{"id":126854,"date":"2021-09-13T17:12:51","date_gmt":"2021-09-13T11:42:51","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126854"},"modified":"2021-09-13T17:12:55","modified_gmt":"2021-09-13T11:42:55","slug":"rd-sharma-class-10-solutions-chapter-8-exercise-8-9","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-9\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.9 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126887\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.9.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.9.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.9-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9:&nbsp;<\/strong>Solving questions with ages is another application of quadratic equations. Many students struggle to comprehend the practical application. That is why we established the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a>&nbsp;at Kopykitab, which includes a step-by-step approach for solving. Students can also download the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-quadratic-equations\/\"><strong>RD Sharma Class 10 Solutions Chapter 8<\/strong><\/a><strong> Exercise 8.9<\/strong>&nbsp;PDF from the link provided 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-69d5687784a41\" 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 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-9\/#download-rd-sharma-class-10-solutions-chapter-8-exercise-89-free-pdf\" title=\"Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9 Free PDF\">Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9 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-8-exercise-8-9\/#access-answers-to-rd-sharma-solutions-class-10-maths-chapter-8-exercise-89-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.9- Important Question with Answers\">Access answers to RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.9- 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-8-exercise-8-9\/#faqs-on-rd-sharma-class-10-solutions-chapter-8-exercise-89\" title=\"FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9\">FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9<\/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-8-exercise-8-9\/#how-is-rd-sharma-class-10-solutions-chapter-8-exercise-89-helpful-for-board-exams\" title=\"How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9 helpful for board exams?\">How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9 helpful for board exams?<\/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-8-exercise-8-9\/#is-the-rd-sharma-class-10-solutions-chapter-8-exercise-89-available-on-the-kopykitab-website\" title=\"Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9 available on the Kopykitab website?\">Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9 available on the Kopykitab website?<\/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-8-exercise-8-9\/#where-can-i-get-rd-sharma-class-10-solutions-chapter-8-exercise-89-free-pdf\" title=\"Where can I get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9 Free PDF?\">Where can I get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9 Free PDF?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-8-exercise-89-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/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-8-Exercise-8.9.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-8-Exercise-8.9.pdf\">RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-8-exercise-89-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.9- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>Ashu is x years old while his mother Mrs. Veena is x\u00b2 years old. Five years hence Mrs. Veena will be three times old as Ashu. Find their present ages.<br>Solution:<br>Present age of Ashu = x years<br>and age of his mother = x\u00b2 years<br>5 years hence,<br>age of Ashu will be = (x + 5) years<br>and age of his mother = (x\u00b2 + 5) years<br>According to the question,<br>x\u00b2 + 5 = 3 (x + 5)<br>\u21d2 x\u00b2 + 5 = 3x + 15<br>\u21d2 x\u00b2 + 5 \u2013 3x \u2013 15 = 0<br>\u21d2 x\u00b2 \u2013 3x \u2013 10 = 0<br>\u21d2 x\u00b2 \u2013 5x + 2x \u2013 10 = 0<br>\u21d2 x (x \u2013 5) + 2 (x \u2013 5) = 0<br>\u21d2 (x \u2013 5) (x + 2) = 0<br>Either x \u2013 5 = 0, then x = 5<br>or x + 2 = 0, then x = -2 which is not possible being negative<br>Present age of Ashu = 5 years<br>and age of his mother = x\u00b2 = (5)\u00b2 = 25 years<\/p>\n<p>Question 2.<br>The sum of the ages of a man and his son is 45 years. Five years ago, the product of their ages was four times the man\u2019s age at that time. Find their present ages.<br>Solution:<br>Sum of ages of a man and his son = 45 years<br>Let the present age of the man = x years<br>Then age of his son = (45 \u2013 x) years<br>5 years ago,<br>Age of the man was = (x \u2013 5) years<br>and age of his son = (45 \u2013 x \u2013 5) years = (40 \u2013 x) years<br>According to the condition,<br>(x \u2013 5) (40 \u2013 x) = 4 (x \u2013 5)<br>\u21d2 40 \u2013 x = 4 [Dividing by (x \u2013 5)]<br>\u21d2 x = 40 \u2013 4 = 36<br>Age of the man = 36 years<br>and age of his son = 45 \u2013 36 = 9 years<\/p>\n<p>Question 3.<br>The product of Shikha\u2019s age five years ago and her age 8 years later is 30, her age at both times being given in years. Find her present age.<br>Solution:<br>Let present age of Shikha = x years<br>5 years ago, her age was = (x \u2013 5) years<br>and 8 years later, her age will be = (x + 8) years<br>According to the condition,<br>(x \u2013 5) (x + 8) = 30<br>\u21d2 x\u00b2 + 3x \u2013 40 = 30<br>\u21d2 x\u00b2 + 3x \u2013 40 \u2013 30 = 0<br>\u21d2 x\u00b2 + 3x \u2013 70 = 0<br>\u21d2 x\u00b2 + 10x \u2013 7x \u2013 70 = 0<br>\u21d2 x (x + 10) \u2013 7 (x + 10) = 0<br>\u21d2 (x + 10)(x \u2013 7) = 0<br>Either x + 10 = 0, then x = -10 which is not possible being negative<br>or x \u2013 7 = 0, then x = 7<br>Her present age = 7 years<\/p>\n<p>Question 4.<br>The product of Ramu\u2019s age (in years) five years ago and his age (in years) nine years later is 15. Determine Ramu\u2019s present age.<br>Solution:<br>Let present age of Ramu = x years<br>5 years ago his age was = (x \u2013 5) years<br>and 9 years later his age will be = (x + 9) years<br>According to the condition,<br>(x \u2013 5) (x + 9) = 15<br>\u21d2 x\u00b2 + 9x \u2013 5x \u2013 45 = 15<br>\u21d2 x\u00b2 + 4x \u2013 45 \u2013 15 = 0<br>\u21d2 x\u00b2 + 4x \u2013 60 = 0<br>\u21d2 x\u00b2 + 10x \u2013 6x \u2013 60 = 0<br>\u21d2 x (x + 10) \u2013 6 (x + 10) = 0<br>\u21d2 (x + 10) (x \u2013 6) = 0<br>Either x + 10 = 0, then x = -10 but it is not possible being negative<br>or x \u2013 6 = 0, then x = 6<br>Present age of Ramu = 6 years<\/p>\n<p>Question 5.<br>Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.<br>Solution:<br>Sum of ages of two friends = 20 years<br>Let present age of one friend = x years<br>Age of second friend = (20 \u2013 x) years<br>4 years ago,<br>Age of first friend = x \u2013 4<br>and age of second friend = 20 \u2013 x \u2013 4 = 16 -x<br>According to the condition,<br>(x \u2013 4) (16 \u2013 x) = 48<br>\u21d2 16x \u2013 x\u00b2 \u2013 64 + 4x = 48<br>\u21d2 \u2013 x\u00b2 + 20x \u2013 64 \u2013 48 = 0<br>\u21d2 \u2013 x\u00b2 + 20x \u2013 112 = 0<br>\u21d2 x\u00b2 \u2013 20x + 112 = 0<br>Here a = 1, b = \u2013 20, c = 112<br>Discriminant(D) = b\u00b2 \u2013 4ac = (-20)\u00b2 \u2013 4 x 1 x 112<br>= 400 \u2013 448 = \u2013 48<br>\u2234 D &lt; 0<br>Roots are not real<br>It is not possible<\/p>\n<p>Question 6.<br>A girl is twice as old as her sister. Four years hence, the product of their ages (in years) will be 160. Find their present ages. [CBSE2010]<br>Solution:<br>Let age of sister = x years<br>Then age of girl = 2x years<br>4 years hence,<br>Girl\u2019s age = 2x + 4<br>and sister\u2019s age = x + 4<br>According to the condition,<br>(2x + 4) (x + 4) = 160<br>\u21d2 2x\u00b2 + 8x + 4x + 16 = 160<br>\u21d2 2x\u00b2 + 12x + 16 \u2013 160 = 0<br>\u21d2 2x\u00b2+ 12x \u2013 144 = 0<br>\u21d2 x\u00b2 + 6x \u2013 12 = 0<br>\u21d2 x\u00b2 + 12x \u2013 6x \u2013 72 = 0<br>\u21d2 x (x + 12) \u2013 6 (x + 12) = 0<br>\u21d2 (x + 12) (x \u2013 6) = 0<br>Either x + 12 = 0, then x = \u2013 12 which is not possible being negative<br>or x \u2013 6 = 0, then x = 6<br>Age of sister = 6 years<br>and age of girl = 2x = 2 x 6 = 12 years<\/p>\n<p>Question 7.<br>The sum of the reciprocals of Rehman\u2019s ages (in years) 3 years ago and 5 years from now is&nbsp;13. Find his present age. [NCERT]<br>Solution:<br>Let the age of Rehman = x years<br>His age 3 years ago = x \u2013 3<br>and age 5 years hence = x + 5<br>According to the condition,<br><img src=\"https:\/\/farm2.staticflickr.com\/1725\/41861605274_110febd861_o.png\" alt=\"RD Sharma Class 10 Chapter 8 Quadratic Equations\" width=\"196\" height=\"194\"><br>\u21d2 x\u00b2 + 2x \u2013 15 = 6x + 6<br>\u21d2 x\u00b2 + 2x \u2013 15 \u2013 6x \u2013 6 = 0<br>\u21d2 x\u00b2 \u2013 4x \u2013 21 =0<br>\u21d2 x\u00b2 \u2013 7x + 3x \u2013 21 = 0<br>\u21d2 x (x \u2013 7) + 3 (x \u2013 7) = 0<br>\u21d2 (x \u2013 7)(x + 3) = 0<br>Either x \u2013 7 = 0, then x = 7<br>or x + 3 = 0 then x = \u2013 3 which is not possible being negative<br>x = 7<br>His present age = 7 years<\/p>\n<p>Question 8.<br>If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than 5 times her actual age. What is her age now? [NCERT Exemplar]<br>Solution:<br>Let the actual age of Zeba = x year.<br>Her age when she was 5 years younger = (x \u2013 5) years<br>Now, by given condition,<br>Square of her age = 11 more than 5 times her actual age<br>(x \u2013 5)\u00b2 = 5 x actual age + 11<br>\u21d2 (x \u2013 5)\u00b2 = 5x + 11<br>\u21d2 x\u00b2 + 25 \u2013 10x = 5x + 11<br>\u21d2 x\u00b2 \u2013 15x + 14 = 0<br>\u21d2 x\u00b2 \u2013 14x \u2013 x + 14 = 0 [by splitting the middle term]<br>\u21d2 x (x \u2013 14) \u2013 1 (x \u2013 14) = 0<br>\u21d2 (x \u2013 1) (x \u2013 14) = 0<br>\u21d2 x = 14<br>[Here, x \u2260 1 because her age is x \u2013 5. So, x \u2013 5 = 1 \u2013 5= -4 i.e., age cannot be negative]<br>Hence, required Zeba\u2019s age now is 14 years.<\/p>\n<p>Question 9.<br>At present Asha\u2019s age (in years) is 2 more than the square of her daughter Nisha\u2019s age. When Nisha grows to her mother\u2019s present age, Asha\u2019s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha. [NCERT Exemplar]<br>Solution:<br>Let Nisha\u2019s present age be x year.<br>Then, Asha\u2019s present age = x\u00b2 + 2 [by given condition]<br>Now, when Nisha grows to her mother\u2019s present age i.e., after {(x\u00b2 + 2) \u2013 x} years.<br>Then, Asha\u2019s age also increased by [(x\u00b2 + 2) \u2013 x] year.<br>Again by given condition,<br>Age of Asha = One years less than 10 times the present age of Nisha<br>(x\u00b2 + 2) + {(x\u00b2 + 2) \u2013 x} = 10x \u2013 1<br>\u21d2 2x\u00b2 \u2013 x + 4 = 10x \u2013 1<br>\u21d2 2x\u00b2 \u2013 11x + 5 = 0<br>\u21d2 2x\u00b2 \u2013 10x \u2013 x + 5 = 0<br>\u21d2 2x (x \u2013 5) \u2013 1(x \u2013 5) = 0<br>\u21d2 (x \u2013 5) (2x \u2013 1) = 0<br>\u2234 x = 5<br>[Here, x =&nbsp;12&nbsp;cannot be possible, because at x =&nbsp;12, Asha\u2019s age is 214&nbsp;years which is not possible]<br>Hence, required age of Nisha = 5 years<br>and required age of Asha = x\u00b2 + 2 = (5)\u00b2 + 2 = 25 + 2 = 27 years.<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>&nbsp;Class 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-8-exercise-89\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9<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-1631525990123\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-is-rd-sharma-class-10-solutions-chapter-8-exercise-89-helpful-for-board-exams\"><\/span>How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9 helpful for board exams?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>For self-evaluation, RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9 provides solutions with thorough descriptions as per term limits specified by the Board. Students will gain valuable experience solving these problems, allowing them to complete the assignment on time.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631526309659\" 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-8-exercise-89-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9 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 8 Exercise 8.9 are available. These solutions are created in a unique method by Kopykitab\u2019s expert faculty.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631526505072\" 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-8-exercise-89-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9 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 8 Exercise 8.9 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 8 Exercise 8.9:&nbsp;Solving questions with ages is another application of quadratic equations. Many students struggle to comprehend the practical application. That is why we established the RD Sharma Class 10 Solutions&nbsp;at Kopykitab, which includes a step-by-step approach for solving. Students can also download the RD Sharma Class 10 Solutions &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.9 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-9\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.9 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":126887,"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\/126854"}],"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=126854"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126854\/revisions"}],"predecessor-version":[{"id":127104,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126854\/revisions\/127104"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126887"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126854"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126854"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126854"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}