{"id":4583,"date":"2023-02-21T09:35:40","date_gmt":"2023-02-21T09:35:40","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4583"},"modified":"2023-02-21T09:39:25","modified_gmt":"2023-02-21T09:39:25","slug":"ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-3\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3"},"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 13.3 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-13-4\/\" 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; 13.3<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Diameter of the base of a cone is 10.5 cm, and its slant height is 10 cm. Find its curved surface area (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>Diameter of the base of the cone, <em>d<\/em> = 10.5 cm<br \/>\nRadius of the base of cone, <em>r <\/em>= diameter\/2 = 10.5\/2 cm = 5.25cm<br \/>\nThe slant height of the cone, <em>l<\/em> = 10 cm<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4654\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.3-Ans01.jpg\" alt=\"NCERT Class 9 Solutions Maths\" width=\"303\" height=\"304\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.3-Ans01.jpg 379w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.3-Ans01-300x300.jpg 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.3-Ans01-150x150.jpg 150w\" sizes=\"auto, (max-width: 303px) 100vw, 303px\" \/><br \/>\nCurved surface area =\u00a0<em>\u03c0rl<br \/>\n<\/em>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 5.25 \u00d7 10 <\/span><br \/>\n<span style=\"color: #000000;\">= 165 cm\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">Thus, curved\u00a0surface area\u00a0of the cone = 165 cm\u00b2.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Find the total surface area of a cone, if its slant height is 21 m and the diameter of its base is 24 m. (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>Diameter of cone, <em>d<\/em> = 24<br \/>\nRadius of cone, <em>r<\/em> = d\/2 = 24\/2 m = 12m<br \/>\nSlant height, <em>l<\/em> = 21 m<br \/>\nTotal Surface area of the cone = \u03c0r(l + r)<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 12 m \u00d7 (12 m + 21 m)<\/span><\/p>\n<p><span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 12 m \u00d7 33 m <\/span><br \/>\n<span style=\"color: #000000;\">= 8712\/7 m\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">= 1244.57 m\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">Thus, total surface area of the cone = 1244.57 m\u00b2.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3.<\/strong>\u00a0<strong>Curved surface area of a cone is 308 cm<sup>2,<\/sup> and its slant height is 14 cm. Find<br \/>\n<\/strong><strong>(i) the radius of the base and<br \/>\n(ii) the total surface area of the cone.<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 the radius of the cone = <em>r<br \/>\n<\/em>The slant height of the cone, <em>l<\/em> = 14 cm<br \/>\nCurved surface area = 308 cm\u00b2<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) the radius of the base\u00a0<\/strong><br \/>\nWe know that the Curved surface area of cone = \u03c0rl<br \/>\n\u03c0rl = 308 cm\u00b2<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/>\u00a0 \u00d7 r \u00d7 14 = 308 cm\u00b2<br \/>\n22 \u00d7 r \u00d7 2 = 308 cm\u00b2<br \/>\n44 r = 308 cm\u00b2<br \/>\nr = 308\/44 = 7 cm<br \/>\nTherefore, the radius of the cone base is 7 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong> Total surface area of cone = Curved surface area of cone + Area of base (\u03c0r<sup>2<\/sup>)<br \/>\nTotal surface area of cone = 308 + <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 7<sup>2<\/sup><br \/>\n= 308 + 154<br \/>\n= 462 cm<sup>2<br \/>\n<\/sup>Therefore, the total surface area of the cone is 462 cm<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. A conical tent is 10 m high, and the radius of its base is 24 m. Find<br \/>\n<\/strong><strong>(i) slant height of the tent.<br \/>\n<\/strong><strong>(ii) cost of the canvas required to make the tent, if the cost of 1 m<sup>2<\/sup> canvas is Rs 70.<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>Let the Slant height of conical tent = <em>l<\/em><br \/>\nHeight of conical tent, <em>h<\/em> = 10 m<br \/>\nRadius of the conical tent, <em>r<\/em> = 24m<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4654\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.3-Ans01.jpg\" alt=\"NCERT Class 9 Solutions Maths\" width=\"264\" height=\"265\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.3-Ans01.jpg 379w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.3-Ans01-300x300.jpg 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.3-Ans01-150x150.jpg 150w\" sizes=\"auto, (max-width: 264px) 100vw, 264px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) slant height of the tent.<br \/>\n<\/strong>Slant height, <em>l<\/em> = \u221ar\u00b2 + h\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">= \u221a(24)\u00b2\u00a0+ (10)\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a576\u00a0+ 100<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a676\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">= 26 m <\/span><br \/>\n<span style=\"color: #000000;\">Therefore, the slant height of the tent is 26 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong> Curved surface area of the tent = \u03c0rl<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 24 \u00d7 26 m<sup>2<br \/>\n<\/sup>Cost of 1 m<sup>2<\/sup> canvas = Rs 70<br \/>\nThe cost of the canvas required to make the tent, at \u20b9 70 per m<sup>\u00b2\u00a0<\/sup>= 70\u00a0\u00d7\u00a0Curved surface area of the cone<br \/>\n= 70 \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 24 \u00d7 26<br \/>\n= 10 \u00d7 22 \u00d7 24 \u00d7 26<br \/>\n= \u20b9 137280<br \/>\nTherefore, the cost of the canvas required to make such a tent is \u20b9 137280.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. What length of tarpaulin 3 m wide will be required to make a conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. [Use \u03c0=3.14]<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Height of conical tent, <em>h<\/em> = 8m<br \/>\nThe radius of the base of the tent, <em>r<\/em> = 6m<br \/>\nThe slant height of the tent = <em>l <\/em><br \/>\n<em>l<\/em> = \u221a(r<sup>2<\/sup>+h<sup>2<\/sup>)<br \/>\n= \u221a(6<sup>2<\/sup>+8<sup>2<\/sup>)<br \/>\n= \u221a(36 + 64)<br \/>\n= \u221a100<br \/>\n= 10 m<br \/>\nTherefore, the curved surface area = \u03c0rl<\/span><br \/>\n<span style=\"color: #000000;\">= 3.14 \u00d7 6m \u00d7 10m<\/span><br \/>\n<span style=\"color: #000000;\">= 188.4 m<sup>2<br \/>\n<\/sup>Now, width of the tarpaulin = 3m<\/span><br \/>\n<span style=\"color: #000000;\">Area of the tarpaulin = 188.4 m<sup>2<br \/>\n<\/sup>So, Area of the tarpaulin =\u00a0width of the tarpaulin\u00a0\u00d7\u00a0length of the tarpaulin<\/span><br \/>\n<span style=\"color: #000000;\">188.4 m<sup>2\u00a0<\/sup>=\u00a03\u00a0\u00d7\u00a0length of the tarpaulin<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 Length of the tarpaulin = 188.4 m<sup>2<\/sup>\/3<\/span><br \/>\n<span style=\"color: #000000;\">= 62.8 m<\/span><br \/>\n<span style=\"color: #000000;\">Extra length of the material = 20cm = 20\/100m = 0.2m<\/span><br \/>\n<span style=\"color: #000000;\">Actual length required = 62.8m + 0.2m = 63m<\/span><br \/>\n<span style=\"color: #000000;\">Thus, the\u00a0required length of the tarpaulin\u00a0is 63 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. The slant height and base diameter of a conical tomb are 25m 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>. (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>The slant height of the conical tomb, <em>l<\/em> = 25m<br \/>\nBase Diameter, d = 14 m<br \/>\nBase Radius, r = diameter\/2 = 14\/2 m = 7m<br \/>\nCurved surface area = \u03c0rl<\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 7 m \u00d7 25 m <\/span><br \/>\n<span style=\"color: #000000;\">= 550m<sup>2<br \/>\n<\/sup>Cost of the whitewashing at \u20b9 210 per 100 m<sup>2<br \/>\n<\/sup>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{210}{100}\" alt=\"\\frac{210}{100}\" align=\"absmiddle\" \/> \u00d7 550 <\/span><br \/>\n<span style=\"color: #000000;\">= \u20b9 1155 <\/span><br \/>\n<span style=\"color: #000000;\">Thus, the cost of whitewashing the conical tomb is \u20b91155.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. A joker\u2019s cap is in the form of a right circular cone with a base radius of 7 cm and a height of 24cm. Find the area of the sheet required to make 10 such caps. (Assume \u03c0 =22\/7)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>The radius of the conical cap, r = 7 cm<br \/>\nHeight of conical cap, h = 24cm<br \/>\nSlant height = <em>l<\/em><br \/>\n<em>l<\/em> = \u221a(r<sup>2 <\/sup>+ h<sup>2<\/sup>)<br \/>\n= \u221a(7<sup>2 <\/sup>+ 24<sup>2<\/sup>)<br \/>\n= \u221a(49 + 576)<br \/>\n= \u221a625<br \/>\n= 25 cm<br \/>\nArea of the sheet required to make each cap = \u03c0rl <\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 7 cm \u00d7 25 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 550 cm\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">Area of the sheet required to make 10 such caps = 10 \u00d7 550 cm<sup>2<\/sup>\u00a0= 5500 cm<sup>2<br \/>\n<\/sup>Thus, the area of the sheet required to make 10 such caps is 5500 cm<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. A bus stop is barricaded from the remaining part of the road by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and a height of 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m<sup>2<\/sup>, what will be the cost of painting all these cones? (Use \u03c0 = 3.14 and take \u221a(1.04) =1.02)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>Diameter, d = 40cm = 40\/100 m = 0.4m<br \/>\nRadius of cone, r = diameter\/2 = 0.4\/2 cm = 0.2 m<br \/>\nHeight of cone, h = 1 m<br \/>\nThe slant height of the cone l = \u221a(r<sup>2 <\/sup>+ h<sup>2<\/sup>)<br \/>\n= \u221a(0.2<sup>2 <\/sup>+ 1<sup>2<\/sup>)<br \/>\n= \u221a1.04<br \/>\n= 1.02 m\u00a0 \u00a0 \u00a0 (given)<br \/>\nThe slant height of the cone is 1.02 m<br \/>\nThe curved surface area = \u03c0rl <\/span><br \/>\n<span style=\"color: #000000;\">= 3.14 \u00d7 0.2m \u00d7 1.02m<\/span><br \/>\n<span style=\"color: #000000;\">= 0.64056 m<sup>2<br \/>\n<\/sup>Curved surface\u00a0area\u00a0of 50 cones = 50 \u00d7 0.64056 m<sup>2<\/sup>\u00a0= 32.028 m<sup>2<br \/>\n<\/sup>Cost of painting of 50 cones at \u20b9 12 per m<sup>2<\/sup>\u00a0= 32.028 \u00d7 12<\/span><br \/>\n<span style=\"color: #000000;\">= \u20b9 384.34 (approx.)<\/span><br \/>\n<span style=\"color: #000000;\">Thus, the cost of painting all the cones is \u20b9 384.34 (approx.)<\/span><\/p>\n<p style=\"text-align: justify;\">\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 13.3 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-4583","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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