{"id":2387,"date":"2020-09-09T09:06:02","date_gmt":"2020-09-09T03:36:02","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=2387"},"modified":"2021-07-02T15:03:36","modified_gmt":"2021-07-02T09:33:36","slug":"kseeb-solutions-class-9-maths-chapter-13-ex-13-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-13-ex-13-3\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3"},"content":{"rendered":"<h2>Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3<\/h2>\n<p>(If values are not given for &#8216;n&#8217; Assume n = \\(\\frac{22}{7}\\).)<\/p>\n<p>Question 1.<br \/>\nDiameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.<br \/>\nSolution:<br \/>\nDiameter of the base of a cone, d = 10.5 cm.<br \/>\nheight h = 10 cm.<br \/>\nCurved Surface Area, C.S.A. = ?<br \/>\nd = 10.5 cm, \\(=10 \\frac{1}{2}=\\frac{21}{2}\\) cm.<br \/>\nh = l = 10 cm.<br \/>\n\u2234 Curved Surface Area of a cone = \u03c0rl<\/p>\n<p>Question 2.<br \/>\nFind the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.<br \/>\nSolution:<br \/>\nDiameter of base of a cone, d = 24 m.<br \/>\nheight, h = 21 m.<br \/>\nTotal Surface Area, T.S.A. = ?<br \/>\nd = 24 m, \u2234 r = \\(\\frac{\\mathrm{d}}{2}=\\frac{24}{2}=12\\)<br \/>\nh = l = 21 m.<br \/>\nT.S.A of a cone = \u03c0r(r + l)<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2401\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.3-Q-2.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3 Q 2\" width=\"162\" height=\"130\" \/><br \/>\n= 1244.57 m<sup>2<\/sup><\/p>\n<p>Question 3.<br \/>\nCurved surface area of a cone is 308 cm<sup>2<\/sup> and its slant height is 14 cm. Find<br \/>\n(i) radius of the base, and<br \/>\n(ii) total surface area of the cone.<br \/>\nSolution:<br \/>\nIf Curved surface area of a cone, \u03c0rl = 308 cm<sup>2<\/sup>.<br \/>\nheight, h = l = 14 cm.<br \/>\nThen, (i) r = ?, (ii) T.S.A. = ?<br \/>\n(i) CSA of a Cone = \u03c0rl<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2402\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.3-Q-3.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3 Q 3\" width=\"202\" height=\"108\" \/><br \/>\n\u2234 r = 7 cm<br \/>\n(ii) TSA of a cone= \u03c0r(r + l)<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2403\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.3-Q-3.1.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3 Q 3.1\" width=\"147\" height=\"87\" \/><br \/>\n= 22 \u00d7 21<br \/>\n= 462 cm<sup>2<\/sup><\/p>\n<p>Question 4.<br \/>\nA conical tent is 10 m high and the radius of its base is 24 m. Find<br \/>\n(i) slant height of the tent.<br \/>\n(ii) cost of the canvas required to make the tent, if the cost of 1 m<sup>2<\/sup> canvas is Rs. 70.<br \/>\nSolution:<br \/>\n(i) ABC is a conical tent,<br \/>\nheight, h = 10 m.<br \/>\nradius, r = 24 m.<br \/>\nSlant height of the tent, l = ?<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2405\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.3-Q-4.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3 Q 4\" width=\"157\" height=\"182\" \/><br \/>\nIn \u22a5\u2206AOC, \u2220O = 90\u00b0<br \/>\nAs per Pythagoras theorem,<br \/>\nAC<sup>2<\/sup> = AO<sup>2<\/sup> + OC<sup>2<\/sup><br \/>\nl<sup>2<\/sup> = (01)<sup>2<\/sup> +(24)<sup>2<\/sup><br \/>\n= 100 + 576<br \/>\nl<sup>2<\/sup> = 676<br \/>\nl = \\(\\sqrt{676}\\)<br \/>\nArea of canvas required to prepare tent, C.S.A. = ?<br \/>\nC.S.A.= \u03c0rl<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2406\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.3-Q-4.1.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3 Q 4.1\" width=\"127\" height=\"88\" \/><br \/>\nCost of 1 m<sup>2<\/sup> canvas is Rs. 70,<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2407\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.3-Q-4.2.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3 Q 4.2\" width=\"287\" height=\"93\" \/><\/p>\n<p>Question 5.<br \/>\nWhat length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use \u03c0 = 3.14.)<br \/>\nSolution:<br \/>\nABC is a conical tent.<br \/>\nHeight, AO = h = 8 m.<br \/>\nbase radius, OC = r = 6 cm<br \/>\nIn \u22a5\u2206AOC,<br \/>\nAC<sup>2<\/sup> = AO<sup>2<\/sup> + OC<sup>2<\/sup><br \/>\nl<sup>2<\/sup> = (8)<sup>2<\/sup> +(6)<sup>2<\/sup><br \/>\nl<sup>2<\/sup> = 64 + 36<br \/>\nl<sup>2<\/sup> = 100<br \/>\n\u2234 l = \\(\\sqrt{100}\\)<br \/>\nl = 10 m<br \/>\nCurved surface Area of cone C.S.A. = \u03c0rl<br \/>\n= 3.14 \u00d7 6 \u00d7 10<br \/>\n= 188.4 m<sup>2<\/sup><br \/>\nLet the length of tarpaulin required be 1 m.<br \/>\nIn that 20 cm. (0.2 m) is wastage means remaining tarpaulin is (1 &#8211; 0.2 m)<br \/>\nbreadth, b = 3m<br \/>\n\u2234 Area of tarpaulin = curved surface area of tent.<br \/>\n(l &#8211; 0.2 m) \u00d7 3 = 188.4 m<sup>2<\/sup><br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2409\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.3-Q-5.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3 Q 5\" width=\"147\" height=\"43\" \/><br \/>\nl &#8211; 0.2 = 62.8<br \/>\n\u2234 l = 62.8 + 0.2<br \/>\n\u2234 l = 63 m.<\/p>\n<p>Question 6.<br \/>\nThe slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m<sup>2<\/sup>.<br \/>\nSolution:<br \/>\nSlant height of a cone, l = 25 m<br \/>\nDiameter, 2r = 14m \u2234 r = 7 m<br \/>\nCurved Surface Area = \u03c0rl<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2410\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.3-Q-6.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3 Q 6\" width=\"130\" height=\"42\" \/><br \/>\n= 22 \u00d7 25<br \/>\n= 550 m<sup>2<\/sup>.<br \/>\nCost of white-washing is Rs. 210 per m<sup>2<\/sup>.<br \/>\nCost of white-washing for 550 m<sup>2<\/sup> &#8230; ? &#8230;<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2413\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.3-Q-6.1.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3 Q 6.1\" width=\"112\" height=\"46\" \/><br \/>\n= Rs. 1155.<\/p>\n<p>Question 7.<br \/>\nA joker&#8217;s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.<br \/>\nSolution:<br \/>\nA joker\u2019s cap is in the form of a right circular cone.<br \/>\nradius of base, r = 7 cm.<br \/>\nheight, h = 24 cm.<br \/>\nslant height, 1 = ?<br \/>\nCurved surface area, C.S.A.= ?<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2415\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.3-Q-7-1.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3 Q 7\" width=\"131\" height=\"162\" \/><br \/>\nIn \u22a5\u2206AOC, \u2220AOC = 90\u00b0<br \/>\n\u2234 AC<sup>2<\/sup> = AO<sup>2<\/sup> + OC<sup>2<\/sup><br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2416\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.3-Q-7.1.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3 Q 7.1\" width=\"126\" height=\"176\" \/><br \/>\n\u2234 Curved surface area= \u03c0rl<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2417\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.3-Q-7.2.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3 Q 7.2\" width=\"127\" height=\"44\" \/><br \/>\n= 550 m<sup>2<\/sup><br \/>\nArea of 1 cap is 550 m<sup>2<\/sup><br \/>\nArea of 10 caps&#8230;,v &#8230; ? &#8230;<br \/>\n= 550 \u00d7 10<br \/>\n= 5500 m<sup>2<\/sup>.<\/p>\n<p>Question 8.<br \/>\nA bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m<sup>2<\/sup>, what will be the cost of painting all these cones ?<br \/>\n(Use \u03c0 = 3. 14 and take \\(\\sqrt{1.04}\\) = 1.02)<br \/>\nSolution:<br \/>\nDiameter of a cone, d = 40 cm = 0.40 m.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2418\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.3-Q-8.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3 Q 8\" width=\"189\" height=\"46\" \/><br \/>\nheight, h = l m.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2419\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.3-Q-8.1.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3 Q 8.1\" width=\"262\" height=\"156\" \/><br \/>\nCurved surface area of cone = \u03c0rl<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-2420\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSEEB-Solutions-for-Class-9-Maths-Chapter-13-Surface-Area-and-Volumes-Ex-13.3-Q-8.2.png\" alt=\"KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3 Q 8.2\" width=\"159\" height=\"47\" \/><br \/>\n= 0.64056 m<sup>2<\/sup><br \/>\nArea of 1 cone = 0.64056 m<sup>2<\/sup><br \/>\nArea of 50 cone = ?<br \/>\n= 50 \u00d7 0.64056<br \/>\n= 32.028 m<sup>2<\/sup><br \/>\nCost of painting for 1 sq.m, is Rs. 12.<br \/>\nCost of painting for 32.028 sq.m. &#8230;? &#8230;<br \/>\n= 32.028 \u00d7 12<br \/>\n= Rs.384.37.<\/p>\n<h5><a href=\"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths\/\">KSEEB Solutions for Class 9 Maths<\/a><\/h5>\n","protected":false},"excerpt":{"rendered":"<p>Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.3 (If values are not given for &#8216;n&#8217; Assume n = .) Question 1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area. Solution: Diameter of the base of a &hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2387"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=2387"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2387\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2387"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2387"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2387"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}