{"id":625011,"date":"2023-09-02T01:59:23","date_gmt":"2023-09-02T01:59:23","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=625011"},"modified":"2023-09-05T02:46:54","modified_gmt":"2023-09-05T02:46:54","slug":"kc-sinha-exercise-14-1-mathematics-solution-class-10-chapter-14-surface-area-and-volumes","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-14-1-mathematics-solution-class-10-chapter-14-surface-area-and-volumes\/","title":{"rendered":"KC Sinha: Exercise 14.1 &#8211; Mathematics Solution Class 10 Chapter 14 Surface Area and Volumes"},"content":{"rendered":"\n\n\n\n\n<h4 class=\"wp-block-heading\" id=\"h-question-1\">Question 1<\/h4>\n\n\n\n<p><strong>Three cubes each of side 5 cm are joined end to end to form a cuboid. Find the surface area of the resulting cuboid.<\/strong><br>Sol :<br><img loading=\"lazy\" decoding=\"async\" height=\"90\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1_1545975392494895.jpg\" width=\"77\"><img loading=\"lazy\" decoding=\"async\" height=\"94\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1545975393333673.jpg\" width=\"85\"><img loading=\"lazy\" decoding=\"async\" height=\"93\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1_1545975394090611.jpg\" width=\"81\"><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1_1545975394903514.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<p>Length of each side of cube = 5cm<br>According to question,<br>Three cubes are joined end to end to form a cuboid<br>So, the length of the resulting cuboid = 5 + 5 + 5 = 15cm<br>But, the breadth and height will remain the same<br>So, Breadth = 5cm<br>and height = 5cm<br>Surface Area of the Cuboid<br>= 2(Length \u00d7 Breadth + Breadth \u00d7 Height + Height \u00d7 Length)<br>= 2(lb + bh + hl)<br>= 2(15\u00d75 + 5\u00d75 + 5\u00d715)<br>= 2(75 + 25 + 75)<br>= 2(175)<br>= 350cm<sup>2<\/sup><br>Hence, the surface area of cuboid is 350cm<sup>2<\/sup>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"h-question-2\">Question 2<\/h4>\n\n\n\n<p><strong>Cardboard boxes of two different sizes are made. The bigger has dimensions 20cm, 15cm and 5cm and the smaller dimensions 16cm, 12cm and 4cm. 5% of the total surface area is required extra for all overlaps. If the cost of the cardboard is Rs. 20 for one square metre, find the cost of the cardboard for supplying 200 boxes of each kind.<\/strong><br>Sol :<br><img loading=\"lazy\" decoding=\"async\" height=\"69\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/1_154597539564216.jpg\" width=\"463\"><\/p>\n\n\n\n<p>For bigger box:<br>Length of the bigger box = 20cm<br>Breadth of the bigger box = 15cm<br>Height of the bigger box = 5cm<br>So, Total Surface Area of bigger box = 2(lb + bh + hl)<br>= 2(20\u00d715 + 15\u00d75 + 5\u00d720)<br>= 2(300 + 75 + 100)<br>= 2(475)<br>= 950cm<sup>2<\/sup><\/p>\n\n\n\n<p>For Smaller box:<br>Length of the smaller box = 16cm<br>Breadth of the smaller box = 12cm<br>Height of the smaller box = 4cm<br>So, Total Surface Area of bigger box = 2(lb + bh + hl)<br>= 2(16\u00d712 + 12\u00d74 + 4\u00d716)<br>= 2(192 + 48 + 64)<br>= 2(304)<br>= 608cm<sup>2<\/sup><\/p>\n\n\n\n<p>Total surface area of 200 boxes of each type = 200(950 +608)<br>= 200 \u00d7 1558<br>= 311600cm<sup>2<\/sup><br>Extra Area Required = 5% of (950 + 608)<br>$=\\frac{5}{100} \\times 1558 \\times 200$<br>= 15580cm<sup>2<\/sup><br>So, Total cardboard Required = 311600 + 15580<br>= 327180cm<sup>2<\/sup><br>$=\\frac{327180}{1000}=327.18 \\mathrm{m}^{2}$<br>Cost of Cardboard for 1m<sup>2<\/sup>&nbsp;= Rs 20<br>Cost of Cardboard for 327.18 m<sup>2<\/sup>&nbsp;= 20 \u00d7 327.18<br>= Rs 654.36<br>Hence, the cost of the cardboard for supplying 200 boxes of each kind is Rs 654.36.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"h-question-3\">Question 3<\/h4>\n\n\n\n<p><strong>The length of cold storage is double its breadth. Its height is 3 metres. The area of its four walls (including doors) is 108 m<sup>2<\/sup>. Find its volume.<\/strong><br>Sol :<br>Given: Area = 108m<sup>2<\/sup>, Height = 3m<br>and Length of a cold storage is double its breadth<br>So, Let breadth of a cold storage = x<br>\u2234&nbsp;length of the cold storage = 2x<br>Area of four walls = 108m<sup>2<\/sup><br>\u2234&nbsp;2(l + b)\u00d7 h = 108<br>\u21d2&nbsp;2 (x + 2x)\u00d73 = 108<br>\u21d2&nbsp;6(3x) = 108<br>$\\Rightarrow \\mathrm{x}=\\frac{108}{18}$<br>\u21d2&nbsp;x = 6<\/p>\n\n\n\n<p>\u2234&nbsp;Breadth of a cold storage = 6m<br>and Length of a cold storage = 2\u00d76 = 12m<br>Hence, Volume of the cold storage = l \u00d7 b \u00d7 h<br>= 12 \u00d7 6 \u00d7 3<br>= 216m<sup>3<\/sup><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"h-question-4\">Question 4<\/h4>\n\n\n\n<p><strong>Find:<\/strong><br><strong>(i) the lateral surface,<\/strong><br><strong>(ii) the whole surface, and<\/strong><br><strong>(iii) the volume of a right circular cylinder whose height is 13.5 cm and radius of the base 7cm<\/strong><br>Sol :<br>Given: Radius of base, r = 7cm<br>Height , h = 13.5cm<\/p>\n\n\n\n<p>(i) Lateral Surface Area of right circular cylinder = 2\u03c0rh<br>$=2 \\times \\frac{22}{7} \\times 7 \\times 13.5$<br>= 2\u00d722\u00d713.5<br>= 594cm<sup>2<\/sup><\/p>\n\n\n\n<p>(ii) Total Surface Area of cylinder = 2\u03c0rh + 2\u03c0r<sup>2<\/sup><br>= 2\u03c0r(h + r)<br>$=2 \\times \\frac{22}{7} \\times 7(13.5+7)$<br>= 2\u00d722\u00d720.5<br>= 902cm<sup>2<\/sup><\/p>\n\n\n\n<p>(iii) Volume of cylinder = \u03c0r<sup>2<\/sup>h<br>$=\\frac{22}{7} \\times 7 \\times 7 \\times 13.5$<br>= 22\u00d77\u00d713.5<br>= 2079cm<sup>3<\/sup><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"h-question-5\">Question 5<\/h4>\n\n\n\n<p><strong>The radius and height of a right circular cone are in the ratio of 5:12. If its volume is 314 cm<sup>3<\/sup>, find its slant height.<\/strong><\/p>\n\n\n\n<p><strong>[Take \u03c0 =3.14]<\/strong><br>Sol :<br>Given: Volume of a right circular cone = 314cm<sup>3<\/sup><br>and r : h = 5 : 12<br>Let r = 5x and h = 12x<br>The volume of a right circular cone$=\\frac{1}{3} \\pi r^{2} h$<br>$\\Rightarrow 314=\\frac{1}{3} \\times 3.14 \\times(5 \\mathrm{x})^{2} \\times 12 \\mathrm{x}$<\/p>\n\n\n\n<p>$\\Rightarrow \\frac{314 \\times 3}{3.14}=25 \\mathrm{x}^{2} \\times 12 \\mathrm{x}$<\/p>\n\n\n\n<p>$\\Rightarrow \\frac{314 \\times 3 \\times 100}{314 \\times 25 \\times 12}=\\mathrm{x}^{3}$<\/p>\n\n\n\n<p>\u21d2&nbsp;x<sup>3<\/sup>&nbsp;= 1<br>Hence, x = 1<br>\u2234&nbsp;r = 5x = 5\u00d71 = 5cm<br>and h = 12x = 12\u00d71 = 12cm<br>Now, we have to find the slant height, l<br>Slant height, l = \u221a(r<sup>2<\/sup>&nbsp;+ h<sup>2<\/sup>)<br>= \u221a{(5)<sup>2<\/sup>&nbsp;+ (12)<sup>2<\/sup>}<br>= \u221a25+144<br>= \u221a169<br>= \u00b113<br>= 13cm<br>[taking positive root, because slant height can\u2019t be negative]<br>\u2234&nbsp;Slant Height = 13cm<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\" id=\"h-question-6-nbsp\">Question 6&nbsp;<\/h4>\n\n\n\n<p><strong>A cylinder, whose height is two-thirds of its diameter, has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.<\/strong><br>Sol :<br>Let height of a cylinder = h<br>and diameter = d<br>According to question,<\/p>\n\n\n\n<p>$\\mathrm{h}=\\frac{2}{3} \\mathrm{d}$<\/p>\n\n\n\n<p>$\\Rightarrow \\mathrm{h}=\\frac{2}{3} \\times 2 \\mathrm{r}$<\/p>\n\n\n\n<p>$\\Rightarrow \\mathrm{h}=\\frac{4}{3} \\mathrm{r}$<\/p>\n\n\n\n<p>Given: Radius of Sphere, R = 4cm<\/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\/kc-sinha-solutions\/\">KC Sinha 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\/kc-sinha-solution-for-class-10\/\">KC Sinha Class 10 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Question 1 Three cubes each of side 5 cm are joined end to end to form a cuboid. Find the surface area of the resulting cuboid.Sol : Length of each side of cube = 5cmAccording to question,Three cubes are joined end to end to form a cuboidSo, the length of the resulting cuboid = 5 [&hellip;]<\/p>\n","protected":false},"author":302,"featured_media":625015,"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":[24],"tags":[],"boards":[],"class_list":["post-625011","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-10","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>KC Sinha: Exercise 14.1 - Mathematics Solution Class 10 Chapter 14 Surface Area and Volumes - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Question 1 Three cubes each of side 5 cm are joined end to end to form a cuboid. 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