. Problems based on Area of sector and segment of a circle Chapter : Areas Related To Circles Website: www.letstute.com Result: Area = Hence, the area of the shaded region is 𝟏𝟒𝟕 𝟒 𝛑 cm2 𝟏𝟒𝟕 𝟒 𝛑cm2 = 𝛑 𝟒 × 𝟏𝟒𝟕 𝐜𝐦 𝟐 = 𝟏𝟒𝟕 𝟒 𝛑cm2 S R O1350 A D C B P Q 14 cm
19. 19. Now we know… 19 Problems based on Area of sector and segment of a circle Please visit www.letstute.com to take a test Chapter : Areas Related To Circles Website: www.letstute.com
20. 20. Next video…. 20 Problems based on Areas of combination of plane figures Please visit www.letstute.com to view the next video Chapter : Areas Related To Circles Website: www.letstute.com