{"id":6097,"date":"2023-10-05T05:12:41","date_gmt":"2023-10-05T05:12:41","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6097"},"modified":"2023-10-05T05:13:40","modified_gmt":"2023-10-05T05:13:40","slug":"ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-4","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-13-surface-areas-and-volumes-ex-13-4\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4"},"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 13.4 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-13-5\" 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; 13.4<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><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.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-6713\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-1-231x300.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"231\" height=\"300\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-1-231x300.png 231w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-1-300x389.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-1.png 341w\" sizes=\"auto, (max-width: 231px) 100vw, 231px\" \/><br \/>\nRadius (r<sub>1<\/sub>) of the upper base = 4\/2 = 2 cm<br \/>\nRadius (r<sub>2<\/sub>) of lower the base = 2\/2 = 1 cm<br \/>\nHeight = 14 cm<br \/>\nNow,<br \/>\nThe capacity of glass = Volume of the frustum of the cone<br \/>\nCapacity of glass = (\u2153) \u00d7 \u03c0 \u00d7 h (r<sub>1<\/sub><sup>2 <\/sup>+ r<sub>2<\/sub><sup>2 <\/sup>+ r<sub>1<\/sub>r<sub>2<\/sub>)<br \/>\n= (\u2153) \u00d7 (22\/7) \u00d7 (14) (2<sup>2 <\/sup>+ 1<sup>2 <\/sup>+\u00a0 (2)(1))<br \/>\n= (44\/3) \u00d7 (4 + 1 + 2)<br \/>\n= 308\/3<br \/>\n= <img decoding=\"async\" id=\"equationview\" title=\"This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.\" src=\"https:\/\/latex.codecogs.com\/gif.latex?102%5Cfrac%7B2%7D%7B3%7D\" name=\"equationview\" \/><br \/>\n\u2234 The capacity of the glass = 102\u00d7(\u2154) cm<sup>3<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><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 surface area of the frustum.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6714\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-2-265x300.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"165\" height=\"187\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-2-265x300.png 265w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-2-300x339.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-2.png 707w\" sizes=\"auto, (max-width: 165px) 100vw, 165px\" \/><br \/>\n<\/strong>Slant height (l) = 4 cm<br \/>\nCircumference of upper circular end of the frustum = 18 cm<br \/>\n\u2234 2\u03c0r<sub>1<\/sub>\u00a0= 18<br \/>\nr<sub>1<\/sub> = 9\/\u03c0<br \/>\nSimilarly, the circumference of the lower end of the frustum = 6 cm<br \/>\n\u2234 2\u03c0r<sub>2<\/sub>\u00a0= 6<br \/>\nr<sub>2<\/sub>\u00a0= 3\/\u03c0<br \/>\nNow, the surface area of the frustum = \u03c0(r<sub>1 <\/sub>+ r<sub>2<\/sub>) \u00d7 l<br \/>\n= \u03c0(9\/\u03c0 + 3\/\u03c0) \u00d7 4<br \/>\n= 12 \u00d7 4<br \/>\n= 48 cm<sup>2<br \/>\n<\/sup>The curved surface area of the frustum is 48 cm<sup>2<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig.). If its radius on the open side is 10 cm, the radius at the upper base is 4 cm, and its slant height is 15 cm, find the area of material used for making it.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6715\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Que-3.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"141\" height=\"183\" \/><br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>Given,<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6716\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-3-243x300.jpg\" alt=\"NCERT Class 10 Maths Solution\" width=\"206\" height=\"254\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-3-243x300.jpg 243w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-3-300x371.jpg 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-3.jpg 326w\" sizes=\"auto, (max-width: 206px) 100vw, 206px\" \/><br \/>\nFor the lower circular end, radius (r<sub>1<\/sub>) = 10 cm<br \/>\nFor the upper circular end, radius (r<sub>2<\/sub>) = 4 cm<br \/>\nSlant height (l) of frustum = 15 cm<br \/>\nNow,<br \/>\nThe area of material to be used for making the fez = CSA of frustum + Area of the upper circular end<br \/>\nCSA of frustum = \u03c0(r<sub>1 <\/sub>+ r<sub>2<\/sub>)\u00d7l<br \/>\n= \u03c0(10<sub>\u00a0<\/sub>+ 4)\u00d715<br \/>\n= 210\u03c0<br \/>\nAnd, the Area of the upper circular end = \u03c0r<sub>2<\/sub><sup>2<br \/>\n<\/sup>= \u03c04<sup>2<\/sup><br \/>\n= 16\u03c0<br \/>\nThe area of material to be used for making the fez = 210\u03c0 + 16\u03c0<br \/>\n= (226 \u00d7 22)\/7<br \/>\n<img decoding=\"async\" title=\"= 710\\frac{2}{7}\" src=\"https:\/\/latex.codecogs.com\/gif.latex?=&amp;space;710\\frac{2}{7}\" \/><br \/>\n= 710.29\u00a0cm\u00b2<br \/>\n710.29 cm\u00b2 of the material used for making Fez.<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><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 Rs. 20 per litre. Also, find the cost of the metal sheet used to make the container if it costs Rs. 8 per 100 cm<sup>2<\/sup>.<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Given,<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-6718\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-4-300x263.jpg\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"263\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-4-300x263.jpg 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-4.jpg 329w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\nr<sub>1<\/sub>\u00a0= 20 cm,<br \/>\nr<sub>2<\/sub>\u00a0= 8 cm and<br \/>\nh = 16 cm<br \/>\n\u2234 Volume of the frustum = (\u2153)\u00d7\u03c0\u00d7h(r<sub>1<\/sub><sup>2<\/sup>+r<sub>2<\/sub><sup>2<\/sup>+r<sub>1<\/sub>r<sub>2<\/sub>)<br \/>\n= 1\/3\u00a0\u00d73.14 \u00d716((20)<sup>2<\/sup>+(8)<sup>2<\/sup>+(20)(8))<br \/>\n=\u00a01\/3\u00a0\u00d73.14 \u00d716(400 + 64 + 160)<br \/>\n= 10449.92 cm<sup>3<\/sup><br \/>\n= 10.45 lit<br \/>\nIt is given that the rate of milk = Rs. 20\/litre<br \/>\nSo, the cost of milk = 20\u00d7volume of the frustum<br \/>\n= 20\u00d710.45<br \/>\n= Rs. 209<br \/>\nNow, the slant height <em>l<\/em> = \u221a[(r\u2081\u00a0&#8211; r\u2082)<sup>2<\/sup>\u00a0+ h<sup>2<\/sup>]<br \/>\n<em>l<\/em> = \u221a[(20 &#8211; 8)<sup>2<\/sup> + (16)<sup>2<\/sup>]<br \/>\n<em>l<\/em> = \u221a[144 + 256]<br \/>\n<em>l<\/em> = \u221a400<br \/>\n<em>l<\/em> = 20 cm<br \/>\nArea of metal sheet required to make the container = CSA of frustum of the cone + Area of lower circular end <\/span><br \/>\n<span style=\"color: #000000;\">= \u03c0 (r\u2081\u00a0+ r\u2082) l + \u03c0r\u2082<sup>2<br \/>\n<\/sup>= \u03c0 [ (r\u2081\u00a0+ r\u2082) l + r\u2082<sup>2<\/sup>] <\/span><br \/>\n<span style=\"color: #000000;\">= 3.14 \u00d7 [ (20 + 8) \u00d7 20 + (8)<sup>2<\/sup>]<\/span><br \/>\n<span style=\"color: #000000;\">= 3.14 \u00d7 [560 + 64 ]<\/span><br \/>\n<span style=\"color: #000000;\">= 3.14 \u00d7 624 cm<sup>2<br \/>\n<\/sup>= 1959.36 cm<sup>2<br \/>\n<\/sup>Cost of 100 cm<sup>2<\/sup> of metal sheet = \u20b9 8 <\/span><br \/>\n<span style=\"color: #000000;\">Therefore, cost of 1959.36 cm<sup>2<\/sup>\u00a0of metal sheet = \u20b9 (8\/100) \u00d7 1959.36<\/span><br \/>\n<span style=\"color: #000000;\">= \u20b9 156.7488<\/span><br \/>\n<span style=\"color: #000000;\">= \u20b9 156.75<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. A metallic right circular cone 20 cm high and whose vertical angle is 60\u00b0 is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained is drawn into a wire of diameter 1\/16 cm, find the length of the wire.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>The diagram will be as follows<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-6721\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-5-300x276.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"300\" height=\"276\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-5-300x276.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-5-768x707.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-5-850x783.png 850w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/10\/NCERT-Class-10-Maths-Ex13.4-Ans-5.png 913w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\nTo get the values of radii of both ends of the frustum formed, compare \u0394ADB and \u0394ADC<br \/>\nTherefore, Volume of the wire =\u00a0Volume of frustum of the cone<\/span><br \/>\n<span style=\"color: #000000;\">We will find the volume of the frustum by using formula; <\/span><br \/>\n<span style=\"color: #000000;\">Volume of frustum of a 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>), where r<sub>1<\/sub>, r<sub>2<\/sub><sub>,<\/sub> and h are the radii and height of the frustum of the cone respectively.\u00a0 <\/span><br \/>\n<span style=\"color: #000000;\">We will find the volume of the wire by using formula; <\/span><br \/>\n<span style=\"color: #000000;\">Volume of cylinder = \u03c0r<sup>2<\/sup>h, where r and h are radius and height of the cylinder <\/span><br \/>\n<span style=\"color: #000000;\">In \u0394ABC , EF parallel to BC and <\/span><br \/>\n<span style=\"color: #000000;\">AD = 20 cm <\/span><br \/>\n<span style=\"color: #000000;\">AG = 10 cm <\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAC = 60\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">To get the values of radii of both ends of the frustum formed compare ADB and ADC <\/span><br \/>\n<span style=\"color: #000000;\">AD = AD (common) <\/span><br \/>\n<span style=\"color: #000000;\">AB = AC (Slant height )<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ADB = \u2220ADC = 90 (Right circular cone)<\/span><br \/>\n<span style=\"color: #000000;\">\u0394ADB \u2245 \u0394ADC (RHS criterion of congruency)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAD = \u2220DAC (CPCT) <\/span><br \/>\n<span style=\"color: #000000;\">Then, <\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAD = \u2220DAC\u00a0 <\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAD = 1\/2 \u00d7 60\u00b0 = 30\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\"><strong>In \u0394ADB <\/strong><\/span><br \/>\n<span style=\"color: #000000;\">BD\/AD = tan 30\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">BD = AD tan 30\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">BD = 20 cm \u00d7 1\/\u221a3<\/span><br \/>\n<span style=\"color: #000000;\">BD = 20\u221a3\/3 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>In \u0394AEG<\/strong><\/span><br \/>\n<span style=\"color: #000000;\">EG\/AG = tan 30\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">EG = AG tan 30\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">EG = 10 cm \u00d7 1\/\u221a3<\/span><br \/>\n<span style=\"color: #000000;\">EG = 10\u221a3\/3 cm<\/span><\/p>\n<p><span style=\"color: #000000;\">Height of the frustum of the cone, h = 10 cm<\/span><br \/>\n<span style=\"color: #000000;\">Radius of lower end, r\u2081\u00a0= (20\u221a3) \/ 3 cm<\/span><br \/>\n<span style=\"color: #000000;\">Radius of upper end, r\u2082\u00a0= (10\u221a3) \/ 3 cm<\/span><br \/>\n<span style=\"color: #000000;\">Diameter of the cylindrical wire, d = 1 \/ 16 cm<\/span><br \/>\n<span style=\"color: #000000;\">Radius of the cylindrical wire, r = 1 \/ 2 \u00d7 1 \/ 16 cm = 1 \/ 32 cm<\/span><br \/>\n<span style=\"color: #000000;\">Let the length of the wire be H<\/span><br \/>\n<span style=\"color: #000000;\">Since the frustum is drawn into wire<\/span><br \/>\n<span style=\"color: #000000;\">Volume of the cylindrical wire = Volume of frustum of the cone <\/span><br \/>\n<span style=\"color: #000000;\">\u03c0r<sup>2<\/sup>H = 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;\">H = [h(r<sub>1<\/sub><sup>2<\/sup> + r<sub>2<\/sub><sup>2<\/sup> + r<sub>1<\/sub>r<sub>2<\/sub>)]\/\u00a03r<sup>2<br \/>\n<\/sup>= [\u00a010 \u00d7\u00a0{(20\u221a3) \/3)<sup>2<\/sup>\u00a0+ (10\u221a3\/ 3)<sup>2<\/sup>\u00a0+ (20\u221a3) \/3 \u00d7\u00a0(10\u221a3) \/3] \/\u00a03 \u00d7 (1\/32)<sup>2<br \/>\n<\/sup>= [10\u00a0\u00d7 {400\/3\u00a0+ 100\/3 \u00a0+ 200\/3}] \/\u00a03 \u00d7 (1\/1024)<\/span><br \/>\n<span style=\"color: #000000;\">= 10240 \u00d7\u00a0700\u00a0\/ 9<\/span><br \/>\n<span style=\"color: #000000;\">= 7168000 \/ 9<\/span><br \/>\n<span style=\"color: #000000;\">= 796444.44 cm<\/span><br \/>\n<span style=\"color: #000000;\">= 7964.4 m<\/span><br \/>\n<span style=\"color: #000000;\">Thus the length of the wire is 7964.4 m.<\/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 13.4 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":[1045,1109,1113,1044,1049,1048],"class_list":["post-6097","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-13-4-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\/ -->\n<title>NCERT 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