{"id":4589,"date":"2023-02-21T09:36:05","date_gmt":"2023-02-21T09:36:05","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4589"},"modified":"2023-02-21T09:37:58","modified_gmt":"2023-02-21T09:37:58","slug":"ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-9","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-9\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9"},"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 13.9 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-13-4\/\" 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<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 13.9<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. A wooden bookshelf has external dimensions as follows: Height = 110cm, Depth = 25cm, <\/strong><strong>Breadth = 85cm (see fig. 13.31). The thickness of the plank is 5cm everywhere. The external faces are to be polished, and the inner faces are to be painted. If the rate of polishing is 20 paise per cm<sup>2<\/sup>\u00a0and the rate of painting is 10 paise per cm<sup>2<\/sup>, find the total expenses required for polishing and painting the surface of the bookshelf.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4673\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.8-Q1.jpeg\" alt=\"NCERT Class 9 Solutions Maths\" width=\"157\" height=\"176\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>The external dimensions of the bookshelf,<br \/>\nBreadth, B = 85 cm<\/span><br \/>\n<span style=\"color: #000000;\">Depth, D = 25 cm<\/span><br \/>\n<span style=\"color: #000000;\">Height, H = 110 cm<\/span><br \/>\n<span style=\"color: #000000;\">The thickness of the plank, t = 5 cm<\/span><br \/>\n<span style=\"color: #000000;\">Internal measures of the bookshelf is,<\/span><br \/>\n<span style=\"color: #000000;\">The breadth of each shelf, b\u00a0= B &#8211; 2t<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 b = 85 cm &#8211; 2 \u00d7 5 cm = 75 cm<\/span><br \/>\n<span style=\"color: #000000;\">Depth of each shelf, d =\u00a0D &#8211; t<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 d = 25 cm &#8211; 5 cm = 20 cm<\/span><br \/>\n<span style=\"color: #000000;\">Height of each shelf, h = H &#8211; 4t <\/span><br \/>\n<span style=\"color: #000000;\">= (110 cm &#8211; 4 \u00d7 5 cm) \u00f7 3 <\/span><br \/>\n<span style=\"color: #000000;\">= 90 cm \/ 3 <\/span><br \/>\n<span style=\"color: #000000;\">= 30 cm<\/span><br \/>\n<span style=\"color: #000000;\">Now,\u00a0Surface area\u00a0to be polished = External 5 surfaces of the bookshelf + border of the shelf<\/span><br \/>\n<span style=\"color: #000000;\">= 2(B\u00a0+ H) D\u00a0+ BH + 2Ht + 4bt<\/span><br \/>\n<span style=\"color: #000000;\">= [2\u00a0\u00d7\u00a0(85 cm + 110 cm) \u00d7 25 cm] + (85 cm \u00d7 110 cm) + (2 \u00d7 110 cm \u00d7 5 cm) + (4 \u00d7 75 cm \u00d7 5 cm)<\/span><br \/>\n<span style=\"color: #000000;\">= 9750 cm<sup>2<\/sup>\u00a0+ 9350 cm<sup>2<\/sup>\u00a0+ 1100 cm<sup>2<\/sup>\u00a0+ 1500 cm<sup>2<br \/>\n<\/sup>= 21700 cm<sup>2<br \/>\n<\/sup>Cost of polishing at the rate of 20 paise per cm<sup>2<\/sup>\u00a0= 21700 cm<sup>2<\/sup>\u00a0\u00d7\u00a0(\u20b9 20\/100) \/ cm<sup>2<\/sup>\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">= \u20b9 4340<\/span><br \/>\n<span style=\"color: #000000;\">Surface area to be painted = Internal 5 surfaces of 3 shelves <\/span><br \/>\n<span style=\"color: #000000;\">= 3 [2(b + h)d + bh] <\/span><br \/>\n<span style=\"color: #000000;\">= 3 [2 \u00d7 (75 cm + 30 cm) \u00d7 20 cm + (75 cm \u00d7 30 cm)]<\/span><br \/>\n<span style=\"color: #000000;\">= 3 [4200cm<sup>2<\/sup>\u00a0+ 2250 cm<sup>2<\/sup>]<\/span><br \/>\n<span style=\"color: #000000;\">= 3 \u00d7 6450 cm<sup>2<br \/>\n<\/sup>= 19350 cm<sup>2<br \/>\n<\/sup>Cost of painting at the rate of 10 paise per cm<sup>2<\/sup>\u00a0= 19350 cm<sup>2<\/sup>\u00a0\u00d7 (\u20b9 10\/100) \/ cm<sup>2<\/sup>\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">= \u20b9 1935<\/span><br \/>\n<span style=\"color: #000000;\">Total expense required for polishing and painting = \u20b9 4340 + \u20b9 1935\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">= \u20b9 6275 <\/span><br \/>\n<span style=\"color: #000000;\">Thus, the total expense required for polishing and painting the surface of the bookshelf is \u20b9 6275.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in fig. 13.32. Eight such spheres are used forth is the purpose and are to be painted silver. Each support is a cylinder of radius 1.5cm and height 7cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm<sup>2,<\/sup>\u00a0and black paint costs 5 paise per cm<sup>2<\/sup>.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4674\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.8-Q2.jpeg\" alt=\"NCERT Class 9 Solutions Maths\" width=\"132\" height=\"117\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Diameter of the wooden sphere = 21 cm<\/span><br \/>\n<span style=\"color: #000000;\">The radius (R) of the wooden sphere = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{21}{2}\" alt=\"\\frac{21}{2}\" align=\"absmiddle\" \/> cm<\/span><br \/>\n<span style=\"color: #000000;\">Surface area\u00a0for wooden 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 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{21}{2}\" alt=\"\\frac{21}{2}\" align=\"absmiddle\" \/> cm \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{21}{2}\" alt=\"\\frac{21}{2}\" align=\"absmiddle\" \/> cm <\/span><br \/>\n<span style=\"color: #000000;\">= 1386 cm<sup>2<br \/>\n<\/sup>Since the support is a cylinder of radius, r = 1.5 cm<\/span><br \/>\n<span style=\"color: #000000;\">Area of the circular end of the cylinder = \u03c0r<sup>2<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 1.5 cm \u00d7 1.5 cm <\/span><br \/>\n<span style=\"color: #000000;\">= 7.07 cm<sup>2<br \/>\n<\/sup>So, the area of each wooden sphere to be painted = 1386 cm<sup>2<\/sup>\u00a0&#8211; 7.07 cm<sup>2<\/sup>\u00a0= 1378.93 cm<sup>2<br \/>\n<\/sup>Total area of the 8 spheres to be painted = 8 \u00d7 1378.93 cm<sup>2<\/sup>\u00a0= 11031.44 cm<sup>2<br \/>\n<\/sup>Cost of silver painting the wooden spheres at the rate of 25 paise per cm<sup>2<br \/>\n<\/sup>= 11031.44 \u00d7 \u20b9 (25\/100) <\/span><br \/>\n<span style=\"color: #000000;\">= \u20b9 2757.86 <\/span><br \/>\n<span style=\"color: #000000;\">Now, <\/span><br \/>\n<span style=\"color: #000000;\">Radius of the cylinder, r = 1.5 cm <\/span><br \/>\n<span style=\"color: #000000;\">Height of the cylinder, h = 7 cm <\/span><br \/>\n<span style=\"color: #000000;\">Curve\u00a0surface area\u00a0of the cylinder = 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 1.5 cm \u00d7 7 cm <\/span><br \/>\n<span style=\"color: #000000;\">= 66 cm<sup>2<br \/>\n<\/sup>CSA of 8 cylindrical support to be painted = 8 \u00d7 66 cm<sup>2<\/sup>\u00a0= 528 cm<sup>2<br \/>\n<\/sup>Cost of black painting the cylindrical support at 5 paise per cm<sup>2<br \/>\n<\/sup>= 528 \u00d7 \u20b9 (5\/100) <\/span><br \/>\n<span style=\"color: #000000;\">= \u20b9 26.40 <\/span><br \/>\n<span style=\"color: #000000;\">Hence the cost of paint required = \u20b9 2757.86 + \u20b9 26.40 <\/span><br \/>\n<span style=\"color: #000000;\">= \u20b9 2784.26 <\/span><br \/>\n<span style=\"color: #000000;\">Thus, the cost of paint required is \u20b9 2784.26 (approx.)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Let the\u00a0radius of the sphere = r<\/span><br \/>\n<span style=\"color: #000000;\">Then its diameter, d= 2r<\/span><br \/>\n<span style=\"color: #000000;\">The\u00a0Surface area of a sphere\u00a0= 4\u03c0r<sup>2<br \/>\n<\/sup>The Curved surface area of the sphere = 4\u03c0r<sup>2<br \/>\n<\/sup>Now it is given in the question that the diameter of the sphere is decreased by 25% hence a new sphere is formed. <\/span><br \/>\n<span style=\"color: #000000;\">Therefore, the diameter of the new sphere can be written as:<\/span><br \/>\n<span style=\"color: #000000;\">= 2r &#8211; (25%) of (2r)<\/span><br \/>\n<span style=\"color: #000000;\">= 2r &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{25}{100}\" alt=\"\\frac{25}{100}\" align=\"absmiddle\" \/> \u00d7 (2r) <\/span><br \/>\n<span style=\"color: #000000;\">= 2r &#8211; (r\/2) <\/span><br \/>\n<span style=\"color: #000000;\">= 3r\/2 <\/span><br \/>\n<span style=\"color: #000000;\">Radius of the new sphere = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3r}{2}\" alt=\"\\frac{3r}{2}\" align=\"absmiddle\" \/>\u00a0 <\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{3r}{4}\" alt=\"\\frac{3r}{4}\" align=\"absmiddle\" \/> <\/span><br \/>\n<span style=\"color: #000000;\">Hence, curved surface area of the new sphere = 4\u03c0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{3r}{4}&amp;space;\\right&amp;space;)^2\" alt=\"\\left ( \\frac{3r}{4} \\right )^2\" align=\"absmiddle\" \/> <\/span><br \/>\n<span style=\"color: #000000;\">= 4\u03c0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;\\frac{9r^2}{16}&amp;space;\\right&amp;space;)\" alt=\"\\left ( \\frac{9r^2}{16} \\right )\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{9\\pi&amp;space;r^2}{4}\" alt=\"\\frac{9\\pi r^2}{4}\" align=\"absmiddle\" \/> <\/span><br \/>\n<span style=\"color: #000000;\">Now, decrease in the original curved surface area = 4\u03c0r &#8211; <img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{9\\pi&amp;space;r^2}{4}\" alt=\"\\frac{9\\pi r^2}{4}\" width=\"36\" height=\"41\" align=\"absmiddle\" \/> <\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{16&amp;space;\\pi&amp;space;r^2&amp;space;-&amp;space;9\\pi&amp;space;r^2}{4}\" alt=\"\\frac{16 \\pi r^2 - 9\\pi r^2}{4}\" align=\"absmiddle\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7\\pi&amp;space;r^2}{4}\" alt=\"\\frac{7\\pi r^2}{4}\" align=\"absmiddle\" \/> <\/span><br \/>\n<span style=\"color: #000000;\">So, the percentage decrease in the curved surface area is,<\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7\\pi&amp;space;r^2}{4}\" alt=\"\\frac{7\\pi r^2}{4}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{4&amp;space;\\pi&amp;space;r^2}\" alt=\"\\frac{1}{4 \\pi r^2}\" align=\"absmiddle\" \/> \u00d7\u00a0100% <\/span><\/p>\n<p><span style=\"color: #000000;\">=\u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7}{16}\" alt=\"\\frac{7}{16}\" align=\"absmiddle\" \/> \u00d7 100% <\/span><br \/>\n<span style=\"color: #000000;\">= 43.75% <\/span><br \/>\n<span style=\"color: #000000;\">Therefore, the percentage decrease in the surface area of the sphere is 43.75% .<\/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 13.9 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-4589","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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