{"id":126852,"date":"2021-09-13T17:12:56","date_gmt":"2021-09-13T11:42:56","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126852"},"modified":"2021-09-14T13:28:45","modified_gmt":"2021-09-14T07:58:45","slug":"rd-sharma-class-10-solutions-chapter-8-exercise-8-11","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-11\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.11 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126889\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.11.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.11.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.11-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11:\u00a0<\/strong>The utilization of quadratic equations to solve problems with mensuration is highlighted in this exercise. Students who are having difficulty solving problems 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>. 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 8.11<\/strong>\u00a0PDF can also be used to clear up any conceptual doubts about this exercise.<\/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-69f8d15c0098a\" 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-8-exercise-8-11\/#download-rd-sharma-class-10-solutions-chapter-8-exercise-811-free-pdf\" title=\"Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11 Free PDF\">Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11 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-8-11\/#access-answers-to-rd-sharma-solutions-class-10-maths-chapter-8-exercise-811-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.11- Important Question with Answers\">Access answers to RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.11- 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-8-11\/#faqs-on-rd-sharma-class-10-solutions-chapter-8-exercise-811\" title=\"FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11\">FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11<\/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-8-11\/#how-is-rd-sharma-class-10-solutions-chapter-8-exercise-811-helpful-for-board-exams\" title=\"How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11 helpful for board exams?\">How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11 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-8-11\/#is-the-rd-sharma-class-10-solutions-chapter-8-exercise-811-available-on-the-kopykitab-website\" title=\"Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11 available on the Kopykitab website?\">Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11 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-8-11\/#where-can-i-get-rd-sharma-class-10-solutions-chapter-8-exercise-811-free-pdf\" title=\"Where can I get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11 Free PDF?\">Where can I get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11 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-811-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11 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-8-Exercise-8.11.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-8.11.pdf\">RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-8-exercise-811-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.11- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 1.<\/strong><br \/>The perimeter of a rectangular field is 82 m and its area is 400 m\u00b2. Find the breadth of the rectangle.<br \/><strong>Solution:<\/strong><br \/>Perimeter of a rectangle field = 82 m<br \/>Length + Breadth =\u00a0822\u00a0= 41 m<br \/>Let breadth = x m<br \/>Length = (41 \u2013 x) m<br \/>According to the condition,<br \/>Area = Length x breadth<br \/>400 = x (41 \u2013 x)<br \/>\u21d2 400 = 4x \u2013 x\u00b2<br \/>\u21d2 x\u00b2 \u2013 41x + 400 = 0<br \/>\u21d2 x\u00b2 \u2013 16x \u2013 25x + 400 = 0<br \/>\u21d2 x (x \u2013 16) \u2013 25 (x \u2013 16) = 0<br \/>\u21d2 (x \u2013 16) (x \u2013 25) = 0<br \/>Either x \u2013 16 = 0, then x = 16<br \/>or x \u2013 25 = 0 then x = 25<br \/>25 &gt; 16 and length &gt; breadth<br \/>Breadth = 16 m<\/p>\n<p><strong>Question 2.<\/strong><br \/>The length of a hall is 5 m more than its breadth. If the area of the floor of the hall is 84 m\u00b2, what are the length and breadth of the hall?<br \/><strong>Solution:<\/strong><br \/>Let breadth of the hall = x m<br \/>Then length = x + 5<br \/>Area of the floor = 84 m2<br \/>Now according to the condition,<br \/>x (x + 5) = 84<br \/>\u21d2 x\u00b2 + 5x \u2013 84 = 0<br \/>\u21d2 x\u00b2 + 12x \u2013 7x \u2013 84 = 0<br \/>\u21d2 x (x + 12) \u2013 7 (x + 12) = 0<br \/>\u21d2 (x + 12) (x \u2013 7) = 0<br \/>Either x + 12 = 0, then x = -12 which is not possible being negative<br \/>or x \u2013 7 = 0, then x = 7<br \/>Breadth of the hall = 7 m and length = 7 + 5 = 12 m<\/p>\n<p><strong>Question 3.<\/strong><br \/>Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 cm\u00b2. Find the sides of the squares.<br \/><strong>Solution:<\/strong><br \/>Side of first square = x cm<br \/>and side of the second square = (x + 4) cm<br \/>According to the condition,<br \/>x\u00b2 + (x + 4)\u00b2 = 656<br \/>\u21d2 x\u00b2 + x\u00b2 + 8x + 16 = 656<br \/>\u21d2 2x\u00b2 + 8x + 16 \u2013 656 = 0<br \/>\u21d2 2x\u00b2 + 8x \u2013 640 = 0<br \/>\u21d2 x\u00b2 + 4x \u2013 320 = 0 (Dividing by 2)<br \/>\u21d2 x\u00b2 + 20x \u2013 16x \u2013 320 = 0<br \/>\u21d2 x (x + 20) \u2013 16 (x + 20) 0<br \/>\u21d2 (x + 20) (x \u2013 16) = 0<br \/>Either x + 20 = 0, then x = -20 which is not possible being negative<br \/>or x \u2013 16 = 0, then x = 16<br \/>Side of first square = 16 cm<br \/>and side of second square = 16 + 4 = 20 cm<\/p>\n<p><strong>Question 4.<\/strong><br \/>The area of a right-angled triangle is 165 m\u00b2. Determine its base and altitude if the latter exceeds the former by 7 m.<br \/><strong>Solution:<\/strong><br \/>Area of a right angled triangle = 165 m\u00b2<br \/>Let its base = x m<br \/>Then altitude = (x + 7) m<br \/>According to the condition,<br \/>12\u00a0x (x + 7) = 165<br \/>\u21d2\u00a012\u00a0(x\u00b2 + 7x) = 165<br \/>\u21d2 x\u00b2 + 7x = 330<br \/>\u21d2 x\u00b2 + 7x \u2013 330 = 0<br \/>\u21d2 x\u00b2 + 22x \u2013 15x \u2013 330 = 0<br \/>\u21d2 x (x + 22) \u2013 15 (x + 22) = 0<br \/>\u21d2 (x + 22) (x \u2013 15) = 0<br \/>Either x + 22 = 0, then x = -22 which is not possible being negative<br \/>or x \u2013 15 = 0, then x = 15<br \/>Base = 15 m<br \/>and altitude = 15 + 7 = 22 m<\/p>\n<p><strong>Question 5.<\/strong><br \/>Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m\u00b2? If so, find its length and breadth.<br \/><strong>Solution:<\/strong><br \/>Area of rectangular mango grove = 800 m\u00b2<br \/>Let breadth = x m<br \/>Then length = 2x m<br \/>According to the condition,<br \/>2x x x = 800<br \/>\u21d2 2x\u00b2 = 800<br \/>\u21d2 x\u00b2 = 400 = (\u00b120)\u00b2<br \/>Yes, it is possible,<br \/>x = 20, -20<br \/>But x = -20 is not possible being negative<br \/>Breadth = 20 m<br \/>and length = 20 x 2 = 40 m<\/p>\n<p><strong>Question 6.<\/strong><br \/>Is it possible to design a rectangular park of perimeter 80 m and area 400 m\u00b2 ? If so, find its length and breadth:<br \/><strong>Solution:<\/strong><br \/>Perimeter of rectangular park = 80 m<br \/>Length + Breadth =\u00a0802\u00a0= 40 m<br \/>Let length = x m<br \/>Them breadth = 40 \u2013 x<br \/>According to the condition,<br \/>Area = Length x Breadth<br \/>x (40 \u2013 x) = 400<br \/>\u21d2 40x \u2013 x\u00b2 = 400<br \/>\u21d2 x\u00b2 \u2013 40x + 400 = 0<br \/>\u21d2 (x \u2013 20)\u00b2 = 0<br \/>\u21d2 x \u2013 20 = 0<br \/>\u21d2 x = 20<br \/>Yes, it is possible<br \/>Length = 20 m<br \/>and breadth = 40 \u2013 x = 40 \u2013 20 = 20 m<\/p>\n<p><strong>Question 7.<\/strong><br \/>Sum of the areas of two squares is 640 m\u00b2. If the difference of their perimeters is 64 m, find the sides of the two squares.\u00a0<strong>[CBSE 2008]<\/strong><br \/><strong>Solution:<\/strong><br \/>Let side of first square = x m<br \/>and of second squares = y m<br \/>According to the given conditions,<br \/>4x \u2013 4y = 64<br \/>\u21d2 x \u2013 y = 16 \u2026.(i)<br \/>and x\u00b2 + y\u00b2 = 640 \u2026.(ii)<br \/>From (i), x = 16 + y<br \/>In (ii)<br \/>(16 + y)\u00b2 + y\u00b2 = 640<br \/>\u21d2 256 + 32y + y\u00b2 + y\u00b2 = 640<br \/>\u21d2 2y\u00b2 + 32y + 256 \u2013 640 = 0<br \/>\u21d2 y\u00b2 + 16y \u2013 192 = 0 (Dividing by 2)<br \/>\u21d2 y\u00b2 + 24y \u2013 8y \u2013 192 = 0<br \/>\u21d2 y (y + 24) \u2013 8 (y + 24) = 0<br \/>\u21d2 (y + 24)(y \u2013 8) = 0<br \/>Either y + 24 = 0, then y = -24, which is not possible as it is negative<br \/>or y \u2013 8 = 0, then y = 8<br \/>x = 16 + y = 16 + 8 = 24<br \/>Side of first square = 24 m<br \/>and side of second square = 8m<\/p>\n<p><strong>Question 8.<\/strong><br \/>Sum of the areas of two squares is 400 cm\u00b2. If the difference of their perimeters is 16 cm, find the sides of two squares.\u00a0<strong>[CBSE 2013]<\/strong><br \/><strong>Solution:<\/strong><br \/>Let perimeter of the first square = x cm<br \/>Then perimeter of second square = (x + 16) cm<br \/>Side of first square =\u00a0x4\u00a0cm<br \/>and side of second square = (x4\u00a0+ 4) cm<br \/>Sum of areas of these two squares = 400 cm\u00b2<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1755\/41710705465_25211b25eb_o.png\" alt=\"RD Sharma Class 10 Chapter 8 Quadratic Equations \" width=\"306\" height=\"334\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1746\/42560334372_605346b0cc_o.png\" alt=\"Quadratic Equations Class 10 RD Sharma \" width=\"354\" height=\"385\" \/><\/p>\n<p><strong>Question 9.<\/strong><br \/>The area of a rectangular plot is 528 m\u00b2. The length of the plot (in metres) is one metre more then twice its breadth. Find the length and the breadth of the plot.<strong>\u00a0[CBSE 2014]<\/strong><br \/><strong>Solution:<\/strong><br \/>Area of a rectangular plot = 528 m\u00b2<br \/>Let breadth = x m<br \/>Then length = (2x + 1) m<br \/>x (2x + 1) = 528 (\u2234 Area = l x b)<br \/>2x\u00b2 + x \u2013 528 = 0<br \/>\u21d2 2x\u00b2 + 33x \u2013 32x\u00b2 \u2013 528 = 0<br \/>\u21d2 x (2x + 33) \u2013 16 (2x + 33) = 0<br \/>\u21d2 (2x + 33) (x \u2013 16) = 0<br \/>Either 2x + 33 = 0 then 2x = \u2013 33 \u21d2 x =\u00a0\u2212332\u00a0but it is not possible being negative<br \/>or x \u2013 16 = 0, then x = 16<br \/>Length = 2x + 1 = 16 x 2 + 1 = 33 m<br \/>and breadth = x = 16 m<\/p>\n<p><strong>Question 10.<\/strong><br \/>In the center of a rectangular lawn of dimensions 50 m x 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m\u00b2. Find the length and breadth of the pond. <strong>[NCERT Exemplar]<\/strong><br \/><strong>Solution:<\/strong><br \/>Given that a rectangular pond has to be constructed in the center of a rectangular lawn of dimensions 50 m x 40 m. So, the distance between the pond and the lawn would be the same around the pond. Say x m.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1727\/42610380211_96d5911040_o.png\" alt=\"RD Sharma Class 10 Solutions Quadratic Equations \" width=\"271\" height=\"194\" \/><br \/>Now, length of rectangular lawn (l<sub>1<\/sub>) = 50 m<br \/>and breadth of rectangular lawn (b<sub>1<\/sub>) = 40 m<br \/>Length of rectangular pond (l<sub>2<\/sub>) = 50 \u2013 (x + x) = 50 \u2013 2x<br \/>Also, area of the grass surrounding the pond = 1184 m\u00b2<br \/>Area of rectangular lawn \u2013 Area of rectangular pond = Area of grass surrounding the pond<br \/>l<sub>1<\/sub>\u00a0x b<sub>1<\/sub>\u00a0\u2013 l<sub>2<\/sub>\u00a0x b<sub>2<\/sub>= 1184 [\u2235 area of rectangle = length x breadth]<br \/>\u21d2 50 x 40 \u2013 (50 \u2013 2x) (40 \u2013 2x) = 1184<br \/>\u21d2 2000 \u2013 (2000 \u2013 80x \u2013 100x + 4x\u00b2) = 1184<br \/>\u21d2 80x + 100x \u2013 4x\u00b2 = 1184<br \/>\u21d2 4x\u00b2 \u2013 180x + 1184 = 0<br \/>\u21d2 x\u00b2 \u2013 45x + 296 = 0<br \/>\u21d2 x\u00b2 \u2013 21x \u2013 8x + 296 = 0 [by splitting the middle term]<br \/>\u21d2 x (x \u2013 37) \u2013 8 (x \u2013 37) = 0<br \/>\u21d2 (x \u2013 37) (x \u2013 8) = 0<br \/>\u2234 x = 8<br \/>[At x = 37, length and breadth, of pond are -24 and -34, respectively but length and breadth cannot be negative. So, x = 37 cannot be possible]<br \/>Length of pond = 50 \u2013 2x = 50 \u2013 2(8) = 50 \u2013 16 = 34 m<br \/>and breadth of pond = 40 \u2013 2x = 40 \u2013 2(8) = 40 \u2013 16 = 24 m<br \/>Hence, required length and .breadth of pond are 34 m and 24 m, respectively.<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11. 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-8-exercise-811\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11<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-1631526011332\" 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-811-helpful-for-board-exams\"><\/span>How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11 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 8.11 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-1631526320696\" 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-811-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11 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 8.11 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-1631526515991\" 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-811-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11 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 8.11 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 8.11:\u00a0The utilization of quadratic equations to solve problems with mensuration is highlighted in this exercise. Students who are having difficulty solving problems can use the RD Sharma Class 10 Solutions. The RD Sharma Class 10 Solutions Chapter 8 Exercise 8.11\u00a0PDF can also be used to clear up &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.11 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-11\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.11 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":126889,"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\/126852"}],"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=126852"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126852\/revisions"}],"predecessor-version":[{"id":127406,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126852\/revisions\/127406"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126889"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126852"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126852"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126852"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}