2<\/sup> \nTherefore, area of major segment is 686.0625 cm\u00b2<\/p>\n <\/p>\n
Question 7. \nA chord of a circle of the radius 12 cm subtends an angle of 120\u00b0 at the centre. Find the area of the corresponding segment of the circle. (Use \u03c0 = 3.14 and \\(\\sqrt{3}\\) = 1.73). \nSolution: \nConstruct a circle C with radius 12 cm and centre O \n \nArea of the corresponding segment = Area of sector – Area of \u2206AOB = 150.72 – 62.28 cm\u00b2 \n= 88.44 cm\u00b2<\/p>\n
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Question 8. \nA 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 figure). Find \n(i) the area of that part of the field in which the horse can graze. \n(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use \u03c0 = 3.14) \n \nSolution: \n(i) Length of the rope = Radius of the sector grazed by horse = 5 m \nHere, angle of the sector = 90\u00b0 \n \nHence, the area of the part of the field which is grazed by horse is 19625 m\u00b2.<\/p>\n
(ii) Now, radius = 10m \nAngle = 90\u00b0 \n <\/p>\n
Question 9. \nA 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 figure. \n \nFind: \n(i) the total length of the silver wire required. \n(ii) the area of each sector of the brooch. \nSolution: \n \nOne diameter of the circle in 35 mm long The length of 5 diameters = 5 x 35 = 175 mm \nThe required length of the silver wire = circumference of circle + length of 5 diameters \n= 110 + 175 = 285 mm<\/p>\n
(ii) Let O be the centre of circle and radius is \n \nThe angle made by the diameters of one part \n \nHence the area of each sector of the brooch is \\(\\frac { 385 }{ 4 }\\) mm\u00b2<\/p>\n
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Question 10. \nAn umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. \n \nSolution: \nLet O be the centre of the umbrella and the radius is 45 cm \nThe angle between two consecutive ribs \\(\\frac { 360\u00b0 }{ 8 }\\) \nArea between two consecutive ribs \n= \\(\\frac{360^{\\circ}}{360^{\\circ} \\times 8} \\times \\frac{22}{7} \\times(45)^{2}\\) \n= \\(\\frac{22275}{28} \\mathrm{~cm}^{2}\\)<\/p>\n
Question 11. \nA car has two wipers which do not overlap. \nEach wiper has a blade of length 25 cm sweeping through an angle of 115\u00b0. Find the total area cleaned at each sweep of the blades. \nSolution: \nGiven: length of blade of wiper = radius of sector sweep by blade = 25 cm \nArea cleaned by each sweep of the blade = area of sector sweep by blade \nAngle of the sector formed by blade of wiper = 115\u00b0 \n \nHence, the total area cleaned at each sweep is \\(\\frac { 158125 }{ 126 }\\) cm\u00b2.<\/p>\n
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Question 12. \nTo warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80\u00b0 to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use \u03c0 = 3.14) \nSolution: \nAngle of the sector = 80\u00b0 \nDistance covered = 16.5 km \nRadius of the sector formed = 16.5 km \n \nHence the area of sea over which the ships are warned is 189.97 km\u00b2.<\/p>\n
Question 13. \nA round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of \u20b9 0.35 per cm2. (Use \\(\\sqrt{3}\\) = 1.7) \n \nSolution: \n \nArea of the design = Area of sector AOB – Area of \u2206AOB \n= 410.67 – 333.20 \n= 77.47 \nThere are six designs in the circle \nArea of six designs = 77.47 x 6 \n= 464.82 cm\u00b2 \nThe cost of making the design at the rate of 0.35 per cm\u00b2 \n\u20b9 (464.82 x 0.35) = \u20b9 162.68 \nHence the required cost of making the design is \u20b9 162.68<\/p>\n
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Question 14. \nTick the correct answer in the following: Area of a sector of angle p (in degrees) of a circle with radius R is \n(a) \\(\\frac{p}{180^{\\circ}}\\) \u00d7 2\u03c0R \n(b) \\(\\frac{p}{180^{\\circ}}\\) \u00d7 \u03c0R2<\/sup> \n(c) \\(\\frac{p}{360^{\\circ}}\\) \u00d7 2\u03c0R \n(d) \\(\\frac{p}{720^{\\circ}}\\) \u00d7 2\u03c0R2<\/sup> \nSolution: \nSector angle is p in degrees \nRadius of the circle = R \nArea of the sector = \\(\\frac{\\pi \\mathrm{R}^{2} p}{361^{6}}\\) = \\(\\frac{\\left(\\pi R^{2} p\\right) 2}{720^{\\circ}}\\) \n= \\(\\frac{p}{720^{\\circ}}\\) \u00d7 2\u03c0R2<\/sup> \nHence (D) is the correct answer.<\/p>\n","protected":false},"excerpt":{"rendered":"These NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Exercise 12.2 Question 1. Find the area of a sector of a circle with radius 6 cm if angle …<\/p>\n
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n