{"id":126425,"date":"2023-03-29T13:27:00","date_gmt":"2023-03-29T07:57:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126425"},"modified":"2023-12-07T10:32:53","modified_gmt":"2023-12-07T05:02:53","slug":"rd-sharma-class-9-solutions-chapter-13-exercise-13-4","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-13-exercise-13-4\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Exercise 13.4 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-127232\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-13-Exercise-13.4.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Exercise 13.4\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-13-Exercise-13.4.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-13-Exercise-13.4-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 13 Exercise 13.4&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\"><strong>RD Sharma Class 9 Solutions Chapter 13 Exercise 13.4<\/strong> are available here. By practicing RD Sharma Class 9 solutions, students can earn higher academic degrees. Our experts solve these solutions with the utmost precision to help students learn. You can download <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-13-linear-equations-in-two-variables\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 13<\/a> Exercise 13.4 from the link provided below.\u00a0<\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-for-maths\/\" target=\"_blank\" rel=\"noopener\"><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Maths Solutions&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Maths Solutions<\/span><\/a><\/li>\n<\/ul>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 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Exercise 13.4 PDF free?\">Can I download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.4 PDF free?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-13-exercise-13-4\/#what-are-the-benefits-of-studying-rd-sharma-class-9-solutions\" title=\"What are the benefits of studying RD Sharma Class 9 Solutions?\">What are the benefits of studying RD Sharma Class 9 Solutions?<\/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-9-solutions-chapter-13-exercise-13-4\/#is-rd-sharma-enough-for-class-12-maths\" title=\"Is RD Sharma enough for Class 12 Maths?\">Is RD Sharma enough for Class 12 Maths?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-13-exercise-134\"><\/span>Download <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 13 Exercise 13.4&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 13 Exercise 13.4<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Go through the RD Sharma Ex 13.4 Class 9 Solutions PDF:<\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<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-9-Maths-Chapter-13-Ex-13.4.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-SHARMA-Solutions-Class-9-Maths-Chapter-13-Ex-13.4.pdf\">RD Sharma Class 9 Solutions Chapter 13 Exercise 13.4<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-exercise-134-class-9th-solutions\"><\/span>Access <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 13 Exercise 13.4&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Exercise 13.4 Class 9th Solutions<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Question 1.<br \/>In an \u2206ABC, D, E, and F are respectively, the mid-points of BC, CA, and AB. If the lengths of side AB, BC, and CA are 7cm, 8cm, and 9cm, respectively, find the perimeter of \u2206DEF.<br \/>Solution:<\/p>\n<p>In \u2206ABC, D, E, and F are the mid-points of sides,<br \/>BC, CA, and AB respectively<br \/>AB = 7cm, BC = 8cm and CA = 9cm<br \/>\u2235 D and E are the midpoints of BC and CA<br \/>\u2234 DE || AB and DE =<span id=\"MathJax-Element-13-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-85\" class=\"math\"><span id=\"MathJax-Span-86\" class=\"mrow\"><span id=\"MathJax-Span-87\" class=\"mfrac\"><span id=\"MathJax-Span-88\" class=\"mn\">1<\/span><span id=\"MathJax-Span-89\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0AB =<span id=\"MathJax-Element-14-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-90\" class=\"math\"><span id=\"MathJax-Span-91\" class=\"mrow\"><span id=\"MathJax-Span-92\" class=\"mfrac\"><span id=\"MathJax-Span-93\" class=\"mn\">1<\/span><span id=\"MathJax-Span-94\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 7 = 3.5cm<br \/>Similarly,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1931\/44730425365_eddfaf5c27_o.png\" alt=\"RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"294\" height=\"346\" \/><\/p>\n<p>Question 2.<br \/>In a triangle \u2220ABC, \u2220A = 50\u00b0, \u2220B = 60\u00b0 and \u2220C = 70\u00b0. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.<br \/>Solution:<\/p>\n<p>In \u2206ABC,<br \/>\u2220A = 50\u00b0, \u2220B = 60\u00b0 and \u2220C = 70\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1969\/44730425025_c87fcbbbe9_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables\" width=\"252\" height=\"170\" \/><br \/>D, E, and F are the midpoints of sides BC, CA, and AB respectively<br \/>DE, EF, and ED are joined<br \/>\u2235 D, E, and F are the midpoints of sides BC, CA, and AB respectively<br \/>\u2234 EF || BC<br \/>DE || AB and FD || AC<br \/>\u2234 BDEF and CDEF are parallelogram<br \/>\u2234 \u2220B = \u2220E = 60\u00b0 and \u2220C = \u2220F = 70\u00b0<br \/>Then \u2220A = \u2220D = 50\u00b0<br \/>Hence \u2220D = 50\u00b0, \u2220E = 60\u00b0 and \u2220F = 70\u00b0<\/p>\n<p>Question 3.<br \/>In a triangle, P, Q, and R are the mid-points of sides BC, CA, and AB respectively. If AC = 21 cm, BC = 29cm, and AB = 30cm, find the perimeter of the quadrilateral ARPQ.<br \/>Solution:<\/p>\n<p>P, Q, and R are the midpoints of sides BC, CA, and AB respectively<br \/>AC = 21 cm, BC = 29 cm and AB = 30\u00b0<br \/>\u2235 P, Q, R, and the midpoints of sides BC, CA, and AB respectively.<br \/>\u2234 PQ || AB and PQ =\u00a0<span id=\"MathJax-Element-15-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-95\" class=\"math\"><span id=\"MathJax-Span-96\" class=\"mrow\"><span id=\"MathJax-Span-97\" class=\"mfrac\"><span id=\"MathJax-Span-98\" class=\"mn\">1<\/span><span id=\"MathJax-Span-99\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0AB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1958\/44730424885_9a0ed779c9_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"265\" height=\"232\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1915\/44730424795_40e17dd689_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 13 Linear Equations in Two Variables\" width=\"302\" height=\"327\" \/><\/p>\n<p>Question 4.<br \/>In an \u2206ABC median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram.<br \/>Solution:<\/p>\n<p>Given: In \u2206ABC, AD is median and AD is produced to X such that DX = AD<br \/>To prove : ABXC is a parallelogram<br \/>Construction: Join BX and CX<br \/>Proof: In \u2206ABD and \u2206CDX<br \/>AD = DX (Given)<br \/>BD = DC (D is midpoints)<br \/>\u2220ADB = \u2220CDX (Vertically opposite angles)<br \/>\u2234 \u2206ABD \u2245 \u2206CDX (SAS criteria)<br \/>\u2234 AB = CX (c.p.c.t.)<br \/>and \u2220ABD = \u2220DCX<\/p>\n<p>But these are alternate angles<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1903\/44730424655_81e6523691_o.png\" alt=\"Linear Equations in Two Variables Class 9 RD Sharma Solutions\" width=\"213\" height=\"254\" \/><br \/>\u2234 AB || CX and AB = CX<br \/>\u2234 ABXC is a parallelogram.<\/p>\n<p>Question 5.<br \/>In\u00a0 \u2206ABC, E, and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.<br \/>Solution:<\/p>\n<p>Given: In \u2206ABC, E, and F are the mid-points of AC and AB respectively.<br \/>EF is joined.<br \/>AP \u22a5 BC is drawn which intersects EF at Q and meets BC at P.<br \/>To prove: AQ = QP<br \/>proof: In \u2206ABC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1914\/44730424515_924d4895f1_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables\" width=\"251\" height=\"175\" \/><br \/>E and F are the midpoints of AC and AB<br \/>\u2234 EF || BC and EF =\u00a0<span id=\"MathJax-Element-16-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-100\" class=\"math\"><span id=\"MathJax-Span-101\" class=\"mrow\"><span id=\"MathJax-Span-102\" class=\"mfrac\"><span id=\"MathJax-Span-103\" class=\"mn\">1<\/span><span id=\"MathJax-Span-104\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>BC<br \/>\u2234 \u2220F = \u2220B<br \/>In \u2206ABP,<br \/>F is midpoint of AB and Q is the midpoint of FE or FQ || BC<br \/>\u2234 Q is midpoint of AP,<br \/>\u2234 AQ = QP<\/p>\n<p>Question 6.<br \/>In an \u2206ABC, BM, and CN are perpendiculars from B and C respectively on any line passing through A. If L is the mid-point of BC, prove that ML NL.<br \/>Solution:<br \/>In \u2206ABC,<br \/>BM and CN are perpendicular on a line drawn from A. L is the midpoint of BC. ML and NL are joined.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1945\/44730424445_004c386a59_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables\" width=\"196\" height=\"261\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1977\/44730424405_55f9c96401_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 - 6a\" width=\"355\" height=\"549\" \/><\/p>\n<p>Question 7.<br \/>In the figure triangle, ABC is right-angled at B. Given that AB = 9cm. AC = 15cm and D, E are the midpoints of the sides AB and AC respectively, calculate.<br \/>(i) The length of BC<br \/>(ii) The area of \u2206ADC<br \/>Solution:<\/p>\n<p>In \u2206ABC, \u2220B = 90\u00b0<br \/>AC =15 cm, AB = 9cm<br \/>D and E are the midpoints of sides AB and AC respectively and D, and E are joined.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1945\/44730424225_79ed4183e6_o.png\" alt=\"Class 9 Maths Chapter 13 Linear Equations in Two Variables RD Sharma Solutions\" width=\"353\" height=\"728\" \/><\/p>\n<p>Question 8.<br \/>In the figure, M, N, and P are the midpoints of AB, AC, and BC respectively. If MN = 3 cm, NP = 3.5cm, and MP = 2.5cm, calculate BC, AB, and AC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1968\/44730423915_6f58122277_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 13 Linear Equations in Two Variables\" width=\"179\" height=\"175\" \/><br \/>Solution:<br \/>In \u2206ABC,<br \/>M, N, and P are the mid points of sides, AB, AC, and BC respectively.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1914\/44730423835_0e7ae9e22a_o.png\" alt=\"RD Sharma Class 9 Book Chapter 13 Linear Equations in Two Variables\" width=\"360\" height=\"606\" \/><\/p>\n<p>Question 9.<br \/>In the figure, AB = AC and CP || BA and AP are the bisector of the exterior \u2220CAD of \u2206ABC. Prove that (i) \u2220PAC = \u2220BCA (ii) ABCP is a parallelogram.<br \/>Solution:<br \/>Given: In ABC, AB = AC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1905\/44730423615_b6c844570b_o.png\" alt=\"Linear Equations in Two Variables With Solutions PDF RD Sharma Class 9 Solutions\" width=\"213\" height=\"197\" \/><br \/>nd CP || BA, AP is the bisector of exterior \u2220CAD of \u2206ABC<br \/>To prove :<\/p>\n<p>(i) \u2220PAC = \u2220BCA<br \/>(ii) ABCP is a ||gm<br \/>Proof : (i) In \u2206ABC,<br \/>\u2235 AB =AC<br \/>\u2234 \u2220C = \u2220B (Angles opposite to equal sides) and ext.<br \/>\u2220CAD = \u2220B + \u2220C<br \/>= \u2220C + \u2220C = 2\u2220C \u2026.(i)<br \/>\u2235 AP is the bisector of \u2220CAD<br \/>\u2234 2\u2220PAC = \u2220CAD \u2026(ii)<br \/>From (i) and (ii)<br \/>\u2220C = 2\u2220PAC<br \/>\u2220C = \u2220CAD or \u2220BCA = \u2220PAC<br \/>Hence \u2220PAC = \u2220BCA<\/p>\n<p>(ii) But there are alternate angles,<br \/>\u2234 AD || BC<br \/>But BA || CP<br \/>\u2234 ABCP is a ||gm.<\/p>\n<p>Question 10.<br \/>ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the mid-points of the sides, in order, is a rectangle.<br \/>Solution:<\/p>\n<p>Given: In the figure, ABCD is a kite in which AB = AD and BC = CD.<br \/>P, Q, R, and S are the mid points of the sides AB, BC, CD, and DA respectively.<br \/>To prove: PQRS is a rectangle.<br \/>Construction: Join AC and BD.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1978\/44730423445_73165e3959_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 13 Linear Equations in Two Variables\" width=\"206\" height=\"264\" \/><br \/>Proof: In \u2206ABD,<br \/>P and S are midpoints of AB and AD<br \/>\u2234 PS || BD and PS =\u00a0<span id=\"MathJax-Element-17-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-105\" class=\"math\"><span id=\"MathJax-Span-106\" class=\"mrow\"><span id=\"MathJax-Span-107\" class=\"mfrac\"><span id=\"MathJax-Span-108\" class=\"mn\">1<\/span><span id=\"MathJax-Span-109\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0BD \u2026(i)<br \/>Similarly in \u2206BCD,<br \/>Q and R the midpoints of BC and CD<br \/>\u2234 QR || BD and<br \/>QR =\u00a0<span id=\"MathJax-Element-18-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-110\" class=\"math\"><span id=\"MathJax-Span-111\" class=\"mrow\"><span id=\"MathJax-Span-112\" class=\"mfrac\"><span id=\"MathJax-Span-113\" class=\"mn\">1<\/span><span id=\"MathJax-Span-114\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0BD \u2026(ii)<\/p>\n<p>\u2234 Similarly, we can prove that PQ || SR and PQ = SR \u2026(iii)<br \/>From (i) and (ii) and (iii)<br \/>PQRS is a parallelogram,<br \/>\u2235 AC and BD intersect each other at right angles.<br \/>\u2234 PQRS is a rectangle.<\/p>\n<p>Question 11.<br \/>Let ABC be an isosceles triangle in which AB = AC. If D, E, and F be the mid-points of the sides BC, CA, and AB respectively, show that the segment AD and EF bisect each other at right angles.<br \/>Solution:<\/p>\n<p>In \u2206ABC, AB = AC<br \/>D, E, and F are the midpoints of the sides BC, CA, and AB respectively,<br \/>AD and EF are joined intersecting at O<br \/>To prove: AD and EF bisect each other at right angles.<br \/>Construction: Join DE and DF.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1916\/43826453580_6d6dd3b19d_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 13 Linear Equations in Two Variables\" width=\"168\" height=\"174\" \/><br \/>Proof : \u2235 D, E, and F are the mid-points of<br \/>the sides BC, CA, and AB respectively<\/p>\n<p>\u2234 AFDE is a ||gm<br \/>\u2234 AF = DE and AE = DF<br \/>But AF = AE<br \/>(\u2235 E and F are mid-points of equal sides AB and AC)<br \/>\u2234 AF = DF = DE = AE<br \/>\u2234AFDE is a rhombus<\/p>\n<p>\u2235 The diagonals of a rhombus bisect each other at a right angle.<br \/>\u2234 AO = OD and EO = OF<br \/>Hence, AD and EF bisect each other at right angles.<\/p>\n<p>Question 12.<br \/>Show that the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other.<br \/>Solution:<\/p>\n<p>Given: In quad. ABCD,<br \/>P, Q, R, and S are the midpoints of sides AB, BC, CD, and DA respectively.<br \/>PR and QS to intersect each other at O<br \/>To prove: PO = OR and QO = OS<br \/>Construction: Join PQ, QR, RS, and SP, and also join AC.<br \/>Proof: In \u2206ABC<\/p>\n<p>P and Q are the mid-points of AB and BC<br \/>\u2234 PQ || AC and PQ =\u00a0<span id=\"MathJax-Element-19-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-115\" class=\"math\"><span id=\"MathJax-Span-116\" class=\"mrow\"><span id=\"MathJax-Span-117\" class=\"mfrac\"><span id=\"MathJax-Span-118\" class=\"mn\">1<\/span><span id=\"MathJax-Span-119\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0AC \u2026(i)<br \/>Similarly\u00a0 \u2206ADC,<br \/>S and R are the mid-points of AD and CD<br \/>\u2234 SR || AC and SR =\u00a0<span id=\"MathJax-Element-20-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-120\" class=\"math\"><span id=\"MathJax-Span-121\" class=\"mrow\"><span id=\"MathJax-Span-122\" class=\"mfrac\"><span id=\"MathJax-Span-123\" class=\"mn\">1<\/span><span id=\"MathJax-Span-124\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0AC ..(ii)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1939\/44730423215_301d3db1da_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"175\" height=\"179\" \/><br \/>from (i) and (ii)<br \/>PQ = SQ and PQ || SR<br \/>PQRS is a ||gm (\u2235 opposite sides are equal area parallel)<br \/>But the diagonals of a ||gm bisect each other.<br \/>\u2234 PR and QS bisect each other.<\/p>\n<p>Question 13.<br \/>Fill in the blanks to make the following statements correct :<br \/>(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is \u2026<br \/>(ii) The triangle formed by joining the mid-points of the sides of a right triangle is \u2026<br \/>(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is \u2026<br \/>Solution:<br \/>(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is an isosceles triangle.<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1969\/43826453400_16ff84798c_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"183\" height=\"167\" \/><br \/>(ii) The triangle formed by joining the mid-points of the sides of a right triangle is a right triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1926\/44730423065_764ffab47f_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 13 Linear Equations in Two Variables\" width=\"132\" height=\"174\" \/><br \/>(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is a parallelogram.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1951\/43826453220_357dc200cc_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"182\" height=\"182\" \/><\/p>\n<p>Question 14.<br \/>ABC is a triangle and through A, B, and C lines are drawn parallel to BC, CA, and AB respectively intersecting at P, Q, and R. Prove that the perimeter of \u2206PQR is double the perimeter of \u2206ABC.<br \/>Solution:<\/p>\n<p>Given: In \u2206ABC,<br \/>Through A, B, and C, lines are drawn parallel to BC, CA, and AB respectively meeting at P, Q, and R.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1960\/44730422895_d17f7caced_o.png\" alt=\"RD Sharma Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"229\" height=\"183\" \/><br \/>To prove : Perimeter of \u2206PQR = 2 x perimeter of \u2206ABC<br \/>Proof : \u2235 PQ || BC and QR || AB<br \/>\u2234 ABCQ is a ||gm<br \/>\u2234 BC = AQ<br \/>Similarly, BCAP is a ||gm<br \/>\u2234 BC = AP \u2026(i)<br \/>\u2234 AQ = AP = BL<br \/>\u21d2 PQ = 2BC<\/p>\n<p>Similarly, we can prove that<br \/>QR = 2AB and PR = 2AC<br \/>Now perimeter of \u2206PQR.<br \/>= PQ + QR + PR = 2AB + 2BC + 2AC<br \/>= 2(AB + BC + AC)<br \/>= 2 perimeters of \u2206ABC.<br \/>Hence proved<\/p>\n<p>Question 15.<br \/>In the figure, BE \u22a5 AC. AD is any line from A to BC intersecting BE in H. P, Q and R are respectively the mid-points of AH, AB, and BC. Prove that PQR = 90\u00b0.<br \/>Solution:<\/p>\n<p>Given: In \u2206ABC, BE \u22a5 AC<br \/>AD is any line from A to BC meeting BC in D and intersecting BE in H. P, Q, and R are respectively midpoints of AH, AB, and BC. PQ and QR are joined B.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1937\/43826453050_1dfb7a3a70_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables\" width=\"220\" height=\"174\" \/><br \/>To prove : \u2220PQR = 90\u00b0<br \/>Proof: In \u2206ABC,<br \/>Q and R the midpoints of AB and BC 1<br \/>\u2234 QR || AC and QR =\u00a0<span id=\"MathJax-Element-21-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-125\" class=\"math\"><span id=\"MathJax-Span-126\" class=\"mrow\"><span id=\"MathJax-Span-127\" class=\"mfrac\"><span id=\"MathJax-Span-128\" class=\"mn\">1<\/span><span id=\"MathJax-Span-129\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0AC<\/p>\n<p>Similarly, in \u2206ABH,<br \/>Q and P are the midpoints of AB and AH<br \/>\u2234 QP || BH or QP || BE<br \/>But AC \u22a5 BE<br \/>\u2234 QP \u22a5 QR<br \/>\u2234 \u2220PQR = 90\u00b0<\/p>\n<p>Question 16.<br \/>ABC is a triangle. D is a point on AB such that AD =\u00a0<span id=\"MathJax-Element-22-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-130\" class=\"math\"><span id=\"MathJax-Span-131\" class=\"mrow\"><span id=\"MathJax-Span-132\" class=\"mfrac\"><span id=\"MathJax-Span-133\" class=\"mn\">1<\/span><span id=\"MathJax-Span-134\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>\u00a0AB and E is a point on AC such that AE =\u00a0<span id=\"MathJax-Element-23-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-135\" class=\"math\"><span id=\"MathJax-Span-136\" class=\"mrow\"><span id=\"MathJax-Span-137\" class=\"mfrac\"><span id=\"MathJax-Span-138\" class=\"mn\">1<\/span><span id=\"MathJax-Span-139\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>\u00a0AC. Prove that DE =\u00a0<span id=\"MathJax-Element-24-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-140\" class=\"math\"><span id=\"MathJax-Span-141\" class=\"mrow\"><span id=\"MathJax-Span-142\" class=\"mfrac\"><span id=\"MathJax-Span-143\" class=\"mn\">1<\/span><span id=\"MathJax-Span-144\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>\u00a0BC.<br \/>Solution:<\/p>\n<p>Given: In \u2206ABC,<br \/>D is a point on AB such that<br \/>AD =\u00a0<span id=\"MathJax-Element-25-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-145\" class=\"math\"><span id=\"MathJax-Span-146\" class=\"mrow\"><span id=\"MathJax-Span-147\" class=\"mfrac\"><span id=\"MathJax-Span-148\" class=\"mn\">1<\/span><span id=\"MathJax-Span-149\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span> AB and E is a point on AC such as 1<br \/>that AE =\u00a0<span id=\"MathJax-Element-26-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-150\" class=\"math\"><span id=\"MathJax-Span-151\" class=\"mrow\"><span id=\"MathJax-Span-152\" class=\"mfrac\"><span id=\"MathJax-Span-153\" class=\"mn\">1<\/span><span id=\"MathJax-Span-154\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>\u00a0AC<br \/>DE is joined.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1975\/44919533644_b8f39aaa22_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 13 Linear Equations in Two Variables\" width=\"247\" height=\"165\" \/><br \/>To prove : DE =\u00a0<span id=\"MathJax-Element-27-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-155\" class=\"math\"><span id=\"MathJax-Span-156\" class=\"mrow\"><span id=\"MathJax-Span-157\" class=\"mfrac\"><span id=\"MathJax-Span-158\" class=\"mn\">1<\/span><span id=\"MathJax-Span-159\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>\u00a0BC<br \/>Construction: Take P and Q the midpoints of AB and AC and join them<br \/>Proof: In \u2206ABC,<\/p>\n<p>\u2235 P and Q are the mid-points of AB and AC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1919\/44919533464_72bb952b2f_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 13 Linear Equations in Two Variables\" width=\"354\" height=\"276\" \/><\/p>\n<p>Question 17.<br \/>In the figure, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ =\u00a0<span id=\"MathJax-Element-28-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-160\" class=\"math\"><span id=\"MathJax-Span-161\" class=\"mrow\"><span id=\"MathJax-Span-162\" class=\"mfrac\"><span id=\"MathJax-Span-163\" class=\"mn\">1<\/span><span id=\"MathJax-Span-164\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>\u00a0AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1915\/43826452710_728564a439_o.png\" alt=\"Linear Equations in Two Variables Class 9 RD Sharma Solutions\" width=\"199\" height=\"161\" \/><br \/>Solution:<\/p>\n<p>Given : In ||gm ABCD,<br \/>P is the mid-point of DC and Q is a point on AC such that CQ =\u00a0<span id=\"MathJax-Element-29-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-165\" class=\"math\"><span id=\"MathJax-Span-166\" class=\"mrow\"><span id=\"MathJax-Span-167\" class=\"mfrac\"><span id=\"MathJax-Span-168\" class=\"mn\">1<\/span><span id=\"MathJax-Span-169\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span> AC. PQ is produced and meets BC at R.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1926\/44919533114_7f0ca04e8c_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 13 Linear Equations in Two Variables\" width=\"201\" height=\"154\" \/><br \/>To prove: R is midpoint of BC<br \/>Construction: Join BD<br \/>Proof : \u2235 In ||gm ABCD,<\/p>\n<p>\u2235 Diagonal AC and BD bisect each other at O<br \/>\u2234 AO = OC =\u00a0<span id=\"MathJax-Element-30-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-170\" class=\"math\"><span id=\"MathJax-Span-171\" class=\"mrow\"><span id=\"MathJax-Span-172\" class=\"mfrac\"><span id=\"MathJax-Span-173\" class=\"mn\">1<\/span><span id=\"MathJax-Span-174\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0AC \u2026(i)<br \/>In \u2206OCD,<br \/>P and Q the mid-points of CD and CO<br \/>\u2234 PQ || OD and PQ =\u00a0<span id=\"MathJax-Element-31-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-175\" class=\"math\"><span id=\"MathJax-Span-176\" class=\"mrow\"><span id=\"MathJax-Span-177\" class=\"mfrac\"><span id=\"MathJax-Span-178\" class=\"mn\">1<\/span><span id=\"MathJax-Span-179\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0OD<br \/>In \u2206BCD,<br \/>P is mid-poiht of DC and PQ || OD (Proved above)<br \/>Or PR || BD<br \/>\u2234 R is mid-point BC.<\/p>\n<p>Question 18.<br \/>In the figure, ABCD and PQRC are rectangles and Q is the mid-point of AC.<br \/>Prove that (i) DP = PC (ii) PR =\u00a0<span id=\"MathJax-Element-32-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-180\" class=\"math\"><span id=\"MathJax-Span-181\" class=\"mrow\"><span id=\"MathJax-Span-182\" class=\"mfrac\"><span id=\"MathJax-Span-183\" class=\"mn\">1<\/span><span id=\"MathJax-Span-184\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0AC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1959\/43826452500_68b2592948_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 13 Linear Equations in Two Variables\" width=\"219\" height=\"168\" \/><br \/>Solution:<br \/>Given: ABCD is PQRC are rectangles and Q is the mid-point of AC.<br \/>To prove : (i) DP = PC (ii) PR =\u00a0<span id=\"MathJax-Element-33-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-185\" class=\"math\"><span id=\"MathJax-Span-186\" class=\"mrow\"><span id=\"MathJax-Span-187\" class=\"mfrac\"><span id=\"MathJax-Span-188\" class=\"mn\">1<\/span><span id=\"MathJax-Span-189\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0AC<br \/>Construction: Join diagonal AC which passes through Q and joins PR.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1943\/44919532684_7f582e57bb_o.png\" alt=\"Class 9 Maths Chapter 13 Linear Equations in Two Variables RD Sharma Solutions\" width=\"209\" height=\"160\" \/><br \/>Proof : (i) In \u2206ACD,<br \/>Q is the mid-point of AC and QP || AD (Sides of rectangles)<br \/>\u2234 P is the mid-point of the CD<br \/>\u2234 DP = PC<br \/>(ii) \u2235PR and QC are the diagonals of rectangle PQRC<br \/>\u2234 PR = QC<br \/>But Q is the mid-point of AC<br \/>\u2234 QC =\u00a0<span id=\"MathJax-Element-34-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-190\" class=\"math\"><span id=\"MathJax-Span-191\" class=\"mrow\"><span id=\"MathJax-Span-192\" class=\"mfrac\"><span id=\"MathJax-Span-193\" class=\"mn\">1<\/span><span id=\"MathJax-Span-194\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0AC<br \/>Hence PR =\u00a0<span id=\"MathJax-Element-35-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-195\" class=\"math\"><span id=\"MathJax-Span-196\" class=\"mrow\"><span id=\"MathJax-Span-197\" class=\"mfrac\"><span id=\"MathJax-Span-198\" class=\"mn\">1<\/span><span id=\"MathJax-Span-199\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0AC<\/p>\n<p>Question 19.<br \/>ABCD is a parallelogram, and E and F are the midpoints AB and CD respectively. GFI is any line intersecting AD, EF and BC at Q P, and H respectively. Prove that GP = PH.<br \/>Solution:<br \/>Given : In ||gm ABCD,<br \/>E and F are mid-points of AB and CD<br \/>GH is any line intersecting AD, EF and BC at GP and H respectively<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1967\/44730422275_995bffb9a9_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 13 Linear Equations in Two Variables\" width=\"229\" height=\"163\" \/><br \/>To prove : GP = PH<br \/>Proof: \u2235 E and F are the mid-points of AB and CD<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1924\/30703633437_7ebbbf3d29_o.png\" alt=\"RD Sharma Class 9 Book Chapter 13 Linear Equations in Two Variables\" width=\"353\" height=\"508\" \/><\/p>\n<p>Question 20.<br \/>BM and CN are perpendicular to a line passing, through the vertex A of a triangle ABC. If L is the mid-point of BC, prove that LM = LN.<br \/>Solution:<br \/>In \u2206ABC,<br \/>BM and CN are perpendicular on a line drawn from A.<br \/>L is the midpoint of BC.<br \/>ML and NL are joined.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1928\/44730422125_915a379f7a_o.png\" alt=\"Linear Equations in Two Variables With Solutions PDF RD Sharma Class 9 Solutions\" width=\"344\" height=\"737\" \/><\/p>\n<p>We have included all the information regarding\u00a0<a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> RD Sharma Class 9 Solutions Chapter 13 Exercise 13.4<span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">. If you have any queries feel free to ask in the comment section.\u00a0<\/span><\/p>\n<h2><span class=\"ez-toc-section\" id=\"faq-rd-sharma-class-9-solutions-chapter-13-exercise-134\"><\/span>FAQ: RD Sharma Class 9 Solutions Chapter 13 Exercise 13.4<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-1631259346686\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-download-rd-sharma-class-9-solutions-chapter-13-exercise-134-pdf-free\"><\/span>Can I download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.4 PDF free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.4 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631259446520\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-studying-rd-sharma-class-9-solutions\"><\/span>What are the benefits of studying RD Sharma Class 9 Solutions?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>By practicing RD Sharma Class 9 Solutions, students can earn higher academic grades. Our experts solve these solutions with utmost accuracy to help students in their studies.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631259447454\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-enough-for-class-12-maths\"><\/span>Is RD Sharma enough for Class 12 Maths?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>RD Sharma is a good book that gives you thousands of questions to practice.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 13 Exercise 13.4 are available here. By practicing RD Sharma Class 9 solutions, students can earn higher academic degrees. Our experts solve these solutions with the utmost precision to help students learn. You can download RD Sharma Class 9 Solutions Chapter 13 Exercise 13.4 from the link provided below.\u00a0 &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Exercise 13.4 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-13-exercise-13-4\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Exercise 13.4 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":127232,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[76786,76845],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126425"}],"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=126425"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126425\/revisions"}],"predecessor-version":[{"id":518352,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126425\/revisions\/518352"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/127232"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126425"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126425"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126425"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}