{"id":6098,"date":"2023-10-05T12:02:24","date_gmt":"2023-10-05T12:02:24","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6098"},"modified":"2023-10-05T12:03:30","modified_gmt":"2023-10-05T12:03:30","slug":"ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-5","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-5\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 13 (Surface Areas and Volumes)\u00a0<\/span><\/strong><\/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 13.5 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-13-1\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 13.1<\/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-13-2\" 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<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 13.5<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><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>.<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong><\/span>A figure is drawn\u00a0to visualize the cylinder.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6726 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-1-1024x464.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"367\" height=\"166\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-1-1024x464.png 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-1-300x136.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-1-768x348.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-1-850x385.png 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-1.png 1146w\" sizes=\"auto, (max-width: 367px) 100vw, 367px\" \/><\/span><br \/>\n<span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\">Diameter of cylinder = 10 cm<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, the radius of the cylinder (r) = 10\/2 cm = 5 cm<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">\u2234 Length of wire in completely one round = 2\u03c0r = 3.14 \u00d7 5 cm = 31.4 cm<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Diameter of wire = 3 mm = 3\/10 cm<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">\u2234 The thickness of the cylinder covered in one round = 3\/10 m<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Hence, the number of turns (rounds) of the wire to cover 12 cm will be\u00a0 = <img decoding=\"async\" title=\"\\frac{12}{\\frac{3}{10}}\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{12}{\\frac{3}{10}}\" \/><br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\"><img decoding=\"async\" title=\"= 12\\times \\frac{10}{3}\" src=\"https:\/\/latex.codecogs.com\/gif.latex?=&amp;space;12\\times&amp;space;\\frac{10}{3}\" \/> <\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\">= 40<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Now, the length of wire required to cover the whole surface = length of wire required to complete 40 rounds = <\/span><span style=\"font-family: Georgia, Palatino;\">40 \u00d7 31.4 cm<br \/>\n= 1256 cm<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Radius of the wire = 0.3\/2 = 0.15 cm<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">The volume of wire = Area of the cross-section of wire \u00d7 Length of wire<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">= \u03c0(0.15)<sup>2 <\/sup>\u00d7 1257.14<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">= 88.898 cm<sup>3<br \/>\n<\/sup><\/span><span style=\"font-family: Georgia, Palatino;\">We know,<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Mass = Volume \u00d7 Density<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">= 88.898 \u00d7 8.88<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">= 789.41 gm<br \/>\nThus, the length of the wire is 12.57 m and the mass is 790 g (approx.)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>2. A right triangle whose sides are 3 cm and 4 cm (other than the hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose the value of \u03c0 as found appropriate)<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">Draw the diagram as follows:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-6727\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-2-300x153.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"153\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-2-300x153.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-2-1024x523.png 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-2-768x392.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-2-850x434.png 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-2.png 1146w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\n<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">From the figure it can be seen that BD \u22a5 AC<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394ABC right-angled at B\u00a0using\u00a0Pythagoras theorem,<\/span><br \/>\n<span style=\"color: #000000;\">AC\u00b2 = AB\u00b2 + BC\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">AC = (3 cm)\u00b2 + (4 cm)\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 AC = 9 cm\u00b2 + 16 cm\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 AC = 25 cm\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 AC = 5 cm<\/span><br \/>\n<span style=\"color: #000000;\">Consider \u0394ABC and \u0394BDC<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ABC = \u2220CDB = 90\u00b0 (BD \u22a5 AC)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BCA = \u2220BCD (common)<\/span><br \/>\n<span style=\"color: #000000;\">By AA criterion of similarity \u0394ABC \u223c \u0394BDC<\/span><br \/>\n<span style=\"color: #000000;\">Therefore,<\/span><br \/>\n<span style=\"color: #000000;\">AB\/BD = AC\/BC (Corresponding sides of similar triangles are in proportion)<\/span><br \/>\n<span style=\"color: #000000;\">BD = (AB \u00d7 BC)\/AC<\/span><br \/>\n<span style=\"color: #000000;\">= (3 cm \u00d7 4 cm)\/5 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 12\/5 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 2.4 cm<\/span><br \/>\n<span style=\"color: #000000;\">We know that,\u00a0Volume of the cone\u00a0= 1\/3\u03c0r\u00b2h<\/span><br \/>\n<span style=\"color: #000000;\">Volume of double cone = Volume of Cone ABB\u2019 + Volume of Cone BCB\u2019<\/span><br \/>\n<span style=\"color: #000000;\">= 1\/3 \u00d7 \u03c0(BD)\u00b2 \u00d7 AD + 1\/3\u03c0 (BD)\u00b2 \u00d7 DC<\/span><br \/>\n<span style=\"color: #000000;\">= 1\/3 \u00d7 \u03c0(BD)\u00b2 [AD + DC]<\/span><br \/>\n<span style=\"color: #000000;\">= 1\/3 \u00d7 \u03c0(BD)\u00b2 \u00d7 AC<\/span><br \/>\n<span style=\"color: #000000;\">= 1\/3 \u00d7 3.14 \u00d7 2.4 cm \u00d7 2.4 cm \u00d7 5 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 30.144 cm\u00b3<\/span><br \/>\n<span style=\"color: #000000;\">We know that,\u00a0CSA of frustum of a cone = \u03c0rl<\/span><br \/>\n<span style=\"color: #000000;\">CSA of double Cone = CSA of cone ABB\u2019 + CSA of cone BCB\u2019<\/span><br \/>\n<span style=\"color: #000000;\">= \u03c0 \u00d7 BD \u00d7 AB + \u03c0 \u00d7 BD \u00d7 BC<\/span><br \/>\n<span style=\"color: #000000;\">= \u03c0 \u00d7 BD [ AB + BC]<\/span><br \/>\n<span style=\"color: #000000;\">= 3.14 \u00d7 2.4 cm \u00d7 [3 cm + 4 cm]<\/span><br \/>\n<span style=\"color: #000000;\">= 3.14 \u00d7 2.4 cm \u00d7 7 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 52.752 cm\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">= 52.75 cm\u00b2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>3. A cistern, internally measuring 150 cm \u00d7 120 cm \u00d7 100 cm, has 129600 cm<sup>3<\/sup>\u00a0of 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 being 22.5 cm \u00d7 7.5 cm \u00d7 6.5 cm?<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">Let&#8217;s construct a diagram according to the given question.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6729 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-3-1024x305.jpg\" alt=\"NCERT Class 10 Maths Solution\" width=\"459\" height=\"137\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-3-1024x305.jpg 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-3-300x89.jpg 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-3-768x229.jpg 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-3-850x253.jpg 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-3.jpg 1107w\" sizes=\"auto, (max-width: 459px) 100vw, 459px\" \/><br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Given that the dimension of the cistern = 150 \u00d7 120 \u00d7 110<br \/>\nSo, volume = 1980000 cm<sup>3<br \/>\n<\/sup><\/span><span style=\"font-family: Georgia, Palatino;\">Volume to be filled in cistern = 1980000 \u2013 129600<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">= 1850400 cm<sup>3<br \/>\n<\/sup><\/span><span style=\"font-family: Georgia, Palatino;\">Now, let the number of bricks placed to be \u201cn\u201d<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">So, the volume of n bricks will be = n\u00d722.5\u00d77.5\u00d76.5<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Now, as each brick absorbs one-seventeenth of its volume, the volume will be<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">= n\/(17)\u00d7(22.5\u00d77.5\u00d76.5)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">For the condition given in the question,<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">The volume of n bricks has to be equal to the volume absorbed by n bricks + the volume to be filled in the cistern<br \/>\n<\/span>1980000 cm\u00b3 = 129600 cm\u00b3 + 16\/17 \u00d7 n \u00d7 1096.875 cm\u00b3<\/span><br \/>\n<span style=\"color: #000000;\">n = [(1980000 cm\u00b3 &#8211; 129600 cm\u00b3) \u00d7 17] \/ (16 \u00d7 1096.875) cm\u00b3<\/span><br \/>\n<span style=\"color: #000000;\">= (1850400 cm\u00b3 \u00d7 17) \/ (16 \u00d7 1096.875 cm\u00b3)<\/span><br \/>\n<span style=\"color: #000000;\">= 1792.41<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><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 7280 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.<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">From the question, it is clear that<br \/>\n<\/span>Volume of the rainfall = Area of the river valley \u00d7\u00a0height of rainfall in the river valley<\/span><br \/>\n<span style=\"color: #000000;\">Volume of the river\u00a0= length \u00d7\u00a0\u00a0breadth \u00d7\u00a0\u00a0height [Assuming the river valley to be cuboidal]<\/span><br \/>\n<span style=\"color: #000000;\">Area of the valley, A = 7280 km\u00b2 = 7280 \u00d7 1000000 m\u00b2 = 7.28 \u00d7 10<sup>9<\/sup>\u00a0m\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">Height of rainfall in a fortnight, h = 10 cm = 10\/100 m = 0.1 m<\/span><br \/>\n<span style=\"color: #000000;\">Volume of the rainfall = Area of the river valley \u00d7 height of rainfall in the river valley<\/span><br \/>\n<span style=\"color: #000000;\">= 7.28 \u00d7 10<sup>9<\/sup>\u00a0m\u00b2 \u00d7 0.1 m<\/span><br \/>\n<span style=\"color: #000000;\">= 7.28 \u00d7 10<sup>8<\/sup>\u00a0m\u00b3<\/span><br \/>\n<span style=\"color: #000000;\">Length of river, l = 1072 km = 1072 \u00d7 1000 m = 1.072 \u00d7 10<sup>6<\/sup>\u00a0m<\/span><br \/>\n<span style=\"color: #000000;\">Width of river, b = 75 m<\/span><br \/>\n<span style=\"color: #000000;\">Depth of river, h = 3 m<\/span><br \/>\n<span style=\"color: #000000;\">Volume of 3 rivers = 3 \u00d7 volume of one river<\/span><br \/>\n<span style=\"color: #000000;\">= 3 lbh<\/span><br \/>\n<span style=\"color: #000000;\">= 3 \u00d7 1.072 \u00d7 10<sup>6<\/sup>\u00a0m \u00d7 75 m \u00d7 3 m<\/span><br \/>\n<span style=\"color: #000000;\">= 723.6 \u00d7 10<sup>6<\/sup>\u00a0m\u00b3<\/span><br \/>\n<span style=\"color: #000000;\">= 7.236 \u00d7 10<sup>8<\/sup> m\u00b3<\/span><br \/>\n<span style=\"color: #000000;\">Since 7.236 \u00d7 10<sup>8<\/sup>\u00a0m\u00b3 is approximately equivalent to 7.28 \u00d7 10<sup>8<\/sup>\u00a0m\u00b3, therefore, we can say that total rainfall in the valley was approximately equivalent to the addition of normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>5. An oil funnel made of a tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, the 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 (see Fig.).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6733\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Que-5.jpg\" alt=\"NCERT Class 10 Maths Solution\" width=\"135\" height=\"134\" \/><br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">Given<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Diameter of the upper circular end of the frustum part = 18 cm<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">radius (r<sub>1<\/sub>) = 9 cm<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">the radius of the lower circular end of the frustum (r<sub>2<\/sub>) will be equal to the radius of the circular end of the cylinder<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">r<sub>2<\/sub> = 8\/2 = 4 cm<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Now, height (h<sub>1<\/sub>) of the frustum section = 22 \u2013 10 = 12 cm<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Height (h<sub>2<\/sub>) of cylindrical section = 10 cm (given)<br \/>\n<\/span><span style=\"font-family: Georgia, Palatino;\">Now, the slant height will be <em>l = = <\/em>\u221ah<sub>1<\/sub><sup>2<\/sup> + (r<sub>1<\/sub> &#8211; r<sub>2<\/sub>)<sup>2<br \/>\n<\/sup><\/span>= \u221a(12)\u00b2 + (9 &#8211; 4)\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a144\u00a0 + 25<\/span><br \/>\n<span style=\"color: #000000;\">= \u221a169 cm\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">= 13 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\">Area of tin sheet required = CSA of frustum part + CSA of the cylindrical part<br \/>\n<\/span>= \u03c0 (r<sub>1<\/sub> + r<sub>2<\/sub>)l + 2\u03c0r<sub>2<\/sub>h<\/span><br \/>\n<span style=\"color: #000000;\">= 22\/7 [(9 + 4) \u00d7 13 + 2 \u00d7 4 \u00d7 10]<\/span><br \/>\n<span style=\"color: #000000;\">= 22\/7 [169 + 80]<\/span><br \/>\n<span style=\"color: #000000;\">= 22\/7 \u00d7 249<\/span><br \/>\n<span style=\"color: #000000;\">= 5478\/7<\/span><br \/>\n<span style=\"color: #000000;\">= 782.57\u00a0cm\u00b2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>6. Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong><\/span>Consider a\u00a0frustum of a cone with h as height, <em>l<\/em> as the slant height, r<sub>1<\/sub> and r<sub>2<\/sub> as radii of the ends where r<sub>1<\/sub> &gt; r<sub>2.<br \/>\n<\/sub>To Prove:<\/span><br \/>\n<span style=\"color: #000000;\">(i)\u00a0CSA of the frustum of the cone = \u03c0l (r<sub>1<\/sub> + r<sub>2<\/sub>)<\/span><br \/>\n<span style=\"color: #000000;\">(ii) TSA of the frustum of the cone = \u03c0l (r<sub>1<\/sub> + r<sub>2<\/sub>) + \u03c0r<sub>1<\/sub>\u00b2 + \u03c0r<sub>2<\/sub>\u00b2<\/span><br \/>\n<span style=\"color: #000000;\">where r<sub>1<\/sub>, r<sub>2<\/sub>, h and l are the radii height and slant height of the frustum of the cone respectively.<\/span><br \/>\n<span style=\"color: #000000;\"><strong>Construction:<br \/>\n<\/strong>Extended side BC and AD of the frustum of cone to meet at O<\/span><br \/>\n<span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6734\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-6-300x256.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"266\" height=\"227\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-6-300x256.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-6-1024x873.png 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-6-768x655.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-6-850x725.png 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-6.png 1146w\" sizes=\"auto, (max-width: 266px) 100vw, 266px\" \/><br \/>\n<\/span><strong>Proof :<br \/>\n<\/strong>The frustum of a cone can be viewed as a difference of two right circular cones OAB and OCD.<\/span><br \/>\n<span style=\"color: #000000;\">Let h<sub>1<\/sub> and l<sub>1<\/sub> be the height and slant height of cone OAB and h<sub>2<\/sub> and l<sub>2<\/sub> be the height and slant height of cone OCD respectively.<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394APO and \u0394DQO<\/span><br \/>\n<span style=\"color: #000000;\">\u2220APO = \u2220DQO = 90\u00b0 (Since both cones are right circular cones)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AOP = \u2220DOQ (Common)<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, \u0394APO \u223c \u0394DQO (AA criterion of similarity)<\/span><br \/>\n<span style=\"color: #000000;\">AP\/DQ = AO\/DO = OP\/OQ (Corresponding sides of similar triangles are proportional)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 r<sub>1<\/sub>\/r<sub>2<\/sub> = l<sub>1<\/sub>\/l<sub>2<\/sub> = h<sub>1<\/sub>\/h<sub>2<br \/>\n<\/sub>\u21d2 r<sub>1<\/sub>\/r<sub>2<\/sub> = l<sub>1<\/sub>\/l<sub>2<\/sub> or \u21d2 r<sub>2<\/sub>\/r<sub>1<\/sub> = l<sub>2<\/sub>\/l<sub>1<br \/>\n<\/sub>Subtracting 1 from both sides we get<\/span><br \/>\n<span style=\"color: #000000;\">r<sub>1<\/sub>\/r<sub>2<\/sub> &#8211; 1 = l<sub>1<\/sub>\/l<sub>2<\/sub> &#8211; 1<\/span><br \/>\n<span style=\"color: #000000;\">(r<sub>1<\/sub> &#8211; r<sub>2<\/sub>)\/r<sub>2<\/sub> = (l<sub>1<\/sub> &#8211; l<sub>2<\/sub>) \/ l<sub>2<br \/>\n<\/sub>(r<sub>1<\/sub> &#8211; r<sub>2<\/sub>)\/r<sub>2<\/sub> = l\/l<sub>2<\/sub> [From diagram, l<sub>1<\/sub> &#8211; l<sub>2<\/sub> = l] <\/span><br \/>\n<span style=\"color: #000000;\">l<sub>2<\/sub> = lr<sub>2<\/sub>\/(r<sub>1<\/sub> &#8211; r<sub>2<\/sub>)\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (i)<\/span><br \/>\n<span style=\"color: #000000;\">r<sub>2 <\/sub>\/ r<sub>1<\/sub> = l<sub>2\u00a0<\/sub>\/ l<sub>1<br \/>\n<\/sub>Subtracting 1 from both sides we get<\/span><br \/>\n<span style=\"color: #000000;\">r<sub>2 <\/sub>\/ r<sub>1<\/sub> &#8211; 1 = l<sub>2 <\/sub>\/ l<sub>1<\/sub> &#8211; 1<\/span><br \/>\n<span style=\"color: #000000;\">(r<sub>2<\/sub> &#8211; r<sub>1<\/sub>) \/ r<sub>1<\/sub>\u00a0= (l<sub>2<\/sub> &#8211; l<sub>1<\/sub>) \/ l<sub>1<br \/>\n<\/sub>(r<sub>1<\/sub>\u00a0&#8211; r<sub>2<\/sub>) \/ r<sub>1<\/sub>\u00a0= (l<sub>1<\/sub>\u00a0&#8211; l<sub>2<\/sub>) \/ l<sub>1<br \/>\n<\/sub>(r<sub>1<\/sub>\u00a0&#8211; r<sub>2<\/sub>) \/ r<sub>1 <\/sub>= l<sub>\u00a0<\/sub>\/ l<sub>1<br \/>\n<\/sub>l<sub>1<\/sub>\u00a0= lr<sub>1<\/sub>\/(r<sub>1<\/sub>\u00a0&#8211; r<sub>2<\/sub>)\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;- (ii)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) CSA of frustum of cone = CSA of cone OAB &#8211; CSA of cone OCD<br \/>\n<\/strong>= \u03c0r<sub>1<\/sub>l<sub>1<\/sub>\u00a0&#8211; \u03c0r<sub>2<\/sub>l<sub>2<br \/>\n<\/sub>= \u03c0 (r<sub>1<\/sub>l<sub>1<\/sub>\u00a0&#8211; r<sub>2<\/sub>l<sub>2<\/sub>)<\/span><br \/>\n<span style=\"color: #000000;\">= \u03c0 [(r<sub>1<\/sub>\u00a0\u00d7 lr<sub>1<\/sub>\/(r<sub>1<\/sub>\u00a0&#8211; r<sub>2<\/sub>) &#8211; r<sub>2<\/sub>\u00a0\u00d7 lr<sub>2<\/sub>\/(r<sub>1<\/sub>\u00a0&#8211; r<sub>2<\/sub>)]\u00a0 \u00a0 \u00a0[Using (i) and (ii)]<\/span><br \/>\n<span style=\"color: #000000;\">= \u03c0 [(lr<sub>1<\/sub><sup>\u00b2<\/sup> &#8211; lr<sub>2<\/sub><sup>\u00b2<\/sup>)\/(r<sub>1<\/sub>\u00a0&#8211; r<sub>2<\/sub>)]<\/span><br \/>\n<span style=\"color: #000000;\">= \u03c0 [l(r<sub>1<\/sub><sup>2<\/sup>\u00a0&#8211; r<sub>2<\/sub><sup>2<\/sup>)\/(r<sub>1<\/sub> &#8211; r<sub>2<\/sub>)]<\/span><br \/>\n<span style=\"color: #000000;\">= \u03c0 [l(r<sub>1<\/sub> &#8211; r<sub>2<\/sub>)(r<sub>1<\/sub> + r<sub>2<\/sub>)\/(r<sub>1<\/sub> &#8211; r<sub>2<\/sub>)]\u00a0 \u00a0 \u00a0[Since, a\u00b2 &#8211; b\u00b2 = (a &#8211; b)(a + b)]<\/span><br \/>\n<span style=\"color: #000000;\">= \u03c0 l(r<sub>1<\/sub> + r<sub>2<\/sub>)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">TSA of frustum of cone = CSA of frustum + Area of lower circular end + Area of top circular end <\/span><br \/>\n<span style=\"color: #000000;\">= \u03c0l (r<sub>1<\/sub> + r<sub>2<\/sub>) + \u03c0r<sub>1<\/sub><sup>\u00b2<\/sup> + \u03c0r<sub>1<\/sub><sup>\u00b2<br \/>\n<\/sup>Therefore, CSA of the frustum of the cone = \u03c0l (r<sub>1<\/sub> + r<sub>2<\/sub>)<\/span><br \/>\n<span style=\"color: #000000;\">TSA of the frustum of the cone = \u03c0l (r<sub>1<\/sub> + r<sub>2<\/sub>) + \u03c0r<sub>1<\/sub><sup>\u00b2<\/sup> + \u03c0r<sub>2<\/sub><sup>\u00b2<br \/>\n<\/sup>Hence Proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><span style=\"font-family: Georgia, Palatino;\"><strong>7. Derive the formula for the volume of the frustum of a cone.<br \/>\n<\/strong><\/span><span style=\"font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"font-family: Georgia, Palatino;\">Consider the same diagram as the previous question.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-6734\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-6-300x256.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"256\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-6-300x256.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-6-1024x873.png 1024w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-6-768x655.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-6-850x725.png 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.5-Ans-6.png 1146w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\n<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>To Prove:<br \/>\n<\/strong>CSA of the frustum of the cone = 1\/3\u03c0h (r<sub>1<\/sub><sup>2<\/sup> + r<sub>2<\/sub><sup>2<\/sup> + r<sub>1<\/sub>r<sub>2<\/sub>) <\/span><br \/>\n<span style=\"color: #000000;\">where r<sub>1<\/sub>, r<sub>2<\/sub>, and h are the radii and height of the frustum of the cone respectively.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Proof :<br \/>\n<\/strong>The\u00a0frustum of a cone\u00a0can be viewed as a difference of two right circular cones OAB and OCD.<\/span><br \/>\n<span style=\"color: #000000;\">Let h<sub>1<\/sub> and l<sub>1<\/sub> be the height and slant height of cone OAB and h\u2082 and l\u2082 be the height and slant height of cone OCD respectively. <\/span><br \/>\n<span style=\"color: #000000;\">In \u0394APO and \u0394DQO <\/span><br \/>\n<span style=\"color: #000000;\">\u2220APO = \u2220DQO = 90\u00b0 (Since both cones are right circular cones) <\/span><br \/>\n<span style=\"color: #000000;\">\u2220AOP = \u2220DOQ (Common) <\/span><br \/>\n<span style=\"color: #000000;\">Therefore, \u0394APO \u223c \u0394DQO (\u00a0A.A criterion of similarity) <\/span><br \/>\n<span style=\"color: #000000;\">AP\/DQ = AO\/DO = OP\/OQ (Corresponding sides of similar triangles are proportional)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 r<sub>1<\/sub>\/r<sub>1<\/sub> = l<sub>1<\/sub>\/l<sub>2<\/sub> = h<sub>1<\/sub>\/h<sub>2<br \/>\n<\/sub>\u21d2 r<sub>1<\/sub>\/r<sub>1<\/sub> = h<sub>1<\/sub>\/h<sub>2<\/sub> <\/span><br \/>\n<span style=\"color: #000000;\">or <\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 r<sub>2<\/sub>\/r<sub>1<\/sub> = h<sub>2<\/sub>\/h<sub>1<br \/>\n<\/sub>Subtracting 1 from both sides <\/span><br \/>\n<span style=\"color: #000000;\">r<sub>1<\/sub>\/r<sub>2<\/sub> &#8211; 1 = h<sub>1<\/sub>\/h<sub>2<\/sub>\u00a0&#8211; 1<\/span><br \/>\n<span style=\"color: #000000;\">(r<sub>1<\/sub>\u00a0&#8211; r<sub>2<\/sub>)\/r<sub>2<\/sub>\u00a0= (h<sub>1<\/sub>\u00a0&#8211; h<sub>2<\/sub>)\/h<sub>2<br \/>\n<\/sub>(r<sub>1<\/sub>\u00a0&#8211; r<sub>2<\/sub>)\/r<sub>2<\/sub> = h\/h<sub>2<\/sub>\u00a0 \u00a0 [From figure h<sub>1<\/sub> &#8211; h<sub>2<\/sub> = h]<\/span><br \/>\n<span style=\"color: #000000;\">h<sub>2<\/sub>\u00a0= hr<sub>2<\/sub>\/(r<sub>1<\/sub>\u00a0&#8211; r<sub>2<\/sub>)\u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (ii)<\/span><br \/>\n<span style=\"color: #000000;\">Now, considering r<sub>2<\/sub>\/r<sub>1<\/sub>\u00a0= h<sub>2<\/sub>\/h<sub>1<br \/>\n<\/sub>Subtracting 1 from both sides we get<\/span><br \/>\n<span style=\"color: #000000;\">r<sub>2<\/sub>\/r<sub>1<\/sub>\u00a0&#8211; 1 = h<sub>2<\/sub>\/h<sub>1<\/sub>\u00a0&#8211; 1<\/span><br \/>\n<span style=\"color: #000000;\">(r<sub>2<\/sub>\u00a0&#8211; r<sub>1<\/sub>)\/r<sub>1<\/sub>\u00a0= (h<sub>2<\/sub>\u00a0&#8211; h<sub>1<\/sub>)\/h<sub>1<br \/>\n<\/sub>(r<sub>1<\/sub>\u00a0&#8211; r<sub>2<\/sub>)\/r<sub>1<\/sub>\u00a0= (h<sub>1<\/sub>\u00a0&#8211; h<sub>2<\/sub>)\/h<sub>1<br \/>\n<\/sub>(r<sub>1<\/sub>\u00a0&#8211; r<sub>2<\/sub>)\/r<sub>1<\/sub>\u00a0= h\/h<sub>1<br \/>\n<\/sub>h<sub>1<\/sub>\u00a0= hr<sub>1<\/sub>\/(r<sub>1<\/sub>\u00a0&#8211; r<sub>2<\/sub>)\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (ii)<\/span><br \/>\n<span style=\"color: #000000;\">Volume of frustum of cone\u00a0= Volume of cone OAB &#8211; Volume of cone OCD<\/span><br \/>\n<span style=\"color: #000000;\">= 1\/3 \u03c0r<sub>1<\/sub><sup>2<\/sup>h<sub>1<\/sub> &#8211; 1\/3 \u03c0r<sub>2<\/sub><sup>2<\/sup>h<sub>2<br \/>\n<\/sub>= 1\/3\u03c0 [r<sub>1<\/sub><sup>2<\/sup>h<sub>1<\/sub> &#8211; r<sub>2<\/sub><sup>2<\/sup>h<sub>2<\/sub>]<\/span><br \/>\n<span style=\"color: #000000;\">= 1\/3\u03c0 [r<sub>1<\/sub><sup>2<\/sup> \u00d7 \u00a0hr<sub>1<\/sub>\/(r<sub>1<\/sub> &#8211; r<sub>2<\/sub>) &#8211; r<sub>2<\/sub><sup>2<\/sup> \u00d7 hr<sub>2<\/sub>(r<sub>1<\/sub> &#8211; r<sub>2<\/sub>)]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [Using (i) and (ii)]<\/span><br \/>\n<span style=\"color: #000000;\">= 1\/3\u03c0 [hr<sub>1<\/sub><sup>3<\/sup>\/(r<sub>1<\/sub> &#8211; r<sub>2<\/sub>) &#8211; hr<sub>2<\/sub><sup>3<\/sup>\/(r<sub>1<\/sub> &#8211; r<sub>2<\/sub>)]<\/span><br \/>\n<span style=\"color: #000000;\">= 1\/3\u03c0h [(r<sub>1<\/sub><sup>3<\/sup> &#8211; r<sub>2<\/sub><sup>3<\/sup>)\/(r<sub>1<\/sub> &#8211; r<sub>2<\/sub>)]<\/span><br \/>\n<span style=\"color: #000000;\">= 1\/3\u03c0h [(r<sub>1<\/sub> &#8211; r<sub>2<\/sub>)(r<sub>1<\/sub><sup>2<\/sup> + r<sub>1<\/sub>r<sub>2<\/sub> + r<sub>2<\/sub><sup>2<\/sup>)\/(r<sub>1<\/sub> &#8211; r<sub>2<\/sub>)]\u00a0 \u00a0 \u00a0[Since, (a\u00b3 &#8211; b\u00b3) = (a &#8211; b)(a\u00b2 + ab + b\u00b2)]<\/span><br \/>\n<span style=\"color: #000000;\">= 1\/3\u03c0h (r<sub>1<\/sub>\u00b2 + r<sub>2<\/sub>\u00b2 + r<sub>1<\/sub>r<sub>2<\/sub>)<\/span><br \/>\n<span style=\"color: #000000;\">Hence proved, volume of the frustum = 1\/3\u03c0h (r<sub>1<\/sub>\u00b2 + r<sub>2<\/sub>\u00b2 + r<sub>1<\/sub>r<sub>2<\/sub>)<\/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 13.5 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":[1045,1109,1114,1044,1049,1048],"class_list":["post-6098","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-13-5-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) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ 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