{"id":126681,"date":"2023-09-12T13:23:00","date_gmt":"2023-09-12T07:53:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126681"},"modified":"2023-11-09T11:34:13","modified_gmt":"2023-11-09T06:04:13","slug":"rd-sharma-class-9-solutions-chapter-11-vsaqs","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-vsaqs\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 11 VSAQs (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126684\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-11-VSAQS.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 11 VSAQs\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-11-VSAQS.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-11-VSAQS-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 9 Solutions Chapter 11 VSAQs:\u00a0<\/strong>If you also find the concepts of Class 9 Maths difficult to understand, this blog is for you. Here at Kopykitab, we believe in providing the best solutions to our readers, and for that, we have <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-for-maths\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a>. All the solutions are easy to understand and very reliable as they are designed by subject matter experts. To know more about the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 11<\/a> VSAQs, read the whole blog.<\/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-69d2c3e4f1634\" 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 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-vsaqs\/#access-answers-of-rd-sharma-class-9-solutions-chapter-11-vsaqs\" title=\"Access answers of RD Sharma Class 9 Solutions Chapter 11 VSAQs\">Access answers of RD Sharma Class 9 Solutions Chapter 11 VSAQs<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-vsaqs\/#faqs-on-rd-sharma-class-9-solutions-chapter-11-vsaqs\" title=\"FAQs on RD Sharma Class 9 Solutions Chapter 11 VSAQs\">FAQs on RD Sharma Class 9 Solutions Chapter 11 VSAQs<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-vsaqs\/#how-many-questions-are-there-in-rd-sharma-class-9-solutions-chapter-11-vsaqs\" title=\"How many questions are there in RD Sharma Class 9 Solutions Chapter 11 VSAQs?\">How many questions are there in RD Sharma Class 9 Solutions Chapter 11 VSAQs?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-vsaqs\/#is-it-even-beneficial-to-study-rd-sharma-for-class-9-solutions-chapter-11-vsaqs\" title=\"Is it even beneficial to study RD Sharma for Class 9 Solutions Chapter 11 VSAQs?\">Is it even beneficial to study RD Sharma for Class 9 Solutions Chapter 11 VSAQs?<\/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-11-vsaqs\/#are-the-solutions-rd-sharma-class-9-solutions-chapter-11-vsaqs-relevant\" title=\"Are the solutions RD Sharma Class 9 Solutions Chapter 11 VSAQs\u00a0relevant?\">Are the solutions RD Sharma Class 9 Solutions Chapter 11 VSAQs\u00a0relevant?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-class-9-solutions-chapter-11-vsaqs\"><\/span><strong>Access answers of RD Sharma Class 9 Solutions Chapter 11 VSAQs<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Question 1.<br \/>Define a triangle.<br \/>Solution:<br \/>A figure bounded by three line segments in a plane is called a triangle.<\/p>\n<p>Question 2.<br \/>Write the sum of the angles of an obtuse triangle.<br \/>Solution:<br \/>The sum of angles of an obtuse triangle is 180\u00b0.<\/p>\n<p>Question 3.<br \/>In \u2206ABC, if \u2220B = 60\u00b0, \u2220C = 80\u00b0, and the bisectors of angles \u2220ABC and \u2220ACB meet at a point O, then find the measure of \u2220BOC.<br \/>Solution:<br \/>In \u2206ABC, \u2220B = 60\u00b0, \u2220C = 80\u00b0<br \/>OB and OC are the bisectors of \u2220B and \u2220C<br \/>\u2235 \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)<br \/>\u21d2 \u2220A + 60\u00b0 + 80\u00b0 = 180\u00b0<br \/>\u21d2 \u2220A + 140\u00b0 = 180\u00b0<br \/>\u2234 \u2220A = 180\u00b0- 140\u00b0 = 40\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1936\/31771518608_97b01672cf_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry\" width=\"287\" height=\"316\" \/><br \/>= 90\u00b0 + \u2013 x 40\u00b0 = 90\u00b0 + 20\u00b0 = 110\u00b0<\/p>\n<p>Question 4.<br \/>If the angles of a triangle are in the ratio 2:1:3. Then find the measure of the smallest angle.<br \/>Solution:<br \/>The sum of angles of a triangle = 180\u00b0<br \/>The ratio in the angles = 2: 1 : 3<br \/>Let first angle = 2x<br \/>Second angle = x<br \/>and third angle = 3x<br \/>\u2234 2x + x + 3x = 180\u00b0 \u21d2 6x = 180\u00b0<br \/>\u2234 x =\u00a0<span id=\"MathJax-Element-30-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-191\" class=\"math\"><span id=\"MathJax-Span-192\" class=\"mrow\"><span id=\"MathJax-Span-193\" class=\"mfrac\"><span id=\"MathJax-Span-194\" class=\"msubsup\"><span id=\"MathJax-Span-195\" class=\"texatom\"><span id=\"MathJax-Span-196\" class=\"mrow\"><span id=\"MathJax-Span-197\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-198\" class=\"texatom\"><span id=\"MathJax-Span-199\" class=\"mrow\"><span id=\"MathJax-Span-200\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-201\" class=\"mn\">6<\/span><\/span><\/span><\/span><\/span>\u00a0 = 30\u00b0<br \/>\u2234 First angle = 2x = 2 x 30\u00b0 = 60\u00b0<br \/>Second angle = x = 30\u00b0<br \/>and third angle = 3x = 3 x 30\u00b0 = 90\u00b0<br \/>Hence angles are 60\u00b0, 30\u00b0, 90\u00b0<\/p>\n<p>Question 5.<br \/>State exterior angle theorem.<br \/>Solution:<br \/>Given: In \u2206ABC, side BC is produced to D<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1972\/45643873851_bd99b68b47_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 11 Coordinate Geometry\" width=\"268\" height=\"207\" \/><br \/>To prove : \u2220ACD = \u2220A + \u2220B<br \/>Proof: In \u2206ABC,<br \/>\u2220A + \u2220B + \u2220ACB = 180\u00b0 \u2026(i) (Sum of angles of a triangle)<br \/>and \u2220ACD + \u2220ACB = 180\u00b0 \u2026(ii) (Linear pair)<br \/>From (i) and (ii)<br \/>\u2220ACD + \u2220ACB = \u2220A + \u2220B + \u2220ACB<br \/>\u2220ACD = \u2220A + \u2220B<br \/>Hence proved.<\/p>\n<p>Question 6.<br \/>The sum of two angles of a triangle is equal to its third angle. Determine the measure of the third angle.<br \/>Solution:<br \/>In \u2206ABC,<br \/>\u2220A + \u2220C = \u2220B<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1945\/45593360382_0f99b8b496_o.png\" alt=\"Coordinate Geometry Class 9 RD Sharma Solutions\" width=\"246\" height=\"189\" \/><br \/>But \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)<br \/>\u2234 \u2220B + \u2220A + \u2220C = 180\u00b0<br \/>\u21d2 \u2220B + \u2220B = 180\u00b0<br \/>\u21d2 2\u2220B = 180\u00b0<br \/>\u21d2 \u2220B =\u00a0<span id=\"MathJax-Element-31-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-202\" class=\"math\"><span id=\"MathJax-Span-203\" class=\"mrow\"><span id=\"MathJax-Span-204\" class=\"mfrac\"><span id=\"MathJax-Span-205\" class=\"msubsup\"><span id=\"MathJax-Span-206\" class=\"texatom\"><span id=\"MathJax-Span-207\" class=\"mrow\"><span id=\"MathJax-Span-208\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-209\" class=\"texatom\"><span id=\"MathJax-Span-210\" class=\"mrow\"><span id=\"MathJax-Span-211\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-212\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0 = 90\u00b0<br \/>\u2234 Third angle = 90\u00b0<\/p>\n<p>Question 7.<br \/>In the figure, if AB || CD, EF || BC, \u2220BAC = 65\u00b0 and \u2220DHF = 35\u00b0, find \u2220AGH.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1953\/31771518268_c234373109_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 11 Coordinate Geometry\" width=\"314\" height=\"196\" \/><br \/>Solution:<br \/>Given : In figure, AB || CD, EF || BC \u2220BAC = 65\u00b0, \u2220DHF = 35\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1930\/45593360252_0f22b72aac_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry\" width=\"310\" height=\"203\" \/><br \/>\u2235 EF || BC<br \/>\u2234 \u2220A = \u2220ACH (Alternate angle)<br \/>\u2234 \u2220ACH = 65\u00b0<br \/>\u2235\u2220GHC = \u2220DHF<br \/>(Vertically opposite angles)<br \/>\u2234 \u2220GHC = 35\u00b0<br \/>Now in \u2206GCH,<br \/>Ext. \u2220AGH = \u2220GCH + \u2220GHC<br \/>= 65\u00b0 + 35\u00b0 = 100\u00b0<\/p>\n<p>Question 8.<br \/>In the figure, if AB || DE and BD || FG such that \u2220FGH = 125\u00b0 and \u2220B = 55\u00b0, find x and y.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1958\/45593360182_d55798124c_o.png\" alt=\"Class 9 Maths Chapter 11 Coordinate Geometry RD Sharma Solutions\" width=\"349\" height=\"262\" \/><br \/>Solution:<br \/>In the figure, AB || DF, BD || FG<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1904\/30703040957_cd16679562_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 11 Coordinate Geometry\" width=\"347\" height=\"257\" \/><br \/>\u2220FGH = 125\u00b0 and \u2220B = 55\u00b0<br \/>\u2220FGH + FGE = 180\u00b0 (Linear pair)<br \/>\u21d2 125\u00b0 + y \u2013 180\u00b0<br \/>\u21d2 y= 180\u00b0- 125\u00b0 = 55\u00b0<br \/>\u2235 BA || FD and BD || FG<br \/>\u2220B = \u2220F = 55\u00b0<br \/>Now in \u2206EFG,<br \/>\u2220F + \u2220FEG + \u2220FGE = 180\u00b0<br \/>(Angles of a triangle)<br \/>\u21d2 55\u00b0 + x + 55\u00b0 = 180\u00b0<br \/>\u21d2 x+ 110\u00b0= 180\u00b0<br \/>\u2234 x= 180\u00b0- 110\u00b0 = 70\u00b0<br \/>Hence x = 70, y = 55\u00b0<\/p>\n<p>Question 9.<br \/>If the angles A, B, and C of \u2206ABC satisfy the relation B \u2013 A = C \u2013 B, then find the measure of \u2220B.<br \/>Solution:<br \/>In \u2206ABC,<br \/>\u2220A + \u2220B + \u2220C= 180\u00b0 \u2026(i)<br \/>and B \u2013 A = C \u2013 B<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1911\/45593360002_b52f470a73_o.png\" alt=\"RD Sharma Class 9 Book Chapter 11 Coordinate Geometry\" width=\"225\" height=\"170\" \/><br \/>\u21d2 B + B = A + C \u21d2 2B = A + C<br \/>From (i),<br \/>B + 2B = 180\u00b0 \u21d2 3B = 180\u00b0<br \/>\u2220B =\u00a0<span id=\"MathJax-Element-32-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-213\" class=\"math\"><span id=\"MathJax-Span-214\" class=\"mrow\"><span id=\"MathJax-Span-215\" class=\"mfrac\"><span id=\"MathJax-Span-216\" class=\"msubsup\"><span id=\"MathJax-Span-217\" class=\"texatom\"><span id=\"MathJax-Span-218\" class=\"mrow\"><span id=\"MathJax-Span-219\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-220\" class=\"texatom\"><span id=\"MathJax-Span-221\" class=\"mrow\"><span id=\"MathJax-Span-222\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-223\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>\u00a0= 60\u00b0<br \/>Hence \u2220B = 60\u00b0<\/p>\n<p>Question 10.<br \/>In \u2206ABC, if bisectors of \u2220ABC and \u2220ACB intersect at O at an angle of 120\u00b0, then find the measure of \u2220A.<br \/>Solution:<br \/>In \u2206ABC, bisectors of \u2220B and \u2220C intersect at O and \u2220BOC = 120\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1967\/30703040777_27f07730c5_o.png\" alt=\"Coordinate Geometry With Solutions PDF RD Sharma Class 9 Solutions\" width=\"282\" height=\"189\" \/><br \/>But \u2220BOC = 90\u00b0+\u00a0<span id=\"MathJax-Element-33-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-224\" class=\"math\"><span id=\"MathJax-Span-225\" class=\"mrow\"><span id=\"MathJax-Span-226\" class=\"mfrac\"><span id=\"MathJax-Span-227\" class=\"mn\">1<\/span><span id=\"MathJax-Span-228\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span><br \/>90\u00b0+\u00a0<span id=\"MathJax-Element-34-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-229\" class=\"math\"><span id=\"MathJax-Span-230\" class=\"mrow\"><span id=\"MathJax-Span-231\" class=\"mfrac\"><span id=\"MathJax-Span-232\" class=\"mn\">1<\/span><span id=\"MathJax-Span-233\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A= 120\u00b0<br \/>\u21d2\u00a0<span id=\"MathJax-Element-35-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-234\" class=\"math\"><span id=\"MathJax-Span-235\" class=\"mrow\"><span id=\"MathJax-Span-236\" class=\"mfrac\"><span id=\"MathJax-Span-237\" class=\"mn\">1<\/span><span id=\"MathJax-Span-238\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A= 120\u00b0-90\u00b0 = 30\u00b0<br \/>\u2234 \u2220A = 2 x 30\u00b0 = 60\u00b0<\/p>\n<p>Question 11.<br \/>If the side BC of \u2206ABC is produced on both sides, then write the difference between the sum of the exterior angles so formed and \u2220A.<br \/>Solution:<br \/>In \u2206ABC, side BC is produced on both sides forming exterior \u2220ABE and \u2220ACD<br \/>Ext. \u2220ABE = \u2220A + \u2220ACB<br \/>and Ext. \u2220ACD = \u2220ABC + \u2220A<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1925\/30703040647_79722b213b_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 11 Coordinate Geometry\" width=\"272\" height=\"165\" \/><br \/>Adding we get,<br \/>\u2220ABE + \u2220ACD = \u2220A + \u2220ACB + \u2220A + \u2220ABC<br \/>\u21d2 \u2220ABE + \u2220ACD \u2013 \u2220A = \u2220A 4- \u2220ACB + \u2220A + \u2220ABC \u2013 \u2220A (Subtracting \u2220A from both sides)<br \/>= \u2220A + \u2220ABC + \u2220ACB = \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)<\/p>\n<p>Question 12.<br \/>In a triangle ABC, if AB = AC and AB is produced to D such that BD = BC, find \u2220ACD: \u2220ADC.<br \/>Solution:<br \/>In \u2206ABC, AB = AC<br \/>AB is produced to D such that BD = BC<br \/>DC is joined<br \/>In \u2206ABC, AB = AC<br \/>\u2234 \u2220ABC = \u2220ACB<br \/>In \u2206 BCD, BD = BC<br \/>\u2234 \u2220BDC = \u2220BCD<br \/>and Ext. \u2220ABC = \u2220BDC + \u2220BCD = 2\u2220BDC (\u2235 \u2220BDC = \u2220BCD)<br \/>\u21d2 \u2220ACB = 2\u2220BCD (\u2235 \u2220ABC = \u2220ACB)<br \/>Adding \u2220BDC to both sides<br \/>\u21d2 \u2220ACB + \u2220BDC = 2\u2220BDC + \u2220BDC<br \/>\u21d2 \u2220ACB + \u2220BCD = 3 \u2220BDC (\u2235 \u2220BDC = \u2220BCD)<br \/>\u21d2 \u2220ACB = 3\u2220BDC<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1916\/30703040537_22366857b0_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 11 Coordinate Geometry\" width=\"335\" height=\"82\" \/><\/p>\n<p>Question 13.<br \/>In the figure, side BC of AABC is produced to point D such that bisectors of \u2220ABC and \u2220ACD meet at point E. If \u2220BAC = 68\u00b0, find \u2220BEC.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1903\/45593359702_c97068d33b_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"318\" height=\"226\" \/><br \/>Solution:<br \/>In the figure,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1945\/30703040377_07031df438_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 11 Coordinate Geometry\" width=\"323\" height=\"238\" \/><br \/>side BC of \u2206ABC is produced to D such that bisectors of \u2220ABC and \u2220ACD meet at E<br \/>\u2220BAC = 68\u00b0<br \/>In \u2206ABC,<br \/>Ext. \u2220ACD = \u2220A + \u2220B<br \/>\u21d2\u00a0<span id=\"MathJax-Element-36-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-239\" class=\"math\"><span id=\"MathJax-Span-240\" class=\"mrow\"><span id=\"MathJax-Span-241\" class=\"mfrac\"><span id=\"MathJax-Span-242\" class=\"mn\">1<\/span><span id=\"MathJax-Span-243\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220ACD =\u00a0<span id=\"MathJax-Element-37-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-244\" class=\"math\"><span id=\"MathJax-Span-245\" class=\"mrow\"><span id=\"MathJax-Span-246\" class=\"mfrac\"><span id=\"MathJax-Span-247\" class=\"mn\">1<\/span><span id=\"MathJax-Span-248\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A +\u00a0<span id=\"MathJax-Element-38-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-249\" class=\"math\"><span id=\"MathJax-Span-250\" class=\"mrow\"><span id=\"MathJax-Span-251\" class=\"mfrac\"><span id=\"MathJax-Span-252\" class=\"mn\">1<\/span><span id=\"MathJax-Span-253\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220B<br \/>\u21d2 \u22202=\u00a0<span id=\"MathJax-Element-39-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-254\" class=\"math\"><span id=\"MathJax-Span-255\" class=\"mrow\"><span id=\"MathJax-Span-256\" class=\"mfrac\"><span id=\"MathJax-Span-257\" class=\"mn\">1<\/span><span id=\"MathJax-Span-258\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A + \u22201 \u2026(i)<br \/>But in \u2206BCE,<br \/>Ext. \u22202 = \u2220E + \u2220l<br \/>\u21d2 \u2220E + \u2220l = \u22202 =\u00a0<span id=\"MathJax-Element-40-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-259\" class=\"math\"><span id=\"MathJax-Span-260\" class=\"mrow\"><span id=\"MathJax-Span-261\" class=\"mfrac\"><span id=\"MathJax-Span-262\" class=\"mn\">1<\/span><span id=\"MathJax-Span-263\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A + \u2220l [From (i)]<br \/>\u21d2 \u2220E =\u00a0<span id=\"MathJax-Element-41-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-264\" class=\"math\"><span id=\"MathJax-Span-265\" class=\"mrow\"><span id=\"MathJax-Span-266\" class=\"mfrac\"><span id=\"MathJax-Span-267\" class=\"mn\">1<\/span><span id=\"MathJax-Span-268\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220A =\u00a0<span id=\"MathJax-Element-42-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-269\" class=\"math\"><span id=\"MathJax-Span-270\" class=\"mrow\"><span id=\"MathJax-Span-271\" class=\"mfrac\"><span id=\"MathJax-Span-272\" class=\"msubsup\"><span id=\"MathJax-Span-273\" class=\"texatom\"><span id=\"MathJax-Span-274\" class=\"mrow\"><span id=\"MathJax-Span-275\" class=\"mn\">68<\/span><\/span><\/span><span id=\"MathJax-Span-276\" class=\"texatom\"><span id=\"MathJax-Span-277\" class=\"mrow\"><span id=\"MathJax-Span-278\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-279\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0 =34\u00b0<\/p>\n<p>This is the whole blog on RD Sharma Class 9 Solutions Chapter 11 VSAQs. To know more about the <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Class 9 Maths, ask in the comments.\u00a0<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-9-solutions-chapter-11-vsaqs\"><\/span><strong>FAQs on RD Sharma Class 9 Solutions Chapter 11 VSAQs<\/strong><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-1631285521913\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-many-questions-are-there-in-rd-sharma-class-9-solutions-chapter-11-vsaqs\"><\/span>How many questions are there in RD Sharma Class 9 Solutions Chapter 11 VSAQs?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are 13 questions in\u00a0RD Sharma Class 9 Solutions for Chapter 11 VSAQs.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631285538818\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-it-even-beneficial-to-study-rd-sharma-for-class-9-solutions-chapter-11-vsaqs\"><\/span>Is it even beneficial to study RD Sharma for Class 9 Solutions Chapter 11 VSAQs?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, your preparation will be strengthened with this amazing help book. All your questions will be answered by this book.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631285560263\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"are-the-solutions-rd-sharma-class-9-solutions-chapter-11-vsaqs-relevant\"><\/span>Are the solutions RD Sharma Class 9 Solutions Chapter 11 VSAQs\u00a0relevant?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>The solutions are relevant as they are designed by the subject matter experts. \u00a0<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 11 VSAQs:\u00a0If you also find the concepts of Class 9 Maths difficult to understand, this blog is for you. Here at Kopykitab, we believe in providing the best solutions to our readers, and for that, we have RD Sharma Solutions Class 9 Maths. All the solutions are easy to &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 11 VSAQs (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-vsaqs\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 11 VSAQs (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":126684,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411],"tags":[3086,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126681"}],"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\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=126681"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126681\/revisions"}],"predecessor-version":[{"id":504788,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126681\/revisions\/504788"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126684"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126681"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126681"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126681"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}