{"id":7116,"date":"2017-08-21T11:26:17","date_gmt":"2017-08-21T05:56:17","guid":{"rendered":"https:\/\/www.cbselabs.com\/?page_id=7116"},"modified":"2021-09-18T15:33:44","modified_gmt":"2021-09-18T10:03:44","slug":"rd-sharma-class-9-solutions-chapter-20-surface-area-volume-right-circular-cone","status":"publish","type":"page","link":"https:\/\/www.cbselabs.com\/rd-sharma-class-9-solutions-chapter-20-surface-area-volume-right-circular-cone\/","title":{"rendered":"RD Sharma Class 9 Solutions Surface Area and Volume of A Right Circular Cone"},"content":{"rendered":"<h2>RD Sharma Class 9 Solutions Surface Area and Volume of A Right Circular Cone<\/h2>\n<p>Summary:<\/p>\n<ol>\n<li>A Right circular cone is a solid generated by revolving a line segment which passes through a fixed point and which makes a constant angle with fixed line. The fixed point is called the vertex of the cone, the fixed line is called the axis of the cone.<\/li>\n<li>For a circular cone of base radius r, slant height l and height h, we have<\/li>\n<li>\n<ul>\n<li>Curved surface area =\u03c0rl<\/li>\n<li>Total surface area =\u00a0\u03c0r(l+r)<\/li>\n<li>Volume = \\(\\frac { 1 }{ 3 } (area of the base)\\times height\\)<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<h3>RD Sharma Class 9 PDF Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1<\/h3>\n<p>Question 1.<br \/>\nFind the curved surface area of a cone, if its slant height is 60 cm and the radius of its base is 21 cm.<br \/>\nSolution:<br \/>\nRadius of the base of the cone = 21 cm<br \/>\nand slant height (l) = 60 cm<br \/>\n<img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1926\/30706412237_13eb67a24a_o.png?resize=156%2C215&#038;ssl=1\" alt=\"RD Sharma Class 9 Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"156\" height=\"215\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 2.<br \/>\nThe radius of a cone is 5 cm and vertical height is 12 cm. Find the area of the curved surface.<br \/>\nSolution:<br \/>\nRadius of the base of a cone = 5 cm<br \/>\nVertical height (h) = 12 cm<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1907\/44733152525_f4f801e3ea_o.png?resize=347%2C446&#038;ssl=1\" alt=\"RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"347\" height=\"446\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 3.<br \/>\nThe radius of a cone is 7 cm and area of curved surface is 176 cm2. Find the slant height.<br \/>\nSolution:<br \/>\nCurved surface area of a cone = 176 cm<sup>2<\/sup><br \/>\nand radius (r) = 1 cm<sup>2<\/sup><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1959\/44733152285_c368e95210_o.png?resize=243%2C111&#038;ssl=1\" alt=\"RD Sharma Solutions Class 9 Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"243\" height=\"111\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 4.<br \/>\nThe height of a cone is 21 cm. Find the area of the base if the slant height is 28 cm.<br \/>\nSolution:<br \/>\nHeight of the cone (h) = 21 cm<br \/>\nSlant height (l) = 28 cm<br \/>\n\u2234 l<sup>2<\/sup> = r<sup>2<\/sup> + h<sup>2<\/sup><br \/>\n\u21d2 r<sup>2<\/sup> = l<sup>2<\/sup>&#8211; h<sup>2<\/sup> = (28)<sup>2<\/sup> &#8211; (21 )<sup>2<\/sup><br \/>\n\u21d2 784 &#8211; 441 = 343 &#8230;(i)<br \/>\nNow area of base = \u03c0r<sup>2<\/sup><br \/>\n= \\(\\frac { 22 }{ 7 }\\) x 343 [From (i)]<br \/>\n= 22 x 49 = 1078 cm<sup>2<\/sup><\/p>\n<p>Question 5.<br \/>\nFind the total surface area of a right circular cone with radius 6 cm and height 8 cm.<br \/>\nSolution:<br \/>\nRadius of base of cone (r) = 6 cm<br \/>\nand height (h) = 8 cm<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1918\/30706411647_1444a4d94d_o.png?resize=333%2C256&#038;ssl=1\" alt=\"RD Sharma Class 9 PDF Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"333\" height=\"256\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 6.<br \/>\nFind the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm. [NCERT]<br \/>\nSolution:<br \/>\nRadius of base of a cone (r) = 5.25 cm<br \/>\nand slant height (l) = 10 cm<br \/>\nCurved surface area = \u03c0rl<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1947\/30706411167_2a16d56fac_o.png?resize=341%2C131&#038;ssl=1\" alt=\"Surface Areas and Volume of A Right Circular Cone Class 9 RD Sharma Solutions\" width=\"341\" height=\"131\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 7.<br \/>\nFind the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. [NCERT]<br \/>\nSolution:<br \/>\nSlant height of a cone (l) = 21 m<br \/>\nand diameter of its base = 24 m<br \/>\n\u2234 Radius (r) = \\(\\frac { 24 }{ 2 }\\) = 12 m<br \/>\nNow total surface area = \u03c0r(l + r)<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1978\/30706410877_266ebb1963_o.png?resize=243%2C153&#038;ssl=1\" alt=\"RD Sharma Class 9 Solution Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"243\" height=\"153\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 8.<br \/>\nThe area of the curved surface of a cone is 60\u03c0 cm<sup>2<\/sup>. If the slant height of the cone be 8 cm, find the radius of the base.<br \/>\nSolution:<br \/>\nCurved surface area of a cone = 6071 cm<sup>2<\/sup><br \/>\nSlant height (l) = 8 cm<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1901\/30706410617_e0af640fd9_o.png?resize=315%2C54&#038;ssl=1\" alt=\"Class 9 RD Sharma Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"315\" height=\"54\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 9.<br \/>\nThe curved surface area of a cone is 4070 cm<sup>2<\/sup> and its diameter is 70 cm. What is the slant height ? (Use \u03c0 = 22\/7).<br \/>\nSolution:<br \/>\nSurface area of a cone = 4070 cm<sup>2<\/sup><br \/>\nDiameter of its base = 70 cm<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1960\/30706410437_381a19d888_o.png?resize=263%2C170&#038;ssl=1\" alt=\"Class 9 Maths Chapter 20 Surface Areas and Volume of A Right Circular Cone RD Sharma Solutions\" width=\"263\" height=\"170\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 10.<br \/>\nThe radius and slant height of a cone are in the ratio of 4 : 7. If its curved surface area is 792 cm<sup>2<\/sup>, find its radius. (Use \u03c0 = 22\/7)<br \/>\nSolution:<br \/>\nCurved surface area of a cone = 792 cm<sup>2<\/sup><br \/>\nRatio in radius and slant height = 4:7<br \/>\nLet radius = 4x<br \/>\nThen slant height = 7x<br \/>\n\u2234 Curved surface area \u03c0rl = 792<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1952\/30706410207_9ccc98f7a1_o.png?resize=277%2C162&#038;ssl=1\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"277\" height=\"162\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 11.<br \/>\nA Joker\u2019s 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. [NCERT]<br \/>\nSolution:<br \/>\nRadius of the base of a conical cap (r) = 7 cm<br \/>\nand height (h) = 24 cm<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1903\/30706410007_02afb3bf66_o.png?resize=324%2C249&#038;ssl=1\" alt=\"RD Sharma Class 9 Book Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"324\" height=\"249\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 12.<br \/>\nFind the ratio of the curved surface areas of two cones if their diameters of the bases are equal and slant heights are in the ratio 4 : 3.<br \/>\nSolution:<br \/>\nLet diameters of each cone = d<br \/>\nThen radius (r) = \\(\\frac { d }{ 2 }\\)<br \/>\nRatio in their slant heights = 4 : 3<br \/>\nLet slant height of first cone = 4x<br \/>\nand height of second cone = 3x<br \/>\nNow curved surface area of the first cone = 2\u03c0rh<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1959\/30706409687_1958a14d58_o.png?resize=298%2C105&#038;ssl=1\" alt=\"Surface Areas and Volume of A Right Circular Cone With Solutions PDF RD Sharma Class 9 Solutions\" width=\"298\" height=\"105\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1956\/44733150755_3b86f9ee05_o.png?resize=281%2C98&#038;ssl=1\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"281\" height=\"98\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 13.<br \/>\nThere are two cones. The curved surface area of one is twice that of the other. The slant height of the later is twice that of the former. Find the ratio of their radii.<br \/>\nSolution:<br \/>\nIn two cones, curved surface of the first cone = 2 x curved surface of the second cone<br \/>\nSlant height of the second cone = 2 x slant height of first cone<br \/>\nLet r<sub>1<\/sub> and r<sub>2<\/sub> be the radii of the two cones<br \/>\nand let height of the first cone = h<br \/>\nThen height of second cone = 2h<br \/>\n\u2234 Curved surface of the first cone = 2\u03c0r<sub>1<\/sub>h<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1973\/30706409467_0af1ff1370_o.png?resize=229%2C205&#038;ssl=1\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"229\" height=\"205\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 14.<br \/>\nThe diameters of two cones are equal. If their slant heights are in the ratio 5 : 4, find the ratio of their curved surfaces.<br \/>\nSolution:<br \/>\nLet diameter of one cone = d<br \/>\nand diameter of second cone = d<br \/>\n\u2234 Radius of the first cone (r) = \\(\\frac { d }{ 2 }\\)<br \/>\nand of second cone (r<sub>2<\/sub>) = \\(\\frac { d }{ 2 }\\)<br \/>\nRatio in their slant heights = 5:4<br \/>\nLet slant height of the first cone = 5x<br \/>\nThen that of second cone = 4x<br \/>\nNow curved surface of the first cone = 2\u03c0rh<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1922\/30706409317_a5904c52a4_o.png?resize=314%2C178&#038;ssl=1\" alt=\"RD Sharma Math Solution Class 9 Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"314\" height=\"178\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 15.<br \/>\nCurved surface area of a cone is 308 cm<sup>2<\/sup> and its slant height is 14 cm. Find the radius of the base and total surface area of the cone. [NCERT]<br \/>\nSolution:<br \/>\nArea of curved surface of a cone = 308 cm<sup>2<\/sup><br \/>\nand slant height (l) = 14 cm<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1919\/44733150285_1fae7812c0_o.png?resize=355%2C233&#038;ssl=1\" alt=\"RD Sharma Class 9 Questions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"355\" height=\"233\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 16.<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>. [NCERT]<br \/>\nSolution:<br \/>\nSlant height of a cone (l) = 25 m<br \/>\nand diameter of base = 14 m<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1946\/30706408997_6765ae16ee_o.png?resize=348%2C201&#038;ssl=1\" alt=\"Maths RD Sharma Class 9 Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"348\" height=\"201\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 17.<br \/>\nA conical tent is 10 m high and the radius of its base is 24 m. Find the slant height of the tent. If the cost of 1 m<sup>2<\/sup> canvas is Rs. 70, find the cost of the canvas required to make the tent. [NCERT]<br \/>\nSolution:<br \/>\nHeight of conical tent (A) = 10 m<br \/>\nRadius of the base (r) = 24 m<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1967\/30706408647_a04bb2d71a_o.png?resize=330%2C516&#038;ssl=1\" alt=\"RD Sharma Class 9 Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"330\" height=\"516\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 18.<br \/>\nThe circumference of the base of a 10 m height conical tent is 44 metres. Calculate the length of canvas used in making the tent if width of canvas is 2 m. (Use \u03c0 = 22\/7)<br \/>\nSolution:<br \/>\nCircumference of the base of a conical tent = 44 m<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1956\/30706408167_3681cee4ba_o.png?resize=330%2C160&#038;ssl=1\" alt=\"RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"330\" height=\"160\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1932\/30706407977_fac10948ba_o.png?resize=329%2C224&#038;ssl=1\" alt=\"RD Sharma Solutions Class 9 Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"329\" height=\"224\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 19.<br \/>\nWhat length of tarpaulin 3 m wide will be required to make a conical tent of height 8 m and base radius 6 m? Assum that the extra length of material will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use \u03c0 = 3.14) [NCERT]<br \/>\nSolution:<br \/>\nHeight of the conical tent (h) = 8 m<br \/>\nand radius of the base (r) = 6 m<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1944\/30706407727_c63859a715_o.png?resize=303%2C97&#038;ssl=1\" alt=\"RD Sharma Class 9 PDF Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"303\" height=\"97\" data-recalc-dims=\"1\" \/><br \/>\nNow curved surface area of the tent = \u03c0rl = 3.14 x 6 x 10 = 188.4 m<br \/>\nWidth of tarpaulin used = 3 m<br \/>\n\u2234 Length = 188.4 , 3 = 62.8 m<br \/>\nExtra length required = 20 cm = 0.2 m<br \/>\n\u2234 Total length of tarpaulin = 62.8 + 0.2 = 63 m<\/p>\n<p>Question 20.<br \/>\nA bus stop is barricated from the remaining part of the road, by using 50 hollow cones made of recycled card-board. 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 \u20b912 per m<sup>2<\/sup>, what will be the cost of painting these cones. (Use \u03c0 = 3.14 and \\(\\sqrt { 1.04 } \\) = 1.02) [NCERT]<br \/>\nSolution:<br \/>\nDiameter of the base of tent = 40 cm<br \/>\n\u2234 Radius of the base of cone (r) = \\(\\frac { 40 }{ 2 }\\)<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1915\/44733149035_ef06e57387_o.png?resize=344%2C374&#038;ssl=1\" alt=\"Surface Areas and Volume of A Right Circular Cone Class 9 RD Sharma Solutions\" width=\"344\" height=\"374\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 21.<br \/>\nA cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, show that the radius of each is to the height of each as 3 : 4.<br \/>\nSolution:<br \/>\nLet radius of cylinder = r<br \/>\nand radius of cone = r<br \/>\nand let height of cylinder = h<br \/>\nand height of cone = h<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1939\/30706407427_6d4446511f_o.png?resize=359%2C383&#038;ssl=1\" alt=\"RD Sharma Class 9 Solution Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"359\" height=\"383\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1953\/30706407087_1aee281133_o.png?resize=361%2C222&#038;ssl=1\" alt=\"Class 9 RD Sharma Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"361\" height=\"222\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 22.<br \/>\nA tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. Find the area of the canvas required for the tent.<br \/>\nSolution:<br \/>\nDiameter of the cylindrical portion = 24 m 24<br \/>\n\u2234 Radius (r) = \\(\\frac { 24 }{ 2 }\\) = 12 m<br \/>\nHeight of cylindrical portion = 11 m<br \/>\nand total height = 16 m<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1929\/30706406877_5b0e9888f9_o.png?resize=149%2C255&#038;ssl=1\" alt=\"Class 9 Maths Chapter 20 Surface Areas and Volume of A Right Circular Cone RD Sharma Solutions\" width=\"149\" height=\"255\" data-recalc-dims=\"1\" \/><br \/>\n\u2234 Height of conical portion = 16 &#8211; 11 = 5 m<br \/>\n\u2234 Slant height of the conical portion (l)<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1927\/30706406757_65f5d90bb3_o.png?resize=249%2C117&#038;ssl=1\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"249\" height=\"117\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1909\/44733148135_160042e1b5_o.png?resize=281%2C158&#038;ssl=1\" alt=\"RD Sharma Class 9 Book Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"281\" height=\"158\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 23.<br \/>\nA circus tent is cylindrical to a height of 3 meters and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.<br \/>\nSolution:<br \/>\nDiameter of the cylindrical tent = 105 m<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1979\/30706406487_c8fa189765_o.png?resize=344%2C667&#038;ssl=1\" alt=\"Surface Areas and Volume of A Right Circular Cone With Solutions PDF RD Sharma Class 9 Solutions\" width=\"344\" height=\"667\" data-recalc-dims=\"1\" \/><\/p>\n<h3>Class 9 RD Sharma Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2<\/h3>\n<p>Question 1.<br \/>\nFind the volume of a right circular cone with:<br \/>\n(i) radius 6 cm, height 7 cm<br \/>\n(ii) radius 3.5 cm, height 12 cm<br \/>\n(iii) height 21 cm and slant height 28 cm. [NCERT]<br \/>\nSolution:<br \/>\n(i) Radius of a cone (r) = 6 cm<br \/>\nand height (h) = 7 cm<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1972\/44733147845_9f35dfc5f5_o.png?resize=356%2C734&#038;ssl=1\" alt=\"RD Sharma Class 9 Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"356\" height=\"734\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 2.<br \/>\nFind the capacity in litres of a conical vessel with:<br \/>\n(i) radius 7 cm, slant height 25 cm,<br \/>\n(ii) height 12 cm, slant height 13 cm. [NCERT]<br \/>\nSolution:<br \/>\n(i) Radius of the conical vessel (r) = 7 cm<br \/>\nSlant height (h) = 25 cm<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1966\/30706406057_5d136bd10c_o.png?resize=355%2C722&#038;ssl=1\" alt=\"RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"355\" height=\"722\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 3.<br \/>\nTwo cones have their heights in the ratio 1 : 3 and the radii of their bases in the ratio 3:1. Find the ratio of their volumes.<br \/>\nSolution:<br \/>\nRatio in the heights of two cones =1:3<br \/>\nand ratio in their radii = 3: 1<br \/>\nLet radius of first cone (r<sub>1<\/sub>) = x<br \/>\nand of second cone (r<sub>2<\/sub>) = 3x<br \/>\nand height of first cone (h<sub>1<\/sub>) = 3y<br \/>\nand of second cone (h<sub>2<\/sub>)<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1973\/44733147555_c438bf2838_o.png?resize=309%2C314&#038;ssl=1\" alt=\"RD Sharma Solutions Class 9 Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"309\" height=\"314\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 4.<br \/>\nThe radius and the height of a right circular cone are in the ratio 5 : 12. If its volume is 314 cubic metre, find the slant height and the radius (Use \u03c0 = 3.14).<br \/>\nSolution:<br \/>\nRatio in the radius and height of a cone = 5 : 12<br \/>\nVolume = 314 cm<sup>3<\/sup><br \/>\nLet radius (r) = 5x<br \/>\nand height (h) = 12x<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1909\/30706405657_34e140f877_o.png?resize=324%2C332&#038;ssl=1\" alt=\"RD Sharma Class 9 PDF Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"324\" height=\"332\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1975\/31774937318_20ffca339e_o.png?resize=261%2C111&#038;ssl=1\" alt=\"Surface Areas and Volume of A Right Circular Cone Class 9 RD Sharma Solutions\" width=\"261\" height=\"111\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 5.<br \/>\nThe radius and height of a right circular cone are in the ratio 5 : 12 and its volume is 2512 cubic cm. Find the slant height and radius of the cone. (Use \u03c0 = 3.14).<br \/>\nSolution:<br \/>\nRatio in the radius and height of a right circular cone = 5 : 12<br \/>\nVolume = 2512 cm<sup>3<\/sup><br \/>\nLet radius (r) = 5x<br \/>\nThen height (h) = 12x<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1928\/44733147225_39134edc2e_o.png?resize=350%2C493&#038;ssl=1\" alt=\"RD Sharma Class 9 Solution Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"350\" height=\"493\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 6.<br \/>\nThe ratio of volumes of two cones is 4 : 5 and the ratio of the radii of their bases is 2:3. Find the ratio of their vertical heights.<br \/>\nSolution:<br \/>\nRatio in volumes of two cones = 4:5<br \/>\nand ratio in radii = 2:3<br \/>\nLet radius of the first cone (r<sub>1<\/sub>) = 2x<br \/>\nThen radius of second cone (r<sub>2<\/sub>) = 3x<br \/>\nLet h<sub>1<\/sub>, h<sub>2<\/sub> be their heights respectively<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1950\/31774937148_7ed1d9b4b0_o.png?resize=362%2C385&#038;ssl=1\" alt=\"Class 9 RD Sharma Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"362\" height=\"385\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 7.<br \/>\nRatio between their vertical heights = 9:5 7. A cylinder and a cone have equal radii of their bases and equal heights. Show that their volumes are in the ratio 3:1.<br \/>\nSolution:<br \/>\nLet r be the radius and h be the height of a cylinder and a cone, then<br \/>\nVolume of cylinder = \u03c0r<sup>2<\/sup>h<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1907\/30706405107_4fbe408fc0_o.png?resize=350%2C193&#038;ssl=1\" alt=\"Class 9 Maths Chapter 20 Surface Areas and Volume of A Right Circular Cone RD Sharma Solutions\" width=\"350\" height=\"193\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 8.<br \/>\nIf the radius of the base of a cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone?<br \/>\nSolution:<br \/>\nLet r be the radius and h be the height of the cone, then<br \/>\nVolume = \\(\\frac { 1 }{ 3 }\\) \u03c0r<sup>2<\/sup>h<br \/>\nBy halving the radius and same height the<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1944\/44733147025_9647a0f080_o.png?resize=347%2C368&#038;ssl=1\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"347\" height=\"368\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 9.<br \/>\nA heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much canvas cloth is required to just cover the heap? (Use \u03c0 = 3.14). [NCERT]<br \/>\nSolution:<br \/>\nDiameter of conical heap of wheat = 9 m<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1930\/30706404887_c39e172208_o.png?resize=353%2C444&#038;ssl=1\" alt=\"RD Sharma Class 9 Book Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"353\" height=\"444\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1910\/45596874872_1f4b65c796_o.png?resize=326%2C54&#038;ssl=1\" alt=\"Surface Areas and Volume of A Right Circular Cone With Solutions PDF RD Sharma Class 9 Solutions\" width=\"326\" height=\"54\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 10.<br \/>\nFind the weight of a solid cone whose base is of diameter 14 cm and vertical height 51 cm, supposing the material of which it is made weighs 10 grams per cubic cm.<br \/>\nSolution:<br \/>\nDiameter of the base of solid cone = 14 cm<br \/>\nand vertical height (h) = 51 cm<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1922\/30706404637_1d63126b23_o.png?resize=345%2C261&#038;ssl=1\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"345\" height=\"261\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 11.<br \/>\nA right angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length, is made to turn round on the longer side. Find the volume of the solid, thus generated. Also, find its curved surface area.<br \/>\nSolution:<br \/>\nLength of sides of a right angled triangle are 6.3 cm and 10 cm<br \/>\nBy turning around the longer side, a cone is formed in which radius (r) = 6.3 cm<br \/>\nand height (h) = 10 cm<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1939\/45596874642_1336611044_o.png?resize=295%2C249&#038;ssl=1\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"295\" height=\"249\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1945\/30706404437_88b860c7e2_o.png?resize=294%2C323&#038;ssl=1\" alt=\"RD Sharma Math Solution Class 9 Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"294\" height=\"323\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 12.<br \/>\nFind the volume of the largest right circular cone that can be fitted in a cube whose edge is 14 cm.<br \/>\nSolution:<br \/>\nSide of cube = 14 cm ,<br \/>\nRadius of the largest cone that can be fitted<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1947\/45596874252_b56cc53178_o.png?resize=354%2C466&#038;ssl=1\" alt=\"RD Sharma Class 9 Questions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"354\" height=\"466\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 13.<br \/>\nThe volume of a right circular cone is 9856 cm<sup>3<\/sup>. If the diameter of the base is 28 cm, find:<br \/>\n(i) height of the cone<br \/>\n(ii) slant height of the cone<br \/>\n(iii) curved surface area of the cone. [NCERT]<br \/>\nSolution:<br \/>\nVolume of a right circular cone = 9856 cm<sup>3<\/sup><br \/>\nDiameter of the base = 28 cm<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1967\/45647408271_1544d8e2af_o.png?resize=330%2C340&#038;ssl=1\" alt=\"RD Sharma Class 9 Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"330\" height=\"340\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 14.<br \/>\nA conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres? [NCERT]<br \/>\nSolution:<br \/>\nDiameter of the top of the conical pit = 3.5 m<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1952\/45596873672_154d29f16d_o.png?resize=273%2C383&#038;ssl=1\" alt=\"RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"273\" height=\"383\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1948\/45647408001_0fbac2a888_o.png?resize=353%2C181&#038;ssl=1\" alt=\"RD Sharma Solutions Class 9 Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"353\" height=\"181\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 15.<br \/>\nMonica has a piece of Canvas whose area is 551 m<sup>2<\/sup>. She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and wastage incurred while cutting, amounts to approximately 1 m<sup>2<\/sup>. Find the volume of the tent that can be made with it. [NCERT]<br \/>\nSolution:<br \/>\nArea of Canvas = 551 m<sup>2<\/sup><br \/>\nArea of wastage = 1 m<sup>2<\/sup><br \/>\nActual area = 551 &#8211; 1 = 550 m<sup>2<\/sup><br \/>\nBase radius of the conical tent = 7 m<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1940\/45647407741_940a640b2e_o.png?resize=181%2C204&#038;ssl=1\" alt=\"RD Sharma Class 9 PDF Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"181\" height=\"204\" data-recalc-dims=\"1\" \/><br \/>\nLet l be the slant height and h be the vertical<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1935\/45647407561_2c24e28cd6_o.png?resize=317%2C267&#038;ssl=1\" alt=\"Surface Areas and Volume of A Right Circular Cone Class 9 RD Sharma Solutions\" width=\"317\" height=\"267\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1969\/45596872472_14d21b698b_o.png?resize=352%2C56&#038;ssl=1\" alt=\"RD Sharma Class 9 Solution Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"352\" height=\"56\" data-recalc-dims=\"1\" \/><\/p>\n<h3>RD Sharma Mathematics Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS<\/h3>\n<p>Question 1.<br \/>\nThe height of a cone is 15 cm. If its volume is 500\u03c0 cm<sup>3<\/sup>, then find the radius of its base.<br \/>\nSolution:<br \/>\nVolume of cone = 500\u03c0 cm<sup>3<\/sup><br \/>\nand height (h) = 15 cm<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1940\/30706413897_0dea2fe491_o.png?resize=292%2C117&#038;ssl=1\" alt=\"Surface Areas and Volume of A Right Circular Cone Class 9 RD Sharma Solutions\" width=\"292\" height=\"117\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 2.<br \/>\nIf the volume of a right circular cone of height 9 cm is 48\u03c0 cm<sup>3<\/sup>, find the diameter of its base.<br \/>\nSolution:<br \/>\nVolume of a cone = 48\u03c0 cm<sup>3<\/sup><br \/>\nHeight (h) = 9 cm<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1947\/44733153485_49861e0187_o.png?resize=281%2C140&#038;ssl=1\" alt=\"RD Sharma Class 9 Solution Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"281\" height=\"140\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 3.<br \/>\nIf the height and slant height of a cone are 21 cm and 28 cm respectively. Find its volume.<br \/>\nSolution:<br \/>\nHeight of a cone (h) = 21 cm<br \/>\nand slant height (l) = 28 cm<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1944\/30706413627_4806e98a1f_o.png?resize=296%2C202&#038;ssl=1\" alt=\"Class 9 RD Sharma Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"296\" height=\"202\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 4.<br \/>\nThe height of a conical vessel is 3.5 cm. If its capacity is 3.3 litres of milk. Find the diameter of its base.<br \/>\nSolution:<br \/>\nCapacity of conical vessel = 3.3 litres<br \/>\nVolume = 3.3 m<sup>3<\/sup><br \/>\n= 3.3 x 1000 = 3300 cm<sup>2<\/sup><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1974\/30706413367_751f1c4c3a_o.png?resize=339%2C192&#038;ssl=1\" alt=\"Class 9 Maths Chapter 20 Surface Areas and Volume of A Right Circular Cone RD Sharma Solutions\" width=\"339\" height=\"192\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 5.<br \/>\nIf the radius and slant height of a cone are in the ratio 7 : 13 and its curved surface area is 286 cm2, find its radius.<br \/>\nSolution:<br \/>\nTwo ratio in radius and slant height of a cone = 7 : 13<br \/>\nLet radius (r) = 7x<br \/>\nand slant height (1) = 13x<br \/>\nCurved surface area = \u03c0rl<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1930\/30706413147_21e2e72bf8_o.png?resize=266%2C181&#038;ssl=1\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"266\" height=\"181\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 6.<br \/>\nFind the area of canvas required for a conical tent of height 24 m and base radius 7 m.<br \/>\nSolution:<br \/>\nBase radius of the closed cone (r) = 7 cm<br \/>\nand vertical height (h) = 24 cm<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1934\/44733152995_acc2275356_o.png?resize=317%2C178&#038;ssl=1\" alt=\"RD Sharma Class 9 Book Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"317\" height=\"178\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 7.<br \/>\nFind the area of metal sheet required in making a closed hollow cone of base radius 7 cm and height 24 cm. making a closed hollow cone of base radius 7 cm and height 24 cm.<br \/>\nSolution:<br \/>\nBase radius of the closed cone (r) = 7 cm<br \/>\nand vertical height (h) = 24 cm<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1913\/30706412867_fd84f5c8d2_o.png?resize=286%2C230&#038;ssl=1\" alt=\"Surface Areas and Volume of A Right Circular Cone With Solutions PDF RD Sharma Class 9 Solutions\" width=\"286\" height=\"230\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 8.<br \/>\nFind the length of cloth used in making a conical pandal of height 100 m and base radius 240 m, if the cloth is 100\u03c0 m wide.<br \/>\nSolution:<br \/>\nHeight of conical pandal (A) = 100 m<br \/>\nBase radius (r) = 240 m<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1967\/30706412597_b6ec99ef55_o.png?resize=318%2C255&#038;ssl=1\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"318\" height=\"255\" data-recalc-dims=\"1\" \/><\/p>\n<h3>RD Sharma Class 9 Solution Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS<\/h3>\n<p>Question 1.<br \/>\nThe number of surfaces of a cone has, is<br \/>\n(a) 1<br \/>\n(b) 2<br \/>\n(c) 3<br \/>\n(d) 4<br \/>\nSolution:<br \/>\nNumber of surfaces of a cone are 2 (b)<\/p>\n<p>Question 2.<br \/>\nThe area of the curved surface of a cone of radius 2r and slant height \\(\\frac { 1 }{ 2 }\\), is<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1974\/44733154205_0b07915046_o.png?resize=293%2C85&#038;ssl=1\" alt=\"RD Sharma Class 9 Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"293\" height=\"85\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nRadius of a cone = 2r<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1932\/44922437144_c7a2e065e3_o.png?resize=327%2C138&#038;ssl=1\" alt=\"RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"327\" height=\"138\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 3.<br \/>\nThe total surface area of a cone of radius \\(\\frac { r }{ 2 }\\) and length 2l, is<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1933\/44733153995_195521b23e_o.png?resize=317%2C97&#038;ssl=1\" alt=\"RD Sharma Solutions Class 9 Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"317\" height=\"97\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1946\/44922436984_7677794241_o.png?resize=351%2C261&#038;ssl=1\" alt=\"RD Sharma Class 9 PDF Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"351\" height=\"261\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 4.<br \/>\nA solid cylinder is melted and cast into a cone of same radius. The heights of the cone and cylinder are in the ratio<br \/>\n(a) 9 : 1<br \/>\n(b) 1 : 9<br \/>\n(c) 3 : 1<br \/>\n(d) 1 : 3<br \/>\nSolution:<br \/>\nLet r be the radius and h be the height of cylinder, then volume = \u03c0r<sup>2<\/sup>h<br \/>\nNow volume of cone = \u03c0r<sup>2<\/sup>h<br \/>\nr is the radius<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1914\/44733153745_55f8e08c98_o.png?resize=310%2C75&#038;ssl=1\" alt=\"Surface Areas and Volume of A Right Circular Cone Class 9 RD Sharma Solutions\" width=\"310\" height=\"75\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1962\/30706414137_b01b32b1d1_o.png?resize=347%2C51&#038;ssl=1\" alt=\"RD Sharma Class 9 Solution Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"347\" height=\"51\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 5.<br \/>\nIf the radius of the base of a right circular cone is 3r and its height is equal to the radius of the base, then its volume is<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1933\/44733153625_7a1e7ffae2_o.png?resize=276%2C85&#038;ssl=1\" alt=\"Class 9 RD Sharma Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"276\" height=\"85\" data-recalc-dims=\"1\" \/><br \/>\nSolution:<br \/>\nRadius of the base of a cone (R) = 3r<br \/>\nand height (H) = 3r<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1977\/31774948598_57f142bfa5_o.png?resize=345%2C106&#038;ssl=1\" alt=\"Class 9 Maths Chapter 20 Surface Areas and Volume of A Right Circular Cone RD Sharma Solutions\" width=\"345\" height=\"106\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 6.<br \/>\nIf the volumes of two cones are in the ratio 1 : 4 and their diameters are in the ratio 4 : 5, then the ratio of their heights, is<br \/>\n(a) 1 : 5<br \/>\n(b) 5 : 4<br \/>\n(c) 5 : 16<br \/>\n(d) 25 : 64<br \/>\nSolution:<br \/>\nRatio in the volumes of two cones =1:4<br \/>\nand ratio in their diameter = 4:5<br \/>\nLet h<sub>1<\/sub>, h<sub>2<\/sub> be their heights<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1970\/44922439454_03a8176583_o.png?resize=334%2C406&#038;ssl=1\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"334\" height=\"406\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1902\/31774948478_c71a78b3d1_o.png?resize=352%2C302&#038;ssl=1\" alt=\"RD Sharma Class 9 Book Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"352\" height=\"302\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 7.<br \/>\nThe curved surface area of one cone is twice that of the other while the slant height of the latter is twice that of the former. The ratio of their radii is<br \/>\n(a) 2 : 1<br \/>\n(b) 4 : 1<br \/>\n(c) 8 : 1<br \/>\n(d) 1 : 1<br \/>\nSolution:<br \/>\nLet r be the radius and l be the slant height<br \/>\n\u2234 Curved surface area of first cone = \u03c0r<sub>1<\/sub>l<sub>1<\/sub><br \/>\nand let curved surface area of second cone = \u03c0r<sub>2<\/sub>l<sub>2<\/sub><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1944\/44922439204_f8e6088009_o.png?resize=347%2C361&#038;ssl=1\" alt=\"Surface Areas and Volume of A Right Circular Cone With Solutions PDF RD Sharma Class 9 Solutions\" width=\"347\" height=\"361\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 8.<br \/>\nIf the height and radius of a cone of volume V are doubled, then the volume of the cone, is<br \/>\n(a) 3V<br \/>\n(b) 4V<br \/>\n(c) 6V<br \/>\n(d) 8V<br \/>\nSolution:<br \/>\nLet r and h be the radius and height of a cone, then<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1975\/31774948358_ea60ae45ca_o.png?resize=325%2C306&#038;ssl=1\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"325\" height=\"306\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 9.<br \/>\nThe ratio of the volume of a right circular cylinder and a right circular cone of the same base and height, is<br \/>\n(a) 1 : 3<br \/>\n(b) 3 : 1<br \/>\n(c) 4 : 3<br \/>\n(d) 3 : 4<br \/>\nSolution:<br \/>\nLet r be the radius and h be the height of a right circular cylinder and a right circular cone, and V<sub>1<\/sub> and V<sub>2<\/sub> are their volumes, the V<sub>1<\/sub> =\u00a0 \u03c0r<sup>2<\/sup>h and<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1940\/31774948048_11c772eb64_o.png?resize=344%2C227&#038;ssl=1\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"344\" height=\"227\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 10.<br \/>\nA right cylinder and a right circular cone have the same radius and same volumes. The ratio of the height of the cylinder to that of the cone is<br \/>\n(a) 3 : 5<br \/>\n(b) 2 : 5<br \/>\n(c) 3 : 1<br \/>\n(d) 1 : 3<br \/>\nSolution:<br \/>\nLet r be the radius of cylinder and cone and volumes are equal<br \/>\nand h<sub>1<\/sub>, and h<sub>2<\/sub> be their have h<sub>2<\/sub> is respectively<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1906\/44922438564_7f7c4e5c06_o.png?resize=349%2C274&#038;ssl=1\" alt=\"RD Sharma Math Solution Class 9 Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"349\" height=\"274\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 11.<br \/>\nThe diameters of two cones are equal. If their slant heights are in the ratio 5 : 4, the ratio of their curved surface areas, is<br \/>\n(a) 4 : 5<br \/>\n(b) 25 : 16<br \/>\n(c) 16 : 25<br \/>\n(d) 5 : 4<br \/>\nSolution:<br \/>\n\u2235 Diameters of two cones are equal<br \/>\n\u2234 Their radii are also be equal<br \/>\nLet r be their radius of each cone,<br \/>\nand ratio in their slant heights = 5:4<br \/>\nLet slant height of first cone (h<sub>1<\/sub>) = 5x<br \/>\nThen height of second cone (h<sub>2<\/sub>) = 4x<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1916\/31774947868_aa239dc666_o.png?resize=344%2C140&#038;ssl=1\" alt=\"RD Sharma Class 9 Questions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"344\" height=\"140\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 12.<br \/>\nIf the heights of two cones are in the ratio of 1 : 4 and the radii of their bases are in the ratio 4 : 1, then the ratio of their volumes is<br \/>\n(a) 1 : 2<br \/>\n(b) 2 : 3<br \/>\n(c) 3 : 4<br \/>\n(d) 4 : 1<br \/>\nSolution:<br \/>\nRatio in the heights of two cones =1 : 4<br \/>\nand ratio in their radii of their bases = 4 : 1<br \/>\nLet height of the first cone = x<br \/>\nand height of the second cone = 4x<br \/>\nRadius of the first cone = 4y<br \/>\nand radius of the second cone = y<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1925\/44733155445_c707c38a2e_o.png?resize=346%2C550&#038;ssl=1\" alt=\"Maths RD Sharma Class 9 Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"346\" height=\"550\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 13.<br \/>\nThe slant height of a cone is increased by 10%. If the radius remains the same, the curved surface area is increased by<br \/>\n(a) 10%<br \/>\n(b) 12.1%<br \/>\n(c) 20%<br \/>\n(d) 21%<br \/>\nSolution:<br \/>\nLet r be radius and l be the slant height of a cone, then curved surface area = \u03c0rl<br \/>\nIf slant height is increased by 10%, then<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1945\/31774947538_e73f918dab_o.png?resize=304%2C186&#038;ssl=1\" alt=\"Maths RD Sharma Class 9 Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"304\" height=\"186\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1935\/44733155215_b068f46be6_o.png?resize=335%2C153&#038;ssl=1\" alt=\"RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS 13a\" width=\"335\" height=\"153\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 14.<br \/>\nThe height of a solid cone is 12 cm and the area of the circular base is 6471 cm<sup>2<\/sup>. A plane parallel to the base of the cone cuts through the cone 9 cm above the vertex of the cone, the area of the base of the new cone so formed is<br \/>\n(a) 9\u03c0 cm<sup>2<\/sup><br \/>\n(b) 16\u03c0 cm<sup>2<\/sup><br \/>\n(c) 25\u03c0 cm<sup>2<\/sup><br \/>\n(d) 36\u03c0 cm<sup>2<\/sup><br \/>\nSolution:<br \/>\nHeight of a solid cone (h) = 12 cm<br \/>\nArea of circular base = 64\u03c0 cm<sup>2<\/sup><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1979\/31774947428_582001c38d_o.png?resize=352%2C579&#038;ssl=1\" alt=\"RD Sharma Class 9 PDF Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"352\" height=\"579\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1934\/44733155115_b760d971b6_o.png?resize=338%2C81&#038;ssl=1\" alt=\"Surface Areas and Volume of A Right Circular Cone Class 9 RD Sharma Solutions\" width=\"338\" height=\"81\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 15.<br \/>\nIf the base radius and the height of a right circular cone are increased by 20%, then the percentage increase in volume is approximately<br \/>\n(a) 60<br \/>\n(b) 68<br \/>\n(c) 73<br \/>\n(d) 78<br \/>\nSolution:<br \/>\nIn first case,<br \/>\nLet r be radius and h be height, in volume<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1912\/45647414801_241d315a92_o.png?resize=348%2C629&#038;ssl=1\" alt=\"Class 9 RD Sharma Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"348\" height=\"629\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 16.<br \/>\nIf h, S and V denote respectively the height, curved surface area and volume of a right circular cone, then 3\u03c0Vh<sup>3<\/sup> &#8211; S2h<sup>2<\/sup> + 9V<sup>2<\/sup> is equal to<br \/>\n(a) 8<br \/>\n(b) 0<br \/>\n(c) 4\u03c0<br \/>\n(d) 32\u03c0<sup>2<\/sup><br \/>\nSolution:<br \/>\nh = height, S = curved surface area<br \/>\nV = volume of a cone<br \/>\nLet r be the radius of the cone, then<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1961\/44733154625_5b6de845f5_o.png?resize=348%2C343&#038;ssl=1\" alt=\"Class 9 Maths Chapter 20 Surface Areas and Volume of A Right Circular Cone RD Sharma Solutions\" width=\"348\" height=\"343\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 17.<br \/>\nIf a cone is cut into two parts by a horizontal plane passing through the mid\u00acpoint of its axis, the ratio of the volumes of upper and lower part is<br \/>\n(a) 1 : 2<br \/>\n(b) 2 : 1<br \/>\n(c) 1 : 7<br \/>\n(d) 1 : 8<br \/>\nSolution:<br \/>\n\u2234 \u2206PDC ~ \u2206PBA (AA axiom)<br \/>\nand O&#8217; is mid point of PO<br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1914\/45647414151_ac8f038d36_o.png?resize=246%2C285&#038;ssl=1\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"246\" height=\"285\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1958\/44733154345_ed7ef1107b_o.png?resize=324%2C274&#038;ssl=1\" alt=\"RD Sharma Class 9 Book Chapter 20 Surface Areas and Volume of A Right Circular Cone\" width=\"324\" height=\"274\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" src=\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1924\/44922437344_f70f765917_o.png?resize=353%2C228&#038;ssl=1\" alt=\"Surface Areas and Volume of A Right Circular Cone With Solutions PDF RD Sharma Class 9 Solutions\" width=\"353\" height=\"228\" data-recalc-dims=\"1\" \/><\/p>\n<p><a title=\"RD Sharma Class 9 solutions Chapter 20 Surface Area and volume of A Right Circular cone Ex 20.1\" href=\"https:\/\/www.cbselabs.com\/rd-sharma-class-9-solutions-chapter-20-surface-area-volume-right-circular-cone-ex-20-1\/\">Chapter 20 Exercise 20.1<\/a><br \/>\n<a title=\"RD Sharma Class 9 solutions Chapter 20 Surface Area and volume of A Right Circular cone Ex 20.2\" href=\"https:\/\/www.cbselabs.com\/rd-sharma-class-9-solutions-chapter-20-surface-area-and-volume-of-a-right-circular-cone-ex-20-2\/\">Chapter 20 Exercise 20.2<\/a><\/p>\n<p><a class=\"button-blue\" title=\"RD Sharma Class 9 Solutions\" href=\"https:\/\/www.cbselabs.com\/rd-sharma-class-9th-solutions\/\">RD Sharma Class 9 Solutions<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Surface Area and Volume of A Right Circular Cone Summary: A Right circular cone is a solid generated by revolving a line segment which passes through a fixed point and which makes a constant angle with fixed line. The fixed point is called the vertex of the cone, the fixed &hellip;<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"open","template":"","meta":{"spay_email":""},"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.2 (Yoast SEO v17.2) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>RD Sharma Class 9 Solutions Surface Area and Volume of A Right Circular Cone - CBSE Labs<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.cbselabs.com\/rd-sharma-class-9-solutions-chapter-20-surface-area-volume-right-circular-cone\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"RD Sharma Class 9 Solutions Surface Area and Volume of A Right Circular Cone\" \/>\n<meta property=\"og:description\" content=\"RD Sharma Class 9 Solutions Surface Area and Volume of A Right Circular Cone Summary: A Right circular cone is a solid generated by revolving a line segment which passes through a fixed point and which makes a constant angle with fixed line. The fixed point is called the vertex of the cone, the fixed &hellip;\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.cbselabs.com\/rd-sharma-class-9-solutions-chapter-20-surface-area-volume-right-circular-cone\/\" \/>\n<meta property=\"og:site_name\" content=\"CBSE Labs\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/ncertsolutionsbooks\/\" \/>\n<meta property=\"article:modified_time\" content=\"2021-09-18T10:03:44+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/farm2.staticflickr.com\/1926\/30706412237_13eb67a24a_o.png\" \/>\n<meta name=\"twitter:card\" content=\"summary\" \/>\n<meta name=\"twitter:site\" content=\"@learncbse\" \/>\n<meta name=\"twitter:label1\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data1\" content=\"17 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Organization\",\"@id\":\"https:\/\/www.cbselabs.com\/#organization\",\"name\":\"LearnCBSE\",\"url\":\"https:\/\/www.cbselabs.com\/\",\"sameAs\":[\"https:\/\/www.facebook.com\/ncertsolutionsbooks\/\",\"https:\/\/www.instagram.com\/learncbse.in\/\",\"https:\/\/www.youtube.com\/user\/CBSEPapers\",\"https:\/\/in.pinterest.com\/learncbsein\/\",\"https:\/\/twitter.com\/learncbse\"],\"logo\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/www.cbselabs.com\/#logo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2017\/01\/LearnCBSE.in_.png?fit=210%2C198&ssl=1\",\"contentUrl\":\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2017\/01\/LearnCBSE.in_.png?fit=210%2C198&ssl=1\",\"width\":210,\"height\":198,\"caption\":\"LearnCBSE\"},\"image\":{\"@id\":\"https:\/\/www.cbselabs.com\/#logo\"}},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/www.cbselabs.com\/#website\",\"url\":\"https:\/\/www.cbselabs.com\/\",\"name\":\"CBSE Labs\",\"description\":\"NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12\",\"publisher\":{\"@id\":\"https:\/\/www.cbselabs.com\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/www.cbselabs.com\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/www.cbselabs.com\/rd-sharma-class-9-solutions-chapter-20-surface-area-volume-right-circular-cone\/#primaryimage\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/farm2.staticflickr.com\/1926\/30706412237_13eb67a24a_o.png\",\"contentUrl\":\"https:\/\/farm2.staticflickr.com\/1926\/30706412237_13eb67a24a_o.png\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.cbselabs.com\/rd-sharma-class-9-solutions-chapter-20-surface-area-volume-right-circular-cone\/#webpage\",\"url\":\"https:\/\/www.cbselabs.com\/rd-sharma-class-9-solutions-chapter-20-surface-area-volume-right-circular-cone\/\",\"name\":\"RD Sharma Class 9 Solutions Surface Area and Volume of A Right Circular Cone - CBSE Labs\",\"isPartOf\":{\"@id\":\"https:\/\/www.cbselabs.com\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/www.cbselabs.com\/rd-sharma-class-9-solutions-chapter-20-surface-area-volume-right-circular-cone\/#primaryimage\"},\"datePublished\":\"2017-08-21T05:56:17+00:00\",\"dateModified\":\"2021-09-18T10:03:44+00:00\",\"breadcrumb\":{\"@id\":\"https:\/\/www.cbselabs.com\/rd-sharma-class-9-solutions-chapter-20-surface-area-volume-right-circular-cone\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/www.cbselabs.com\/rd-sharma-class-9-solutions-chapter-20-surface-area-volume-right-circular-cone\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/www.cbselabs.com\/rd-sharma-class-9-solutions-chapter-20-surface-area-volume-right-circular-cone\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/www.cbselabs.com\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"RD Sharma Class 9 Solutions Surface Area and Volume of A Right Circular Cone\"}]}]}<\/script>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"RD Sharma Class 9 Solutions Surface Area and Volume of A Right Circular Cone - CBSE Labs","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/www.cbselabs.com\/rd-sharma-class-9-solutions-chapter-20-surface-area-volume-right-circular-cone\/","og_locale":"en_US","og_type":"article","og_title":"RD Sharma Class 9 Solutions Surface Area and Volume of A Right Circular Cone","og_description":"RD Sharma Class 9 Solutions Surface Area and Volume of A Right Circular Cone Summary: A Right circular cone is a solid generated by revolving a line segment which passes through a fixed point and which makes a constant angle with fixed line. 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