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Q1. |
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Given: Radius of circle OA = OB = 6 cm Angle of the sector \(\theta\) = \(60^{0}\) Let OAMB be a sector of a circle. We knot that Area of sector (A) is: $$A=\frac{\theta}{360^{0}}\times\pi r^{2}$$ Putting value $$A=\frac{60^{0}}{360^{0}}\times\frac{22}{7}\times (6)^{2}cm^{2}$$ $$A=\frac{1}{6}\times\frac{22}{7}\times 36 cm^{2}$$ $$A=\frac{22}{7}\times 6=\frac{132}{7} cm^{2}$$ |
Q2. | ||
Given: Circumference of circle = 22 cm We know that Circumference of circle (C): \[C=2\pi r\] putting value \[22=2 \pi r\] \[r = \frac {22}{2\pi}=\frac{11}{\pi}\] We know that Quadrant of a circle will subtend \(\frac{360^{0}}{4}=90^{0}\) angle with centre of circle. so, by Area of sector (A) formula: $$A=\frac{\theta}{360^{0}}\times\pi r^{2}$$ Putting value \[A=\frac{90^{0}}{360^{0}}\times\pi\times \left(\frac{11}{\pi}\right)^{2}\] \[A=\frac{1}{4}\times\pi\times \left(\frac{121}{\pi\times\pi}\right)\] \[A=\frac{121}{4\pi}=\frac{121\times7}{4\times22}=\frac{847}{88}=\frac{77}{8}cm^{2}\] Alternative method: We know that the area (A) of a quadrant is one-fourth of a circle. So we can write as \[\frac {1}{4}\pi r^{2}\] Putting value \[\frac {1}{4}\times\pi\times \left(\frac{11}{\pi}\right)^{2}\] \[A=\frac{1}{4}\times\pi\times \left(\frac{121}{\pi\times\pi}\right)\] \[A=\frac{121}{4\pi}=\frac{121\times7}{4\times22}=\frac{847}{88}=\frac{77}{8}cm^{2}\] |
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 \(\Rightarrow 360^{0}\) So in 5 minutes, minute hand will rotate\(=\frac{360^{0}}{60}\times 5 = 30^{0}\) So, we can say that- Area swept by minute hand in 5 minutes = the area of a sector of \(30^{0}\) in a circle of 14 cm radius. We know that Area of a sector of angle \(\theta\) is \[A=\frac{\theta}{360^{0}}\times \pi r^{2}\] Putting value \[A=\frac{30^{0}}{360^{0}}\times\frac{22}{7}\times 14\times 14\] \[A=\frac{1}{12}\times\frac{22}{7}\times 14\times 14\] \[A=\frac{11}{6}\times 2\times 14\] \[A=\frac{11}{3}\times 14=\frac{154}{3}cm^{2}\] |
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 \(\Pi\) = 3.14) | |
So, Area of segment PMQ (Minor segment)= Area of sector OPMQ — Area of △POQ Area of sector OPMQ : Area of △POQ:\(=\frac{\theta}{360^{0}}\times\pi r^{2}\)\(=\frac{90^{0}}{360^{0}}\times\frac{22}{7} \times 10\times 10\)\(=\frac{1}{4}\times\frac{22}{7} \times 100\)\(=\frac{11}{14}\times 100\)\(=\frac{1100}{14}=78.5 cm^{2}\) Area of segment PMQ (Minor segment) = Area of sector OPMQ — Area of △POQ\(=\frac{1}{2}\times base \times height\)\(=\frac{1}{2}\times OP \times OQ\)\(=\frac{1}{2}\times 10 \times 10\)\(=\frac{1}{2}\times 100=50 cm^{2}\) Area of segment PMQ (Minor segment) = \(78.5 - 50 = 28.5 cm^{2}\) Area of Major segment = Area of circle — Area of miner sector OPMQ Area of Major segment = \(=\frac{360^{0}}{360^{0}}\times \pi r^{2}-\frac{90^{0}}{360^{0}}\times \pi r^{2}\)\(=\left(\frac{360^{0}}{360^{0}}-\frac{90^{0}}{360^{0}}\right)\times \pi r^{2}\)\(=\left(\frac{360^{0}-90^{0}}{360^{0}}\right)\times \pi r^{2}\)\(=\left(\frac{270^{0}}{360^{0}}\right)\times \pi r^{2}\)\(=\frac{3}{4}\times 3.14\times10\times10=\frac{942}{4}=235.5 cm^{2}\) |
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 = \(60^{0}\) (i) Length of Arc ACB = \(\frac {\theta}{360^{0}}\times 2\pi r\) Putting value (ii) Area of sector OACB = \(\frac {\theta}{360^{0}}\times \pi r^{2}\)\(=\frac {60}{360}\times 2\pi \times 21\)\(=\frac {1}{6}\times 2\pi \times 21\)\(=\frac {\pi}{3}\times 21\)\(=7\pi = 7\times\frac{22}{7}=22cm\) Putting value (iii) In \(triangle OAB\)\(=\frac {60}{360}\times \frac{22}{7} \times 21\times 21\)\(=\frac {1}{6}\times \frac{22}{7} \times 21\times 21\)\(=\frac {1}{2}\times \frac{22}{7} \times 7\times 21\)\(=11\times 21=231cm^{2}\) thus we can say that \(\triangle OAB\) is an equilateral traiangle.OA = OB so,\(\angle A = \angle B\)\(\angle A + \angle O + \angle B = 180^{0}\)\(2\angle A + 60^{0} = 180^{0}\)So \(\angle A = 60^{0}\) We know that area of equilateral traiangle \((\triangle OAB) = \frac{\sqrt{3}}{4}\times \left(side\right)^{2}\) Putting value Thus Area of segment ACB = Area of sector OACB — Area of \(triangle OAB\)\(=\frac{\sqrt{3}}{4} \times 21 \times 21=\frac{441\sqrt{3}}{4} cm^{2}\) \(=\left(231-\frac{441\sqrt{3}}{4}\right)cm^{2}\) |
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 \(\pi = 3.14\) and \(\sqrt {3}=1.73\). | |
Radius of circle = OA = OB = 15 cm Chord of circle subtends an angle of 60° \(\pi = 3.14\) and \(\sqrt {3}=1.73\) Area of Sector OACB : In \(\triangle AOB\)\(=\frac{60}{360}\times \pi r^{2}\)\(=\frac{60}{360}\times 3.14 \times 15 \times 15\)\(=\frac{1}{6}\times 3.14 \times 15 \times 15\)\(=\frac{706.5}{6}=117.75 cm^{2}\) We know that area of a equilateral triangle is: \(=\frac{\sqrt{3}}{4} \times (side)^{2}\)Since \(OA=OB\)So, \(\angle A = \angle B\)\(\angle A + \angle B + \angle O = 180^{0}\)\(2\angle A + 60^{0} = 180^{0}\)\(2\angle A = 180^{0}-60^{0}\)\(2\angle A = 120^{0}\)\(\angle A = 60^{0}\)So, \(\triangle AOB\) is a equilateral triangle. Putting value, we get Area of ACB = Area of sector OACB - Area of \(triangle AOB\)\(=\frac{\sqrt{3}}{4} \times 15 \times 15=\frac{225\times {1.73}}{4} = 97.3125 cm^{2}\) Area of Major Segment = Area of circle — Area of minor segment (ACB)\(= 117.75 - 97.3125 = 20.4375 cm^{2}\) \(=\pi r^{2}-20.4375\)\(=3.14 \times 15 \times 15 - 20.4375\)\(=3.14 \times 225 - 20.4375\)\(=706.5 - 20.4375\)\(=686.0625 cm^{2}\) |
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 \(\pi = 3.14\) and \(\sqrt {3}=1.73\). | |
Radius of circle = OA = OB = 12 cm Chord of circle subtends an angle of 120° \(\pi = 3.14\) and \(\sqrt {3}=1.73\) 1. Area of the Sector : \[\text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2\] putting given values: \[\text{Area of sector} = \frac{120^\circ}{360^\circ} \times 3.14 \times (12)^2\] \[\text{Area of sector} = \frac{1}{3} \times 3.14 \times 144\] \[\text{Area of sector} = 150.72 \, \text{cm}^2\] 2. Area of the Triangle: The triangle is an equilateral triangle because the central angle is 120 degrees. \[\text{Area of triangle} = \frac{\sqrt{3}}{4} \times a^2\] In this case, the side length is the radius \(r = 12\) cm: \[\text{Area of triangle} = \frac{\sqrt{3}}{4} \times (12)^2\] \[\text{Area of triangle} = \frac{\sqrt{3}}{4} \times 144\] \[\text{Area of triangle} = 36 \sqrt{3} \, \text{cm}^2\] 3. Area of the Segment: \[\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}\] \[\text{Area of segment} = 150.72 - 36 \sqrt{3} \, \text{cm}^2\] Using the given approximation \(\sqrt{3} \approx 1.73\): \[\text{Area of segment} \approx 150.72 - 36 \times 1.73 \, \text{cm}^2\] \[\text{Area of segment} \approx 150.72 - 62.28 \, \text{cm}^2\] \[\text{Area of segment} \approx 88.44 \, \text{cm}^2\] So, the area of the corresponding segment of the circle is approximately \(88.44 \, \text{cm}^2\). |
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 \((\pi\=3.14)\) | |
Radius of circle = length of rope = OA = OB = 5 m As per construction \(\theta = 120^{0}\) \(\pi = 3.14\) \) (i) Part of the area where horse can graze = Area of sector OACB \[\text{Area of sector OACB} = \frac{\theta}{360^\circ} \times \pi r^2\] putting given values: \[\text{Area of sector OACB} = \frac{90^\circ}{360^\circ} \times 3.14 \times (5)^2\] \[\text{Area of sector OACB} = \frac{1}{4} \times 3.14 \times 25\] \[\text{Area of sector OACB} = 19.625 \, \text{m}^2\] (ii) Part of the area where horse can graze if rope length is 10 m long = Area of sector OACB when r = 10 m \[\text{Area of sector OACB} = \frac{\theta}{360^\circ} \times \pi r^2\] putting given values: \[\text{Area of sector OACB} = \frac{90^\circ}{360^\circ} \times 3.14 \times (10)^2\] \[\text{Area of sector OACB} = \frac{1}{4} \times 3.14 \times 100\] \[\text{Area of sector OACB} = 78.5\, \text{m}^2\] Change in graze are when rope length changed 5 m to 10 m = \(78.5-19.625 = 58.875 m^{2}\) |
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 \(=\frac {35}{2}\text {mm}\) We know that circumference of circle \(=2\pi r\) putting value we get \(\text{Length of wire required} = 5 \times \text {Diameter} + \text {circumference}\)\(=2\times \frac {22}{7} \times \left(\frac{35}{2}\right) \)\(=2\times \times 11 \times 5 = 110 mm\) Putting value \(\text{Length of wire required} = \text {5} \times \text {35} + \text {110}\) \(\text{Length of wire required} = \text {175} + \text {110}=\text {285 mm}\) Each of 10 sector of circle subtending \(\frac{360}{10}=36^{0}\) at center of circle \(\text {So Area of each sector of the brooch}\) = \(\frac {36^{0}}{360^{0}} \times \pi r^{2}\) Putting value \(= \frac {36^{0}}{360^{0}} \times \frac {22}{7} \times\left(\frac{35}{2}\right)\times\left(\frac{35}{2}\right)=\frac {385}{4} mm^{2}\) |
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 \((\theta) = \frac {360^{0}}{8}=45^{0}\) So, the area of two consecutive ribs \(=\frac {\theta}{360} \times \pi r^{2}\) Putting value \(=\frac {45}{360} \times \frac{22}{7} \times 45 \times 45\)\(=\frac {1}{8} \times \frac{22}{7} \times 45 \times 45\)\(=\frac {11}{28} \times 2025\)\(=\frac {22275}{28}=795.53 cm^{2}\) |
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 \((\theta) = 115^{0}\) Area cleaned at each sweep of the blades = Area of sector So, The area cleaned at each sweep of the blade \(=\frac {\theta}{360} \times \pi r^{2}\) Putting value \(\text {Total area swept by both blade} = 2 \times \frac {158125}{252} =\frac {158125}{126} = 1254.96 cm^{2}\)\(=\frac {115}{360} \times \frac{22}{7} \times 25 \times 25\)\(=\frac {23}{72} \times \frac{22}{7} \times 25 \times 25\)\(=\frac {158125}{252} cm^{2}\) |
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 \(\pi = 3.14)\) | |
Distance of light spread = radius = 16.5 km Sector of Angle \((\theta) = 80^{0}\) Area of sector = Area of the sea over which the ships are warned So, The area of sector \(=\frac {\theta}{360} \times \pi r^{2}\) Putting value \(=\frac {80}{360} \times 3.14 \times 16.5 \times 16.5\)\(=\frac {2}{9} \times 3.14 \times 16.5 \times 16.5\)\(=189.97 km^{2}\) |
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 \(cm^{2}\).  | |
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 \(\frac {360}{6} = 60^{0}\) at the centre of circle. In \(\triangle OBC\) We know that area of a equilateral triangle is: \(=\frac{\sqrt{3}}{4} \times (side)^{2}\)Since \(OB=OC\)So, \(\angle B = \angle C\)\(\angle B + \angle C + \angle O = 180^{0}\)\(2\angle B + 60^{0} = 180^{0}\)\(2\angle B = 180^{0}-60^{0}\)\(2\angle B = 120^{0}\)\(\angle B = 60^{0}\)So, \(\triangle OBC\) is a equilateral triangle. Putting value, we get \(=\frac{\sqrt{3}}{4} \times 28 \times 28\)
Area of sector OBPC \(=\frac {\theta}{360} \times \pi r^{2}\) Putting value
Area of segment BPC = Area of sector OBPC - Area of \(\triangle OBC\) \(\text {Area of segment BPC} = \text {410.6667 - 333.2 = 77.4667} cm^{2}\) \(\text {Total area of design} = 6 \times 77.4667 = 464.8 cm^{2}\) \(\text {Cost for making } 1 cm^{2} \text { designs = Rs. 0.36}\) \(\text {Cost for making } 464.8 cm^{2} \text { designs =} 464.8 \times 0.35 = Rs. 162.68\) 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 \(\text {(A)} = \frac {p}{180} \times 2 \pi R\) \(\text { }\) \(\text {(B)} = \frac {p}{180} \times \pi R^{2}\) \(\text {(C)} = \frac {p}{360} \times 2 \pi R\) \(\text { }\) \(\text {(D)} = \frac {p}{720} \times 2 \pi R^{2}\) | |
Angle with Center \(\theta = p\) We know that area of sector = \(\frac {\theta}{360^{0}} \times \pi R^{2}\) Putting value By multiplying numerator and denominator by 2, we get\(=\frac {p^{0}}{360^{0}} \times \pi R^{2}\) Hence, right answer is option (D).\( = \frac {p}{720} \times 2 \pi R^{2}\) |
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