{"id":4584,"date":"2023-02-21T09:35:43","date_gmt":"2023-02-21T09:35:43","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4584"},"modified":"2023-02-21T09:39:11","modified_gmt":"2023-02-21T09:39:11","slug":"ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-4","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-4\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4"},"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 13.4 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-13-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 13.2<\/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-13-5\/\" 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; 13.4<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Find the surface area of a sphere of radius:<br \/>\n<\/strong><strong>(i) 10.5cm<br \/>\n(ii) 5.6cm<br \/>\n(iii) 14cm<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><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)<\/strong> Radius of the sphere, <em>r<\/em> = 10.5 cm<br \/>\nSurface area\u00a0of the\u00a0sphere\u00a0= 4\u03c0r<sup>2<br \/>\n<\/sup>= 4 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 10.5 cm<sup>2<br \/>\n<\/sup>= 1386 cm<sup>2<br \/>\n<\/sup>Surface area of the sphere is 1386 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong> Radius of the sphere, <em>r<\/em> = 5.6cm<br \/>\nSurface area of the sphere = 4\u03c0r<sup>2<br \/>\n<\/sup>= 4 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 5.6 cm<sup>2<br \/>\n<\/sup>= 394.24 cm<sup>2<\/sup><\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(iii)<\/strong> Radius, <em>r<\/em> = 14 cm<\/span><br \/>\n<span style=\"color: #000000;\">Surface area of the sphere = 4\u03c0r<sup>2<br \/>\n<\/sup>= 4 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 14 cm<sup>2<br \/>\n<\/sup>= 2464 cm<sup>2<br \/>\n<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Find the surface area of a sphere of diameter:<br \/>\n<\/strong><strong>(i) 14 cm<br \/>\n(ii) 21 cm<br \/>\n(iii) 3.5 cm<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><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)<\/strong> Radius of sphere, <em>r<\/em> = diameter\/2 = 14\/2 cm = 7 cm<br \/>\nSurface area of a\u00a0sphere\u00a0= 4\u03c0r<sup>2<br \/>\n<\/sup>= 4 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 7cm \u00d7 7cm<\/span><br \/>\n<span style=\"color: #000000;\">= 616 cm<sup>2<br \/>\n<\/sup>Surface area of the sphere is 616 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">(ii) Radius of the sphere, <em>r <\/em>= diameter\/2 = 21\/2 = 10.5 cm<br \/>\nSurface area of a sphere = 4\u03c0r<sup>2<br \/>\n<\/sup>= 4 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 10.5<sup>2<br \/>\n<\/sup>= 1386<br \/>\nSurface area of the sphere is 1386 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii)<\/strong> Radius of sphere, <em>r <\/em>= diameter\/2 = 3.5\/2 = 1.75 cm<br \/>\nSurface area of a sphere = 4\u03c0r<sup>2<br \/>\n<\/sup>= 4 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 1.75<sup>2<\/sup><br \/>\n= 38.5<br \/>\nSurface area of the sphere is 38.5 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Find the total surface area of a hemisphere of radius 10 cm. [Use \u03c0=3.14]<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Radius of the hemisphere, <em>r<\/em> = 10 cm<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4656\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.4-Ans3.jpg\" alt=\"NCERT Class 9 Solutions Maths\" width=\"302\" height=\"232\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.4-Ans3.jpg 302w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.4-Ans3-300x230.jpg 300w\" sizes=\"auto, (max-width: 302px) 100vw, 302px\" \/><br \/>\nThe total surface area of a hemisphere = (4\u03c0r<sup>2<\/sup>)\/2 + \u03c0r<sup>2<\/sup>\u00a0= 3\u03c0r<sup>2<br \/>\n<\/sup>= 3 \u00d7 3.14 \u00d7 10<sup>2<\/sup><br \/>\n= 942<br \/>\nThe total surface area of the given hemisphere is 942 cm<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>The radius of the balloon before pumping air, r<sub>\u2081<\/sub>\u00a0= 7cm<\/span><br \/>\n<span style=\"color: #000000;\">The radius of the balloon after pumping air, r<sub>\u2082<\/sub>\u00a0= 14cm<\/span><br \/>\n<span style=\"color: #000000;\">The surface area of the balloon before pumping air, SA<sub>\u2081<\/sub>\u00a0= 4\u03c0(r<sub>\u2081<\/sub>)<sup>2<br \/>\n<\/sup>The surface area of the balloon after pumping air, SA<sub>\u2082<\/sub>\u00a0= 4\u03c0(r<sub>\u2082<\/sub>)<sup>2<br \/>\n<\/sup>The ratio of the surface areas of the balloon, = SA<sub>\u2081<\/sub>\/CSA<sub>\u2082<br \/>\n<\/sub>= 4\u03c0(r<sub>\u2081<\/sub>)<sup>2<\/sup>\/4\u03c0(r<sub>\u2082<\/sub>)<sup>2<br \/>\n<\/sup>= (r<sub>\u2081<\/sub>)<sup>2<\/sup>\/(r<sub>\u2082<\/sub>)<sup>2<br \/>\n<\/sup>= (r<sub>\u2081<\/sub>\/r<sub>\u2082<\/sub>)<sup>2<br \/>\n<\/sup>= (7\/14)<sup>2<br \/>\n<\/sup>=\u00a0(1\/2)<sup>2<br \/>\n<\/sup>= (1\/4) <\/span><br \/>\n<span style=\"color: #000000;\">The ratio of the surface areas of the balloons = 1: 4<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm<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;<br \/>\n<\/strong>Inner diameter, d = 10.5 cm<\/span><br \/>\n<span style=\"color: #000000;\">Inner radius, r = d\/2 = 10.5\/2cm = 5.25 cm\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">CSA of hemispherical bowl = 2\u03c0r<sup>2<br \/>\n<\/sup>= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 5.25 cm \u00d7 5.25 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 173.25 cm<sup>2<br \/>\n<\/sup>The cost of tin-plating 100 cm<sup>2<\/sup>\u00a0of the bowl =\u00a0\u20b916<\/span><br \/>\n<span style=\"color: #000000;\">The cost of tin-plating 1 cm<sup>2<\/sup> of the bowl =\u00a0\u20b916\/100 <\/span><br \/>\n<span style=\"color: #000000;\">The cost of tin-plating 173.25 cm<sup>2\u00a0<\/sup>area of the bowl = (\u20b916\/100) \u00d7 173.25 = 27.72<\/span><br \/>\n<span style=\"color: #000000;\">Thus, the cost of tin-plating is \u20b9 27.72.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. Find the radius of a sphere whose surface area is 154 cm<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\" \/>)<br \/>\n<\/strong><strong>Answer &#8211; <\/strong>Let the radius of the sphere = <em>r<\/em><br \/>\nThe surface area of the sphere = 154 cm<sup>2<br \/>\n<\/sup>4\u03c0r<sup>2<\/sup>\u00a0= 154 cm<sup>2<br \/>\n<\/sup>r<sup>2<\/sup> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{154}{4\\Pi}\" alt=\"\\frac{154}{4\\Pi}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\">r<sup>2<\/sup> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{154}{4\\times&amp;space;\\frac{22}{7}}\" alt=\"\\frac{154}{4\\times \\frac{22}{7}}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\">r<sup>2<\/sup> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{154\\times&amp;space;7}{4\\times&amp;space;22}\" alt=\"\\frac{154\\times 7}{4\\times 22}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\">r<sup>2<\/sup> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{49}{4}\" alt=\"\\frac{49}{4}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\">r = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7}{2}\" alt=\"\\frac{7}{2}\" align=\"absmiddle\" \/> = 3.5 cm<\/span><br \/>\n<span style=\"color: #000000;\">Thus, the radius of the sphere whose surface area is 154 cm<sup>2<\/sup>\u00a0is\u00a03.5 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. The diameter of the moon is approximately one fourth of the diameter of the earth. <\/strong><strong>Find the ratio of their surface areas. <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Let the\u00a0radius of the earth = <em>R<\/em><br \/>\nthe radius of the moon = <em>r<\/em><br \/>\nDiameter of the moon = 1\/4 \u00d7 diameter of the earth<br \/>\nThus, the radius of the moon = 1\/4 \u00d7 radius of the earth\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Since, radius = 2 \u00d7 Diameter]<\/span><br \/>\n<span style=\"color: #000000;\">r = 1\/4 \u00d7 R<\/span><br \/>\n<span style=\"color: #000000;\">r\/R = 1\/4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212; (1)<\/span><br \/>\n<span style=\"color: #000000;\">Now, the\u00a0surface area\u00a0of\u00a0earth = 4\u03c0R<sup>2<br \/>\n<\/sup>The surface\u00a0area\u00a0of moon = 4\u03c0r<sup>2<br \/>\n<\/sup>The ratio of their surface areas = 4\u03c0r<sup>2<\/sup>\/4\u03c0R<sup>2<br \/>\n<\/sup>= r<sup>2<\/sup>\/R<sup>2<br \/>\n<\/sup>= (r\/R)<sup>2<br \/>\n<\/sup>= (1\/4)<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[From equation (1)]<\/span><br \/>\n<span style=\"color: #000000;\">= 1\/16<\/span><br \/>\n<span style=\"color: #000000;\">The ratio of their surface areas = 1 : 16<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. (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>The inner radius of the bowl, <em>r<\/em> = 5 cm<\/span><br \/>\n<span style=\"color: #000000;\">Thickness of steel = 0.25 cm<\/span><br \/>\n<span style=\"color: #000000;\">Outer radius of the bowl, <em>R<\/em> = 5 cm + 0.25 cm = 5.25 cm<\/span><br \/>\n<span style=\"color: #000000;\">Outer CSA of the hemisphere = 2\u03c0R<sup>2<br \/>\n<\/sup>= 2 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 5.25 cm \u00d7 5.25 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 173.25 cm<sup>2<br \/>\n<\/sup>Thus, the outer curved surface area of the hemisphere\u00a0is\u00a0173.25 cm<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find<br \/>\n<\/strong><strong>(i) surface area of the sphere,<br \/>\n<\/strong><strong>(ii) curved surface area of the cylinder,<br \/>\n<\/strong><strong>(iii) ratio of the areas obtained in(i) and (ii).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4657\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.4-Ans09.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"213\" height=\"282\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>The radius of the sphere = Radius of the cylinder = <em>r<\/em><\/span><br \/>\n<span style=\"color: #000000;\">Height of the cylinder, <em>h<\/em> = diameter of the sphere = 2<em>r<br \/>\n<\/em>Thus, h = 2r<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)<\/strong> Surface area of sphere = 4\u03c0r<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong> Height of cylinder, h = r + r = 2r<br \/>\nRadius of cylinder = r<br \/>\nCSA of cylinder formula = 2\u03c0rh = = 2\u03c0r \u00d7 2r<br \/>\n= 4\u03c0r<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii)<\/strong> Ratio between areas = (Surface area of sphere)\/(CSA of Cylinder)<br \/>\n= 4\u03c0r<sup>2<\/sup>\/4\u03c0r<sup>2\u00a0<\/sup>= 1\/1<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Thus, the surface area of the sphere and the curved surface area of the cylinder is\u00a04\u03c0r<sup>2<\/sup> and the ratio between these areas is 1 : 1.<\/span><\/p>\n<p>&nbsp;<\/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 13.4 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-4584","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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