{"id":55457,"date":"2020-11-26T13:50:11","date_gmt":"2020-11-26T13:50:11","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=55457"},"modified":"2023-09-19T02:48:05","modified_gmt":"2023-09-19T02:48:05","slug":"ncert-solutions-for-chapter-13-surface-areas-and-volumes","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-13-surface-areas-and-volumes\/","title":{"rendered":"NCERT Solutions for Maths: Chapter 13 &#8211; Surface Areas and Volumes"},"content":{"rendered":"\n<p>Class 10: Maths Chapter 13 solutions. Complete Class 10 Maths Chapter 13 Notes.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-ncert-solutions-for-maths-chapter-13-surface-areas-and-volumes\"><strong>NCERT Solutions for Maths: Chapter 13 &#8211; Surface Areas and Volumes<\/strong><\/h2>\n\n\n\n<p>NCERT 10th Maths Chapter 13, class 10 Maths Chapter 13 solutions<\/p>\n\n\n\n<p>Page No: 244<\/p>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"h-exercise-13-1\">Exercise 13.1<\/h4>\n\n\n\n<p><strong>Unless stated otherwise, take \u03c0 = 22\/7.<\/strong><\/p>\n\n\n\n<p><strong>1. 2 cubes each of volume 64 cm<sup>3<\/sup> are joined end to end. Find the surface area of the resulting cuboid.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class-10-ncert-maths-ch13-surface-areas-and-volumes-1.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.1 Que. 1.2\"\/><\/figure>\n\n\n\n<p>&nbsp;Volume of each cube(a<sup>3<\/sup>) = 64 cm<sup>3<\/sup><br>\u21d2 a<sup>3<\/sup> = 64 cm<sup>3<\/sup><br>\u21d2 a = 4 cm<br>Side of the cube = 4 cm<br>Length of the resulting cuboid = 4 cm<br>Breadth of the resulting cuboid = 4 cm<br>Height of the resulting cuboid = 8 cm<br>\u2234 Surface area of the cuboid = 2(lb + bh + lh)<br>= 2(8\u00d74 + 4\u00d74 + 4\u00d78)&nbsp;cm<sup>2<\/sup><br>= 2(32 + 16 + 32) cm<sup>2<\/sup><br>= (2 \u00d7 80) cm<sup>2<\/sup> = 160 cm<sup>2<\/sup><\/p>\n\n\n\n<p><strong>2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class-10-ncert-maths-ch13-surface-areas-and-volumes-2.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.1 Que. 2\"\/><\/figure>\n\n\n\n<p>Diameter of the hemisphere = 14 cm<br>Radius of the hemisphere(r) = 7 cm<br>Height of the cylinder(h) = 13 &#8211; 7 = 6 cm<br>Also, radius of the hollow hemisphere = 7 cm<br>Inner surface area of the vessel = CSA of the cylindrical part + CSA of hemispherical part<br>= (2\u03c0rh+2\u03c0r<sup>2<\/sup>) cm<sup>2<\/sup>= 2\u03c0r(h+r) cm<sup>2<\/sup><br>= 2 \u00d7 22\/7 \u00d7 7 (6+7) cm<sup>2<\/sup>= 572 cm<sup>2<\/sup><\/p>\n\n\n\n<p><strong>3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class-10-ncert-maths-ch13-surface-areas-and-volumes-3.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.1 Que. 3\"\/><\/figure>\n\n\n\n<p>Radius of the cone and the hemisphere(r) = 3.5 = 7\/2 cm<br>Total height of the toy = 15.5 cm<br>Height of the cone(h) = 15.5 &#8211; 3.5 = 12 cm<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class-10-maths-surface-area-and-volume-1.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<p>Curved Surface Area of cone = \u03c0rl = 22\/7 \u00d7 7\/2 \u00d7&nbsp;25\/2<br>= 275\/2 cm<sup>2<\/sup><\/p>\n\n\n\n<p>Curved Surface Area of hemisphere = 2\u03c0r<sup>2<\/sup><br>= 2 \u00d7 22\/7 \u00d7 (7\/2)<sup>2<\/sup><br>= 77 cm<sup>2<\/sup><\/p>\n\n\n\n<p>Total surface area of the toy = CSA of cone + CSA of hemisphere<br>= (275\/2&nbsp;+ 77) cm<sup>2<\/sup><br>= (275+154)\/2 cm<sup>2&nbsp;<\/sup><br>= 429\/2 cm<sup>2<\/sup> = 214.5cm<sup>2<\/sup><br>The total surface area of the toy is 214.5cm<sup>2<\/sup><\/p>\n\n\n\n<p>NCERT 10th Maths Chapter 13, class 10 Maths Chapter 13 solutions<\/p>\n\n\n\n<p><strong>4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class-10-maths-surface-area-and-volume-2.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.1 Que. 4\"\/><\/figure>\n\n\n\n<p>Each side of cube = 7 cm<br>\u2234 Radius of the hemisphere = 7\/2 cm<br>Total surface area of solid = Surface area of cubical block + CSA of hemisphere &#8211; Area of base of hemisphere<\/p>\n\n\n\n<p>TSA of solid = 6\u00d7(side)<sup>2&nbsp;<\/sup>+ 2\u03c0r<sup>2&nbsp;<\/sup>-\u03c0r<sup>2<\/sup><br>= 6\u00d7(side)<sup>2&nbsp;<\/sup>+ \u03c0r<sup>2<\/sup><br>= 6\u00d7(7)<sup>2&nbsp;<\/sup>+ (22\/7 \u00d7 7\/2 \u00d7 7\/2)<\/p>\n\n\n\n<p>= (6\u00d749)&nbsp;+ (77\/2)<br>= 294&nbsp;+ 38.5 = 332.5 cm<sup>2<\/sup><\/p>\n\n\n\n<p>The surface area of the solid is 332.5 cm<sup>2<\/sup><\/p>\n\n\n\n<p><strong>5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class-10-maths-surface-area-and-volume-4.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.1 Que. 5\"\/><\/figure>\n\n\n\n<p>Diameter of hemisphere = Edge of cube = l<br>Radius of hemisphere = l\/2<br>Total surface area of solid = Surface area of cube + CSA of hemisphere &#8211; Area of base of hemisphere<br>TSA of remaining solid = 6 (edge)<sup>2&nbsp;<\/sup>+ 2\u03c0r<sup>2&nbsp;<\/sup>&#8211; \u03c0r<sup>2<\/sup><br>= 6l<sup>2&nbsp;<\/sup>+ \u03c0r<sup>2<\/sup><br>= 6l<sup>2&nbsp;<\/sup>+ \u03c0(l\/2)<sup>2<\/sup><br>= 6l<sup>2&nbsp;<\/sup>+ \u03c0l<sup>2<\/sup>\/4<br>= l<sup>2<\/sup>\/4(24&nbsp;+ \u03c0) sq units<\/p>\n\n\n\n<p>NCERT Solutions for Maths Chapter 13<\/p>\n\n\n\n<p><strong>6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class-10-maths-surface-area-and-volume-5.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.1 Que. 6\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class-10-maths-surface-area-and-volume-6.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.1 Que. 6\"\/><\/figure>\n\n\n\n<p>Two hemisphere and one cylinder are given in the figure.&nbsp;<\/p>\n\n\n\n<p>Diameter of the capsule = 5 mm<\/p>\n\n\n\n<p>\u2234&nbsp;Radius = 5\/2 = 2.5 mm<\/p>\n\n\n\n<p>Length of the capsule = 14 mm<\/p>\n\n\n\n<p>\u2234 Length of the cylinder = 14 &#8211; (2.5 +&nbsp;2.5) = 9mm<\/p>\n\n\n\n<p>Surface area of a hemisphere = 2\u03c0r<sup>2<\/sup> = 2 \u00d7 22\/7 \u00d7&nbsp;2.5 \u00d7&nbsp;2.5<\/p>\n\n\n\n<p>= 275\/7 mm<sup>2<\/sup>&nbsp;<\/p>\n\n\n\n<p>Surface area of the cylinder = 2\u03c0rh<br>= 2 \u00d7&nbsp;22\/7 \u00d7&nbsp;2.5 x 9<br>= 22\/7 \u00d7&nbsp;45<br>990\/7 mm<sup>2<\/sup><\/p>\n\n\n\n<p>\u2234 Required surface area of medicine capsule<br>= 2 \u00d7 Surface area of hemisphere + Surface area of cylinder<br>= (2 \u00d7 275\/7) \u00d7 990\/7<br>= 550\/7&nbsp;+ 990\/7<br>= 1540\/7 = 220 mm<sup>2<\/sup><\/p>\n\n\n\n<p>NCERT Solutions for Maths Chapter 13<\/p>\n\n\n\n<p><strong>7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m<sup>2<\/sup>. (Note that the base of the tent will not be covered with canvas.)<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Tent is combination of cylinder and cone.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-area-and-volume-ncert-solutions-1.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.1 Que. 7\"\/><\/figure>\n\n\n\n<p>Diameter = 4 m<br>Slant height of the cone (l) = 2.8 m<br>Radius of the cone (r) = Radius of cylinder = 4\/2 = 2 m<br>Height of the cylinder (h) = 2.1 m<\/p>\n\n\n\n<p>\u2234 Required surface area of tent = Surface area of cone+Surface area of cylinder<br>= \u03c0rl&nbsp;+ 2\u03c0rh<br>= \u03c0r(l+2h)<br>= 22\/7 \u00d7 2 (2.8&nbsp;+ 2\u00d72.1)<br>= 44\/7 (2.8&nbsp;+ 4.2)<br>= 44\/7 \u00d7 7 = 44 m<sup>2<\/sup><\/p>\n\n\n\n<p>Cost of the canvas of the tent at the rate of \u20b9500 per m<sup>2<\/sup><br>= Surface area \u00d7 cost per m<sup>2<\/sup><br>= 44 \u00d7 500 = \u20b922000<\/p>\n\n\n\n<p><strong>8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm<sup>2<\/sup>.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-area-and-volume-ncert-solutions-2.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.1 Que. 8\"\/><\/figure>\n\n\n\n<p>Diameter of cylinder = diameter of conical cavity = 1.4 cm<\/p>\n\n\n\n<p>\u2234 Radius of cylinder = Radius of conical cavity = 1.4\/2 = 0.7<\/p>\n\n\n\n<p>Height of cylinder = Height of conical cavity = 2.4 cm<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-area-and-volume-ncert-solutions-3.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.1 Que. 8\"\/><\/figure>\n\n\n\n<p>TSA of remaining solid = Surface area of conical cavity+TSA of cylinder<\/p>\n\n\n\n<p>= \u03c0rl&nbsp;+ (2\u03c0rh&nbsp;+&nbsp;\u03c0r<sup>2<\/sup>)<\/p>\n\n\n\n<p>= \u03c0r (l&nbsp;+ 2h&nbsp;+ r)<\/p>\n\n\n\n<p>= 22\/7 \u00d7 0.7 (2.5&nbsp;+ 4.8&nbsp;+ 0.7)<\/p>\n\n\n\n<p>= 2.2 \u00d7 8 = 17.6 cm<sup>2<\/sup><\/p>\n\n\n\n<p>NCERT Solutions for Maths Chapter 13<\/p>\n\n\n\n<p><strong>9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-chapter13-mathematics-ncert-solutions-exercise-13-1-1.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.1 Que. 9\"\/><\/figure>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>Height of the cylinder, h = 10cm and radius of base of cylinder = Radius of hemisphere (r) = 3.5 cm<\/p>\n\n\n\n<p>Now, required total surface area of the article = 2 \u00d7 Surface area of hemisphere + Lateral surface area of cylinder<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-chapter13-mathematics-ncert-solutions-exercise-13-1-2.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.1 Que. 9\"\/><\/figure>\n\n\n\n<h2>Exercise 13.2<\/h2>\n\n\n\n<p><strong>1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of \u03c0.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Given, solid is a combination of a cone and a hemisphere.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-chapter13-mathematics-ncert-solutions-exercise-13-1-4.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.2 Que. 1\"\/><\/figure>\n\n\n\n<p>Also, we have radius of the cone (r) = Radius of the hemisphere = 1cm and height of the cone (h) = 1cm<\/p>\n\n\n\n<p>\u2234 Required volume of the solid = Volume of the cone + Volume of the hemisphere<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-chapter13-mathematics-ncert-solutions-exercise-13-1-3.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.2 Que. 1\"\/><\/figure>\n\n\n\n<p><strong>2. Rachel, an engineering, student was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel mode. (Assume the outer and inner dimensions of the model to be nearly the same.)<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Given, model is a combination of a cylinder and two cones. Also, we have, diameter of the model,BC = ED =&nbsp;3 cm.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-chapter13-mathematics-ncert-solutions-exercise-13-1-5.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.2 Que. 2\"\/><\/figure>\n\n\n\n<p><strong>3. A gulab jamun, contains sugar syrup upto about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure).<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-chapter13-mathematics-ncert-solutions-exercise-13-1-6.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.2 Que. 3\"\/><\/figure>\n\n\n\n<p>Let r be the radius of the hemisphere and cylinder both. h1 be the height of the hemisphere which is equal to its radius and h2 be the height of the cylinder.&nbsp;<\/p>\n\n\n\n<p>Given, length = 5 cm, diameter = 2.8 cm<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-chapter13-mathematics-ncert-solutions-exercise-13-1-7.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.2 Que. 3\"\/><\/figure>\n\n\n\n<p>NCERT Solutions for Maths Chapter 13<\/p>\n\n\n\n<p><strong>4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see figure).<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-chapter13-mathematics-ncert-solutions-exercise-13-1-8.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.2 Que. 4\"\/><\/figure>\n\n\n\n<p>Given, length of cuboid (l) = 15 cm<\/p>\n\n\n\n<p>Breadth of cuboid (b) = 10cm<\/p>\n\n\n\n<p>and height of cuboid (h) = 3.5 cm<\/p>\n\n\n\n<p>\u2234 Volume of cuboid = l\u00d7b\u00d7h = 15 \u00d710 \u00d7 3.5 = 525<sup>3<\/sup><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-chapter13-mathematics-ncert-solutions-exercise-13-1-9.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.2 Que. 4\"\/><\/figure>\n\n\n\n<p><strong>5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water upto the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel,one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-chapter13-mathematics-ncert-solutions-exercise-13-1-10.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.2 Que. 5\"\/><\/figure>\n\n\n\n<p><strong>6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm<sup>3<\/sup>&nbsp;of iron has approximately 8 g mass. (Use \u03c0 = 3.14)<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-chapter13-mathematics-ncert-solutions-exercise-13-1-11.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.2 Que. 6\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-chapter13-mathematics-ncert-solutions-exercise-13-1-12.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.2 Que. 6\"\/><\/figure>\n\n\n\n<p>NCERT Solutions for Maths Chapter 13<\/p>\n\n\n\n<p><strong>7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-chapter13-mathematics-ncert-solutions-exercise-13-1-13.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.2 Que. 7\"\/><\/figure>\n\n\n\n<p><strong>8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm<sup>3<\/sup>. Check whether she is correct, taking the above as the inside measurements and \u03c0 = 3.14.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>She is not correct.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-chapter13-mathematics-ncert-solutions-exercise-13-1-14.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.2 Que. 8\"\/><\/figure>\n\n\n\n<h5>Exercise 13.3<\/h5>\n\n\n\n<p><strong>1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Given, the radius of the sphere (r) = 4.2 cm<\/p>\n\n\n\n<p>Radius of the cylinder (r1) = 6 cm<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-15.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.3 Que. 1\"\/><\/figure>\n\n\n\n<p><strong>2. Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let r1, r2 and r3 be the radius of given three spheres and R be the radius of a single solid sphere.<\/p>\n\n\n\n<p>Given, r1 = 6 cm , r2 = 8 cm and r3 =10 cm<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-16.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.3 Que. 2\"\/><\/figure>\n\n\n\n<p><strong>3. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Given, the height of deep well which form a cylinder (h1) = 20 m<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-17.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.3 Que. 3\"\/><\/figure>\n\n\n\n<p><strong>4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4m to form an embankment. Find the height of the embankment.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Given, the height of deep well which form a cylinder (h1) = 14 m<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-19.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.3 Que. 4\"\/><\/figure>\n\n\n\n<p>NCERT 10th Maths Chapter 13, class 10 Maths Chapter 13 solutions<\/p>\n\n\n\n<p><strong>5. A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let the height and radius of ice cream container (cylinder) be h1 and r1.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-20.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.3 Que. 5\"\/><\/figure>\n\n\n\n<p><strong>6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 10 3 5 . . cm cm cm&nbsp; \u00d7 ?<\/strong><\/p>\n\n\n\n<p>Answer<\/p>\n\n\n\n<p>We know that, every coin has a shape of cylinder. Let radius and height of the coin are r1 and h1 respectively.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-20-1.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.3 Que. 6\"\/><\/figure>\n\n\n\n<p>NCERT 10th Maths Chapter 13, class 10 Maths Chapter 13 solutions<\/p>\n\n\n\n<p><strong>7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let the radius and slant height of the heap of sand are r and l.<\/p>\n\n\n\n<p>Given, the height of the heap of sand h = 24 cm.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-22.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.3 Que. 7\"\/><\/figure>\n\n\n\n<p><strong>8. Water in a canal, 6 m wide and 1.5 m deep is flowing with a speed of 10 kmh<sup>\u22121<\/sup>. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Given, speed of flow of water (l) = 10 kmh-1 = 10 \u00d7 1000 mh<sup>\u22121<\/sup><\/p>\n\n\n\n<p>Area of canal = 6 \u00d7 1.5 = 9m<sup>2<\/sup><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-23.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.3 Que. 8\"\/><\/figure>\n\n\n\n<p><strong>9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 kmh<sup>\u22121<\/sup>, in how much time will the tank be filled?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Given, speed of flow of water = 3 kmh<sup>\u22121<\/sup> = 3\u00d71000 mh<sup>\u22121<\/sup><\/p>\n\n\n\n<p>\u2234 Length of water in 1 h = 3000 m<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-24.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.3 Que. 9\"\/><\/figure>\n\n\n\n<h5>Exercise 13.4<\/h5>\n\n\n\n<p>NCERT 10th Maths Chapter 13, class 10 Maths Chapter 13 solutions<\/p>\n\n\n\n<p><strong>1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let the height of the frustum of a cone be h.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-25.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.4 Que. 1\"\/><\/figure>\n\n\n\n<p><strong>2. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let the slant height of the frustum be l and radius of the both ends of the frustum be r1 and r2.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-26.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.4 Que. 2\"\/><\/figure>\n\n\n\n<p><strong>3. A fez, the cap used by the turks, is shaped like the frustum of a cone (see figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-27.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.4 Que. 3\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let the slant height of fez be l and the radius of upper end which is closed be r1 and the other end which is open be r2.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-28.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.4 Que. 4\"\/><\/figure>\n\n\n\n<p>NCERT 10th Maths Chapter 13, class 10 Maths Chapter 13 solutions<\/p>\n\n\n\n<p><strong>4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ` 20 per litre. Also find the cost of metal sheet used to make the container. If it costs \u20b98 per 100 cm<sup>2<\/sup>. (Take \u03c0 = 3.14)<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let h be the height of the container, which is in the form of a frustum of a cone whose lower end is closed and upper end is opened. Also, let the radius of its lower end be r1 and upper end be r2.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-29.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.4 Que. 4\"\/><\/figure>\n\n\n\n<p><strong>5. A metallic right circular cone 20 cm high and whose vertical angle is 60\u00ba is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1\/16 cm, find the length of the wire.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let r1 and r2 be the radii of the frustum of upper and lower ends cut by a plane. Given, height of the cone = 20 cm.<\/p>\n\n\n\n<p>\u2234 Height of the frustum = 10 cm<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-30.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.4 Que. 5\"\/><\/figure>\n\n\n\n<h5>Exercise 13.5 (Optional)<\/h5>\n\n\n\n<p>NCERT 10th Maths Chapter 13, class 10 Maths Chapter 13 solutions<\/p>\n\n\n\n<p><strong>1. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm<sup>3<\/sup>.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Since, the diameter of the wire is 3 mm.<\/p>\n\n\n\n<p>When a wire is one round wound about a cylinder, it covers a 3 mm of length of the cylinder.<\/p>\n\n\n\n<p>Given, length of the cylinder = 12 cm = 120 mm<\/p>\n\n\n\n<p>\u2234 Number of rounds to cover 120 mm = 120\/3 = 40<\/p>\n\n\n\n<p>Given, diameter of a cylinder is d = 10 cm<\/p>\n\n\n\n<p>\u2234 Radius, r = 10\/2 = 5 cm<\/p>\n\n\n\n<p>\u2234 Length of wire required to complete one round = 2\u03c0r = 2\u03c0(5) = 10\u03c0 cm<\/p>\n\n\n\n<p>\u2234 Length of the wire in covering the whole surface<\/p>\n\n\n\n<p>= Length of the wire in completing 40 rounds<\/p>\n\n\n\n<p>= 10\u03c0 \u00d7 40 = 400\u03c0 cm<\/p>\n\n\n\n<p>= 400 \u00d7 3.14 = 1256 cm<\/p>\n\n\n\n<p>Now, radius of copper wire = 3\/2 mm = 3\/20 cm<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-31.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.5 Que. 1\"\/><\/figure>\n\n\n\n<p><strong>2. A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of \u03c0 as found appropriate).<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Here, ABC is a right angled triangle at A and BC is the hypotenuse.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-32.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.5 Que. 2\"\/><\/figure>\n\n\n\n<p><strong>3. A cistern, internally measuring 150 cm \u00d7 120 cm\u00d7 110 cm, has 129600 cm<sup>3<\/sup>&nbsp;of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. how many bricks can be put in without overflowing the water, each brick being 22.5 cm \u00d7 7.5 cm \u00d7 6.5 cm?<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Given, internally dimensions of cistern = 150 cm\u00d7120 cm\u00d7110 cm<\/p>\n\n\n\n<p>\u2234 Volume of cistern = 150\u00d7120\u00d7110&nbsp;<\/p>\n\n\n\n<p>= 1980000 cm<sup>3<\/sup><\/p>\n\n\n\n<p>Volume of water = 129600 cm<sup>3<\/sup><\/p>\n\n\n\n<p>\u2234 Volume of cistern to be filled = 1980000 &#8211; 129600 = 1850400 cm<sup>3<\/sup><\/p>\n\n\n\n<p>Let required number of bricks = n<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-33.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.5 Que. 3\"\/><\/figure>\n\n\n\n<p>NCERT 10th Maths Chapter 13, class 10 Maths Chapter 13 solutions<\/p>\n\n\n\n<p><strong>4. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km<sup>2<\/sup>, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Given, area of the valley = 97280 km<sup>2<\/sup><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-34.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.5 Que. 4\"\/><\/figure>\n\n\n\n<p><strong>5. An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-35.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.5 Que. 5\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Given, oil funnel is a combination of a cylinder and a frustum of a cone.<\/p>\n\n\n\n<p>Also, given height of cylindrical portion h = 10 cm<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-36.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.5 Que. 5\"\/><\/figure>\n\n\n\n<p>NCERT 10th Maths Chapter 13, class 10 Maths Chapter 13 solutions<\/p>\n\n\n\n<p><strong>6. Derive the formula for the curved surface area and total surface area of the frustum of a cone. Using the symbols as explained.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Leth be the height, l be the slant height and r<sub>1<\/sub> and r<sub>2<\/sub>&nbsp;be the radii of the bases (r<sub>1<\/sub>&gt;r<sub>2<\/sub>) of the frustum of a cone. We complete the conical part OCD.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-37.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.5 Que. 6\"\/><\/figure>\n\n\n\n<p>The frustum of the right circular cone can be viewed as the difference of the two right circular cones OAB and OCD. Let slant height of the cone OAB be l<sub>1<\/sub>&nbsp;and its height be h<sub>1<\/sub>&nbsp;i.e.,OB = OA = l<sub>1<\/sub>&nbsp;and OP = h<sub>1<\/sub><\/p>\n\n\n\n<p>The frustum of the right circular cone can be viewed as the difference of the two right circular cones OAB and OCD. Let slant height of the cone OAB be l<sub>1<\/sub>&nbsp;and its height be h<sub>1<\/sub>&nbsp;i.e., OB = OA = l<sub>1<\/sub>&nbsp;and OP = h<sub>1<\/sub><\/p>\n\n\n\n<p>Then in \u2206ACE,<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-38.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.5 Que. 6\"\/><\/figure>\n\n\n\n<p><strong>7. Derive the formula for the volume of the frustum of a cone given to you in the section 13.5 using the symbols as explained.<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Leth be the height, l be the slant height and r<sub>1<\/sub>&nbsp;and r<sub>2<\/sub>&nbsp;be the radii of the bases (r<sub>1<\/sub>&gt;r<sub>2<\/sub>) of the frustum of a cone. We complete the conical part OCD.<\/p>\n\n\n\n<p>The frustum of the right circular cone can be viewed as the difference of the two right circular cones OAB and OCD. Let the height of the cone OAB be h<sub>1<\/sub>&nbsp;and its slant height be l<sub>1<\/sub>.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-37-1.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.5 Que. 7\"\/><\/figure>\n\n\n\n<p>i.e., OP = h<sub>1<\/sub>&nbsp;and OA = OB = l<sub>1<\/sub><\/p>\n\n\n\n<p>Then, height of the cone OCD = h<sub>1<\/sub>&nbsp;\u22121<\/p>\n\n\n\n<p>\u2206OQD ~ \u2206OPB (AA similarity criterion)<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/class10-ch13-surface-areas-and-volume-ncert-solutions-39.jpg\" alt=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes Ex. 13.5 Que. 7\"\/><\/figure>\n\n\n\n<p>NCERT 10th Maths Chapter 13, class 10 Maths Chapter 13 solutions<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-ncert-solutions-for-maths-chapter-13-nbsp-download-pdf\">NCERT Solutions for Maths: Chapter 13:&nbsp;<strong>Download PDF<\/strong><\/h2>\n\n\n\n<p>NCERT Solutions for Maths: Chapter 13 &#8211; Surface Areas and Volumes<\/p>\n\n\n\n<p><a href=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/08\/NCERT-Solutions-for-Maths_-Chapter-13-Surface-Areas-and-Volumes.pdf\"><strong>Download PDF<\/strong>: NCERT Solutions for Maths: Chapter 13 &#8211; Surface Areas and Volumes PDF<\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-chapterwise-ncert-solutions-for-class-10-maths\"><strong>Chapterwise NCERT Solutions for Class 10 Maths<\/strong>:<\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-10th-mathematics-chapter-1-real-numbers\/\">Chapter 1 Real Numbers<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-10th-mathematics-chapter-2-polynomials\/\">Chapter 2 Polynomials<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables\/\">Chapter 3 Pair of Linear Equations in Two Variables<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-10th-mathematics-chapter-4-quadratic-equations\/\">Chapter 4 Quadratic Equations<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-10th-mathematics-chapter-5-arithmetic-progressions\/\">Chapter 5 Arithmetic Progressions<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-10th-mathematics-chapter-6-triangles\/\">Chapter 6 Triangles<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-10th-mathematics-chapter-7-coordinate-geometry\/\">Chapter 7 Coordinate Geometry<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-8-introduction-to-trigonometry\/\">Chapter 8 Introduction to Trigonometry<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-9-some-applications-of-trigonometry\/\">Chapter 9 Applications of Trigonometry<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-10-circles\/\">Chapter 10 Circle<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-11-constructions\/\">Chapter 11 Constructions<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-12-areas-related-to-circles\/\">Chapter 12 Areas related to Circles<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-13-surface-areas-and-volumes\/\">Chapter 13 Surface Areas and Volumes<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-14-statistics\/\">Chapter 14 Statistics<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-15-probability\/\">Chapter 15 Probability<\/a><\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-about-ncert\">About NCERT<\/h2>\n\n\n\n<p>The National Council of Educational Research and Training is an autonomous organization of the Government of India which was established in 1961 as a literary, scientific, and charitable Society under the Societies Registration Act. Its headquarters are located at Sri Aurbindo Marg in New Delhi. <a href=\"https:\/\/ncert.nic.in\/\" target=\"_blank\" rel=\"noreferrer noopener\">Visit the Official NCERT website<\/a> to learn more. <\/p>\n\n\n\n<div class=\"wp-block-buttons is-layout-flex wp-block-buttons-is-layout-flex\">\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link has-primary-background-color has-background wp-element-button\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions\/\">NCERT Solutions<\/a><\/div>\n\n\n\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link has-primary-background-color has-background wp-element-button\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-class-10\/\">NCERT Solutions for Class 10<\/a><\/div>\n\n\n\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link has-primary-background-color has-background wp-element-button\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-10th-mathematics\/\">NCERT Solutions for Class 10 Mathematics<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Class 10: Maths Chapter 13 solutions. Complete Class 10 Maths Chapter 13 Notes. NCERT Solutions for Maths: Chapter 13 &#8211; Surface Areas and Volumes NCERT 10th Maths Chapter 13, class 10 Maths Chapter 13 solutions Page No: 244 Exercise 13.1 Unless stated otherwise, take \u03c0 = 22\/7. 1. 2 cubes each of volume 64 cm3 [&hellip;]<\/p>\n","protected":false},"author":294,"featured_media":628034,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,24],"tags":[1443],"boards":[1180],"class_list":["post-55457","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-10","tag-ncert-maths-class-10","boards-ncert","entry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.0 (Yoast SEO v27.1.1) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions for Class 10, Maths Chapter 13 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes | Browse all Class 10 Maths Chapters NCERT books - IndCareer Schools\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-13-surface-areas-and-volumes\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes\" \/>\n<meta property=\"og:description\" content=\"Class 10: Maths Chapter 13 solutions. 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NCERT Solutions for Maths: Chapter 13 - Surface Areas and Volumes NCERT\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-13-surface-areas-and-volumes\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" content=\"2020-11-26T13:50:11+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-09-19T02:48:05+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2020\/11\/NCERT-Solutions-43-scaled.jpg\" \/>\n\t<meta property=\"og:image:width\" content=\"1600\" \/>\n\t<meta property=\"og:image:height\" content=\"900\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\n<meta name=\"author\" content=\"Mukesh Kaple\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@indcareer\" \/>\n<meta name=\"twitter:site\" content=\"@indcareer\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Mukesh Kaple\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"28 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-13-surface-areas-and-volumes\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-13-surface-areas-and-volumes\/\"},\"author\":{\"name\":\"Mukesh Kaple\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/c9fbb187e0abc227936e80a406bdbb2b\"},\"headline\":\"NCERT Solutions for Maths: Chapter 13 &#8211; 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