{"id":125883,"date":"2023-09-11T16:24:00","date_gmt":"2023-09-11T10:54:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125883"},"modified":"2023-11-09T12:04:27","modified_gmt":"2023-11-09T06:34:27","slug":"rd-sharma-class-10-solutions-chapter-4-exercise-4-7","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-4-exercise-4-7\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125914\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-Exercise-4.7.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-Exercise-4.7.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-Exercise-4.7-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7:\u00a0<\/strong>The fundamental topic in this exercise is the Baudhayan theorem, also known as the Pythagoras theorem. When students solve the exercise&#8217;s difficulties, they will gain a clear understanding of the idea. In order to lead students in the right route, Kopykitab&#8217;s <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> is developed by subject experts. In addition, you can download 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 Triangles<\/strong><\/a> Exercise 4.7 PDF.<\/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-69e9160be1842\" 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-4-7\/#download-rd-sharma-class-10-solutions-chapter-4-exercise-47-free-pdf\" title=\"Download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7 Free PDF\">Download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7 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-4-7\/#access-answers-to-rd-sharma-solutions-class-10-maths-chapter-4-exercise-47-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 10 Maths Chapter 4 Exercise 4.7- Important Question with Answers\">Access answers to RD Sharma Solutions Class 10 Maths Chapter 4 Exercise 4.7- 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-4-7\/#where-can-i-download-rd-sharma-class-10-solutions-chapter-4-exercise-47-free-pdf\" title=\"Where can I download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7 free PDF?\">Where can I download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7 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-4-7\/#what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-4-exercise-47\" title=\"What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7?\">What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-4-exercise-47-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7 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-4.7.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-4.7.pdf\">RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-4-exercise-47-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 4 Exercise 4.7- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>If the sides of a triangle are 3 cm, 4 cm, and 6 cm long, determine whether the triangle is right-angled. (C.B.S.E. 1992)<br \/>Solution:<br \/>We know that if the square of the hypotenuse (longest side) is equal to the sum of squares of the other two sides then it is a right triangle<br \/>Now the sides of a triangle are 3 cm, 4 cm, and 6 cm<br \/>(Longest side)\u00b2 = (6)\u00b2 = 36<br \/>and sum of two smaller sides = (3)\u00b2 + (4)\u00b2 = 9 + 16 = 25<br \/>36 \u2260 25<br \/>It is not a right-angled triangle<\/p>\n<p>Question 2.<br \/>The sides of certain triangles are given below. Determine which of them are right triangles :<br \/>(i) a = 1 cm, b = 24 cm and c = 25 cm<br \/>(ii) a = 9 cm, b = 16 cm and c = 18 cm<br \/>(iii) a = 1.6 cm, b = 3.8 cm and c = 4 cm<br \/>(iv) a = 8 cm, b = 10 cm and c = 6 cm (C.B.S.E. 1992)<br \/>Solution:<br \/>We know that if the square of the hypotenuse is equal to the sum of squares of the other two sides, then it is a right triangle<br \/>(i) Sides of a triangle are a = 7 cm, b = 5.24 cm and c = 25 cm<br \/>(Longest side)\u00b2 = (25)\u00b2 = 625<br \/>Sum of square of shorter sides = (7)\u00b2 + (24)\u00b2 = 49 + 576 = 625<br \/>625 = 625<br \/>This is right triangle<br \/>(ii) Sides of the triangle are a = 9 cm, b = 16 cm, c = 18 cm<br \/>(Longest side)\u00b2 = (18)\u00b2 = 324<br \/>and sum of squares of shorter sides = (9)\u00b2 + (16)\u00b2 = 81 + 256 = 337<br \/>324 \u2260 337<br \/>It is not a right-angled triangle<br \/>(iii) Sides of the triangle are a = 1.6 cm, 6 = 3.8 cm, c = 4 cm<br \/>(Longest side)\u00b2 = (4)\u00b2 =16<br \/>Sum of squares of shorter two sides + (1.6)\u00b2 + (3.8)\u00b2 = 2.56 + 14.44 = 17.00<br \/>16 \u2260 17<br \/>It is not a right triangle<br \/>(iv) Sides of the triangle are a = 8 cm, b = 10 cm, c = 6 cm<br \/>(Longest side)\u00b2 = (10)\u00b2 = 100<br \/>Sum of squares of shorter sides = (8)\u00b2 + (6)\u00b2 = 64 + 36 = 100<br \/>100 = 100<br \/>It is a right triangle.<\/p>\n<p>Question 3.<br \/>A man goes 15 meters due west and then 8 meters due north. How far is he from the starting point?<br \/>Solution:<br \/>Let a man starts from O, the starting point to west 15 m at A and then from A, 8 m due north at B<br \/>Join OB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1722\/27900390287_4f04a0fced_o.png\" alt=\"RD Sharma Class 10 Chapter 4 Triangles \" width=\"306\" height=\"285\" \/><br \/>Now in right \u2206OAB<br \/>OB\u00b2 = OA\u00b2 + AB\u00b2 (Pythagoras Theorem)<br \/>OB\u00b2 = (15)\u00b2 + (8)\u00b2 = 225 + 64 = 289 = (17)\u00b2<br \/>OB = 17<br \/>The man is 17 m away from the starting point.<\/p>\n<p>Question 4.<br \/>A ladder 17 m long reaches a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.<br \/>Solution:<br \/>Length of ladder = 17 m<br \/>Height of window = 15 m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1757\/27900390337_6fb0e82737_o.png\" alt=\"Triangles Class 10 RD Sharma \" width=\"183\" height=\"172\" \/><br \/>Let the distance of the foot of ladder from the building = x<br \/>Using Pythagoras Theorem<br \/>AC\u00b2 = AB\u00b2 + BC\u00b2<br \/>=&gt; (17)\u00b2 = (15)\u00b2 + x\u00b2<br \/>=&gt; 289 = 225 + x\u00b2<br \/>=&gt; x\u00b2 = 289 \u2013 225<br \/>=&gt; x\u00b2 = 64 = (8)\u00b2<br \/>x = 8<br \/>Distance of the foot of the ladder from the building = 8m<\/p>\n<p>Question 5.<br \/>Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops. (C.B.S.E. 1996C, 2002C)<br \/>Solution:<br \/>Two poles AB and CD which are 6 m and 11 m long respectively are standing on the ground 12 m apart<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1759\/27900390397_e5ecac277c_o.png\" alt=\"RD Sharma Class 10 Solutions Triangles \" width=\"239\" height=\"241\" \/><br \/>Draw AE || BD so that AE = BD = 12 m and ED = AB = 6 m<br \/>Then CE = CD \u2013 ED = 11 \u2013 6 = 5 m<br \/>Now in right \u2206ACE<br \/>Using Pythagoras Theorem,<br \/>AC\u00b2 = AE\u00b2 + EC\u00b2 = (12)\u00b2 + (5)\u00b2 = 144 + 25 = 169 = (13)\u00b2<br \/>AC = 13<br \/>Distance between their tops = 13 m<\/p>\n<p>Question 6.<br \/>In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC. (C.B.S.E. 1994)<br \/>Solution:<br \/>\u2206ABC is an isosceles triangle in which AB = AC = 25 cm .<br \/>AD \u22a5 BC BC = 14 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1759\/27900390737_f719e94c8c_o.png\" alt=\"RD Sharma Class 10 Solutions Chapter 4 Triangles \" width=\"226\" height=\"249\" \/><br \/>Perpendicular AD bisects the base i.e . BD = DC = 7 cm<br \/>Let perpendicular AD = x<br \/>In right \u2206ABD,<br \/>AB\u00b2 = AD\u00b2 + BD\u00b2 (Pythagoras Theorem)<br \/>=&gt; (25)\u00b2 = AD\u00b2 + (7)\u00b2<br \/>=&gt; 625 = AD\u00b2 + 49<br \/>=&gt; AD\u00b2 = 625 \u2013 49<br \/>=&gt; AD\u00b2 = 576 = (24)\u00b2<br \/>=&gt; AD = 24<br \/>Perpendicular AD = 24 cm<\/p>\n<p>Question 7.<br \/>The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach?<br \/>Solution:<br \/>In the first case,<br \/>The foot of the ladder is 6 m away from the wall and its top reaches a window 8 m high<br \/>Let AC be ladder and BC = 6 m, AB = 8 m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1741\/42719859562_4bbff95778_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 4 Triangles \" width=\"301\" height=\"262\" \/><br \/>Now in right \u2206ABC,<br \/>Using Pythagoras Theorem<br \/>AC\u00b2 = BC\u00b2 + AB\u00b2 = (6)\u00b2 + (8)\u00b2 = 36 + 64 = 100 = (10)\u00b2<br \/>AC = 10 m<br \/>In second case,<br \/>ED = AC = 10 m<br \/>BD = 8 m, let ED = x<br \/>ED\u00b2 = BD\u00b2 + EB\u00b2<br \/>=&gt; (10)\u00b2 = (8)\u00b2 + x\u00b2<br \/>=&gt; 100 = 64 + x\u00b2<br \/>=&gt; x\u00b2 = 100 \u2013 64 = 36 = (6)\u00b2<br \/>x = 6<br \/>Height of the ladder on the wall = 6 m<\/p>\n<p>Question 8.<br \/>Two poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.<br \/>Solution:<br \/>Let CD and AB be two poles which are 12 m apart<br \/>AB = 14 m, CD = 9 m and BD = 12 m<br \/>From C, draw CE || DB<br \/>CB = DB = 12 m<br \/>EB = CD = 9 m<br \/>and AE = 14 \u2013 9 = 5 m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1745\/42719859822_d2c4f7a669_o.png\" alt=\"RD Sharma Solutions Class 10 Chapter 4 Triangles \" width=\"234\" height=\"264\" \/><br \/>Now in right \u2206ACE,<br \/>AC\u00b2 = AE\u00b2 + CE\u00b2 (Pythagoras Theorem)<br \/>= (5)\u00b2 + (12)\u00b2<br \/>= 25 + 144 = 169 = (13)\u00b2<br \/>AC = 13<br \/>Distance between their tops = 13 m<\/p>\n<p>Question 9.<br \/>Using the Pythagoras theorem, determine the length of AD in terms of b and c shown in the figure. (C.B.S.E. 1997C)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1731\/42720148462_86259e993c_o.png\" alt=\"Learncbse.In Class 10 Chapter 4 Triangles \" width=\"252\" height=\"182\" \/><br \/>Solution:<br \/>In right \u2206ABC, \u2220A = 90\u00b0<br \/>AB = c, AC = b<br \/>AD \u22a5 BC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1724\/42719860142_6530298544_o.png\" alt=\"Class 10 RD Sharma Solutions Chapter 4 Triangles \" width=\"350\" height=\"560\" \/><br \/>Question 10.<br \/>A triangle has sides 5 cm, 12 cm, and 13 cm. Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is 13 cm. (C.B.S.E. 1992C)<br \/>Solution:<br \/>A triangle has sides 5 cm, 12 cm, and 13 cm<br \/>(Longest side)\u00b2 = (13)\u00b2 = 169<br \/>Sum of squares of shorter sides = (5)\u00b2 + (12)\u00b2 = 25 + 144= 169<br \/>169 = 169<br \/>It is a right triangle whose hypotenuse is 13 cm<br \/><img src=\"https:\/\/farm1.staticflickr.com\/875\/42719860352_10728fb20a_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 4 Triangles \" width=\"311\" height=\"603\" \/><br \/>BD = 4.6 cm<\/p>\n<p>Question 11.<br \/>ABCD is a square, F is the midpoint of AB. BE is one-third of BC. If the area of \u2206FBE = 108 cm\u00b2 finds the length of AC. (C.B.S.E. 1995)<br \/>Solution:<br \/>In square ABCD, F is the midpoint of AB i.e.,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1724\/42719860702_697f151b4c_o.png\" alt=\"RD Sharma Class 10 Solution Chapter 4 Triangles \" width=\"345\" height=\"513\" \/><br \/><img src=\"https:\/\/farm1.staticflickr.com\/880\/42719860532_8ee7e9ff46_o.png\" alt=\"RD Sharma Class 10 Solution Chapter 4 Triangles \" width=\"269\" height=\"253\" \/><\/p>\n<p>Question 12.<br \/>In an isosceles triangle ABC, if AB = AC = 13 cm and the altitude from A on BC is 5 cm, find BC. (C.B.S.E. 2000)<br \/>Solution:<br \/>In \u2206ABC, AB = AC<br \/>AD \u22a5 BC<br \/>AB = AC = 13 cm, AD = 5 cm<br \/>AD \u22a5 BC<br \/>AD bisects BC at D<br \/>BD =\u00a012\u00a0BC<br \/>=&gt; BC = 2BD<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1741\/27900391657_aceb0dccd4_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 4 Triangles \" width=\"353\" height=\"403\" \/><\/p>\n<p>Question 13.<br \/>In a \u2206ABC, AB = BC = CA = 2a and AD \u22a5 BC. Prove that<br \/>(i) AD = a \u221a3<br \/>(ii) area (\u2206ABC) = \u221a3 a\u00b2 (C.B.S.E. 1991)<br \/>Solution:<br \/>In \u2206ABC, AB = BC = AC = 2a<br \/>AD \u22a5 BC<br \/>AD bisects BC at D<br \/>BD = DC =\u00a012\u00a0BC = a<br \/><img src=\"https:\/\/farm1.staticflickr.com\/899\/42719861902_e484d6393e_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Pdf Free Download Chapter 4 Triangles \" width=\"261\" height=\"231\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1736\/27900391777_38d8d16cdc_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Pdf Free Download Chapter 4 Triangles \" width=\"358\" height=\"308\" \/><\/p>\n<p>Question 14.<br \/>The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus. (C.B.S.E. 1993C)<br \/>Solution:<br \/>ABCD is a rhombus whose diagonals AC = 24 cm and BD = 10 cm<br \/>The diagonals of a rhombus bisect each other at right angles<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1752\/42719862022_a4f2c69777_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 4 Triangles \" width=\"239\" height=\"175\" \/><br \/>AO = OC =\u00a0242\u00a0= 12 cm<br \/>and BO = OD =\u00a0102\u00a0= 5 cm<br \/>Now in right \u2206AOB,<br \/>AB\u00b2 = AO\u00b2 + BO\u00b2 (Pythagoras Theorem)<br \/>(12)\u00b2 + (5)\u00b2 = 144 + 25 = 169 = (13)\u00b2<br \/>AB = 13<br \/>Each side of rhombus = 13 cm<\/p>\n<p>Question 15.<br \/>Each side of a rhombus is 10 cm. If one of its diagonals is 16 cm, find the length of the other diagonal.<br \/>Solution:<br \/>In rhombus ABCD, diagonals AC and BD bisect each other at O at right angles<br \/>Each side = 10 cm and one diagonal AC = 16 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1741\/42719862252_3633f13bdc_o.png\" alt=\"Class 10 RD Sharma Chapter 4 Triangles \" width=\"236\" height=\"184\" \/><br \/>AO = OC =\u00a0162\u00a0= 8 cm<br \/>Now in right angled triangle AOB,<br \/>AB\u00b2 = AO\u00b2 + OB\u00b2 (Pythagoras Theorem)<br \/>(10)\u00b2 = (8)\u00b2 + (BO)\u00b2<br \/>=&gt; 100 = 64 + BO\u00b2<br \/>=&gt; BO\u00b2 = 100 \u2013 64 = 36 = (6)\u00b2<br \/>BO = 6<br \/>BD = 2BO = 2 x 6 = 12 cm<\/p>\n<p>Question 16.<br \/>Calculate the height of an equilateral triangle each of whose sides measures 12 cm.<br \/>Solution:<br \/>Each side of the equilateral \u2206ABC = 12 cm<br \/>AD \u22a5 BC which bisects BC at D<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1749\/42719862452_27c879a2bc_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 4 Triangles \" width=\"224\" height=\"214\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1738\/42719862372_4548baeb3b_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 4 Triangles \" width=\"351\" height=\"347\" \/><br \/>BD = DC =\u00a0122\u00a0= 6 cm<\/p>\n<p>Question 17.<br \/>In the figure, \u2220B &lt; 90\u00b0 and segment AD \u22a5 BC. Show that:<br \/>(i) b\u00b2 = h\u00b2 + a\u00b2 + x\u00b2 \u2013 2ax<br \/>(ii) b\u00b2 = a\u00b2 + c\u00b2 \u2013 2ax<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1757\/42719862642_2b6bd08d84_o.png\" alt=\"RD Sharma 10 Class Solutions Chapter 4 Triangles \" width=\"255\" height=\"229\" \/><br \/>Solution:<br \/>Given : In \u2206ABC, \u2220B &lt; 90\u00b0<br \/>AD \u22a5 BC<br \/>AD = c, BC = a, CA = b AD = h, BD = x, DC = a \u2013 x<br \/>To prove: (i) b\u00b2 = h\u00b2 + a\u00b2 + x\u00b2 \u2013 2ax<br \/>(ii) b\u00b2 = a\u00b2 + c\u00b2 \u2013 2ax<br \/>Proof: (i) In right \u2206ADC, AC\u00b2 = AD\u00b2 + DC\u00b2 (Pythagoras Theorem)<br \/>=&gt; b\u00b2 = h\u00b2 + (a \u2013 x)\u00b2 = h\u00b2 + a\u00b2 + x\u00b2 \u2013 2ax<br \/>(ii) Similarly in right \u2206ADB<br \/>AB\u00b2 = AD\u00b2 + BD\u00b2<br \/>c\u00b2 = h\u00b2 + x\u00b2 \u2026.(i)<br \/>b\u00b2 = h\u00b2 + a\u00b2 + x\u00b2 \u2013 2ax = h\u00b2 + x\u00b2 + a\u00b2 \u2013 2ax<br \/>= c\u00b2 + a\u00b2 \u2013 2ax {From (i)}<br \/>= a\u00b2 + c\u00b2 \u2013 2ax<br \/>Hence proved.<\/p>\n<p>Question 18.<br \/>In an equilateral \u2206ABC, AD \u22a5 BC, prove that AD\u00b2 = 3 BD\u00b2. (C.B.S.E. 2002C)<br \/>Solution:<br \/>Given : \u2206ABC is an equilateral in which AB = BC = CA .<br \/>AD \u22a5 BC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1749\/42719863022_0072a1bc3e_o.png\" alt=\"RD Sharma Class 10 Textbook PDF Chapter 4 Triangles \" width=\"347\" height=\"534\" \/><\/p>\n<p>Question 19.<br \/>\u2206ABD is a right triangle right-angled at A and AC \u22a5 BD. Show that<br \/>(i) AB\u00b2 = BC.BD<br \/>(ii) AC\u00b2 = BC.DC<br \/>(iii) AD\u00b2 = BD.CD<br \/>(iv)\u00a0AB2AC2\u00a0=\u00a0BDDC\u00a0[NCERT]<br \/>Solution:<br \/>In \u2206ABD, \u2220A = 90\u00b0<br \/>AC \u22a5 BD<br \/><img src=\"https:\/\/farm1.staticflickr.com\/895\/28895892718_717743f88e_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 4 Triangles \" width=\"366\" height=\"563\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1741\/27900393157_53b154ba00_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 4 Triangles \" width=\"272\" height=\"163\" \/><br \/><img src=\"https:\/\/farm1.staticflickr.com\/887\/27900393337_70fed0e13e_o.png\" alt=\"10th Maths Solution Book Pdf Chapter 4 Triangles \" width=\"368\" height=\"596\" \/><\/p>\n<p>Question 20.<br \/>A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?<br \/>Solution:<br \/>Let AB be the vertical pole whose height = 18 m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1721\/40958363040_c92f68a437_o.png\" alt=\"RD Sharma 10 Chapter 4 Triangles \" width=\"239\" height=\"200\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1741\/28895893058_5af1000668_o.png\" alt=\"RD Sharma 10 Chapter 4 Triangles \" width=\"354\" height=\"271\" \/><\/p>\n<p>Question 21.<br \/>Determine whether the triangle having sides (a \u2013 1) cm, 2 \u221aa cm and (a + 1) cm is a right angled triangle. [CBSE 2010]<br \/>Solution:<br \/>Sides of a triangle are (a \u2013 1) cm, 2 \u221aa cm and (a + 1) cm<br \/>Let AB = (a \u2013 1) cm BC = (a + 1) cm<br \/>and AC = 2 \u221aa<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1727\/40958363430_86b0eb1e2b_o.png\" alt=\"RD Sharma Class 10 Book Pdf Chapter 4 Triangles \" width=\"344\" height=\"402\" \/><\/p>\n<p>Question 22.<br \/>In an acute-angled triangle, express a median in terms of its sides.<br \/>Solution:<br \/>In acute-angled \u2206ABC,<br \/>AD is median and AL \u22a5 BC<br \/>Using a result of a theorem, a sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side<br \/><img src=\"https:\/\/farm1.staticflickr.com\/883\/40958363590_433b51e292_o.png\" alt=\"Solution Of RD Sharma Class 10 Chapter 4 Triangles \" width=\"311\" height=\"626\" \/><\/p>\n<p>Question 23.<br \/>In the right-angled triangle ABC in which \u2220C = 90\u00b0, if D is the midpoint of BC, prove that AB\u00b2 = 4AD\u00b2 \u2013 3AC\u00b2.<br \/>Solution:<br \/>Given : In right \u2206ABC, \u2220C = 90\u00b0<br \/>D is the midpoint of BC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1725\/28895894098_495ba325ca_o.png\" alt=\"RD Sharma 10 Solutions Chapter 4 Triangles \" width=\"356\" height=\"582\" \/><\/p>\n<p>Question 24.<br \/>In the figure, D is the mid-point of side BC and AE \u22a5 BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that :<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1723\/40958364670_9cfcd00aa4_o.png\" alt=\"RD Sharma 10 Solutions Chapter 4 Triangles \" width=\"286\" height=\"282\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1736\/40958364050_41b4f5023b_o.png\" alt=\"RD Sharma Maths Book For Class 10 Solution Chapter 4 Triangles \" width=\"193\" height=\"120\" \/><br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1759\/40958364280_6e5f1389d0_o.png\" alt=\"RD Sharma Maths Book For Class 10 Solution Chapter 4 Triangles \" width=\"366\" height=\"485\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1724\/40958364430_58eeec5773_o.png\" alt=\"RD Sharma Maths Book For Class 10 Solution Chapter 4 Triangles \" width=\"354\" height=\"339\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1744\/40958364550_4c94d21e0b_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 4 Triangles \" width=\"368\" height=\"366\" \/><\/p>\n<p>Question 25.<br \/>In \u2206ABC, \u2220A is obtuse, PB x AC and QC x AB. Prove that:<br \/>(i) AB x AQ = AC x AP<br \/>(ii) BC\u00b2 = (AC x CP + AB x BQ)<br \/>Solution:<br \/>Given: In \u2206ABC, \u2220A is an obtuse angle PB x AC and QC x AB on producing them<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1726\/28895895648_901afc0fe1_o.png\" alt=\"RD Sharma Mathematics Class 10 Pdf Download Free Chapter 4 Triangles \" width=\"298\" height=\"405\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1749\/42719866112_b77cd27f88_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 4 Triangles \" width=\"359\" height=\"381\" \/><\/p>\n<p>Question 26.<br \/>In a right \u2206ABC right-angled at C, if D is the mid-point of BC, prove that BC\u00b2 = 4 (AD\u00b2 \u2013 AC\u00b2).<br \/>Solution:<br \/>Given : In \u2206ABC, \u2220C = 90\u00b0<br \/>D is mid-point of BC<br \/>AD is joined<br \/><img src=\"https:\/\/farm1.staticflickr.com\/892\/28895896038_37d5363a7d_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 4 Triangles \" width=\"351\" height=\"441\" \/><br \/>=&gt; 4AD\u00b2 = 4AC\u00b2 + BC\u00b2<br \/>=&gt; BC\u00b2 = 4AD\u00b2 \u2013 4AC\u00b2<br \/>=&gt; BC\u00b2 = 4 (AD\u00b2 \u2013 AC\u00b2)<br \/>Hence proved.<\/p>\n<p>Question 27.<br \/>In a quadrilateral ABCD, \u2220B = 90\u00b0, AD\u00b2 = AB\u00b2 + BC\u00b2 + CD\u00b2, prove that \u2220ACD = 90\u00b0.<br \/>Solution:<br \/>Given : ABCD is a quadrilateral in which \u2220B = 90\u00b0 and AD\u00b2 = AB\u00b2 + BC\u00b2 + CD\u00b2<br \/>To prove : \u2220ACD = 90\u00b0<br \/>Construction: Join AC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1758\/28895896198_391bb8f92d_o.png\" alt=\"Class 10 RD Sharma Pdf Chapter 4 Triangles \" width=\"210\" height=\"222\" \/><br \/>proof : In right \u2206ABC, \u2220B = 90\u00b0<br \/>AC\u00b2 = AB\u00b2 + BC\u00b2 \u2026.(i) (Pythagoras Theorem)<br \/>But AD\u00b2 = AB\u00b2 + BC\u00b2 + CD\u00b2 (given)<br \/>=&gt; AD\u00b2 = AC\u00b2 + CD\u00b2 {From (i)}<br \/>\u2206ACD is a right angle with right angle ACD<br \/>Hence proved.<\/p>\n<p>Question 28.<br \/>An airplane leaves an airport and flies due north at a speed of 1000 km\/hr. At the same time, another airplane leaves the same airport and flies due west at a speed of 1200 km\/ hr. How far apart will be the two planes after 112 hours?<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1753\/28895896808_e6fd1ee29f_o.png\" alt=\"RD Sharma Class 10 Solutions Pdf Free Download Chapter 4 Triangles \" width=\"307\" height=\"308\" \/><br \/>Speed of the first plane = 1000 km\/hr<br \/>Distance travelled in 112\u00a0hour due north = 1000 x\u00a032\u00a0= 1500 km<br \/>Speed of the second plane = 1200 km\/hr<br \/>Distance traveled in 112 hours due west<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1752\/28895896528_6d1d7a00fb_o.png\" alt=\"RD Sharma Class 10 Solutions Pdf Free Download Chapter 4 Triangles \" width=\"353\" height=\"301\" \/><\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7. 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-47\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7<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-1631178093074\" 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-47\"><\/span>Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7?<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 4.7 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-1631178225213\" 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-47-free-pdf\"><\/span>Where can I download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7 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 4.7 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631178365182\" 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-47\"><\/span>What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7?<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 4.7 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 4.7:\u00a0The fundamental topic in this exercise is the Baudhayan theorem, also known as the Pythagoras theorem. When students solve the exercise&#8217;s difficulties, they will gain a clear understanding of the idea. In order to lead students in the right route, Kopykitab&#8217;s RD Sharma Class 10 Solutions is &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-4-exercise-4-7\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 4 Exercise 4.7 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":125914,"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\/125883"}],"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=125883"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125883\/revisions"}],"predecessor-version":[{"id":504840,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125883\/revisions\/504840"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125914"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125883"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125883"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125883"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}