{"id":125338,"date":"2023-09-05T09:08:00","date_gmt":"2023-09-05T03:38:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125338"},"modified":"2023-11-02T10:03:40","modified_gmt":"2023-11-02T04:33:40","slug":"rd-sharma-class-9-solutions-chapter-15-vsaqs","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-15-vsaqs\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 15 VSAQS (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-127699\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-15-VSAQS.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 15 VSAQS\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-15-VSAQS.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-15-VSAQS-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 15 VSAQS&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:268732,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;14&quot;:[null,2,0],&quot;15&quot;:&quot;Arial&quot;,&quot;21&quot;:1}\"><strong>RD Sharma Class 9 Solutions Chapter 15 VSAQS<\/strong>: These solutions are created by our experts in a detailed manner. This exercise helps students to revise all the concepts they have studied in this chapter. They will help you during the class 9 exam preparation. You can get direct access to <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 15<\/a>&nbsp;VSAQS from the below article.&nbsp;<\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-for-maths\/\" target=\"_blank\" rel=\"noopener\"><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Maths Solutions&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Maths Solutions<\/span><\/a><\/li>\n<\/ul>\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-69da25b89a4bb\" 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-69da25b89a4bb\"><\/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-9-solutions-chapter-15-vsaqs\/#access-rd-sharma-class-9-solutions-chapter-15-vsaqs\" title=\"Access RD Sharma Class 9 Solutions Chapter 15 VSAQS\">Access RD Sharma Class 9 Solutions Chapter 15 VSAQS<\/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-9-solutions-chapter-15-vsaqs\/#faq-rd-sharma-class-9-solutions-chapter-15-vsaqs\" title=\"FAQ: RD Sharma Class 9 Solutions Chapter 15 VSAQS\">FAQ: RD Sharma Class 9 Solutions Chapter 15 VSAQS<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-15-vsaqs\/#can-i-access-rd-sharma-class-9-solutions-chapter-15-vsaqs-on-my-smartphone\" title=\"Can I access RD Sharma Class 9 Solutions Chapter 15 VSAQS on my smartphone?\">Can I access RD Sharma Class 9 Solutions Chapter 15 VSAQS on my smartphone?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-15-vsaqs\/#what-are-the-benefits-of-studying-rd-sharma-class-9-solutions\" title=\"What are the benefits of studying RD Sharma Class 9 Solutions?\">What are the benefits of studying RD Sharma Class 9 Solutions?<\/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-9-solutions-chapter-15-vsaqs\/#is-rd-sharma-enough-for-class-12-maths\" title=\"Is RD Sharma enough for Class 12 Maths?\">Is RD Sharma enough for Class 12 Maths?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-class-9-solutions-chapter-15-vsaqs\"><\/span>Access <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 15 VSAQS&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:268732,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;14&quot;:[null,2,0],&quot;15&quot;:&quot;Arial&quot;,&quot;21&quot;:1}\">RD Sharma Class 9 Solutions Chapter 15 VSAQS<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1: If ABC and BDE are two equilateral triangles such that D is the mid-point of BC, then find ar(\u25b3ABC): ar(\u25b3BDE).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: ABC and BDE are two equilateral triangles.<\/p>\n<p>We know, the area of an equilateral triangle = \u221a3\/4 (side)<sup>2<\/sup><\/p>\n<p>Let \u201ca\u201d be the side measure of the given triangle.<\/p>\n<p>Find ar(\u25b3ABC):<\/p>\n<p>ar(\u25b3ABC) = \u221a3\/4 (a)<sup>&nbsp;2<\/sup><\/p>\n<p>Find ar(\u25b3BDE):<\/p>\n<p>ar(\u25b3BDE) = \u221a3\/4 (a\/2)<sup>&nbsp;2<\/sup><\/p>\n<p>(D is the mid-point of BC)<\/p>\n<p>Now,<\/p>\n<p>ar(\u25b3ABC) : ar(\u25b3BDE)<\/p>\n<p>or \u221a3\/4 (a)<sup>&nbsp;2<\/sup>&nbsp;: \u221a3\/4 (a\/2)<sup>&nbsp;2<\/sup><\/p>\n<p>or 1 : 1\/4<\/p>\n<p>or 4:1<\/p>\n<p>This implies, ar(\u25b3ABC) : ar(\u25b3BDE) = 4:1<\/p>\n<p><strong>Question 2: In the figure, ABCD is a rectangle in which CD = 6 cm, AD = 8 cm. Find the area of parallelogram CDEF.<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 15 ex vsaqs question 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-15-ex-vsaqs-questi.png\" alt=\"RD sharma class 9 maths chapter 15 ex vsaqs question 2\" width=\"282\" height=\"151\"><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>ABCD is a rectangle, where CD = 6 cm and AD = 8 cm (Given)<\/p>\n<p>From figure: Parallelogram CDEF and rectangle ABCD on the same base and between the same parallels, which means both have equal areas.<\/p>\n<p>Area of parallelogram CDEF = Area of rectangle ABCD \u2026.(1)<\/p>\n<p>Area of rectangle ABCD = CD x AD = 6 x 8 cm<sup>2&nbsp;<\/sup>= 48 cm<sup>2<\/sup><\/p>\n<p>Equation (1) =&gt; Area of parallelogram CDEF = 48 cm<sup>2<\/sup>.<\/p>\n<p><strong>Question 3: In the figure, find the area of \u0394GEF.<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 15 ex vsaqs question 3\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-15-ex-vsaqs-questi-1.png\" alt=\"RD sharma class 9 maths chapter 15 ex vsaqs question 3\" width=\"288\" height=\"154\"><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>From figure:<\/p>\n<p>Parallelogram CDEF and rectangle ABCD are on the same base and between the same parallels, which means both have equal areas.<\/p>\n<p>Area of CDEF = Area of ABCD = 8 x 6 cm<sup>2<\/sup>= 48 cm<sup>2<\/sup><\/p>\n<p>Again,<\/p>\n<p>Parallelogram CDEF and triangle EFG are on the same base and between the same parallels, then<\/p>\n<p>Area of a triangle = \u00bd(Area of parallelogram)<\/p>\n<p>In this case,<\/p>\n<p>Area of a triangle EFG = \u00bd(Area of parallelogram CDEF) = 1\/2(48 cm<sup>2<\/sup>) = 24 cm<sup>2<\/sup>.<\/p>\n<p><strong>Question 4: In the figure, ABCD is a rectangle with sides AB = 10 cm and AD = 5 cm. Find the area of \u0394EFG.<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 15 ex vsaqs question 4\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-15-ex-vsaqs-questi-2.png\" alt=\"RD sharma class 9 maths chapter 15 ex vsaqs question 4\" width=\"309\" height=\"165\"><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>From figure:<\/p>\n<p>Parallelogram ABEF and rectangle ABCD are on the same base and between the same parallels, which means both have equal areas.<\/p>\n<p>Area of ABEF = Area of ABCD = 10 x 5 cm<sup>2<\/sup>= 50 cm<sup>2<\/sup><\/p>\n<p>Again,<\/p>\n<p>Parallelogram ABEF and triangle EFG are on the same base and between the same parallels, then<\/p>\n<p>Area of a triangle = \u00bd(Area of parallelogram)<\/p>\n<p>In this case,<\/p>\n<p>Area of a triangle EFG = \u00bd(Area of parallelogram ABEF) = 1\/2(50 cm<sup>2<\/sup>) = 25 cm<sup>2<\/sup>.<\/p>\n<p><strong>Question 5: PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then find ar(\u25b3RAS).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>PQRS is a rectangle with PS = 5 cm and PR = 13 cm (Given)<\/p>\n<p>In \u25b3PSR:<\/p>\n<p>Using Pythagoras&#8217; theorem,<\/p>\n<p>SR<sup>2<\/sup>&nbsp;= PR<sup>2<\/sup>&nbsp;\u2013 PS<sup>2<\/sup>&nbsp;= (13)<sup>&nbsp;2<\/sup>&nbsp;\u2013 (5)<sup>&nbsp;2<\/sup>&nbsp;= 169 \u2013 25 = 114<\/p>\n<p>or SR = 12<\/p>\n<p>Now,<\/p>\n<p>Area of \u25b3RAS = 1\/2 x SR x PS<\/p>\n<p>= 1\/2 x 12 x 5<\/p>\n<p>= 30<\/p>\n<p>Therefore, the Area of \u25b3RAS is 30 cm<sup>2<\/sup>.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"faq-rd-sharma-class-9-solutions-chapter-15-vsaqs\"><\/span>FAQ: <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 15 VSAQS&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:268732,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;14&quot;:[null,2,0],&quot;15&quot;:&quot;Arial&quot;,&quot;21&quot;:1}\">RD Sharma Class 9 Solutions Chapter 15 VSAQS<\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631088249794\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-rd-sharma-class-9-solutions-chapter-15-vsaqs-on-my-smartphone\"><\/span>Can I access RD Sharma Class 9 Solutions Chapter 15 VSAQS on my smartphone?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can access RD Sharma Class 9 Solutions Chapter 15 VSAQS on any device.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631088364811\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-studying-rd-sharma-class-9-solutions\"><\/span>What are the benefits of studying RD Sharma Class 9 Solutions?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>By practising RD Sharma Class 9 Solutions, students can earn higher academic grades. Our experts solve these solutions with utmost accuracy to help students in their studies.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631088366130\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-enough-for-class-12-maths\"><\/span>Is RD Sharma enough for Class 12 Maths?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>RD Sharma is a good book that gives you thousands of questions to practice.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 15 VSAQS: These solutions are created by our experts in a detailed manner. This exercise helps students to revise all the concepts they have studied in this chapter. They will help you during the class 9 exam preparation. You can get direct access to RD Sharma Class 9 Solutions &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 15 VSAQS (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-15-vsaqs\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 15 VSAQS (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":127699,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[76786,76825],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125338"}],"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\/244"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=125338"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125338\/revisions"}],"predecessor-version":[{"id":501110,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125338\/revisions\/501110"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/127699"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125338"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125338"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125338"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}