{"id":128520,"date":"2021-09-16T17:30:03","date_gmt":"2021-09-16T12:00:03","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=128520"},"modified":"2021-09-16T17:30:06","modified_gmt":"2021-09-16T12:00:06","slug":"rd-sharma-class-10-solutions-chapter-16-exercise-16-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-exercise-16-2\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 16 Surface Areas and Volumes Exercise 16.2 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-128530\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-16-Exercise-16.2.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-16-Exercise-16.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-16-Exercise-16.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2:\u00a0<\/strong>This exercise focuses on determining the surface areas and volumes of solid combinations. Students can use the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> to clear up any doubts they have about this chapter. Students can also download the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-surface-areas-and-volumes\/\"><strong>RD Sharma Solutions for Class 10 Maths Chapter 16<\/strong><\/a> Surface Areas And Volumes Exercise 16.2 PDF.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69ea5a1740685\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69ea5a1740685\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-exercise-16-2\/#download-rd-sharma-class-10-solutions-chapter-16-exercise-162-free-pdf\" title=\"Download RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2 Free PDF\">Download RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2 Free PDF<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-exercise-16-2\/#access-answers-to-rd-sharma-solutions-class-10-maths-chapter-16-exercise-162-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 10 Maths Chapter 16 Exercise 16.2- Important Question with Answers\">Access answers to RD Sharma Solutions Class 10 Maths Chapter 16 Exercise 16.2- Important Question with Answers<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-exercise-16-2\/#faqs-on-rd-sharma-class-10-solutions-chapter-16-exercise-162\" title=\"FAQs on RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2\">FAQs on RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-exercise-16-2\/#where-can-i-get-rd-sharma-class-10-solutions-chapter-16-exercise-162-free-pdf\" title=\"Where can I get RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2 Free PDF?\">Where can I get RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2 Free PDF?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-exercise-16-2\/#where-can-i-get-the-correct-rd-sharma-solutions-for-class-10-maths-chapter-16-exercise-162\" title=\"Where can I get the correct RD Sharma Solutions for Class 10 Maths Chapter 16 Exercise 16.2?\">Where can I get the correct RD Sharma Solutions for Class 10 Maths Chapter 16 Exercise 16.2?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-exercise-16-2\/#is-the-rd-sharma-class-10-solutions-chapter-16-exercise-162-available-on-the-kopykitab-website\" title=\"Is the RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2 available on the Kopykitab website?\">Is the RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2 available on the Kopykitab website?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-16-exercise-162-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-16-Exercise-16.2.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-16-Exercise-16.2.pdf\">RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-16-exercise-162-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 16 Exercise 16.2- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. Find the area of canvas required for the tent.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The diameter of the cylinder (also the same for cone) = 24 m.<\/p>\n<p>So, its radius (R) =\u00a024\/2 = 12 m<\/p>\n<p>The height of the Cylindrical part (H<sub>1<\/sub>) = 11m<\/p>\n<p>So, Height of the cone part (H<sub>2<\/sub>) = 16 \u2013 11 = 5 m<\/p>\n<p>Now,<\/p>\n<p>Vertex of the cone above the ground = 11 + 5 = 16 m<\/p>\n<p>Curved Surface area of the Cone (S<sub>1<\/sub>) = \u03c0RL =\u00a022\/7 \u00d7 12 \u00d7 L<\/p>\n<p>The slant height (L) is given by,<\/p>\n<p>L = \u221a(R<sup>2<\/sup>\u00a0+ H<sub>2<\/sub><sup>2<\/sup>) = \u221a(12<sup>2<\/sup>\u00a0+ 5<sup>2<\/sup>) = \u221a169<\/p>\n<p>L = 13 m<\/p>\n<p>So,<\/p>\n<p>Curved Surface Area of Cone (S<sub>1<\/sub>) =\u00a022\/7 \u00d7 12 \u00d7 13<\/p>\n<p>And,<\/p>\n<p>Curved Surface Area of Cylinder (S<sub>2<\/sub>) = 2\u03c0RH<sub>1<\/sub><\/p>\n<p>S<sub>2\u00a0<\/sub>= 2\u03c0(12)(11) m<sup>2 <\/sup><\/p>\n<p>Thus, the area of Canvas required for tent<\/p>\n<p>S = S<sub>1<\/sub>\u00a0+ S<sub>2\u00a0<\/sub>= (22\/7 \u00d7 12 \u00d7 13) + (2 \u00d7 22\/7 \u00d7 12 \u00d7 11)<\/p>\n<p>S = 490 + 829.38<\/p>\n<p>S = 1319.8 m<sup>2<\/sup><\/p>\n<p>S = 1320 m<sup>2<\/sup><\/p>\n<p>Therefore, the area of canvas required for the tent is 1320 m<sup>2<\/sup><\/p>\n<p><strong>2. A rocket is in the form of a circular cylinder closed at the lower end with a cone of the same radius attached to the top. The cylinder is of radius 2.5 m and height 21 m and the cone has the slant height 8 m. Calculate the total surface area and the volume of the rocket.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Radius of the cylindrical portion of the rocket (R) = 2.5 m<\/p>\n<p>Height of the cylindrical portion of the rocket (H) = 21 m<\/p>\n<p>Slant Height of the Conical surface of the rocket (L) = 8 m<\/p>\n<p>Curved Surface Area of the Cone (S<sub>1<\/sub>) = \u03c0RL = \u03c0(2.5)(8)= 20\u03c0<\/p>\n<p>And,<\/p>\n<p>Curved Surface Area of the Cone (S<sub>2<\/sub>) = 2\u03c0RH + \u03c0R<sup>2<\/sup><\/p>\n<p>S<sub>2\u00a0<\/sub>= (2\u03c0 \u00d7 2.5\u00a0\u00d7 21) + \u03c0 (2.5)<sup>2<\/sup><\/p>\n<p>S<sub>2<\/sub>\u00a0= (\u03c0 \u00d7 105) + (\u03c0 \u00d7 6.25)<\/p>\n<p>Thus, the total curved surface area S is<\/p>\n<p>S = S<sub>1<\/sub>\u00a0+ S<sub>2<\/sub><\/p>\n<p>S = (\u03c020) + (\u03c0105) + (\u03c06.25)<\/p>\n<p>S = (22\/7)(20 + 105 + 6.25) = 22\/7 x 131.25<\/p>\n<p>S = 412.5 m<sup>2<\/sup><\/p>\n<p>Therefore, the total Surface Area of the Conical Surface = 412.5 m<sup>2<\/sup><\/p>\n<p>Now, calculating the volume of the rocket<\/p>\n<p>Volume of the conical part of the rocket (V<sub>1<\/sub>) =\u00a01\/3 \u00d7 22\/7 \u00d7 R<sup>2<\/sup>\u00a0\u00d7 h<\/p>\n<p>V<sub>1<\/sub>\u00a0=\u00a01\/3 \u00d7 22\/7 \u00d7 (2.5)<sup>2<\/sup>\u00a0\u00d7 h<\/p>\n<p>Let, h be the height of the conical portion in the rocket.<\/p>\n<p>We know that,<\/p>\n<p>L<sup>2<\/sup>\u00a0= R<sup>2\u00a0<\/sup>+ h<sup>2<\/sup><\/p>\n<p>h<sup>2<\/sup>\u00a0= L<sup>2\u00a0<\/sup>\u2013 R<sup>2<\/sup>\u00a0= 8<sup>2<\/sup>\u00a0\u2013 2.5<sup>2<\/sup><\/p>\n<p>h = 7.6 m<\/p>\n<p>Using the value of h, we will get<\/p>\n<p>Volume of the conical part (V<sub>1<\/sub>) =\u00a01\/3 \u00d7 22\/7 \u00d7 2.5<sup>2<\/sup>\u00a0\u00d7 7.6\u00a0m<sup>2 <\/sup>\u00a0= 49.67 m<sup>2 <\/sup><\/p>\n<p>Next,<\/p>\n<p>Volume of the Cylindrical Portion (V<sub>2<\/sub>) = \u03c0R<sup>2<\/sup>h<\/p>\n<p>V<sub>2<\/sub>\u00a0=\u00a022\/7 \u00d7 2.5<sup>2<\/sup>\u00a0\u00d7 21 = 412.5 m<sup>2<\/sup><\/p>\n<p>Thus, the total volume of the rocket = V<sub>1<\/sub>\u00a0+ V<sub>2<\/sub><\/p>\n<p>V = 412.5 + 49.67 = 462.17 m<sup>2<\/sup><\/p>\n<p>Hence, the total volume of the Rocket is 462.17 m<sup>2\u00a0<\/sup><\/p>\n<p><strong>3. A tent of height 77 dm is in the form of a right circular cylinder of diameter 36 m and height 44 dm surmounted by a right circular cone. Find the cost of the canvas at Rs. 3.50 per m<sup>2<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Height of the tent = 77 dm<\/p>\n<p>Height of a surmounted cone = 44 dm<\/p>\n<p>Height of the Cylindrical Portion = Height of the tent \u2013 Height of the surmounted Cone<\/p>\n<p>= 77 \u2013 44<\/p>\n<p>= 33 dm = 3.3 m<\/p>\n<p>And, given diameter of the cylinder (d) = 36 m<\/p>\n<p>So, its radius (r) of the cylinder =\u00a036\/2 = 18 m<\/p>\n<p>Let\u2019s consider L as the slant height of the cone.<\/p>\n<p>Then, we know that<\/p>\n<p>L<sup>2<\/sup>\u00a0= r<sup>2<\/sup>\u00a0+ h<sup>2<\/sup><\/p>\n<p>L<sup>2<\/sup>\u00a0= 18<sup>2<\/sup>\u00a0+ 3.3<sup>2<\/sup><\/p>\n<p>L<sup>2\u00a0<\/sup>= 324 + 10.89<\/p>\n<p>L<sup>2<\/sup>\u00a0= 334.89<\/p>\n<p>L = 18.3 m<\/p>\n<p>Thus, slant height of the cone (L) = 18.3 m<\/p>\n<p>Now, the Curved Surface area of the Cylinder (S<sub>1<\/sub>) = 2\u03c0rh<\/p>\n<p>S<sub>1<\/sub>\u00a0= 2\u03c0 (184.4) m<sup>2\u00a0<\/sup><\/p>\n<p>And, the Curved Surface area of the cone (S<sub>2<\/sub>) = \u03c0rL<\/p>\n<p>S<sub>2\u00a0<\/sub>= \u03c0 \u00d7 18 \u00d7 18.3 m<sup>2\u00a0<\/sup><\/p>\n<p>So, the total curved surface of the tent (S) = S<sub>1\u00a0<\/sub>+ S<sub>2<\/sub><\/p>\n<p>S = S<sub>1<\/sub>\u00a0+ S<sub>2<\/sub><\/p>\n<p>S = (2\u03c018 \u00d7\u00a04.4) + (\u03c018 \u00d7\u00a018.3)<\/p>\n<p>S = 1533.08 m<sup>2<\/sup><\/p>\n<p>Hence, the total Curved Surface Area (S) = 1533.08 m<sup>2<\/sup><\/p>\n<p>Next,<\/p>\n<p>The cost of 1 m<sup>2<\/sup>\u00a0canvas = Rs 3.50<\/p>\n<p>So, 1533.08 m<sup>2<\/sup>\u00a0of canvas will cost = Rs (3.50 x 1533.08)<\/p>\n<p>= Rs 5365.8<\/p>\n<p><strong>4. A toy is in the form of a cone surmounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm, respectively. Determine the surface area of the toy.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given that,<\/p>\n<p>The height of the cone (h) = 4 cm<\/p>\n<p>Diameter of the cone (d) = 6 cm<\/p>\n<p>So, its radius (r) = 3<\/p>\n<p>Let, \u2018l\u2019 be the slant height of cone.<\/p>\n<p>Then, we know that<\/p>\n<p>l<sup>2<\/sup>\u00a0= r<sup>2<\/sup>\u00a0+ h<sup>2<\/sup><\/p>\n<p>l<sup>2<\/sup>\u00a0= 3<sup>2<\/sup>\u00a0+ 4<sup>2<\/sup>\u00a0= 9 + 16 = 25<\/p>\n<p>l = 5 cm<\/p>\n<p>Hence, the slant height of the cone (l) = 5 cm<\/p>\n<p>So, the curved surface area of the cone (S<sub>1<\/sub>) =\u00a0\u03c0rl<\/p>\n<p>S<sub>1<\/sub>\u00a0=\u00a0\u03c0(3)(5)<\/p>\n<p>S<sub>1<\/sub>\u00a0= 47.1 cm<sup>2<\/sup><\/p>\n<p>And, the curved surface area of the hemisphere (S<sub>2<\/sub>) =\u00a02\u03c0r<sup>2<\/sup><\/p>\n<p>S<sub>2<\/sub>\u00a0=\u00a02\u03c0(3)<sup>2<\/sup><\/p>\n<p>S<sub>2<\/sub>\u00a0= 56.23 cm<sup>2<\/sup><\/p>\n<p>So, the total surface area (S) = S<sub>1\u00a0<\/sub>+ S<sub>2<\/sub><\/p>\n<p>S = 47.1 + 56.23<\/p>\n<p>S = 103.62 cm<sup>2<\/sup><\/p>\n<p>Therefore, the curved surface area of the toy is 103.62 cm<sup>2<\/sup><\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-16-exercise-162\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631792782880\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-rd-sharma-class-10-solutions-chapter-16-exercise-162-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2 Free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can get RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631792916619\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-the-correct-rd-sharma-solutions-for-class-10-maths-chapter-16-exercise-162\"><\/span>Where can I get the correct RD Sharma Solutions for Class 10 Maths Chapter 16 Exercise 16.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>On the Kopykitab website, students can find appropriate and accurate RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2. The best reference materials are RD Sharma textbooks, which give students and teachers a wide choice of sample questions to answer. The solutions are carefully crafted to ensure that students grasp the concepts and perform well on their board exams.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631792999692\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-the-rd-sharma-class-10-solutions-chapter-16-exercise-162-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2 available on the Kopykitab website?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, the PDFs of RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2 are available. These solutions are created in a unique method by Kopykitab\u2019s expert faculty.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 16 Exercise 16.2:\u00a0This exercise focuses on determining the surface areas and volumes of solid combinations. Students can use the RD Sharma Class 10 Solutions to clear up any doubts they have about this chapter. Students can also download the RD Sharma Solutions for Class 10 Maths Chapter 16 Surface &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 16 Surface Areas and Volumes Exercise 16.2 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-exercise-16-2\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 16 Surface Areas and Volumes Exercise 16.2 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":128530,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/128520"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=128520"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/128520\/revisions"}],"predecessor-version":[{"id":128576,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/128520\/revisions\/128576"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/128530"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=128520"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=128520"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=128520"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}