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Given: Radius of circle OA = OB = 6 cm Angle of the sector θ = 600 Let OAMB be a sector of a circle. We knot that Area of sector (A) is: A=θ3600×πr2 Putting value A=6003600×227×(6)2cm2 A=16×227×36cm2 A=227×6=1327cm2 |
Q2. | ||
Given: Circumference of circle = 22 cm We know that Circumference of circle (C): C=2πr putting value 22=2πr r=222π=11π We know that Quadrant of a circle will subtend 36004=900 angle with centre of circle. so, by Area of sector (A) formula: A=θ3600×πr2 Putting value A=9003600×π×(11π)2 A=14×π×(121π×π) A=1214π=121×74×22=84788=778cm2 Alternative method: We know that the area (A) of a quadrant is one-fourth of a circle. So we can write as 14πr2 Putting value 14×π×(11π)2 A=14×π×(121π×π) A=1214π=121×74×22=84788=778cm2 |
Q3. | The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. | |
Minute hand rotates in 1 hour = 60 minutes ⇒3600 So in 5 minutes, minute hand will rotate=360060×5=300 So, we can say that- Area swept by minute hand in 5 minutes = the area of a sector of 300 in a circle of 14 cm radius. We know that Area of a sector of angle θ is A=θ3600×πr2 Putting value A=3003600×227×14×14 A=112×227×14×14 A=116×2×14 A=113×14=1543cm2 |
Q4. | 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) | |
So, Area of segment PMQ (Minor segment)= Area of sector OPMQ — Area of △POQ Area of sector OPMQ : Area of △POQ:=θ3600×πr2=9003600×227×10×10=14×227×100=1114×100=110014=78.5cm2 Area of segment PMQ (Minor segment) = Area of sector OPMQ — Area of △POQ=12×base×height=12×OP×OQ=12×10×10=12×100=50cm2 Area of segment PMQ (Minor segment) = 78.5−50=28.5cm2 Area of Major segment = Area of circle — Area of miner sector OPMQ Area of Major segment = =36003600×πr2−9003600×πr2=(36003600−9003600)×πr2=(3600−9003600)×πr2=(27003600)×πr2=34×3.14×10×10=9424=235.5cm2 |
Q5. | 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 | |
Angle sbtended by arc with centre = 600 (i) Length of Arc ACB = θ3600×2πr Putting value (ii) Area of sector OACB = θ3600×πr2=60360×2π×21=16×2π×21=π3×21=7π=7×227=22cm Putting value (iii) In triangleOAB=60360×227×21×21=16×227×21×21=12×227×7×21=11×21=231cm2 thus we can say that △OAB is an equilateral traiangle.OA = OB so,∠A=∠B∠A+∠O+∠B=18002∠A+600=1800So ∠A=600 We know that area of equilateral traiangle (△OAB)=√34×(side)2 Putting value Thus Area of segment ACB = Area of sector OACB — Area of triangleOAB=√34×21×21=441√34cm2 =(231−441√34)cm2 |
Q6. | 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. | |
Radius of circle = OA = OB = 15 cm Chord of circle subtends an angle of 60° π=3.14 and √3=1.73 Area of Sector OACB : In △AOB=60360×πr2=60360×3.14×15×15=16×3.14×15×15=706.56=117.75cm2 We know that area of a equilateral triangle is: =√34×(side)2Since OA=OBSo, ∠A=∠B∠A+∠B+∠O=18002∠A+600=18002∠A=1800−6002∠A=1200∠A=600So, △AOB is a equilateral triangle. Putting value, we get Area of ACB = Area of sector OACB - Area of triangleAOB=√34×15×15=225×1.734=97.3125cm2 Area of Major Segment = Area of circle — Area of minor segment (ACB)=117.75−97.3125=20.4375cm2 =πr2−20.4375=3.14×15×15−20.4375=3.14×225−20.4375=706.5−20.4375=686.0625cm2 |
Q7. | 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. | |
Radius of circle = OA = OB = 12 cm Chord of circle subtends an angle of 120° π=3.14 and √3=1.73 1. Area of the Sector : Area of sector=θ360∘×πr2 putting given values: Area of sector=120∘360∘×3.14×(12)2 Area of sector=13×3.14×144 Area of sector=150.72cm2 2. Area of the Triangle: The triangle is an equilateral triangle because the central angle is 120 degrees. Area of triangle=√34×a2 In this case, the side length is the radius r=12 cm: Area of triangle=√34×(12)2 Area of triangle=√34×144 Area of triangle=36√3cm2 3. Area of the Segment: Area of segment=Area of sector−Area of triangle Area of segment=150.72−36√3cm2 Using the given approximation √3≈1.73: Area of segment≈150.72−36×1.73cm2 Area of segment≈150.72−62.28cm2 Area of segment≈88.44cm2 So, the area of the corresponding segment of the circle is approximately 88.44cm2. |
Q8. | 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 (see Fig. 11.8). 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) | |
Radius of circle = length of rope = OA = OB = 5 m As per construction θ=1200 π=3.14 \) (i) Part of the area where horse can graze = Area of sector OACB Area of sector OACB=θ360∘×πr2 putting given values: Area of sector OACB=90∘360∘×3.14×(5)2 Area of sector OACB=14×3.14×25 Area of sector OACB=19.625m2 (ii) Part of the area where horse can graze if rope length is 10 m long = Area of sector OACB when r = 10 m Area of sector OACB=θ360∘×πr2 putting given values: Area of sector OACB=90∘360∘×3.14×(10)2 Area of sector OACB=14×3.14×100 Area of sector OACB=78.5m2 Change in graze are when rope length changed 5 m to 10 m = 78.5−19.625=58.875m2 |
Q9. | 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 Fig. 11.9. Find : (i) the total length of the silver wire required. (ii) the area of each sector of the brooch. | |
Diameter of circle = 35 mm So, Radius of circle =352mm We know that circumference of circle =2πr putting value we get Length of wire required=5×Diameter+circumference=2×227×(352)=2××11×5=110mm Putting value Length of wire required=5×35+110 Length of wire required=175+110=285 mm Each of 10 sector of circle subtending 36010=360 at center of circle So Area of each sector of the brooch = 3603600×πr2 Putting value =3603600×227×(352)×(352)=3854mm2 |
Q10. | An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. | |
Ribs in Umbrella = 8 So, Radius of circle = 45 cm We have to find Area between two consecutive ribs. Angle from the centre (θ)=36008=450 So, the area of two consecutive ribs =θ360×πr2 Putting value =45360×227×45×45=18×227×45×45=1128×2025=2227528=795.53cm2 |
Q11. | A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. | |
Length of wiper blade = radius = 25 cm Angle from the centre (θ)=1150 Area cleaned at each sweep of the blades = Area of sector So, The area cleaned at each sweep of the blade =θ360×πr2 Putting value Total area swept by both blade=2×158125252=158125126=1254.96cm2=115360×227×25×25=2372×227×25×25=158125252cm2 |
Q12. | 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) | |
Distance of light spread = radius = 16.5 km Sector of Angle (θ)=800 Area of sector = Area of the sea over which the ships are warned So, The area of sector =θ360×πr2 Putting value =80360×3.14×16.5×16.5=29×3.14×16.5×16.5=189.97km2 |
Q13. | A round table cover has six equal designs as shown in Fig. 11.11. 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. | |
radius = 28 cm As per construction we can say table design in hexagon shape and design are segment of a circle. Lets take a segment PBC. Chord BC is a side of hexagon. Each chord will substitute 3606=600 at the centre of circle. In △OBC We know that area of a equilateral triangle is: =√34×(side)2Since OB=OCSo, ∠B=∠C∠B+∠C+∠O=18002∠B+600=18002∠B=1800−6002∠B=1200∠B=600So, △OBC is a equilateral triangle. Putting value, we get =√34×28×28
Area of sector OBPC =θ360×πr2 Putting value
Area of segment BPC = Area of sector OBPC - Area of △OBC Area of segment BPC=410.6667 - 333.2 = 77.4667cm2 Total area of design=6×77.4667=464.8cm2 Cost for making 1cm2 designs = Rs. 0.36 Cost for making 464.8cm2 designs =464.8×0.35=Rs.162.68 Hence, The cost of making such type designs is Rs. 162.68 |
Q14. | Tick the correct answer in the following : Area of a sector of angle p (in degrees) of a circle with radius R is (A)=p180×2πR (B)=p180×πR2 (C)=p360×2πR (D)=p720×2πR2 | |
Angle with Center θ=p We know that area of sector = θ3600×πR2 Putting value By multiplying numerator and denominator by 2, we get=p03600×πR2 Hence, right answer is option (D).=p720×2πR2 |
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