{"id":4587,"date":"2023-02-21T09:35:51","date_gmt":"2023-02-21T09:35:51","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4587"},"modified":"2023-02-21T09:38:29","modified_gmt":"2023-02-21T09:38:29","slug":"ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-7","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-13-surface-areas-and-volumes-ex-13-7\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7"},"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 13.7 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-13-8\/\" 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; 13.7<\/span><\/h2>\n<p style=\"text-align: center;\"><span style=\"color: #000000;\"><strong>(Assume \u03c0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{22}{7}}\" alt=\"\\mathbf{\\frac{22}{7}}\" align=\"absmiddle\" \/>, <span class=\"fontstyle0\">unless stated otherwise<\/span>)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Find the volume of the right circular cone with<br \/>\n<\/strong><strong>(i) radius 6 cm, height 7 cm<br \/>\n(ii) radius 3.5 cm, height 12 cm\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong><strong>(i)<\/strong> Radius of the cone, <em>r<\/em> = 6 cm<br \/>\nHeight of cone, <em>h<\/em> = 7 cm<br \/>\nVolume of the\u00a0cone = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/>\u03c0r\u00b2h<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 36 \u00d7 7<br \/>\n= 12 \u00d7 22<br \/>\n= 264<br \/>\nThe volume of the cone is 264 cm<sup>3<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)<\/strong> Radius of the cone, <em>r<\/em> = 3.5 cm<br \/>\nHeight of the cone, <em>h<\/em> = 12 cm<br \/>\nVolume of the\u00a0cone = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/>\u03c0r\u00b2h<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 3.5<sup>2 <\/sup>\u00d7 7<br \/>\n= 154<br \/>\nThe volume of the cone is 154 cm<sup>3<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Find the capacity in litres of a conical vessel with<br \/>\n<\/strong><strong>(i) radius 7cm, slant height 25 cm<br \/>\n(ii) height 12 cm, slant height 13 cm<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong><strong>(i)<\/strong> Radius of the cone, <em>r<\/em> =7 cm<br \/>\nSlant height of the cone, <em>l<\/em> = 25 cm<br \/>\nHeight of the\u00a0conical vessel, h = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{l^2&amp;space;-&amp;space;r^2}\" alt=\"\\sqrt{l^2 - r^2}\" align=\"absmiddle\" \/> <\/span><br \/>\n<span style=\"color: #000000;\">= \u221a(25)\u00b2 &#8211; (7)\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a625\u00a0&#8211; 49<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a576<\/span><br \/>\n<span style=\"color: #000000;\">h = 24 cm<\/span><br \/>\n<span style=\"color: #000000;\">Capacity of the conical vessel = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/>\u03c0r\u00b2h <\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 7 cm \u00d7 7 cm \u00d7 24 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 1232 cm\u00b3<\/span><br \/>\n<span style=\"color: #000000;\">= 1232 \u00d7 (1\/1000L)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[\u2235 1000 cm\u00b3 = 1 litre]<\/span><br \/>\n<span style=\"color: #000000;\">= 1.232 litres<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(ii)<\/strong> Height of the conical vessel, h = 7 cm<\/span><br \/>\n<span style=\"color: #000000;\">Slant height of the conical vessel, l = 13 cm<\/span><\/p>\n<p><span style=\"color: #000000;\">Radius of the conical vessel, r = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\sqrt{l^2&amp;space;-&amp;space;h^2}\" alt=\"\\sqrt{l^2 - h^2}\" align=\"absmiddle\" \/> <\/span><br \/>\n<span style=\"color: #000000;\">= \u221a(13)\u00b2 &#8211; (12)\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a169 -144<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a25<\/span><br \/>\n<span style=\"color: #000000;\">r = 5 cm<\/span><br \/>\n<span style=\"color: #000000;\">Capacity of the conical vessel = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/>\u03c0r\u00b2h <\/span><br \/>\n<span style=\"color: #000000;\">= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/>\u00d7 5 cm \u00d7 5 cm \u00d7 12 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 2200\/7 cm\u00b3 <\/span><br \/>\n<span style=\"color: #000000;\">= 2200\/7 \u00d7 1\/1000 <em>l<\/em>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [\u2235 1000 cm\u00b3 = 1 litre] <\/span><br \/>\n<span style=\"color: #000000;\">= 11\/35 litres<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. The height of a cone is 15cm. If its volume is 1570cm<sup>3<\/sup>, find the diameter of its base. (Use \u03c0 = 3.14)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Let the radius of the cone = <em>r<\/em><br \/>\nHeight of the cone, <em>h<\/em> = 15 cm<br \/>\nVolume of the cone = 1570 cm<sup>3<br \/>\n<\/sup><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/>\u03c0r\u00b2h = 1570<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/> \u00d7 3.14 \u00d7 r<sup>2\u00a0<\/sup>\u00d7 15 = 1570<br \/>\nr<sup>2<\/sup> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1570\\times&amp;space;3}{3.14\\times&amp;space;15}\" alt=\"\\frac{1570\\times 3}{3.14\\times 15}\" align=\"absmiddle\" \/><br \/>\nr<sup>2<\/sup> = 100<br \/>\nr = 10<br \/>\nRadius of the base of the cone is 10 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4.<\/strong>\u00a0<strong>If the volume of a right circular cone of height 9cm is 48\u03c0cm<sup>3<\/sup>, find the diameter of its base.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Let the radius of the cone = <em>r<\/em><br \/>\nHeight of cone, <em>h<\/em> = 9 cm<br \/>\nVolume of cone = 48\u03c0 cm<sup>3<br \/>\n<\/sup><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/>\u03c0r\u00b2h = 48\u03c0 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/>\u03c0r<sup>2 <\/sup>\u00d7 9 = 48 \u03c0<br \/>\nr<sup>2<\/sup>\u00a0= 16<br \/>\nr = 4<br \/>\nRadius of the cone is 4 cm.<br \/>\nSo, diameter = 2\u00d7Radius = 8<br \/>\nThus, diameter of the base is 8 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. A conical pit of top diameter 3.5m is 12m deep. What is its capacity in kiloliters?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Diameter of the conical pit = 3.5 m<br \/>\nRadius of the conical pit, r = diameter\/ 2 = (3.5\/2)m = 1.75m<br \/>\nHeight of the pit, h = Depth of the pit = 12m<br \/>\nVolume of the cone, V = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/>\u03c0r<sup>2<\/sup>h<br \/>\nV = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/> \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>\u00d7 12<br \/>\n= 38.5<br \/>\nVolume of the cone is 38.5 m<sup>3<br \/>\n<\/sup>Hence, capacity of the pit = (38.5 \u00d7 1) kiloliters = 38.5 kiloliters.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. The volume of a right circular cone is 9856cm<sup>3<\/sup>. If the diameter of the base is 28cm, find<br \/>\n<\/strong><strong>(i) height of the cone<br \/>\n<\/strong><strong>(ii) slant height of the cone<br \/>\n<\/strong><strong>(iii) curved surface area of the cone<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Diameter of the cone, d\u00a0= 28 cm<\/span><br \/>\n<span style=\"color: #000000;\">Radius of the cone, r = 28\/2 cm = 14 cm<\/span><br \/>\n<span style=\"color: #000000;\">Volume of a right circular cone = 9856 cm<sup>3<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) <\/strong>Let the height of the cone be h<br \/>\nVolume of the cone, V = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/>\u03c0r<sup>2<\/sup>h<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/>\u03c0r<sup>2<\/sup>h = 9856<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 14 \u00d7 14 \u00d7 h = 9856<br \/>\nh = 48<br \/>\nThe height of the cone is 48 cm.<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(ii)<\/strong> Slant height of the cone = <em>l<br \/>\n<\/em>l\u00a0= \u221ar\u00b2 + h\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a(14)\u00b2 + (48)\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a(196 + 2304)<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a(2500)<\/span><br \/>\n<span style=\"color: #000000;\">= 50 cm<\/span><br \/>\n<span style=\"color: #000000;\">Slant height of the cone is 50 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii)<\/strong> curved surface area of the cone = \u03c0rl<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 14 \u00d7 50<br \/>\n= 2200<br \/>\nThe curved surface area of the cone is 2200 cm<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. A right triangle ABC with sides 5cm, 12cm and 13cm is revolved about the side 12 cm. Find the volume of the solid so obtained.<br \/>\n<\/strong><strong>Answer &#8211;<br \/>\n<\/strong>Height (h)= 12 cm<br \/>\nRadius (r) = 5 cm, and<br \/>\nSlant height (l) = 13 cm<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4668\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.7-Ans7.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"201\" height=\"212\" \/><br \/>\nVolume of cone, V = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/>\u03c0r<sup>2<\/sup>h<br \/>\nV = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/> \u00d7 \u03c0 \u00d7 5<sup>2 <\/sup>\u00d7 12<br \/>\n= 100\u03c0<br \/>\nThe volume of the cone so formed is 100\u03c0 cm<sup>3<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. If the triangle ABC in the Question 7 is revolved about the side 5cm, then find the volume of the solids so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>Radius of the cone, r =12 cm<\/span><br \/>\n<span style=\"color: #000000;\">Height of the cone, h = 5 cm<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4669\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-13.7-Ans8.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"254\" height=\"261\" \/><\/span><\/p>\n<p><span style=\"color: #000000;\">Volume of the cone = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/>\u03c0r<sup>2<\/sup>h<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/> \u00d7 \u03c0 \u00d7 12cm \u00d7 12cm \u00d7 5cm<\/span><br \/>\n<span style=\"color: #000000;\">= 240\u03c0 cm\u00b3 <\/span><br \/>\n<span style=\"color: #000000;\">Volume of the cone in question 7 = 100\u03c0 cm <\/span><br \/>\n<span style=\"color: #000000;\">Ratio = Volume of the cone in question 7 : Volume of the cone in question 8 <\/span><br \/>\n<span style=\"color: #000000;\">= 100\u03c0 : 240\u03c0 \u00a0<\/span><br \/>\n<span style=\"color: #000000;\">= 5 : 12<\/span><br \/>\n<span style=\"color: #000000;\">The volume of the cone is 240\u03c0 cm\u00b3 and the required ratio is\u00a0 5 : 12.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong>Diameter of the conical heap, d = 10.5 m<\/span><br \/>\n<span style=\"color: #000000;\">Radius of the conical heap, r = 10.5\/2 m = 5.25 m<\/span><br \/>\n<span style=\"color: #000000;\">Height of the conical heap, h = 3 m<\/span><br \/>\n<span style=\"color: #000000;\">Volume of heap = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/>\u03c0r<sup>2<\/sup>h<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/> \u00d7 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{22}{7}\" alt=\"\\frac{22}{7}\" align=\"absmiddle\" \/> \u00d7 5.25 \u00d7 5.25 \u00d7 3<br \/>\n= 86.625<br \/>\nThe volume of the heap of wheat is 86.625 m<sup>3<\/sup>.<br \/>\nSlant height, l = \u221ar\u00b2 + h\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">= \u221a(5.25)\u00b2 + (3)\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">= \u221a(27.5625 + 9)<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a(36.5625)<\/span><br \/>\n<span style=\"color: #000000;\">= 6.046\u00a0m (approx.)<\/span><br \/>\n<span style=\"color: #000000;\">The area of the canvas required to cover the heap of wheat = \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 5.25 m \u00d7 6.046 m<\/span><br \/>\n<span style=\"color: #000000;\">= 99.759 m\u00b2 <\/span><br \/>\n<span style=\"color: #000000;\">The volume of the conical heap is 86.625 m\u00b3 and the area of the canvas required is 99.759\u00a0m\u00b2.<\/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 13.7 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-4587","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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