{"id":125884,"date":"2023-09-12T16:24:00","date_gmt":"2023-09-12T10:54:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125884"},"modified":"2023-11-28T11:09:47","modified_gmt":"2023-11-28T05:39:47","slug":"rd-sharma-class-10-solutions-chapter-4-exercise-4-6","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-4-exercise-4-6\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125913\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-Exercise-4.6.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-Exercise-4.6.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-Exercise-4.6-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6:&nbsp;<\/strong>In this exercise, you&#8217;ll learn more about triangle similarity. The essential ideas covered in the exercise problems are other distinguishing features and the areas of two identical triangles. <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> has all of the solutions needed for quick reference and exam preparation. Students can also refer to the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-4-triangles\/\"><strong>RD Sharma Solutions for Class 10 Maths Chapter 4<\/strong><\/a> Exercise 4.6 PDF, which is available 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-6a038fce420f5\" 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\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-4-exercise-4-6\/#download-rd-sharma-class-10-solutions-chapter-4-exercise-46-free-pdf\" title=\"Download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6 Free PDF\">Download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6 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-4-exercise-4-6\/#access-answers-to-rd-sharma-solutions-class-10-maths-chapter-4-exercise-46-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 10 Maths Chapter 4 Exercise 4.6- Important Question with Answers\">Access answers to RD Sharma Solutions Class 10 Maths Chapter 4 Exercise 4.6- 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\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-4-exercise-4-6\/#where-can-i-download-rd-sharma-class-10-solutions-chapter-4-exercise-46-free-pdf\" title=\"Where can I download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6 free PDF?\">Where can I download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6 free PDF?<\/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-4-exercise-4-6\/#what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-4-exercise-46\" title=\"What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6?\">What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-4-exercise-46-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6 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-Solutions-Class-10-Maths-Chapter-4-Ex-4.6.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-SHARMA-Solutions-Class-10-Maths-Chapter-4-Ex-4.6.pdf\">RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-4-exercise-46-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 4 Exercise 4.6- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Triangles ABC and DEF are similar.<\/strong><\/p>\n<p><strong>(i) If area of (\u0394ABC) = 16 cm<sup>2<\/sup>, area (\u0394DEF) = 25 cm<sup>2&nbsp;<\/sup>and BC = 2.3 cm, find EF.<\/strong><\/p>\n<p><strong>(ii) If area (\u0394ABC) = 9 cm<sup>2<\/sup>, area (\u0394DEF) = 64 cm<sup>2&nbsp;<\/sup>and DE = 5.1 cm, find AB.<\/strong><\/p>\n<p><strong>(iii) If AC = 19 cm and DF = 8 cm, find the ratio of the area of two triangles.<\/strong><\/p>\n<p><strong>(iv) If area of (\u0394ABC) = 36 cm<sup>2<\/sup>, area (\u0394DEF) = 64 cm<sup>2&nbsp;<\/sup>and DE = 6.2 cm, find AB.<\/strong><\/p>\n<p><strong>(v) If AB = 1.2 cm and DE = 1.4 cm, find the ratio of the area of two triangles.<\/strong><\/p>\n<p><strong>Solutions:<\/strong><\/p>\n<p>As we know, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, we get<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.6 - 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-4-21.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.6 - 1\"><\/p>\n<p><strong>2. In the fig 4.178,&nbsp;\u0394ACB \u223c \u0394APQ. If BC = 10 cm, PQ = 5 cm, BA = 6.5 cm, AP = 2.8 cm, find CA and AQ. Also, find the area (\u0394ACB): area&nbsp;(\u0394APQ).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>\u0394ACB&nbsp;is similar to&nbsp;\u0394APQ<\/p>\n<p>BC = 10 cm<\/p>\n<p>PQ = 5 cm<\/p>\n<p>BA = 6.5 cm<\/p>\n<p>AP = 2.8 cm<\/p>\n<p>Required to Find: CA, AQ and that the area (\u0394ACB): area&nbsp;(\u0394APQ).<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.6 - 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-4-22.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.6 - 2\"><\/p>\n<p>Since \u0394ACB \u223c \u0394APQ<\/p>\n<p>We know that,<\/p>\n<p>AB\/ AQ = BC\/ PQ = AC\/ AP&nbsp;[Corresponding Parts of Similar Triangles]<\/p>\n<p>AB\/ AQ = BC\/ PQ<\/p>\n<p>6.5\/ AQ = 10\/5<\/p>\n<p>\u21d2 AQ = 3.25 cm<\/p>\n<p>Similarly,<\/p>\n<p>BC\/ PQ = CA\/ AP<\/p>\n<p>CA\/ 2.8 = 10\/5<\/p>\n<p>\u21d2 CA = 5.6 cm<\/p>\n<p>Next,<\/p>\n<p>Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, we have,<\/p>\n<p>ar(\u0394ACQ): ar(\u0394APQ) = (BC\/ PQ)2<\/p>\n<p>=&nbsp;(10\/5)2<\/p>\n<p>=&nbsp;(2\/1)2<\/p>\n<p>=&nbsp;4\/1<\/p>\n<p>Therefore, the ratio is 4:1.<\/p>\n<p><strong>3. The areas of two similar triangles are 81 cm\u00b2 and 49 cm\u00b2 respectively. Find the ration of their corresponding heights. What is the ratio of their corresponding medians?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: The areas of two similar triangles are 81cm<sup>2<\/sup>&nbsp;and 49cm<sup>2<\/sup>.<\/p>\n<p>Required to find: The ratio of their corresponding heights and the ratio of their corresponding medians.<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.6 - 3\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-4-23.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.6 - 3\">Let\u2019s consider the two similar triangles as \u0394ABC and \u0394PQR, AD and PS be the altitudes of \u0394ABC and \u0394PQR respectively.<\/p>\n<p>So,<\/p>\n<p>By area of similar triangle theorem, we have<\/p>\n<p>ar(\u0394ABC)\/ ar(\u0394PQR) = AB<sup>2<\/sup>\/ PQ<sup>2<\/sup><\/p>\n<p>\u21d2 81\/ 49 = AB<sup>2<\/sup>\/ PQ<sup>2<\/sup><\/p>\n<p>\u21d2 9\/ 7 = AB\/ PQ<\/p>\n<p>In \u0394ABD and \u0394PQS<\/p>\n<p>\u2220B&nbsp;<strong>=&nbsp;<\/strong>\u2220Q [Since \u0394ABC \u223c \u0394PQR]<\/p>\n<p>\u2220ABD&nbsp;<strong>=&nbsp;<\/strong>\u2220PSQ = 90<sup>o<\/sup><\/p>\n<p>\u21d2 \u0394ABD \u223c \u0394PQS [By AA similarity]<strong>&nbsp; &nbsp;<\/strong><\/p>\n<p>Hence, as the corresponding parts of similar triangles are proportional, we have<\/p>\n<p>AB\/ PQ = AD\/ PS<\/p>\n<p>Therefore,<\/p>\n<p>AD\/ PS = 9\/7 (Ratio of altitudes)<\/p>\n<p>Similarly,<\/p>\n<p>The ratio of two similar triangles is equal to the ratio of the squares of their corresponding medians also.<\/p>\n<p>Thus, ratio of altitudes = Ratio of medians = 9\/7<\/p>\n<p><strong>4. The areas of two similar triangles are 169 cm<sup>2<\/sup>and 121 cm<sup>2&nbsp;<\/sup>respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>The area of two similar triangles is 169cm<sup>2<\/sup>&nbsp;and 121cm<sup>2<\/sup>.<\/p>\n<p>The longest side of the larger triangle is 26cm.<\/p>\n<p>Required to find: the longest side of the smaller triangle<\/p>\n<p>Let the longer side of the smaller triangle = x<\/p>\n<p>We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, we have<\/p>\n<p>ar(larger triangle)\/ ar(smaller triangle) = (side of the larger triangle\/ side of the smaller triangle)<sup>2<\/sup><\/p>\n<p>=&nbsp;169\/ 121<\/p>\n<p>Taking the square roots of LHS and RHS, we get<\/p>\n<p>= 13\/ 11<\/p>\n<p>Since sides of similar triangles are propositional, we can say<\/p>\n<p>3\/ 11 = (longer side of the larger triangle)\/ (longer side of the smaller triangle)<\/p>\n<p>\u21d2 13\/ 11 = 26\/ x<\/p>\n<p>x = 22<\/p>\n<p>Therefore, the longest side of the smaller triangle is 22 cm.<\/p>\n<p><strong>5. The area of two similar triangles are 25 cm<sup>2<\/sup>&nbsp;and 36cm<sup>2<\/sup>&nbsp;respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: The area of two similar triangles are 25 cm<sup>2<\/sup>&nbsp;and 36cm<sup>2<\/sup>&nbsp;respectively, the altitude of the first triangle is 2.4 cm<\/p>\n<p>Required to find: the altitude of the second triangle<\/p>\n<p>We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes, we have<\/p>\n<p>\u21d2 ar(triangle1)\/ar(triangle2) = (altitude1\/ altitude2)<sup>2<\/sup><\/p>\n<p>\u21d2 25\/ 36 = (2.4)<sup>2<\/sup>\/ (altitude2)<sup>2<\/sup><\/p>\n<p>Taking the square roots of LHS and RHS, we get<\/p>\n<p>5\/ 6 = 2.4\/ altitude2<\/p>\n<p>\u21d2 altitude2 = (2.4 x 6)\/5 = 2.88cm<\/p>\n<p>Therefore, the altitude of the second triangle is 2.88cm.<\/p>\n<p><strong>6. The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>The corresponding altitudes of two similar triangles are 6 cm and 9 cm.<\/p>\n<p>Required to find the ratio of areas of the two similar triangles<\/p>\n<p>We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes, we have<\/p>\n<p>ar(triangle1)\/ar(triangle2) = (altitude1\/ altitude2)<sup>2<\/sup>&nbsp;= (6\/9)<sup>2<\/sup><\/p>\n<p>= 36\/ 81<\/p>\n<p>= 4\/9<\/p>\n<p>Therefore, the ratio of the areas of two triangles = 4: 9.<\/p>\n<p><strong>7.<\/strong>&nbsp;<strong>ABC is a triangle in which&nbsp;\u2220 A = 90\u00b0, AN&nbsp;\u22a5 BC, BC = 12 cm and AC = 5 cm. Find the ratio of the areas of&nbsp;<\/strong>\u0394ANC<strong>&nbsp;and&nbsp;<\/strong>\u0394ABC<strong>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>Given:<\/strong><\/p>\n<p>Given,<\/p>\n<p>\u0394ABC, \u2220A = 90<sup>\u2218<\/sup>, AN&nbsp;\u22a5&nbsp;BC<\/p>\n<p>BC= 12 cm<\/p>\n<p>AC = 5 cm.<\/p>\n<p>Required to find: ar(\u0394ANC)\/ ar(\u0394ABC).<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.6 - 4\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-4-24.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.6 - 4\"><\/p>\n<p>We have,<\/p>\n<p>In&nbsp;\u0394ANC and \u0394ABC,<\/p>\n<p>\u2220ACN = \u2220ACB&nbsp; [Common]<\/p>\n<p>\u2220A = \u2220ANC&nbsp; [each 90<sup>\u2218<\/sup>]<\/p>\n<p>\u21d2 \u0394ANC \u223c \u0394ABC&nbsp; [AA similarity]<\/p>\n<p>Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, we get have<\/p>\n<p>ar(\u0394ANC)\/ ar(\u0394ABC) = (AC\/ BC)<sup>2<\/sup>&nbsp;= (5\/12)<sup>2<\/sup>&nbsp;= 25\/ 144<\/p>\n<p>Therefore, ar(\u0394ANC)\/ ar(\u0394ABC) = 25:144<\/p>\n<p><strong>&nbsp;<\/strong><\/p>\n<p><strong>8. In Fig 4.179, DE || BC<\/strong><\/p>\n<p><strong>(i) If DE = 4m, BC = 6 cm and Area&nbsp;(\u0394ADE) = 16cm<sup>2<\/sup>, find the area of&nbsp;\u0394ABC.<\/strong><\/p>\n<p><strong>(ii) If DE = 4cm, BC = 8 cm and Area&nbsp;(\u0394ADE) = 25cm<sup>2<\/sup>, find the area of&nbsp;\u0394ABC.<\/strong><\/p>\n<p><strong>(iii) If DE: BC = 3: 5. Calculate the ratio of the areas of&nbsp;\u0394ADE&nbsp;and the trapezium BCED.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>DE&nbsp;\u2225&nbsp;BC.<\/p>\n<p>In&nbsp;\u0394ADE and \u0394ABC<\/p>\n<p>We know that,<\/p>\n<p>\u2220ADE = \u2220B&nbsp; [Corresponding angles]<\/p>\n<p>\u2220DAE = \u2220BAC&nbsp; [Common]<\/p>\n<p>Hence,&nbsp;\u0394ADE ~ \u0394ABC&nbsp;(AA Similarity)<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.6 - 5\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-4-25.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.6 - 5\"><\/p>\n<p>(i)&nbsp;Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, we have,<\/p>\n<p>Ar(\u0394ADE)\/ Ar(\u0394ABC) = DE<sup>2<\/sup>\/ BC<sup>2<\/sup><\/p>\n<p>16\/ Ar(\u0394ABC) = 4<sup>2<\/sup>\/ 6<sup>2<\/sup><\/p>\n<p>\u21d2 Ar(\u0394ABC) = (6<sup>2<\/sup>&nbsp;\u00d7 16)\/ 4<sup>2<\/sup><\/p>\n<p>\u21d2 Ar(\u0394ABC)&nbsp;= 36 cm<sup>2<\/sup><\/p>\n<p>(ii)&nbsp;Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, we have,<\/p>\n<p>Ar(\u0394ADE)\/ Ar(\u0394ABC) = DE<sup>2<\/sup>\/ BC<sup>2<\/sup><\/p>\n<p>25\/ Ar(\u0394ABC) = 4<sup>2<\/sup>\/ 8<sup>2<\/sup><\/p>\n<p>\u21d2 Ar(\u0394ABC) = (8<sup>2<\/sup>&nbsp;\u00d7 25)\/ 4<sup>2<\/sup><\/p>\n<p>\u21d2 Ar(\u0394ABC)&nbsp;= 100 cm<sup>2<\/sup><\/p>\n<p>(iii) According to the question,<\/p>\n<p>Ar(\u0394ADE)\/ Ar(\u0394ABC) = DE<sup>2<\/sup>\/ BC<sup>2<\/sup><\/p>\n<p>Ar(\u0394ADE)\/ Ar(\u0394ABC) = 3<sup>2<\/sup>\/ 5<sup>2<\/sup><\/p>\n<p>Ar(\u0394ADE)\/ Ar(\u0394ABC) = 9\/25<\/p>\n<p>Assume that the area of&nbsp;\u0394ADE&nbsp;= 9x sq units<\/p>\n<p>And, area of&nbsp;\u0394ABC&nbsp;= 25x sq units<\/p>\n<p>So,<\/p>\n<p>Area of trapezium BCED = Area of \u0394ABC \u2013 Area of \u0394ADE<\/p>\n<p>= 25x \u2013 9x<\/p>\n<p>= 16x<\/p>\n<p>Now,&nbsp; Ar(\u0394ADE)\/ Ar(trap BCED) = 9x\/ 16x<\/p>\n<p>Ar(\u0394ADE)\/ Ar(trapBCED) = 9\/16<\/p>\n<p><strong>9. In \u0394ABC, D and E are the mid-points of AB and AC respectively. Find the ratio of the areas \u0394ADE and \u0394ABC.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>In \u0394ABC, D and E are the midpoints of AB and AC, respectively.<\/p>\n<p>Required to find: Ratio of the areas of&nbsp;\u0394ADE and \u0394ABC<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.6 - 6\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-4-26.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.6 - 6\"><\/p>\n<p>Since D and E are the midpoints of AB and AC, respectively.<\/p>\n<p>We can say,<\/p>\n<p>DE || BC (By converse of mid-point theorem)<\/p>\n<p>Also, DE =&nbsp;(1\/2) BC<\/p>\n<p>In&nbsp;\u0394ADE and \u0394ABC,<\/p>\n<p>\u2220ADE = \u2220B (Corresponding angles)<\/p>\n<p>\u2220DAE = \u2220BAC&nbsp; (common)<\/p>\n<p>Thus,&nbsp;\u0394ADE ~ \u0394ABC&nbsp; (AA Similarity)<\/p>\n<p>Now, we know that<\/p>\n<p>The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, so<\/p>\n<p>Ar(\u0394ADE)\/ Ar(\u0394ABC) = AD<sup>2<\/sup>\/ AB<sup>2<\/sup><\/p>\n<p>Ar(\u0394ADE)\/ Ar(\u0394ABC) = 1<sup>2<\/sup>\/ 2<sup>2<\/sup><\/p>\n<p>Ar(\u0394ADE)\/ Ar(\u0394ABC) = 1\/4<\/p>\n<p>Therefore, the ratio of the areas&nbsp;\u0394ADE and \u0394ABC is 1:4<\/p>\n<p><strong>10. The areas of two similar triangles are 100 cm<sup>2<\/sup>&nbsp;and 49 cm<sup>2<\/sup>&nbsp;respectively. If the altitude of the bigger triangles is 5 cm, find the corresponding altitude of the other.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: The area of the two similar triangles is&nbsp;100cm<sup>2<\/sup>&nbsp;and&nbsp;49cm<sup>2<\/sup>. And the altitude of the bigger triangle is 5cm.<\/p>\n<p>Required to find: The corresponding altitude of the other triangle<\/p>\n<p>We know that,<\/p>\n<p>The ratio of the areas of the two similar triangles is equal to the ratio of squares of their corresponding altitudes.<\/p>\n<p>ar(bigger triangle)\/ ar(smaller triangle) = (altitude of the bigger triangle\/ altitude of the smaller triangle)<sup>2<\/sup><\/p>\n<p>(100\/ 49) = (5\/ altitude of the smaller triangle)<sup>2<\/sup><\/p>\n<p>Taking square root on LHS and RHS, we get<\/p>\n<p>(10\/ 7) = (5\/ altitude of the smaller triangle) = 7\/2<\/p>\n<p>Therefore, the altitude of the smaller triangle = 3.5cm<\/p>\n<p><strong>11. The areas of two similar triangles are 121 cm<sup>2<\/sup>&nbsp;and 64 cm<sup>2<\/sup>&nbsp;respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: the area of the two triangles is&nbsp;121cm<sup>2<\/sup>&nbsp;and&nbsp;64cm<sup>2&nbsp;<\/sup>respectively and the median of the first triangle is 12.1cm<\/p>\n<p>Required to find: the corresponding median of the other triangle<\/p>\n<p>We know that,<\/p>\n<p>The ratio of the areas of the two similar triangles are equal to the ratio of the squares of their medians.<\/p>\n<p>ar(triangle1)\/ ar(triangle2) = (median of triangle 1\/median of triangle 2)<sup>2<\/sup><\/p>\n<p>121\/ 64 = (12.1\/ median of triangle 2)<sup>2<\/sup><\/p>\n<p>Taking the square roots on both LHS and RHS, we have<\/p>\n<p>11\/8 = (12.1\/ median of triangle 2) = (12.1 x 8)\/ 11<\/p>\n<p>Therefore, the Median of the other triangle = 8.8cm<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6. 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-4-exercise-46\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6<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-1631178084853\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-it-important-to-study-all-of-the-concepts-included-in-rd-sharma-class-10-solutions-chapter-4-exercise-46\"><\/span>Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, learning all of the concepts included in RD Sharma Solutions for Class 10 Maths Chapter 4 Exercise 4.6 is required in order to achieve high scores on the Class 10 board exams. These solutions were created by subject matter specialists who compiled model questions that covered all of the textbook&#8217;s exercise questions.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631178220332\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-download-rd-sharma-class-10-solutions-chapter-4-exercise-46-free-pdf\"><\/span>Where can I download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6 free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631178356451\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-4-exercise-46\"><\/span>What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>1. Correct answers according to the latest CBSE guidelines and syllabus.<br \/>2. The RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6 is written in simple language to assist students in their board examination, &amp; competitive examination preparation.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6:&nbsp;In this exercise, you&#8217;ll learn more about triangle similarity. The essential ideas covered in the exercise problems are other distinguishing features and the areas of two identical triangles. RD Sharma Class 10 Solutions has all of the solutions needed for quick reference and exam preparation. Students can &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-4-exercise-4-6\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 4 Exercise 4.6 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":125913,"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\/125884"}],"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=125884"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125884\/revisions"}],"predecessor-version":[{"id":513221,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125884\/revisions\/513221"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125913"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125884"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125884"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125884"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}