{"id":6094,"date":"2023-09-26T14:44:49","date_gmt":"2023-09-26T14:44:49","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6094"},"modified":"2023-10-03T04:44:46","modified_gmt":"2023-10-03T04:44:46","slug":"ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-1\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 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 10 Mathematics <strong>Chapter &#8211; 13 Surface Areas and Volumes <\/strong>Exercise 13.1 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><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-2\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 13.2<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-3\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 13.3<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-4\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 13.4<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-5\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 13.5<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 13.1<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Unless stated otherwise, take \u03c0 = 22\/7<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><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.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>The diagram is given as:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6666 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-1-195x300.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"140\" height=\"215\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-1-195x300.png 195w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-1-666x1024.png 666w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-1-768x1181.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-1-300x461.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-1.png 815w\" sizes=\"auto, (max-width: 140px) 100vw, 140px\" \/><br \/>\n<strong>Given-<br \/>\n<\/strong>The Volume (V) of each cube is = 64 cm<sup>3<br \/>\n<\/sup>This implies that a<sup>3<\/sup>\u00a0= 64 cm<sup>3<br \/>\n<\/sup>\u2234 a = 4 cm<br \/>\nNow, the side of the cube = a = 4 cm<br \/>\nAlso, the length and breadth of the resulting cuboid will be 4 cm each, while its height will be 8 cm.<br \/>\nSo, the surface area of the cuboid = 2(lb+bh+lh)<br \/>\n= 2(8 \u00d7 4 + 4 \u00d7 4 + 4 \u00d7 8) cm<sup>2<br \/>\n<\/sup>= 2(32 + 16 + 32) cm<sup>2<br \/>\n<\/sup>= (2 \u00d7 80) cm<sup>2<\/sup><br \/>\n= 160 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><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.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>The diagram is as follows:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6667 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-2-300x262.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"177\" height=\"155\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-2-300x262.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-2.png 742w\" sizes=\"auto, (max-width: 177px) 100vw, 177px\" \/><br \/>\nNow, the given parameters are:<br \/>\nThe diameter of the hemisphere = D = 14 cm<br \/>\nThe radius of the hemisphere = r = 7 cm<br \/>\nAlso, the height of the cylinder = h = (13 &#8211; 7) = 6 cm<br \/>\nAnd the radius of the hollow hemisphere = 7 cm<br \/>\nNow, the inner surface area of the vessel = CSA of the cylindrical part + CSA of the hemispherical part<br \/>\n(2\u03c0rh + 2\u03c0r<sup>2<\/sup>) cm<sup>2<\/sup><br \/>\n= 2\u03c0r(h + r) cm<sup>2<br \/>\n<\/sup>= 2 \u00d7 (22\/7) \u00d7 7(6 + 7) cm<sup>2<\/sup><br \/>\n= 572 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>The diagram is as follows:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6668 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-3-285x300.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"208\" height=\"219\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-3-285x300.png 285w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-3-300x316.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-3.png 722w\" sizes=\"auto, (max-width: 208px) 100vw, 208px\" \/><br \/>\nGiven that the radius of the cone and the hemisphere (r) = 3.5 cm = 7\/2 cm<br \/>\nThe total height of the toy is given as 15.5 cm.<br \/>\nSo, the height of the cone (h) = 15.5-3.5 = 12 cm<br \/>\nSlant height of the cone, l = \u221a(r<sup>2<\/sup>\u00a0+ h<sup>2<\/sup>) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">l = \u221a[(3.5 cm)<sup>2<\/sup>\u00a0+ (12 cm)<sup>2<\/sup>] <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">l = \u221a[12.25 cm<sup>2<\/sup>\u00a0+ 144 cm<sup>2<\/sup>] <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">l = \u221a\u00a056.25 cm<sup>2<br \/>\n<\/sup>l = 12.5 cm\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">l = 25\/2 cm <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 The curved surface area of the cone = \u03c0rl<br \/>\n= (22\/7) \u00d7 (7\/2) \u00d7 (25\/2)<br \/>\n= 275\/2 cm<sup>2<br \/>\n<\/sup>Also, the curved surface area of the hemisphere = 2\u03c0r<sup>2<br \/>\n<\/sup>= 2 \u00d7 (22\/7) \u00d7 (7\/2)<sup>2<br \/>\n<\/sup>= 77 cm<sup>2<br \/>\n<\/sup>Now, the Total surface area of the toy = CSA of the cone + CSA of the hemisphere<br \/>\n= (275\/2) + 77 cm<sup>2<br \/>\n<\/sup>= (275 + 154)\/2 cm<sup>2<br \/>\n<\/sup>= 429\/2 cm<sup>2<\/sup>\u00a0= 214.5cm<sup>2<br \/>\n<\/sup>So, the total surface area (TSA) of the toy is 214.5cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><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.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>It is given that each side of the cube is 7 cm. So, the radius will be 7\/2 cm.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6669 size-medium\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-4-300x170.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"170\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-4-300x170.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-4-1024x579.png 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-4-768x434.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-4-850x481.png 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-4.png 1146w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\nThe total surface area of solid (TSA) = surface area of the cubical block + CSA of the hemisphere \u2013 Area of the base of the hemisphere<br \/>\n\u2234 TSA of solid = 6 \u00d7 (side)<sup>2 <\/sup>+ 2\u03c0r<sup>2 <\/sup>&#8211; \u03c0r<sup>2<br \/>\n<\/sup>= 6 \u00d7 (side)<sup>2 <\/sup>+ \u03c0r<sup>2<br \/>\n<\/sup>= 6 \u00d7 (7)<sup>2 <\/sup>+ (22\/7) \u00d7 (7\/2) \u00d7 (7\/2)<br \/>\n= (6 \u00d7 49) + (77\/2)<br \/>\n= 294 + 38.5<br \/>\n= 332.5 cm<sup>2<br \/>\n<\/sup>So, the surface area of the solid is 332.5 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><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.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>The diagram is as follows:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6670 size-medium\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-5-300x150.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"150\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-5-300x150.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-5-1024x511.png 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-5-768x383.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-5-850x424.png 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-5.png 1146w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\nNow, the diameter of the hemisphere = Edge of the cube = l<br \/>\nSo, the radius of the hemisphere = l\/2<br \/>\n\u2234 The total surface area of solid = surface area of cube + CSA of the hemisphere \u2013 Area of the base of the hemisphere<br \/>\nThe surface area of the\u00a0remaining solid = 6 (edge)<sup>2 <\/sup>+ 2\u03c0r<sup>2 <\/sup>&#8211; \u03c0r<sup>2<br \/>\n<\/sup>= 6l<sup>2<\/sup>\u00a0+ \u03c0r<sup>2<br \/>\n<\/sup>= 6l<sup>2 <\/sup>+ \u03c0(l\/2)<sup>2<br \/>\n<\/sup>= 6l<sup>2 <\/sup>+ \u03c0l<sup>2<\/sup>\/4<br \/>\n= l<sup>2<\/sup>\/4(24 + \u03c0) sq. units <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><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.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6671 size-medium\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Que-6-300x134.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"134\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Que-6-300x134.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Que-6.png 362w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>Two hemispheres and one cylinder are shown in the figure given below.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6672\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-6.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"141\" \/><br \/>\nHere, the diameter of the capsule = 5 mm<br \/>\n\u2234 Radius = 5\/2 = 2.5 mm<br \/>\nNow, the length of the capsule = 14 mm<br \/>\nSo, the length of the cylinder = 14 &#8211; (2.5 + 2.5) = 9 mm<br \/>\n\u2234 The surface area of a hemisphere = 2\u03c0r<sup>2<\/sup><br \/>\n= 2 \u00d7 (22\/7) \u00d7 2.5 \u00d7 2.5<br \/>\n= 275\/7 mm<sup>2<br \/>\n<\/sup>Now, the surface area of the cylinder = 2\u03c0rh<br \/>\n= 2 \u00d7 (22\/7) \u00d7 2.5 \u00d7 9<br \/>\n= (22\/7) \u00d7 45<br \/>\n= 990\/7 mm<sup>2<br \/>\n<\/sup>Thus, the required surface area of the medicine capsule will be = 2 \u00d7 surface area of hemisphere + surface area of the cylinder<br \/>\n= (2 \u00d7 275\/7) \u00d7 990\/7<br \/>\n= (550\/7) + (990\/7)<br \/>\n= 1540\/7<br \/>\n= 220 mm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><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.)<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>It is known that a tent is a combination of a cylinder and a cone.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6673 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-7-287x300.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"218\" height=\"228\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-7-287x300.png 287w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-7-300x314.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-7.png 748w\" sizes=\"auto, (max-width: 218px) 100vw, 218px\" \/><br \/>\nDiameter = 4 m<br \/>\nThe slant height of the cone (l) = 2.8 m<br \/>\nRadius of the cone (r) = Radius of cylinder = 4\/2 = 2 m<br \/>\nHeight of the cylinder (h) = 2.1 m<br \/>\nSo, the required surface area of the tent = surface area of the cone + surface area of the cylinder<br \/>\n= \u03c0rl + 2\u03c0rh<br \/>\n= \u03c0r(l + 2h)<br \/>\n= (22\/7) \u00d7 2(2.8 + 2 \u00d7 2.1)<br \/>\n= (44\/7)(2.8 + 4.2)<br \/>\n= (44\/7) \u00d7 7<br \/>\n= 44 m<sup>2<br \/>\n<\/sup>\u2234 The cost of the canvas of the tent at the rate of \u20b9500 per m<sup>2\u00a0<\/sup>will be<br \/>\n= Surface area \u00d7 cost per m<sup>2<br \/>\n<\/sup>= 44 \u00d7 500<br \/>\n= \u20b922000<br \/>\nSo, Rs. 22000 will be the total cost of the canvas.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>8. From a solid cylinder whose height is 2.4 cm and diameter is 1.4 cm, a conical cavity of the <\/strong><strong>same height and same diameter is hollowed out. Find the total surface area of the <\/strong><strong>remaining solid to the nearest cm<sup>2<\/sup>.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>The diagram for the question is as follows:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6674 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-8-218x300.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"148\" height=\"204\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-8-218x300.png 218w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-8-743x1024.png 743w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-8-768x1058.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-8-300x413.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-8-850x1171.png 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Ans-8.png 967w\" sizes=\"auto, (max-width: 148px) 100vw, 148px\" \/><br \/>\nFrom the question, we know the following:<br \/>\nThe diameter of the cylinder = diameter of conical cavity = 1.4 cm<br \/>\nSo, the radius of the cylinder = radius of the conical cavity = 1.4\/2 = 0.7<br \/>\nAlso, the height of the cylinder = height of the conical cavity = 2.4 cm<br \/>\nSlant height of the\u00a0cone, l = \u221a[r<sup>2<\/sup>\u00a0+ h<sup>2<\/sup>]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">l = \u221a[(0.7 cm)<sup>2<\/sup>\u00a0+ (2.4 cm)<sup>2<\/sup>]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u221a[0.49 cm<sup>2<\/sup>\u00a0+ 5.76 cm<sup>2<\/sup>]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= \u221a[6.25 cm<sup>2<\/sup>]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 2.5 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Now, the TSA of the remaining solid = surface area of conical cavity + TSA of the cylinder<br \/>\n= \u03c0rl + (2\u03c0rh + \u03c0r<sup>2<\/sup>)<br \/>\n= \u03c0r(l + 2h + r)<br \/>\n= (22\/7) \u00d7 0.7(2.5 + 4.8 + 0.7)<br \/>\n= 2.2 \u00d7 8<br \/>\n= 17.6 cm<sup>2<br \/>\n<\/sup>So, the total surface area of the remaining solid is 17.6 cm<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 13.11. 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.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6677\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/09\/NCERT-Class-10-Maths-Ex13.1-Que-9.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"164\" height=\"230\" \/><br \/>\n<span style=\"font-family: Georgia, Palatino;\">Solution &#8211;<\/span><\/strong><span style=\"font-family: Georgia, Palatino;\"> \u00a0<\/span>Radius (<i>r<\/i>) of cylindrical part = Radius (<i>r<\/i>) of hemispherical part = 3.5 cm <\/span><br \/>\n<span style=\"color: #000000;\">Height of cylindrical part (<i>h<\/i>) = 10 cm\u00a0 <\/span><br \/>\n<span style=\"color: #000000;\">TSA of the article = 2 \u00d7 CSA of the hemispherical part + CSA of the cylindrical part <\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 2\u03c0r<sup>2<\/sup>\u00a0+ 2\u03c0rh<\/span><br \/>\n<span style=\"color: #000000;\">= 2\u03c0r (2r + h)<\/span><br \/>\n<span style=\"color: #000000;\">= 2 \u00d7 22\/7 \u00d7 3.5 cm \u00d7 (2 \u00d7 3.5 cm + 10 cm)<\/span><br \/>\n<span style=\"color: #000000;\">= 22 cm \u00d7 17 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 374 cm<sup>2<br \/>\n<\/sup>Thus, the\u00a0total surface area of the article\u00a0is 374 cm<sup>2<\/sup>.<\/span><\/p>\n<p><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\" target=\"_blank\" rel=\"noopener\"><b><i>Go Back To Chapters<\/i><\/b><\/a><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 13 (Surface Areas and Volumes)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 13 Surface Areas and Volumes Exercise 13.1 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 13 Surface Areas and Volumes NCERT Class [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1109,1110,1044,1049,1048],"class_list":["post-6094","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-13-surface-areas-and-volumes-solutions","tag-ncert-class-10-mathematics-exercise-13-1-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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