{"id":125885,"date":"2021-09-09T16:23:33","date_gmt":"2021-09-09T10:53:33","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125885"},"modified":"2021-09-09T16:30:46","modified_gmt":"2021-09-09T11:00:46","slug":"rd-sharma-class-10-solutions-chapter-4-exercise-4-5","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-4-exercise-4-5\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125912\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-Exercise-4.5.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-Exercise-4.5.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-Exercise-4.5-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5:\u00a0<\/strong>The resemblance of triangles can be demonstrated using a variety of criteria. In this exercise, all of these criteria are explored. Students who wish to learn more about these criteria 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 get high or full marks in their exams. For additional help, check 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 4.5 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-6a03a06fdac88\" 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-4-5\/#where-can-i-download-rd-sharma-class-10-solutions-chapter-4-exercise-45-free-pdf\" title=\"Where can I download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5 free PDF?\">Where can I download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5 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-4-5\/#what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-4-exercise-45\" title=\"What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5?\">What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-4-exercise-45-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5 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-4.5.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-4.5.pdf\">RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-4-exercise-45-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 4 Exercise 4.5- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>In the figure, \u2206ACB ~ \u2206APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ. (C.B.S.E. 1991)<br \/>Solution:<br \/>In the figure,<br \/>\u2206ACB ~ \u2206APQ<br \/>BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm<br \/><img src=\"https:\/\/farm1.staticflickr.com\/876\/40940567570_b73b13a7ac_o.png\" alt=\"RD Sharma Class 10 Chapter 4 Triangles \" width=\"350\" height=\"549\" \/><\/p>\n<p>Question 2.<br \/>In the figure, AB || QR. Find the length of PB. (C.B.S.E. 1994)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1739\/40940567810_0e733f5735_o.png\" alt=\"Triangles Class 10 RD Sharma \" width=\"261\" height=\"228\" \/><br \/>Solution:<br \/>In the figure,<br \/>In \u2206PQR, AB || QR<br \/>AB = 3 cm, QR =9 cm, PR = 6 cm<br \/>In \u2206PAB and \u2206PQR<br \/>\u2220P = \u2220P (common)<br \/>\u2220PAB = \u2220PQR (corresponding angles)<br \/>\u2220PBA = \u2220PRQ (corresponding angles)<br \/>\u2220PAB = \u2220PQR (AAA axiom)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1721\/40940567690_cb56a86408_o.png\" alt=\"RD Sharma Class 10 Solutions Triangles \" width=\"224\" height=\"138\" \/><\/p>\n<p>Question 3.<br \/>In the figure, XY || BC. Find the length of XY. (C.B.S.E. 1994C)<br \/>Solution:<br \/>In the figure<br \/>In \u2206ABC XY || BC<br \/>AX = 1 cm, BC = 6 cm<br \/><img src=\"https:\/\/farm1.staticflickr.com\/888\/40940568320_1116f7bc68_o.png\" alt=\"RD Sharma Class 10 Solutions Chapter 4 Triangles \" width=\"250\" height=\"224\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1735\/40940568040_cf6b99f5a1_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 4 Triangles \" width=\"358\" height=\"297\" \/><\/p>\n<p>Question 4.<br \/>In a right-angled triangle with sides a and b and hypotenuse c, the altitude drawn on the hypotenuse is x. Prove that ab = cx.<br \/>Solution:<br \/>Given : In right \u2206ABC, \u2220B is right angle BD \u22a5 AC<br \/>Now AB = a, BC = b, AC = c and BD = x<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1760\/40940568930_8bc9930d47_o.png\" alt=\"RD Sharma Solutions Class 10 Chapter 4 Triangles \" width=\"334\" height=\"488\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1751\/40940568530_467464906d_o.png\" alt=\"Learncbse.In Class 10 Chapter 4 Triangles \" width=\"324\" height=\"117\" \/><\/p>\n<p>Question 5.<br \/>In the figure, \u2220ABC = 90\u00b0 and BD \u22a5 AC. If BD = 8 cm and AD = 4 cm, find CD.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1722\/40940569370_1d9bb88f31_o.png\" alt=\"Class 10 RD Sharma Solutions Chapter 4 Triangles \" width=\"285\" height=\"410\" \/><\/p>\n<p>Question 6.<br \/>In the figure, \u2220ABC = 90\u00b0 and BD \u22a5 AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.<br \/>Solution:<br \/>In right \u2206ABC, \u2220B = 90\u00b0<br \/>BD \u22a5 AC<br \/>AB = 5.7 cm, BD = 3.8 and CD = 5.4 cm<br \/><img src=\"https:\/\/farm1.staticflickr.com\/891\/40940569710_21a027b7d1_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 4 Triangles \" width=\"269\" height=\"197\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1722\/40940569570_738d1f94ed_o.png\" alt=\"RD Sharma Class 10 Solution Chapter 4 Triangles \" width=\"247\" height=\"291\" \/><\/p>\n<p>Question 7.<br \/>In the figure, DE || BC such that AE =\u00a014\u00a0AC. If AB = 6 cm, find AD.<br \/>Solution:<br \/>In the figure, in \u2206ABC, DE || BC<br \/>AE =\u00a014\u00a0AC, AB = 6 cm<br \/><img src=\"https:\/\/farm1.staticflickr.com\/887\/40940569900_6e3a9beff5_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 4 Triangles \" width=\"303\" height=\"432\" \/><\/p>\n<p>Question 8.<br \/>In the figure, if AB \u22a5 BC, DC \u22a5 BC, and DE \u22a5 AC, prove that \u2206CED ~ \u2206ABC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1737\/40940570100_1fcc21e456_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Pdf Free Download Chapter 4 Triangles \" width=\"297\" height=\"217\" \/><br \/>Solution:<br \/>Given : In the figure AB \u22a5 BC, DC \u22a5 BC and DE \u22a5 AC<br \/>To prove : \u2206CED ~ \u2206ABC.<br \/>Proof: AB \u22a5 BC<br \/>\u2220B = 90\u00b0<br \/>and \u2220A + \u2220ACB = 90\u00b0 \u2026.(i)<br \/>DC \u22a5 BC<br \/>\u2220DCB = 90\u00b0<br \/>=&gt; \u2220ACB + \u2220DCA = 90\u00b0 \u2026.(ii)<br \/>From (i) and (ii)<br \/>\u2220A = \u2220DCA<br \/>Now in \u2206CED and \u2206ABC,<br \/>\u2220E = \u2220B (each 90\u00b0)<br \/>\u2220DEA or \u2220DCE = \u2220A (proved)<br \/>\u2206CED ~ \u2206ABC (AA axiom)<br \/>Hence proved.<\/p>\n<p>Question 9.<br \/>Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using similarity criterion<br \/>for two triangles, show that\u00a0OAOC\u00a0=\u00a0OBOD<br \/>Solution:<br \/>Given: ABCD is a trapezium in which AB || DC and diagonals AC and BD intersect each other at O<br \/>To Prove :\u00a0OAOC\u00a0=\u00a0OBOD<br \/><img src=\"https:\/\/farm1.staticflickr.com\/884\/40940570490_06d9e576bc_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 4 Triangles \" width=\"344\" height=\"403\" \/><\/p>\n<p>Question 10.<br \/>In \u2206ABC and \u2206AMP are two right triangles, right-angled at B and M respectively such that \u2220MAP = \u2220BAC. Prove that<br \/>(i) \u2206ABC ~ \u2206AMP<br \/>(ii)\u00a0CAPA\u00a0=\u00a0BCMP.<br \/>Solution:<br \/>Given : In \u2206ABC and \u2206AMP,<br \/>\u2220B = \u2220M = 90\u00b0<br \/>\u2220MAP = \u2220BAC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1758\/40940571050_9434cc364a_o.png\" alt=\"Class 10 RD Sharma Chapter 4 Triangles \" width=\"289\" height=\"271\" \/><br \/><img src=\"https:\/\/farm1.staticflickr.com\/880\/40940570800_01ce650199_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 4 Triangles \" width=\"354\" height=\"301\" \/><\/p>\n<p>Question 11.<br \/>A vertical stick 10 cm long casts a shadow of 8 cm long. At the same time, a tower casts a shadow 30 m long. Determine the height of the tower. (CB.S.E. 1991)<br \/>Solution:<br \/>The shadows are cast by a vertical stick and a tower at the same time<br \/>Their angles will be equal<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1736\/40940571290_2ac48e6eb0_o.png\" alt=\"RD Sharma 10 Class Solutions Chapter 4 Triangles \" width=\"355\" height=\"461\" \/><\/p>\n<p>Question 12.<br \/>The figure, A = CED, proves that \u2220CAB ~ \u2220CED. Also, find the value of x.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1730\/40940571600_617e1908d5_o.png\" alt=\"RD Sharma Class 10 Textbook PDF Chapter 4 Triangles \" width=\"253\" height=\"217\" \/><br \/>Solution:<br \/>Given : In \u2206ABC,<br \/>\u2220CED = \u2220A<br \/>AB = 9 cm, BE = 2 cm, EC = 10 cm, AD = 7 cm and DC = 8 cm<br \/>To prove :<br \/>(i) \u2206CAB ~ \u2206CED<br \/>(ii) Find the value of x<br \/>Proof: BC = BE + EC = 2 + 10 = 12 cm<br \/>AC = AD + DC = 7 + 8 = 15 cm<br \/>(i) Now in \u2206CAB and \u2206CED,<br \/>\u2220A = \u2220CED (given)<br \/>\u2220C = \u2220C (common)<br \/>\u2206CAB ~ \u2206CED (AA axiom)<br \/><img src=\"https:\/\/farm1.staticflickr.com\/876\/40940571460_18c2b584c2_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 4 Triangles \" width=\"240\" height=\"198\" \/><\/p>\n<p>Question 13.<br \/>The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of the first triangle is 9 cm, what is the corresponding side of the other triangle? (C.B.S.E. 2002C)<br \/>Solution:<br \/>Let perimeter of \u2206ABC = 25 cm<br \/>and perimeter of \u2206DEF = 15 cm<br \/>and side BC of \u2206ABC = 9 cm<br \/>Now we have to find the side EF of \u2206DEF<br \/>\u2206ABC ~ \u2206DEF (given)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1760\/40940571840_55d3c9bb9b_o.png\" alt=\"10th Maths Solution Book Pdf Chapter 4 Triangles \" width=\"361\" height=\"324\" \/><\/p>\n<p>Question 14.<br \/>In \u2206ABC and \u2206DEF, it is being given that: AB = 5 cm, BC = 4 cm and CA = 4.2 cm; DE = 10 cm, EF = 8 cm and FD = 8.4 cm. If AL \u22a5 BC and DM \u22a5 EF, find AL : DM.<br \/>Solution:<br \/>In \u2206ABC and \u2206DEF,<br \/>AB = 5 cm, BC = 4 cm, CA = 4.2 cm, DE = 10 cm, EF = 8 cm and FD = 8.4 cm<br \/>AL \u22a5 BC and DM \u22a5 EF<br \/><img src=\"https:\/\/farm1.staticflickr.com\/899\/40940572210_9f4c1d6a5a_o.png\" alt=\"RD Sharma 10 Chapter 4 Triangles \" width=\"262\" height=\"329\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1736\/27882386097_4359eca0f9_o.png\" alt=\"RD Sharma Class 10 Book Pdf Chapter 4 Triangles \" width=\"358\" height=\"480\" \/><\/p>\n<p>Question 15.<br \/>D and E are the points on the sides AB and AC respectively of a \u2206ABC such that: AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm. Prove that BC =\u00a052\u00a0DE.<br \/>Solution:<br \/>Given : In \u2206ABC, points D and E are on the sides AB and AC respectively<br \/>and AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1738\/40940572590_53610bc291_o.png\" alt=\"Solution Of RD Sharma Class 10 Chapter 4 Triangles \" width=\"295\" height=\"239\" \/><br \/><img src=\"https:\/\/farm1.staticflickr.com\/899\/40940572410_d58aab836d_o.png\" alt=\"RD Sharma 10 Solutions Chapter 4 Triangles \" width=\"285\" height=\"507\" \/><br \/>Hence proved<\/p>\n<p>Question 16.<br \/>D is the mid-point of side BC of an \u2206ABC. AD is bisected at point E and BE produced cuts AC at point X. Prove that BE: EX = 3: 1.<br \/>Solution:<br \/>Given: In \u2206ABC, D is the midpoint of BC, and E is the midpoint of AD<br \/>BE is joined and produced to meet AC at X<br \/>To prove : BE : EX = 3 : 1<br \/>Construction: From D, draw DY || BX meeting AC at Y<br \/><img src=\"https:\/\/farm1.staticflickr.com\/876\/27882387027_42010cf627_o.png\" alt=\"RD Sharma Maths Book For Class 10 Solution Chapter 4 Triangles \" width=\"300\" height=\"187\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1752\/27882386677_f1d23eb285_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 4 Triangles \" width=\"360\" height=\"623\" \/><br \/>Hence proved.<\/p>\n<p>Question 17.<br \/>ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q. Prove that the rectangle obtained by BP and DQ is equal to the rectangle contained AB and BC.<br \/>Solution:<br \/>Given: ABCD is a parallelogram.<br \/>APQ is a straight line that meets BC at P and DC on producing at Q<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1737\/27882387507_8eafb699fb_o.png\" alt=\"RD Sharma Mathematics Class 10 Pdf Download Free Chapter 4 Triangles \" width=\"317\" height=\"132\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1738\/27882387317_04b86f51c9_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 4 Triangles \" width=\"351\" height=\"283\" \/><\/p>\n<p>Question 18.<br \/>In \u2206ABC, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that :<br \/>(i) \u2206OMA ~ \u2206OLC<br \/>(ii)\u00a0OAOC\u00a0=\u00a0OMOL<br \/>Solution:<br \/>Given: In \u2206ABC, AL \u22a5 BC, CM \u22a5 AB which intersect each other at O<br \/><img src=\"https:\/\/farm1.staticflickr.com\/884\/27882387797_f5328db2da_o.png\" alt=\"Class 10 RD Sharma Pdf Chapter 4 Triangles \" width=\"365\" height=\"437\" \/><br \/>(ii)\u00a0OAOC\u00a0=\u00a0OMOL<br \/>Hence proved.<\/p>\n<p>Question 19.<br \/>ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the mid-points of AB, AC, CD, and BD respectively, show that PQRS is a rhombus.<br \/>Solution:<br \/>Given : In quadrilateral ABCD, AD = BC<br \/>P, Q, R, and S are the midpoints of AB, AC, CD, and AD respectively<br \/>PQ, QR, RS, SP are joined<br \/><img src=\"https:\/\/farm1.staticflickr.com\/900\/27882388087_fa9ba0a559_o.png\" alt=\"RD Sharma Class 10 Solutions Pdf Free Download Chapter 4 Triangles \" width=\"347\" height=\"587\" \/><\/p>\n<p>Question 20.<br \/>In an isosceles \u2206ABC, the base AB has produced both the ways to P and Q such that AP x BQ = AC\u00b2. Prove that \u2206APC ~ \u2206BCQ.<br \/>Solution:<br \/>Given : In \u2206ABC, AC = BC<br \/>Base AB is produced to both sides and points P and Q are taken in such a way that<br \/>AP x BQ = AC\u00b2<br \/>CP and CQ are joined<br \/>To prove: \u2206APC ~ \u2206BCQ<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1729\/42751507131_40a8f09be0_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 4 Triangles \" width=\"366\" height=\"508\" \/><\/p>\n<p>Question 21.<br \/>A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m\/sec. If the lamp is 3.6 m above ground, find the length of her shadow after 4 seconds.<br \/>Solution:<br \/>Let AB be the lamp post and CD be the girl and AB = 3.6 m, CD =\u00a090100\u00a0= 9 m<br \/>Distance covered in 4 seconds = 1.2 m x 4 = 4.8 m<br \/>BD = 4.8 m<br \/><img src=\"https:\/\/farm1.staticflickr.com\/875\/42751507341_30eef4c453_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 4 Triangles \" width=\"357\" height=\"556\" \/><br \/>x = 1.6<br \/>Length of her shadow = 1.6 m<\/p>\n<p>Question 22.<br \/>A vertical stick of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.<br \/>Solution:<br \/>Let AB be stick and DE be a tower.<br \/>A stick 6 m long casts a shadow of 4 m i.e., AB = 6 m and BC = 4 m<br \/>Let DE casts a shadow at the same time which is EF = 28 m<br \/>Let the height of tower DE = x<br \/>Now in \u2206ABC and \u2206DEF,<br \/>\u2220B = \u2220E (each 90\u00b0)<br \/>\u2220C = \u2220F (shadows at the same time)<br \/>\u2206ABC ~ \u2206DEF (AA criterion)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1743\/27882388857_f1dde1936e_o.png\" alt=\"Solution Of RD Sharma Class 10 Chapter 4 Triangles \" width=\"347\" height=\"387\" \/><\/p>\n<p>Question 23.<br \/>In the figure, \u2206ABC is right-angled at C and DE \u22a5 AB. Prove that \u2206ABC ~ \u2206ADE and hence find the lengths of AE and DE.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1754\/27882389887_4eed7dc6b6_o.png\" alt=\"RD Sharma 10 Chapter 4 Triangles \" width=\"209\" height=\"149\" \/><br \/>Solution:<br \/>Given: In the figure, \u2206ABC is a right-angled triangle right angle at C.<br \/>DE \u22a5 AB<br \/>To prove:<br \/>(i) \u2206ABC ~ \u2206ADE<br \/>(ii) Find the length of AE and DE<br \/>Proof: In \u2206ABC and \u2206ADE,<br \/>\u2220ACB = \u2220AED (each 90\u00b0)<br \/>\u2220BAC = \u2220DAE (common)<br \/>\u2206ABC ~ \u2206ADE (AA axiom)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1722\/42033682624_5fd6659a5d_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 4 Triangles \" width=\"352\" height=\"94\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1750\/42033682994_446470bc3f_o.png\" alt=\"RD Sharma 10 Class Solutions Chapter 4 Triangles \" width=\"361\" height=\"324\" \/><\/p>\n<p>Question 24.<br \/>In the figure, PA, QB, and RC are each perpendicular to AC. Prove that<br \/><img src=\"https:\/\/farm1.staticflickr.com\/895\/27882391107_416b135b00_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 4 Triangles \" width=\"324\" height=\"292\" \/><br \/>Solution:<br \/>Given: In the figure, PA, QB, and RC are perpendicular on AC and PA = x, QB = z, and RC = y<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1749\/27882390147_16480bb4d1_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 4 Triangles \" width=\"345\" height=\"193\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1748\/27882390647_4b4ced68b3_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 4 Triangles \" width=\"355\" height=\"397\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1751\/27882390807_4f4b9f55c8_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 4 Triangles \" width=\"218\" height=\"311\" \/><br \/>Hence proved.<\/p>\n<p>Question 25.<br \/>In the figure, we have AB || CD || EF. If AB = 6 cm, CD = x cm, EF = 10 cm, BD = 4 cm and DE = y cm, calculate the values of x and y.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1749\/27882391737_6a637702c1_o.png\" alt=\"Learncbse.In Class 10 Chapter 4 Triangles \" width=\"337\" height=\"203\" \/><br \/>Solution:<br \/>In the figure, AB || CD || EF<br \/>AB = 6 cm, EF = 10 cm, BD = 4 cm, CD = x cm and DE = y cm<br \/>In \u2206ABE, CE || AB<br \/>\u2206CED ~ \u2206AEB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1745\/27882391457_caeb767bee_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 4 Triangles \" width=\"353\" height=\"578\" \/><\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5. 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-45\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5<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-1631178077721\" 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-45\"><\/span>Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5?<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 4.5 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-1631178214968\" 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-45-free-pdf\"><\/span>Where can I download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5 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 4.5 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631178349550\" 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-45\"><\/span>What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5?<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 4.5 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 4.5:\u00a0The resemblance of triangles can be demonstrated using a variety of criteria. In this exercise, all of these criteria are explored. Students who wish to learn more about these criteria can use the RD Sharma Class 10 Solutions to get high or full marks in their exams. &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-4-exercise-4-5\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":125912,"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\/125885"}],"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=125885"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125885\/revisions"}],"predecessor-version":[{"id":126013,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125885\/revisions\/126013"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125912"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125885"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125885"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125885"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}