{"id":125888,"date":"2023-09-12T16:21:00","date_gmt":"2023-09-12T10:51:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125888"},"modified":"2023-11-10T10:24:51","modified_gmt":"2023-11-10T04:54:51","slug":"rd-sharma-class-10-solutions-chapter-4-exercise-4-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-4-exercise-4-2\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 4 Exercise 4.2 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125909\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-Exercise-4.2.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 4 Exercise 4.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-Exercise-4.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-Exercise-4.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 4 Exercise 4.2:\u00a0<\/strong>In this exercise, the main focus is on similar triangles and their attributes. In the exercise problems, basic conclusions on proportionality and theorems are examined. 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> to clarify their problems and as a study material for exams. Students can also use 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 4.2 PDF 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-69de712765f12\" 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 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-4-exercise-4-2\/#where-can-i-download-rd-sharma-class-10-solutions-chapter-4-exercise-42-free-pdf\" title=\"Where can I download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.2 free PDF?\">Where can I download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.2 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-4-2\/#what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-4-exercise-42\" title=\"What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 Exercise 4.2?\">What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 Exercise 4.2?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-4-exercise-42-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.2 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-4-Exercise-4.2.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-4-Exercise-4.2.pdf\">RD Sharma Class 10 Solutions Chapter 4 Exercise 4.2<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-4-exercise-42-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 4 Exercise 4.2- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>i) If AD = 6 cm, DB = 9 cm and AE = 8 cm, Find AC.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:\u00a0\u0394\u00a0ABC, DE\u00a0\u2225\u00a0BC, AD = 6 cm, DB = 9 cm and AE = 8 cm.<\/p>\n<p>Required to find AC.<\/p>\n<p>By using Thales Theorem, [As DE\u00a0\u2225\u00a0BC]<\/p>\n<p>AD\/BD = AE\/CE<\/p>\n<p>Let CE = x.<\/p>\n<p>So then,<\/p>\n<p>6\/9 = 8\/x<\/p>\n<p>6x = 72 cm<\/p>\n<p>x = 72\/6 cm<\/p>\n<p>x = 12 cm<\/p>\n<p>\u2234 AC = AE + CE = 12 + 8 = 20.<\/p>\n<p><strong>ii) If\u00a0AD\/DB = 3\/4\u00a0and AC = 15 cm, Find AE.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:\u00a0AD\/BD = 3\/4\u00a0and AC = 15 cm [As DE\u00a0\u2225\u00a0BC]<\/p>\n<p>Required to find AE.<\/p>\n<p>By using Thales Theorem, [As DE\u00a0\u2225\u00a0BC]<\/p>\n<p>AD\/BD = AE\/CE<\/p>\n<p>Let, AE = x, then CE = 15-x.<\/p>\n<p>\u21d2 3\/4 = x\/ (15\u2013x)<\/p>\n<p>45 \u2013 3x = 4x<\/p>\n<p>-3x \u2013 4x = \u2013 45<\/p>\n<p>7x = 45<\/p>\n<p>x = 45\/7<\/p>\n<p>x = 6.43 cm<\/p>\n<p>\u2234 AE= 6.43cm<\/p>\n<p><strong>iii) If\u00a0AD\/DB = 2\/3\u00a0and AC = 18 cm, Find AE.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:\u00a0AD\/BD = 2\/3\u00a0and AC = 18 cm<\/p>\n<p>Required to find AE.<\/p>\n<p>By using Thales Theorem, [As DE\u00a0\u2225\u00a0BC]<\/p>\n<p>AD\/BD = AE\/CE<\/p>\n<p>Let, AE = x and CE = 18 \u2013 x<\/p>\n<p>\u21d2 23 = x\/ (18\u2013x)<\/p>\n<p>3x = 36 \u2013 2x<\/p>\n<p>5x = 36 cm<\/p>\n<p>x = 36\/5 cm<\/p>\n<p>x = 7.2 cm<\/p>\n<p>\u2234 AE = 7.2 cm<\/p>\n<p><strong>iv) If AD = 4 cm, AE = 8 cm, DB = x \u2013 4 cm and EC = 3x \u2013 19, find x.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: AD = 4 cm, AE = 8 cm, DB = x \u2013 4 and EC = 3x \u2013 19<\/p>\n<p>Required to find x.<\/p>\n<p>By using Thales Theorem, [As DE\u00a0\u2225\u00a0BC]<\/p>\n<p>AD\/BD = AE\/CE<\/p>\n<p>Then,\u00a04\/ (x \u2013 4) = 8\/ (3x \u2013 19)<\/p>\n<p>4(3x \u2013 19) = 8(x \u2013 4)<\/p>\n<p>12x \u2013 76 = 8(x \u2013 4)<\/p>\n<p>12x \u2013 8x = \u2013 32 + 76<\/p>\n<p>4x = 44 cm<\/p>\n<p>x = 11 cm<\/p>\n<p><strong>v) If AD = 8 cm, AB = 12 cm and AE = 12 cm, find CE.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: AD = 8 cm, AB = 12 cm, and AE = 12 cm.<\/p>\n<p>Required to find CE,<\/p>\n<p>By using Thales Theorem, [As DE\u00a0\u2225\u00a0BC]<\/p>\n<p>AD\/BD = AE\/CE<\/p>\n<p>8\/4 = 12\/CE<\/p>\n<p>8 x CE = 4 x 12 cm<\/p>\n<p>CE = (4 x 12)\/8 cm<\/p>\n<p>CE = 48\/8 cm<\/p>\n<p>\u2234 CE = 6 cm<\/p>\n<p><strong>vi) If AD = 4 cm, DB = 4.5 cm and AE = 8 cm, find AC.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: AD = 4 cm, DB = 4.5 cm, AE = 8 cm<\/p>\n<p>Required to find AC.<\/p>\n<p>By using Thales Theorem, [As DE\u00a0\u2225\u00a0BC]<\/p>\n<p>AD\/BD = AE\/CE<\/p>\n<p>4\/4.5 = 8\/AC<\/p>\n<p>AC = (4.5 \u00d7 8)\/4\u00a0cm<\/p>\n<p>\u2234AC = 9 cm<\/p>\n<p><strong>vii) If AD = 2 cm, AB = 6 cm and AC = 9 cm, find AE.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: AD = 2 cm, AB = 6 cm and AC = 9 cm<\/p>\n<p>Required to find AE.<\/p>\n<p>DB = AB \u2013 AD = 6 \u2013 2 = 4 cm<\/p>\n<p>By using Thales Theorem, [As DE\u00a0\u2225\u00a0BC]<\/p>\n<p>AD\/BD = AE\/CE<\/p>\n<p>2\/4 = x\/ (9\u2013x)<\/p>\n<p>4x = 18 \u2013 2x<\/p>\n<p>6x = 18<\/p>\n<p>x = 3 cm<\/p>\n<p>\u2234 AE= 3cm<\/p>\n<p><strong>viii) If\u00a0AD\/BD = 4\/5\u00a0and EC = 2.5 cm, Find AE.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:\u00a0AD\/BD = 4\/5\u00a0and EC = 2.5 cm<\/p>\n<p>Required to find AE.<\/p>\n<p>By using Thales Theorem, [As DE\u00a0\u2225\u00a0BC]<\/p>\n<p>AD\/BD = AE\/CE<\/p>\n<p>Then,\u00a04\/5 = AE\/2.5<\/p>\n<p>\u2234 AE =\u00a04 \u00d7 2.55\u00a0= 2 cm<\/p>\n<p><strong>ix) If AD = x cm, DB = x \u2013 2 cm, AE = x + 2 cm, and EC = x \u2013 1 cm, find the value of x.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: AD = x, DB = x \u2013 2, AE = x + 2 and EC = x \u2013 1<\/p>\n<p>Required to find the value of x.<\/p>\n<p>By using Thales Theorem, [As DE\u00a0\u2225\u00a0BC]<\/p>\n<p>AD\/BD = AE\/CE<\/p>\n<p>So, x\/ (x\u20132) = (x+2)\/ (x\u20131)<\/p>\n<p>x(x \u2013 1) = (x \u2013 2)(x + 2)<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 x \u2013 x<sup>2<\/sup>\u00a0+ 4 = 0<\/p>\n<p>x = 4<\/p>\n<p><strong>x) If AD = 8x \u2013 7 cm, DB = 5x \u2013 3 cm, AE = 4x \u2013 3 cm, and EC = (3x \u2013 1) cm, Find the value of x.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: AD = 8x \u2013 7, DB = 5x \u2013 3, AER = 4x \u2013 3 and EC = 3x -1<\/p>\n<p>Required to find x.<\/p>\n<p>By using Thales Theorem, [As DE\u00a0\u2225\u00a0BC]<\/p>\n<p>AD\/BD = AE\/CE<\/p>\n<p>(8x\u20137)\/ (5x\u20133) = (4x\u20133)\/ (3x\u20131)<\/p>\n<p>(8x \u2013 7)(3x \u2013 1) = (5x \u2013 3)(4x \u2013 3)<\/p>\n<p>24x<sup>2<\/sup>\u00a0\u2013 29x + 7 = 20x<sup>2\u00a0<\/sup>\u2013 27x + 9<\/p>\n<p>4x<sup>2<\/sup>\u00a0\u2013 2x \u2013 2 = 0<\/p>\n<p>2(2x<sup>2<\/sup>\u00a0\u2013 x \u2013 1) = 0<\/p>\n<p>2x<sup>2<\/sup>\u00a0\u2013 x \u2013 1 = 0<\/p>\n<p>2x<sup>2<\/sup>\u00a0\u2013 2x + x \u2013 1 = 0<\/p>\n<p>2x(x \u2013 1) + 1(x \u2013 1) = 0<\/p>\n<p>(x \u2013 1)(2x + 1) = 0<\/p>\n<p>\u21d2 x = 1 or x = -1\/2<\/p>\n<p>We know that the side of a triangle can never be negative. Therefore, we take the positive value.<\/p>\n<p>\u2234 x = 1.<\/p>\n<p><strong>xi) If AD = 4x \u2013 3, AE = 8x \u2013 7, BD = 3x \u2013 1, and CE = 5x \u2013 3, find the value of x.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: AD = 4x \u2013 3, BD = 3x \u2013 1, AE = 8x \u2013 7 and EC = 5x \u2013 3<\/p>\n<p>Required to find x.<\/p>\n<p>By using Thales Theorem, [As DE\u00a0\u2225\u00a0BC]<\/p>\n<p>AD\/BD = AE\/CE<\/p>\n<p>So,\u00a0(4x\u20133)\/ (3x\u20131) = (8x\u20137)\/ (5x\u20133)<\/p>\n<p>(4x \u2013 3)(5x \u2013 3) = (3x \u2013 1)(8x \u2013 7)<\/p>\n<p>4x(5x \u2013 3) -3(5x \u2013 3) = 3x(8x \u2013 7) -1(8x \u2013 7)<\/p>\n<p>20x<sup>2<\/sup>\u00a0\u2013 12x \u2013 15x + 9 = 24x<sup>2<\/sup>\u00a0\u2013 29x + 7<\/p>\n<p>20x<sup>2<\/sup>\u00a0-27x + 9 = 24<sup>2<\/sup>\u00a0-29x + 7<\/p>\n<p>\u21d2 -4x<sup>2<\/sup>\u00a0+ 2x + 2 = 0<\/p>\n<p>4x<sup>2<\/sup>\u00a0\u2013 2x \u2013 2 = 0<\/p>\n<p>4x<sup>2<\/sup>\u00a0\u2013 4x + 2x \u2013 2 = 0<\/p>\n<p>4x(x \u2013 1) + 2(x \u2013 1) = 0<\/p>\n<p>(4x + 2)(x \u2013 1) = 0<\/p>\n<p>\u21d2 x = 1 or x = -2\/4<\/p>\n<p>We know that the side of a triangle can never be negative. Therefore, we take the positive value.<\/p>\n<p>\u2234 x = 1<\/p>\n<p><strong>xii) If AD = 2.5 cm, BD = 3.0 cm, and AE = 3.75 cm, find the length of AC.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: AD = 2.5 cm, AE = 3.75 cm and BD = 3 cm<\/p>\n<p>Required to find AC.<\/p>\n<p>By using Thales Theorem, [As DE\u00a0\u2225\u00a0BC]<\/p>\n<p>AD\/BD = AE\/CE<\/p>\n<p>2.5\/ 3 = 3.75\/ CE<\/p>\n<p>2.5 x CE = 3.75 x 3<\/p>\n<p>CE = 3.75\u00d732.5<\/p>\n<p>CE = 11.252.5<\/p>\n<p>CE = 4.5<\/p>\n<p>Now, AC = 3.75 + 4.5<\/p>\n<p>\u2234 AC = 8.25 cm.<\/p>\n<p><strong>2. In a\u00a0<\/strong>\u0394<strong>\u00a0ABC, D and E are points on the sides AB and AC, respectively. For each of the following cases show that DE\u00a0<\/strong>\u2225<strong>\u00a0BC:<\/strong><\/p>\n<p><strong>i) AB = 12 cm, AD = 8 cm, AE = 12 cm, and AC = 18 cm.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Required to prove DE\u00a0\u2225\u00a0BC.<\/p>\n<p>We have,<\/p>\n<p>AB = 12 cm, AD = 8 cm, AE = 12 cm, and AC = 18 cm<strong>.\u00a0<\/strong>(Given)<\/p>\n<p>So,<\/p>\n<p>BD = AB \u2013 AD = 12 \u2013 8 = 4 cm<\/p>\n<p>And,<\/p>\n<p>CE = AC \u2013 AE = 18 \u2013 12 = 6 cm<\/p>\n<p>It\u2019s seen that,<\/p>\n<p>AD\/BD = 8\/4 = 1\/2<\/p>\n<p>AE\/CE = 12\/6 = 1\/2<\/p>\n<p>Thus,<\/p>\n<p>AD\/BD = AE\/CE<\/p>\n<p>So, by the converse of Thale\u2019s Theorem<\/p>\n<p>We have,<\/p>\n<p>DE\u00a0\u2225\u00a0BC.<\/p>\n<p>Hence Proved.<\/p>\n<p><strong>ii) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, and AE = 1.8 cm.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Required to prove DE\u00a0\u2225\u00a0BC.<\/p>\n<p>We have,<\/p>\n<p>AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, and AE = 1.8 cm<strong>.\u00a0<\/strong>(Given)<\/p>\n<p>So,<\/p>\n<p>BD = AB \u2013 AD = 5.6 \u2013 1.4 = 4.2 cm<\/p>\n<p>And,<\/p>\n<p>CE = AC \u2013 AE = 7.2 \u2013 1.8 = 5.4 cm<\/p>\n<p>It\u2019s seen that,<\/p>\n<p>AD\/BD = 1.4\/4.2 = 1\/3<\/p>\n<p>AE\/CE = 1.8\/5.4 =1\/3<\/p>\n<p>Thus,<\/p>\n<p>AD\/BD = AE\/CE<\/p>\n<p>So, by the converse of Thale\u2019s Theorem<\/p>\n<p>We have,<\/p>\n<p>DE\u00a0\u2225\u00a0BC.<\/p>\n<p>Hence Proved.<\/p>\n<p><strong>iii) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm, and AE = 2.8 cm.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Required to prove DE\u00a0\u2225\u00a0BC.<\/p>\n<p>We have<\/p>\n<p>AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm, and AE = 2.8 cm.<\/p>\n<p>So,<\/p>\n<p>AD = AB \u2013 DB = 10.8 \u2013 4.5 = 6.3<\/p>\n<p>And,<\/p>\n<p>CE = AC \u2013 AE = 4.8\u00a0\u2013 2.8 = 2<\/p>\n<p>It\u2019s seen that,<\/p>\n<p>AD\/BD = 6.3\/ 4.5 = 2.8\/ 2.0 = AE\/CE = 7\/5<\/p>\n<p>So, by the converse of Thale\u2019s Theorem<\/p>\n<p>We have,<\/p>\n<p>DE\u00a0\u2225\u00a0BC.<\/p>\n<p>Hence Proved.<\/p>\n<p><strong>iv) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm, and EC = 5.5 cm.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Required to prove DE\u00a0\u2225\u00a0BC.<\/p>\n<p>We have<\/p>\n<p>AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm, and EC = 5.5 cm<\/p>\n<p>Now,<\/p>\n<p>AD\/BD = 5.7\/9.5 =3\/5<\/p>\n<p>And,<\/p>\n<p>AE\/CE = 3.3\/5.5 = 3\/5<\/p>\n<p>Thus,<\/p>\n<p>AD\/BD = AE\/CE<\/p>\n<p>So, by the converse of Thale\u2019s Theorem<\/p>\n<p>We have,<\/p>\n<p>DE\u00a0\u2225\u00a0BC.<\/p>\n<p>Hence Proved.<\/p>\n<p><strong>3. In a\u00a0<\/strong>\u0394<strong>\u00a0ABC, P and Q are the points on sides AB and AC, respectively, such that PQ\u00a0<\/strong>\u2225<strong>\u00a0BC. If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm. Find AB and PQ.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: \u0394\u00a0ABC, AP = 2.4 cm, AQ = 2 cm, QC = 3 cm, and BC = 6 cm. Also, PQ\u00a0\u2225\u00a0BC.<\/p>\n<p>Required to find: AB and PQ.<\/p>\n<p><img title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.2 - 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-4.jpeg\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.2 - 1\" \/><\/p>\n<p>By using Thales Theorem, we have [As it\u2019s given that PQ\u00a0\u2225\u00a0BC]<\/p>\n<p>AP\/PB = AQ\/ QC<\/p>\n<p>2.4\/PB = 2\/3<\/p>\n<p>2 x PB = 2.4 x 3<\/p>\n<p>PB = (2.4 \u00d7 3)\/2\u00a0cm<\/p>\n<p>\u21d2 PB = 3.6 cm<\/p>\n<p>Now finding, AB = AP + PB<\/p>\n<p>AB = 2.4 + 3.6<\/p>\n<p>\u21d2 AB = 6 cm<\/p>\n<p>Now, considering \u0394\u00a0APQ and \u0394\u00a0ABC<\/p>\n<p>We have,<\/p>\n<p>\u2220A = \u2220A<\/p>\n<p>\u2220APQ = \u2220ABC (Corresponding angles are equal, PQ||BC and AB being a transversal)<\/p>\n<p>Thus, \u0394\u00a0APQ and \u0394\u00a0ABC are similar to each other by AA criteria.<\/p>\n<p>Now, we know that<\/p>\n<p>Corresponding parts of similar triangles are propositional.<\/p>\n<p>\u21d2 AP\/AB = PQ\/ BC<\/p>\n<p>\u21d2 PQ = (AP\/AB) x BC<\/p>\n<p>= (2.4\/6) x 6 = 2.4<\/p>\n<p>\u2234 PQ = 2.4 cm.<\/p>\n<p><strong>4. In a\u00a0<\/strong>\u0394<strong>\u00a0ABC, D and E are points on AB and AC, respectively, such that DE\u00a0<\/strong>\u2225<strong>\u00a0BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm. Find BD and CE.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: \u0394\u00a0ABC such that AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BE = 5 cm. Also DE\u00a0\u2225\u00a0BC.<\/p>\n<p>Required to find: BD and CE.<\/p>\n<p>As DE\u00a0\u2225\u00a0BC, AB is transversal,<\/p>\n<p>\u2220APQ = \u2220ABC (corresponding angles)<\/p>\n<p>As DE\u00a0\u2225\u00a0BC, AC is transversal,<\/p>\n<p>\u2220AED = \u2220ACB (corresponding angles)<\/p>\n<p>In\u00a0\u0394\u00a0ADE and\u00a0\u0394\u00a0ABC,<\/p>\n<p>\u2220ADE=\u2220ABC<\/p>\n<p>\u2220AED=\u2220ACB<\/p>\n<p>\u2234\u00a0\u0394\u00a0ADE =\u00a0\u0394\u00a0ABC (AA similarity criteria)<\/p>\n<p>Now, we know that<\/p>\n<p>Corresponding parts of similar triangles are propositional.<\/p>\n<p>\u21d2 AD\/AB = AE\/AC = DE\/BC<\/p>\n<p>AD\/AB = DE\/BC<\/p>\n<p>2.4\/ (2.4 + DB) = 2\/5 [Since, AB = AD + DB]<\/p>\n<p>2.4 + DB = 6<\/p>\n<p>DB = 6 \u2013 2.4<\/p>\n<p>DB = 3.6 cm<\/p>\n<p>In the same way,<\/p>\n<p>\u21d2 AE\/AC = DE\/BC<\/p>\n<p>3.2\/ (3.2 + EC) = 2\/5 [Since AC = AE + EC]<\/p>\n<p>3.2 + EC = 8<\/p>\n<p>EC = 8 \u2013 3.2<\/p>\n<p>EC = 4.8 cm<\/p>\n<p>\u2234 BD = 3.6 cm and CE = 4.8 cm.<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 4 Exercise 4.2. 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-4-exercise-42\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 4 Exercise 4.2<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-1631178055664\" 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-42\"><\/span>Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 4 Exercise 4.2?<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 4.2 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-1631178196403\" 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-42-free-pdf\"><\/span>Where can I download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.2 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 4.2 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631178321217\" 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-42\"><\/span>What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 Exercise 4.2?<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 4.2 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 4.2:\u00a0In this exercise, the main focus is on similar triangles and their attributes. In the exercise problems, basic conclusions on proportionality and theorems are examined. Students can use the RD Sharma Class 10 Solutions to clarify their problems and as a study material for exams. Students can &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 4 Exercise 4.2 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-4-exercise-4-2\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 4 Exercise 4.2 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":125909,"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\/125888"}],"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=125888"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125888\/revisions"}],"predecessor-version":[{"id":505680,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125888\/revisions\/505680"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125909"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125888"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125888"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125888"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}