{"id":126853,"date":"2021-09-13T17:12:54","date_gmt":"2021-09-13T11:42:54","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126853"},"modified":"2021-09-13T17:12:57","modified_gmt":"2021-09-13T11:42:57","slug":"rd-sharma-class-10-solutions-chapter-8-exercise-8-10","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-10\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.10 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126888\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.10.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.10.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.10-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10:&nbsp;<\/strong>In this exercise, you will learn how to use quadratic equations to solve geometry problems. 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>&nbsp;to gain a better understanding of how to solve questions in this chapter. Students can also utilise the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-quadratic-equations\/\"><strong>RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations<\/strong><\/a> Exercise 8.10 PDF as a reference when tackling the problems.<\/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-69d7a1987809f\" 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-8-10\/#download-rd-sharma-class-10-solutions-chapter-8-exercise-810-free-pdf\" title=\"Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10 Free PDF\">Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10 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-8-10\/#access-answers-to-rd-sharma-solutions-class-10-maths-chapter-8-exercise-810-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.10- Important Question with Answers\">Access answers to RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.10- 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-8-10\/#faqs-on-rd-sharma-class-10-solutions-chapter-8-exercise-810\" title=\"FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10\">FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10<\/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-8-10\/#how-is-rd-sharma-class-10-solutions-chapter-8-exercise-810-helpful-for-board-exams\" title=\"How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10 helpful for board exams?\">How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10 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-8-10\/#is-the-rd-sharma-class-10-solutions-chapter-8-exercise-810-available-on-the-kopykitab-website\" title=\"Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10 available on the Kopykitab website?\">Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10 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-8-10\/#where-can-i-get-rd-sharma-class-10-solutions-chapter-8-exercise-810-free-pdf\" title=\"Where can I get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10 Free PDF?\">Where can I get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10 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-810-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10 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-8.10.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-8.10.pdf\">RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-8-exercise-810-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.10- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides.<br>Solution:<br>Length of the hypotenuse of a let right \u2206ABC = 25 cm<br>Let the length of one of the other two sides = x cm<br>Then other side = x + 5 cm<br><img src=\"https:\/\/farm2.staticflickr.com\/1758\/41709657135_7751294f7f_o.png\" alt=\"RD Sharma Class 10 Chapter 8 Quadratic Equations \" width=\"279\" height=\"193\"><br>According to the condition,<br>(x)\u00b2 + (x + 5)\u00b2 = (25)\u00b2 (Using Pythagoras Theorem)<br>\u21d2 x\u00b2 + x\u00b2 + 10x + 25 = 625<br>\u21d2 2x\u00b2 + 10x + 25 \u2013 625 = 0<br>\u21d2 2x\u00b2 + 10x \u2013 600 = 0<br>\u21d2 x\u00b2 + 5x \u2013 300 = 0 (Dividing by 2)<br>\u21d2 x\u00b2 + 20x \u2013 15x \u2013 300 = 0<br>\u21d2 x (x + 20) \u2013 15 (x + 20) = 0<br>\u21d2 (x + 20) (x \u2013 15) = 0<br>Either x + 20 = 0, then x = -20, which is not possible being negative<br>or x \u2013 15 = 0, then x = 15<br>One side = 15 cm<br>and second side = 15 + 5 = 20 cm<\/p>\n<p>Question 2.<br>The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.<br>Solution:<br>Let shorter side of the rectangular field = x m<br>Then diagonal = (x + 60) m<br>and longer side = (x + 30) m<br>According to the condition,<br>(Diagonal)\u00b2 = Sum of squares of the two sides<br><img src=\"https:\/\/farm2.staticflickr.com\/1739\/40802597830_20271c7205_o.png\" alt=\"Quadratic Equations Class 10 RD Sharma \" width=\"236\" height=\"151\"><br>\u21d2 (x + 60)\u00b2 = x\u00b2 + (x + 30)\u00b2<br>\u21d2 x\u00b2 + 120x + 3600 = x\u00b2 + x\u00b2 + 60x + 900<br>\u21d2 2x\u00b2 + 60x + 900 \u2013 x\u00b2 \u2013 120x \u2013 3600 = 0<br>\u21d2 x\u00b2 \u2013 60x \u2013 2700 = 0<br>\u21d2 x\u00b2 \u2013 90x + 30x \u2013 2700 = 0<br>\u21d2 x (x \u2013 90) + 30 (x \u2013 90) = 0<br>\u21d2 (x \u2013 90) (x + 30) = 0<br>Either x \u2013 90 = 0, then x = 90<br>or x + 30 = 0, then x = \u2013 30 which is not possible being negative<br>Longer side (length) = x + 30 = 90 + 30= 120<br>and breadth = x = 90 m<\/p>\n<p>Question 3.<br>The hypotenuse of a right triangle is 3\u221a10 cm. If the smaller leg is tripled and the longer leg doubled, the new hypotenuse will be 9\u221a5 cm. How long are the legs of the triangle?<br>Solution:<br>Let the smaller leg of right triangle = x cm<br>and larger leg = y cm<br>Then x\u00b2 + y\u00b2 = (3\u221a10)\u00b2 (Using Pythagoras Theorem)<br>x\u00b2 + y\u00b2 = 90 \u2026.(i)<br>According to the second condition,<br>(3x)\u00b2 + (2y)\u00b2 = (9\u221a5)\u00b2<br>\u21d2 9x\u00b2 + 4y\u00b2 = 405 \u2026.(ii)<br>Multiplying (i) by 9 and (ii) by 1<br><img src=\"https:\/\/farm2.staticflickr.com\/1728\/40802598000_1d8a0e768c_o.png\" alt=\"RD Sharma Class 10 Solutions Quadratic Equations \" width=\"259\" height=\"165\"><br>y = 9<br>Substituting the value of y in (i)<br>x\u00b2 + (9)\u00b2 = 90<br>\u21d2 x\u00b2 + 81 = 90<br>\u21d2 x\u00b2 = 90 \u2013 81 = 9 = (3)\u00b2<br>x = 3<br>Length of smaller leg = 3 cm<br>and length of longer leg = 9 cm<\/p>\n<p>Question 4.<br>A pole has to be erected at a point on the boundary of a circular park of diameter 13 meters in such a way that. the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 meters. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?<br>Solution:<br>In a circle, AB is the diameters and AB = 13 m<br>Let P be the pole on the circle Let PB = x m,<br>then PA = (x + 7) m<br>Now in right \u2206APB (P is in a semi-circle)<br>AB\u00b2 = AB\u00b2 + AP\u00b2 (Pythagoras Theorem)<br><img src=\"https:\/\/farm2.staticflickr.com\/1749\/41709657365_9ca4759b67_o.png\" alt=\"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations \" width=\"187\" height=\"157\"><br>(13)\u00b2 = x\u00b2 + (x + 7)\u00b2<br>\u21d2 x\u00b2 + x\u00b2 + 14x + 49 = 169<br>\u21d2 2x\u00b2 + 14x + 49 \u2013 169 = 0<br>\u21d2 2x\u00b2+ 14x \u2013 120 = 0<br>\u21d2 x2 + 7x \u2013 60 = 0 (Dividing by 2)<br>\u21d2 x\u00b2 + 12x \u2013 5x \u2013 60 = 0<br>\u21d2 x (x + 12) \u2013 5 (x + 12) = 0<br>\u21d2 (x + 12) (x \u2013 5) = 0<br>Either x + 12 = 0, then x = -12 which is not possible being negative<br>or x \u2013 5 = 0, then x = 5<br>P is at a distance of 5 m from B and 5 + 7 = 12 m from A.<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10. 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-810\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10<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-1631525997903\" 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-810-helpful-for-board-exams\"><\/span>How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10 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 8.10 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-1631526315062\" 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-810-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10 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 8.10 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-1631526510324\" 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-810-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.10 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 8.10 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 8.10:&nbsp;In this exercise, you will learn how to use quadratic equations to solve geometry problems. Students can use the RD Sharma Class 10 Solutions&nbsp;to gain a better understanding of how to solve questions in this chapter. Students can also utilise the RD Sharma Solutions for Class 10 &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.10 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-10\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.10 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":126888,"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\/126853"}],"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=126853"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126853\/revisions"}],"predecessor-version":[{"id":127105,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126853\/revisions\/127105"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126888"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126853"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126853"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126853"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}