{"id":128519,"date":"2021-09-16T17:30:07","date_gmt":"2021-09-16T12:00:07","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=128519"},"modified":"2021-09-16T17:30:11","modified_gmt":"2021-09-16T12:00:11","slug":"rd-sharma-class-10-solutions-chapter-16-exercise-16-3","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-exercise-16-3\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 16 Surface Areas and Volumes Exercise 16.3 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-128531\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-16-Exercise-16.3.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 16 Exercise 16.3\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-16-Exercise-16.3.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-16-Exercise-16.3-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 16 Exercise 16.3:\u00a0<\/strong>This exercise includes problems on the volume and surface area of a frustum of a right circular cone. Students can get the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> prepared by Kopykitab&#8217;s experts for the primary objective of reinforcing ideas and problem-solving skills. Students can also get 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 Surface Areas And Volumes<\/strong><\/a> Exercise 16.3 PDF by clicking on the link below.<\/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-69d095d2b09ed\" 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\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-exercise-16-3\/#download-rd-sharma-class-10-solutions-chapter-16-exercise-163-free-pdf\" title=\"Download RD Sharma Class 10 Solutions Chapter 16 Exercise 16.3 Free PDF\">Download RD Sharma Class 10 Solutions Chapter 16 Exercise 16.3 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-16-3\/#access-answers-to-rd-sharma-solutions-class-10-maths-chapter-16-exercise-163-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 10 Maths Chapter 16 Exercise 16.3- Important Question with Answers\">Access answers to RD Sharma Solutions Class 10 Maths Chapter 16 Exercise 16.3- 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-16-3\/#faqs-on-rd-sharma-class-10-solutions-chapter-16-exercise-163\" title=\"FAQs on RD Sharma Class 10 Solutions Chapter 16 Exercise 16.3\">FAQs on RD Sharma Class 10 Solutions Chapter 16 Exercise 16.3<\/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-16-3\/#where-can-i-get-rd-sharma-class-10-solutions-chapter-16-exercise-163-free-pdf\" title=\"Where can I get RD Sharma Class 10 Solutions Chapter 16 Exercise 16.3 Free PDF?\">Where can I get RD Sharma Class 10 Solutions Chapter 16 Exercise 16.3 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-16-3\/#where-can-i-get-the-correct-rd-sharma-solutions-for-class-10-maths-chapter-16-exercise-163\" title=\"Where can I get the correct RD Sharma Solutions for Class 10 Maths Chapter 16 Exercise 16.3?\">Where can I get the correct RD Sharma Solutions for Class 10 Maths Chapter 16 Exercise 16.3?<\/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-16-3\/#is-the-rd-sharma-class-10-solutions-chapter-16-exercise-163-available-on-the-kopykitab-website\" title=\"Is the RD Sharma Class 10 Solutions Chapter 16 Exercise 16.3 available on the Kopykitab website?\">Is the RD Sharma Class 10 Solutions Chapter 16 Exercise 16.3 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-163-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 16 Exercise 16.3 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-16.3.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-16.3.pdf\">RD Sharma Class 10 Solutions Chapter 16 Exercise 16.3<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-16-exercise-163-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 16 Exercise 16.3- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. A bucket has top and bottom diameters of 40 cm and 20 cm respectively. Find the volume of the bucket if its depth is 12 cm.\u00a0Also, find the cost of tin sheet used for making the bucket at the rate of Rs 1.20 per\u00a0dm<sup>2<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Diameter to top of bucket = 40 cm<\/p>\n<p>So, the radius (r<sub>1<\/sub>) =\u00a040\/2 = 20 cm<\/p>\n<p>Diameter of bottom part of the bucket = 20 cm<\/p>\n<p>So, the radius (r<sub>2<\/sub>) = 30\/2 = 10cm<\/p>\n<p>Depth of the bucket (h) = 12 cm<\/p>\n<p>Volume of the bucket = 1\/3 \u03c0(r<sub>2<\/sub><sup>2<\/sup>+ r<sub>1<\/sub><sup>2<\/sup>\u00a0+ r<sub>1\u00a0<\/sub>r<sub>2\u00a0<\/sub>)h<\/p>\n<p>=\u00a0\u03c0\/3(20<sup>2<\/sup>\u00a0+ 10<sup>2<\/sup>\u00a0+ 20 \u00d7 10)12<\/p>\n<p>=\u00a08800 cm<sup>3<\/sup><\/p>\n<p>Now,<\/p>\n<p><img class=\"alignnone size-full wp-image-128567\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a2-1.png\" alt=\"a2\" width=\"470\" height=\"264\" \/><\/p>\n<p>Given that the cost of tin sheet used for making bucket per\u00a0dm<sup>2<\/sup>\u00a0= Rs 1.20<\/p>\n<p>So, the total cost for\u00a017.87dm<sup>2<\/sup>\u00a0=\u00a01.20 \u00d7 17.87\u00a0= Rs 21.40<\/p>\n<p>Therefore, the cost of tin sheet used for making the bucket is Rs 21.40<\/p>\n<p><strong>2. A frustum of a right circular cone has a diameter of base 20 cm, of top 12 cm and height 3 cm. Find the area of its whole surface and volume.\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Base diameter of cone\u00a0(d<sub>1<\/sub>) = 20 cm<\/p>\n<p>So the radius\u00a0(r<sub>1<\/sub>)\u00a0=\u00a020\/2 cm = 10 cm<\/p>\n<p>Top diameter of Cone\u00a0(d<sub>2<\/sub>)\u00a0= 12 cm<\/p>\n<p>So, the radius\u00a0(r<sub>2<\/sub>)\u00a0=\u00a012\/2 cm = 6 cm<\/p>\n<p>Height of the cone (h) = 3 cm<\/p>\n<p>Volume of the frustum of a right circular cone = 1\/3 \u03c0(r<sub>2<\/sub><sup>2<\/sup>+ r<sub>1<\/sub><sup>2<\/sup>\u00a0+ r<sub>1\u00a0<\/sub>r<sub>2\u00a0<\/sub>)h<\/p>\n<p>=\u00a0\u03c0\/3(10<sup>2<\/sup>\u00a0+ 6<sup>2\u00a0<\/sup>+ 10 \u00d7 6)<sup>3<\/sup><\/p>\n<p>= 616\u00a0cm<sup>3<\/sup><\/p>\n<p>Let \u2018L\u2019 be the slant height of cone, then we know that<\/p>\n<p>L = \u221a(r<sub>1<\/sub>\u00a0\u2013 r2<sub>1<\/sub>)<sup>2<\/sup>\u00a0+ h<sup>2<\/sup><\/p>\n<p>L = \u221a(10 \u2013 6)<sup>2<\/sup>\u00a0+ 3<sup>2<\/sup><\/p>\n<p>L = \u221a(16 + 9)<\/p>\n<p>L = 5cm<\/p>\n<p>So, the slant height of cone\u00a0(L) = 5 cm<\/p>\n<p>Thus,<\/p>\n<p>Total surface area of the frustum = \u03c0(r<sub>1<\/sub>\u00a0+ r<sub>2<\/sub>) x L + \u03c0 r<sub>1<\/sub><sup>2<\/sup>\u00a0+ \u03c0 r<sub>2<\/sub><sup>2<\/sup><\/p>\n<p>=\u00a0\u03c0(10 + 6) \u00d7 5 + \u03c0 \u00d7 10<sup>2<\/sup>\u00a0+ \u03c0 \u00d7 6<sup>2<\/sup><\/p>\n<p>=\u00a0\u03c0(80 + 100 + 36)<\/p>\n<p>=\u00a0\u03c0(216)<\/p>\n<p>=\u00a0678.85 cm<sup>2<\/sup><\/p>\n<p><strong>3. The slant height of the frustum of a cone is 4 cm and the perimeters of its circular ends are 18 cm and 6 cm.\u00a0Find the curved surface of the frustum.\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Slant height of frustum of cone (l) = 4 cm<\/p>\n<p>Let ratio of the top and bottom circles be\u00a0r<sub>1<\/sub>\u00a0and\u00a0r<sub>2<\/sub><\/p>\n<p>And given perimeters of its circular ends as 18 cm and 6 cm<\/p>\n<p>\u27f9\u00a02\u03c0r<sub>1<\/sub>\u00a0= 18 cm; 2\u03c0r<sub>2<\/sub>\u00a0= 6 cm<\/p>\n<p>\u27f9\u00a0\u03c0r<sub>1<\/sub>= 9 cm and \u03c0r<sub>2<\/sub>\u00a0= 3 cm<\/p>\n<p>We know that,<\/p>\n<p>Curved surface area of frustum of a cone =\u00a0\u03c0(r<sub>1<\/sub>\u00a0+ r<sub>2<\/sub>)l<\/p>\n<p>=\u00a0\u03c0(r<sub>1\u00a0<\/sub>+ r<sub>2<\/sub>)l<\/p>\n<p>=\u00a0(\u03c0r<sub>1<\/sub>+\u03c0r<sub>2<\/sub>)l\u00a0= (9 + 3)\u00a0\u00d7 4\u00a0= (12)\u00a0\u00d7 4\u00a0=\u00a048 cm<sup>2<\/sup><\/p>\n<p>Therefore, the curved surface area of the frustum =\u00a048 cm<sup>2<\/sup><\/p>\n<p><strong>4. The perimeters of the ends of a frustum of a right circular cone are 44 cm and 33 cm. If the height of the frustum be 16 cm, find its volume, the slant surface and the total surface.\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Perimeter of the upper end = 44 cm<\/p>\n<p>2 \u03c0 r<sub>1<\/sub>\u00a0= 44<\/p>\n<p>2(22\/7) r<sub>1<\/sub>\u00a0= 44<\/p>\n<p>r<sub>1\u00a0<\/sub>= 7 cm<\/p>\n<p>Perimeter of the lower end = 33 cm<\/p>\n<p>2 \u03c0 r<sub>2<\/sub>\u00a0= 33<\/p>\n<p>2(22\/7) r<sub>2<\/sub>\u00a0= 33<\/p>\n<p>r<sub>2\u00a0<\/sub>= 21\/4 cm<\/p>\n<p>Now,<\/p>\n<p>Let the slant height of the frustum of a right circular cone be L<\/p>\n<p><img class=\"alignnone size-full wp-image-128566\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a1-1.png\" alt=\"a1\" width=\"185\" height=\"68\" \/><\/p>\n<p>L = 16.1 cm<\/p>\n<p>So, the curved surface area of the frustum cone =\u00a0\u03c0(r<sub>1<\/sub>\u00a0+ r<sub>2<\/sub>)l<\/p>\n<p>=\u00a0\u03c0(7 + 5.25)16.1<\/p>\n<p>Curved surface area of the frustum cone = 619.65\u00a0cm<sup>3<\/sup><\/p>\n<p>Next,<\/p>\n<p>The volume of the frustum cone = 1\/3 \u03c0(r<sub>2<\/sub><sup>2<\/sup>+ r<sub>1<\/sub><sup>2<\/sup>\u00a0+ r<sub>1\u00a0<\/sub>r<sub>2\u00a0<\/sub>)h<\/p>\n<p>= 1\/3 \u03c0(7<sup>2<\/sup>+ 5.25<sup>2<\/sup>\u00a0+ (7) (5.25)) x 16<\/p>\n<p>= 1898.56\u00a0cm<sup>3<\/sup><\/p>\n<p>Thus, volume of the cone = 1898.56\u00a0cm<sup>3<\/sup><\/p>\n<p>Finally, the total surface area of the frustum cone<\/p>\n<p>= \u03c0(r<sub>1<\/sub>\u00a0+ r<sub>2<\/sub>) x L + \u03c0 r<sub>1<\/sub><sup>2<\/sup>\u00a0+ \u03c0 r<sub>2<\/sub><sup>2<\/sup><\/p>\n<p>= \u03c0(7 + 5.25) \u00d7 16.1 + \u03c07<sup>2\u00a0<\/sup>+ \u03c05.25<sup>2<\/sup><\/p>\n<p>=\u00a0\u03c0(7 + 5.25) \u00d7 16.1 + \u03c0(7<sup>2<\/sup>\u00a0+ 5.25<sup>2<\/sup>)\u00a0= 860.27\u00a0cm<sup>2<\/sup><\/p>\n<p>Therefore, the total surface area of the frustum cone is 860.27\u00a0cm<sup>2<\/sup><\/p>\n<p><strong>5. If the radii of the circular ends of a conical bucket which is 45 cm high be 28 cm and 7 cm, find the capacity of the bucket.\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Height of the conical bucket = 45 cm<\/p>\n<p>Radii of the 2 circular ends of the conical bucket are 28 cm and 7 cm<\/p>\n<p>So, r<sub>1<\/sub>\u00a0= 28 cm\u00a0r<sub>2<\/sub>\u00a0= 7 cm<\/p>\n<p>Volume of the conical bucket = 1\/3 \u03c0(r<sub>1<\/sub><sup>2<\/sup>+ r<sub>2<\/sub><sup>2<\/sup>\u00a0+ r<sub>1\u00a0<\/sub>r<sub>2\u00a0<\/sub>)h<\/p>\n<p>=\u00a01\/3 \u03c0(28<sup>2<\/sup>\u00a0+ 7<sup>2<\/sup>\u00a0+ 28 \u00d7 7)45\u00a0=\u00a015435\u03c0<\/p>\n<p>Therefore, the volume\/ capacity of the bucket is 48510\u00a0cm<sup>3<\/sup>.<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 16 Exercise 16.3. 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-163\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 16 Exercise 16.3<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-1631792790072\" 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-163-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 16 Exercise 16.3 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 16.3 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631792921942\" 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-163\"><\/span>Where can I get the correct RD Sharma Solutions for Class 10 Maths Chapter 16 Exercise 16.3?<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 Solutions for Class 10 Maths Chapter 16 Exercise 16.3. 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-1631793004873\" 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-163-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 16 Exercise 16.3 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 16.3 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 16.3:\u00a0This exercise includes problems on the volume and surface area of a frustum of a right circular cone. Students can get the RD Sharma Class 10 Solutions prepared by Kopykitab&#8217;s experts for the primary objective of reinforcing ideas and problem-solving skills. Students can also get the RD &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 16 Surface Areas and Volumes Exercise 16.3 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-16-exercise-16-3\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 16 Surface Areas and Volumes Exercise 16.3 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":128531,"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\/128519"}],"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=128519"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/128519\/revisions"}],"predecessor-version":[{"id":128580,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/128519\/revisions\/128580"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/128531"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=128519"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=128519"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=128519"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}