{"id":4582,"date":"2023-02-21T09:35:37","date_gmt":"2023-02-21T09:35:37","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4582"},"modified":"2023-02-21T09:39:39","modified_gmt":"2023-02-21T09:39:39","slug":"ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-2\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 9 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 13 (Surface Areas and Volumes)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 9 Mathematics\u00a0<strong>Chapter &#8211; 13 Surface Areas and Volumes <\/strong>Exercise 13.2 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter 13: Surface Areas and Volumes<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.1<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.3<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-4\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.4<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-5\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.5<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-6\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.6<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-7\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.7<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-8\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.8<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-9\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.9<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 13.2<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm<sup>2<\/sup>. Find the diameter of the base of the cylinder. (Assume \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;\u00a0<\/strong>Let the\u00a0radius of the cylinder = r<br \/>\nHeight of cylinder, h = 14 cm<br \/>\nCurved surface area of cylinder = 88 cm<sup>2<\/sup><br \/>\nCurved Surface Area\u00a0of a\u00a0right circular cylinder = 2\u03c0rh<br \/>\n2\u03c0rh = 88 cm\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 r \u00d7 14 cm = 88 cm\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">r = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{88\\times&amp;space;7}{2\\times&amp;space;22\\times&amp;space;14}\" alt=\"\\frac{88\\times 7}{2\\times 22\\times 14}\" align=\"absmiddle\" \/> = 1 cm<\/span><\/p>\n<p><span style=\"color: #000000;\">Diameter = 2 \u00d7 radius <\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 1 cm <\/span><br \/>\n<span style=\"color: #000000;\">= 2 cm <\/span><br \/>\n<span style=\"color: #000000;\">Thus, the diameter of the base of the cylinder is 2 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? Assume \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>Height of the tank, h = 1 m<br \/>\nDiameter of cylindrical tank, d = 140 cm<br \/>\nRadius = half of diameter = (140\/2) cm = 70cm = 0.7m<br \/>\nTotal surface area of the closed cylindrical tank = 2\u03c0r(r + h)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 0.7 m \u00d7 (0.7 m + 1 m)<\/span><br \/>\n<span style=\"color: #000000;\">= 4.4 m \u00d7 1.7 m<\/span><br \/>\n<span style=\"color: #000000;\">= 7.48 m\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">7.48 m\u00b2 of the sheet is required for making the cylindrical tank.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4cm. (see fig. 13.11). Find its<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4651\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.1-Q3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"83\" height=\"157\" \/><br \/>\n<\/strong><strong>(i) inner curved surface area,<br \/>\n<\/strong><strong>(ii) outer curved surface area<br \/>\n<\/strong><strong>(iii) total surface area<br \/>\n<\/strong><strong>(Assume \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>Let r and R be the inner and outer radii of the cylindrical pipe<br \/>\nr<sub>\u00a0<\/sub>= 4\/2 cm = 2 cm<br \/>\nR = 4.4\/2 cm = 2.2 cm<br \/>\nHeight of cylindrical pipe, h = length of cylindrical pipe = 77 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)<\/strong> Inner curved surface area = 2\u03c0rh<br \/>\n= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 2 \u00d7 77 cm<sup>2<br \/>\n<\/sup>= 968 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong> Outer curved surface area = 2\u03c0Rh<br \/>\n= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 2.2 \u00d7 77 cm<sup>2<br \/>\n<\/sup>= (22 \u00d7 22 \u00d7 2.2) cm<sup>2<br \/>\n<\/sup>= 1064.8 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii)<\/strong> Total surface area of pipe = inner curved surface area+ outer curved surface area+ Area of both circular ends of pipe.<br \/>\n= 2\u03c0rh + 2\u03c0Rh + 2\u03c0(R\u00b2 &#8211; r\u00b2)<br \/>\n= 9668 + 1064.8 + 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 (2.2<sup>2 <\/sup>&#8211; 2<sup>2<\/sup>) cm<sup>2<\/sup><br \/>\n= 2031.8 + 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 5.28 cm<sup>2<br \/>\n<\/sup>= 2031.8 + 5.28 cm<sup>2<\/sup><br \/>\n= 2038.08 cm<sup>2<br \/>\n<\/sup>Therefore, the total surface area of the cylindrical pipe is 2038.08 cm<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to <\/strong><strong>move once over to level a playground. Find the area of the playground in m<sup>2<\/sup>? (Assume \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>) <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>A roller is shaped like a cylinder.<br \/>\nThe Height of the roller, h = Length of roller = 120 cm<br \/>\nThe Diameter of the roller, d = 84 cm<br \/>\nThe Radius of the roller, r = Diameter\/2 = 84\/2 cm = 42 cm<br \/>\nCurved Surface Area of the roller = 2\u03c0rh <\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 42 cm \u00d7 120 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 31680 cm\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">Area of the playground = Area leveled by the cylinder in 500 revolutions <\/span><br \/>\n<span style=\"color: #000000;\">= 500 \u00d7 31680 cm\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">= 15840000 cm\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">= 15840000\/10000 m\u00b2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Since 1cm\u00b2 = 1\/10000 m\u00b2]<\/span><br \/>\n<span style=\"color: #000000;\">= 1584 m\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">Thus, area of the playground = 1584 m\u00b2.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m<sup>2<\/sup>.<br \/>\n<\/strong><strong>(Assume \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>) <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>The Height of a cylindrical pillar, h = 3.5 m<br \/>\nThe Diameter of a cylindrical pillar, d = 50 cm = 50\/100 m = 0.50 m<br \/>\nThe Radius of a cylindrical pillar, r = Diameter\/2 = 0.50\/2 =0.25 m<br \/>\nCurved surface area of the pillar = 2\u03c0rh<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 0.25 m \u00d7 3.5 m<\/span><br \/>\n<span style=\"color: #000000;\">= 5.5 m\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">Cost of painting the curved surface area at \u20b912.50 per m<sup>\u00b2<\/sup> = 12.50 \u00d7 5.5 <\/span><br \/>\n<span style=\"color: #000000;\">= \u20b9 68.75 <\/span><br \/>\n<span style=\"color: #000000;\">Thus, cost of painting the curved surface of the pillar is \u20b9 68.75.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. Curved surface area of a right circular cylinder is 4.4 m<sup>2<\/sup>. If the radius of the base of the base of the cylinder is 0.7 m, find its height. (Assume \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>) <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>The curved surface area of the pillar = 4.4 m\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">Radius\u00a0of the cylinder, r = 0.7 m<\/span><br \/>\n<span style=\"color: #000000;\">Height of the cylinder = h<\/span><br \/>\n<span style=\"color: #000000;\">2\u03c0rh = 4.4 m\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 0.7 m \u00d7 h = 4.4 m\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">h = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7\\times&amp;space;4.4}{2\\times22\\times0.7}\" alt=\"\\frac{7\\times 4.4}{2\\times22\\times0.7}\" align=\"absmiddle\" \/><\/span><br \/>\n<span style=\"color: #000000;\">h = 1 m<\/span><br \/>\n<span style=\"color: #000000;\">The height of the\u00a0right circular cylinder is 1 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. The inner diameter of a circular well is 3.5m. It is 10m deep. Find<br \/>\n<\/strong><strong>(i) its inner curved surface area,<br \/>\n<\/strong><strong>(ii) the cost of plastering this curved surface at the rate of Rs. 40 per m<sup>2<\/sup>.<br \/>\n<\/strong><strong>(Assume \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Inner Diameter of the circular well, d = 3.5 m<br \/>\nInner Radius of the circular well, r = d\/2 = 3.5\/2 m = 1.75 m<br \/>\nDepth of circular well, say h = 10 m<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)<\/strong> Inner curved surface area = 2\u03c0rh<br \/>\n= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 1.75 \u00d7 10<br \/>\n= 110\u00a0m<sup>2<br \/>\n<\/sup>Therefore, the inner curved surface area of the circular well is 110 m<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong> Cost of plastering 1 m<sup>2<\/sup> area = Rs. 40<br \/>\nCost of plastering 110 m<sup>2<\/sup> area = Rs (110 \u00d7 40)<br \/>\n= Rs. 4400<br \/>\nTherefore, the cost of plastering the curved surface of the well is Rs. 4400.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find <\/strong><strong>the total radiating surface in the system. (Assume \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Diameter of the pipe, d = 5 cm<br \/>\nRadius\u00a0of the pipe, r = d\/2 = 5\/2 cm = 2.5 cm = 2.5\/100 m = 0.025 m<\/span><br \/>\n<span style=\"color: #000000;\">Length of the pipe, h = 28 m<\/span><br \/>\n<span style=\"color: #000000;\">The total radiating surface area of the pipe = 2\u03c0rh <\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 0.025 m \u00d7 28 m <\/span><br \/>\n<span style=\"color: #000000;\">= 4.4 m\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">Thus, the total radiating surface is 4.4 m\u00b2.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. Find<br \/>\n<\/strong><strong>(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in <\/strong><strong>diameter and 4.5m high.<br \/>\n<\/strong><strong>(ii) How much steel was actually used, if <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{12}}\" alt=\"\\mathbf{\\frac{1}{12}}\" align=\"absmiddle\" \/> of the steel actually used was wasted in making the tank. (Assume \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Diameter of the tank, d = 4.2 m<\/span><br \/>\n<span style=\"color: #000000;\">Radius\u00a0of the tank, r = d\/2 = 4.2\/2 m = 2.1 m<\/span><br \/>\n<span style=\"color: #000000;\">Height of the tank, h = 4.5 m<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)<\/strong> the lateral or curved surface area of the cylindrical tank = 2\u03c0rh<br \/>\n= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 2.1 \u00d7 4.5 m<sup>2<br \/>\n<\/sup>= 44 \u00d7 0.3 \u00d7 4.5 m<sup>2<br \/>\n<\/sup>= 59.4 m<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong> Total surface area of tank = 2\u03c0r(r+h)<br \/>\n= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 2.1 \u00d7 (2.1 + 4.5)<br \/>\n= 44 \u00d7 0.3 \u00d7 6.6<br \/>\n= 87.12 m<sup>2<br \/>\n<\/sup>Let the amount of steel required to make the tank be &#8216;x&#8217;. <\/span><br \/>\n<span style=\"color: #000000;\">Amount of steel required &#8211; Amount of steel wasted =\u00a0Total surface area of the tank<\/span><br \/>\n<span style=\"color: #000000;\">x &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{x}{12}\" alt=\"\\frac{x}{12}\" align=\"absmiddle\" \/> = 87.12 m\u00b2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [Since, 1\/12 of the steel was wasted]<\/span><br \/>\n<span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{11x}{12}\" alt=\"\\frac{11x}{12}\" align=\"absmiddle\" \/> = 87.12 m\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{12}{11}\" alt=\"\\frac{12}{11}\" align=\"absmiddle\" \/> \u00d7 87.12 m\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">x =\u00a095.04 m\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">Thus, Curve surface area = 59.4m\u00b2, Steel actually used to make the tank = 95.04m\u00b2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>10. In fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. <\/strong><strong>The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. (Assume \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4652\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.1-Q10.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"86\" height=\"120\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>Diameter of the frame, d = 20 cm<br \/>\nRadius of the frame, r = d\/2 = 20\/2 cm = 10 cm<br \/>\nThe frame has a base height, h = 30 cm<br \/>\nHeight of the lampshade, h = 30 cm + 2.5 cm + 2.5 cm = 35 cm<br \/>\nThus, cloth required for covering the lampshade = 2\u03c0rh <\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 10 cm \u00d7 35 cm <\/span><br \/>\n<span style=\"color: #000000;\">= 2200 cm\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">Thus, the area of the cloth required is 2200 cm\u00b2.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>11. The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? (Assume \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Radius\u00a0of the penholder, r = 3 cm<\/span><br \/>\n<span style=\"color: #000000;\">Height of the penholder, h = 10.5 cm<\/span><br \/>\n<span style=\"color: #000000;\">Area of cardboard required for each penholder = 2\u03c0rh + \u03c0r\u00b2 = \u03c0r(2h + r)<\/span><br \/>\n<span style=\"color: #000000;\">Area of cardboard required for 35 penholders = 35 \u00d7 \u03c0r (2h + r )<\/span><br \/>\n<span style=\"color: #000000;\">= 35 \u00d7 \u03c0r(2h + r )<\/span><br \/>\n<span style=\"color: #000000;\">= 35 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 3 cm \u00d7 (2 \u00d7 10.5 cm + 3 cm)<\/span><br \/>\n<span style=\"color: #000000;\">= 330 cm \u00d7 24 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 7920 cm\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">Therefore, 7920 cm<sup>2<\/sup>\u00a0cardboard sheet will be needed for the competition.<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 13 (Surface Areas and Volumes)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 13 Surface Areas and Volumes Exercise 13.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 13: Surface Areas and Volumes NCERT Solution Class 9 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[698],"tags":[725,702,186,726,701,5],"class_list":["post-4582","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-13-surface-areas-and-volumes-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-13-in-english","tag-ncert-solutions-class-9-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.4) - 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