{"id":6050,"date":"2023-08-16T08:18:54","date_gmt":"2023-08-16T08:18:54","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6050"},"modified":"2023-08-16T08:18:54","modified_gmt":"2023-08-16T08:18:54","slug":"ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-2\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 6 Triangles Ex 6.2"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 6 (Triangles)\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 6 Triangles <\/strong>Exercise 6.2 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter : 6 Triangles<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-1\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 6.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-3\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 6.3<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-4\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 6.4<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-5\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 6.5<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-6\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 6.6<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 6.2<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6329\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.2-Que1.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"648\" height=\"222\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.2-Que1.png 648w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.2-Que1-300x103.png 300w\" sizes=\"auto, (max-width: 648px) 100vw, 648px\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong>(i) Given, in \u25b3 ABC, DE || BC<br \/>\n\u2234 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{AD}{DB}&amp;space;=&amp;space;\\frac{AE}{EC}\" alt=\"\\frac{AD}{DB} = \\frac{AE}{EC}\" align=\"absmiddle\" \/> [Using Basic proportionality theorem]<\/span><\/p>\n<p>\u21d2 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1.5}{3}&amp;space;=&amp;space;\\frac{1}{EC}\" alt=\"\\frac{1.5}{3} = \\frac{1}{EC}\" align=\"absmiddle\" \/><\/p>\n<p>\u21d2 EC = 3\/1.5<br \/>\n\u21d2 EC = 2 cm<br \/>\nHence, EC = 2 cm.<\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">(ii) Given, in \u25b3 ABC, DE || BC<br \/>\n\u2234 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{AD}{DB}&amp;space;=&amp;space;\\frac{AE}{EC}\" alt=\"\\frac{AD}{DB} = \\frac{AE}{EC}\" align=\"absmiddle\" \/>\u00a0 [Using Basic proportionality theorem]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{AD}{7.2}&amp;space;=&amp;space;\\frac{1.8}{5.4}\" alt=\"\\frac{AD}{7.2} = \\frac{1.8}{5.4}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 AD = 1.8 \u00d7 7.2\/5.4<br \/>\n\u21d2 AD = 2.4<br \/>\nHence, AD = 2.4 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>2. E and F are points on the sides PQ and PR, respectively of a \u0394PQR. For each of the following cases, state whether EF || QR.<br \/>\n<\/strong>(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm<br \/>\n(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm<br \/>\n(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong>Given, in \u0394PQR, E and F are two points on side PQ and PR, respectively. See the figure below;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6330\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.2-Ans2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"192\" height=\"192\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.2-Ans2.png 192w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.2-Ans2-150x150.png 150w\" sizes=\"auto, (max-width: 192px) 100vw, 192px\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm<br \/>\n<\/strong>Therefore, by using Basic proportionality theorem, we get,<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{PE}{EQ}&amp;space;=&amp;space;\\frac{3.9}{3}\" alt=\"\\frac{PE}{EQ} = \\frac{3.9}{3}\" align=\"absmiddle\" \/> = 1.3<br \/>\nAnd<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{PE}{FR}&amp;space;=&amp;space;\\frac{3.6}{2.4}\" alt=\"\\frac{PE}{FR} = \\frac{3.6}{2.4}\" align=\"absmiddle\" \/> = 1.5<br \/>\nSo, we get, PE\/EQ \u2260 PF\/FR<br \/>\nHence, EF is not parallel to QR.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) Given, PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm<br \/>\n<\/strong>Therefore, by using Basic proportionality theorem, we get,<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{PE}{QE}=\\frac{4}{4.5}\" alt=\"\\frac{PE}{QE}=\\frac{4}{4.5}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{8}{9}\" alt=\"\\frac{8}{9}\" align=\"absmiddle\" \/><br \/>\nAnd,<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{PF}{RF}&amp;space;=&amp;space;\\frac{8}{9}\" alt=\"\\frac{PF}{RF} = \\frac{8}{9}\" align=\"absmiddle\" \/><br \/>\nSo, we get here,<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{PE}{QE}=\\frac{PF}{RF}\" alt=\"\\frac{PE}{QE}=\\frac{PF}{RF}\" align=\"absmiddle\" \/><br \/>\nHence, EF is parallel to QR.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii) Given, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm<br \/>\n<\/strong>EQ = PQ \u2013 PE = 1.28 \u2013 0.18 = 1.10 cm<br \/>\nAnd,<br \/>\nFR = PR \u2013 PF = 2.56 \u2013 0.36 = 2.20 cm<br \/>\nSo,<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{PE}{EQ}&amp;space;=\\frac{0.18}{1.10}\" alt=\"\\frac{PE}{EQ} =\\frac{0.18}{1.10}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{9}{55}\" alt=\"\\frac{9}{55}\" align=\"absmiddle\" \/> &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (i)<br \/>\nAnd,<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{PE}{FR}&amp;space;=\\frac{0.36}{2.20}=\\frac{9}{55}\" alt=\"\\frac{PE}{FR} =\\frac{0.36}{2.20}=\\frac{9}{55}\" align=\"absmiddle\" \/>\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (ii)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">So, we get here,<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{PE}{EQ}=\\frac{PF}{FR}\" alt=\"\\frac{PE}{EQ}=\\frac{PF}{FR}\" align=\"absmiddle\" \/> <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, EF is parallel to QR.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. In the figure, if LM || CB and LN || CD, prove that AM\/AB = AN\/AD<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6331\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.2-Que3.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"403\" height=\"166\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.2-Que3.png 403w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.2-Que3-300x124.png 300w\" sizes=\"auto, (max-width: 403px) 100vw, 403px\" \/><br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>In the given figure, LM || CB <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">By using basic proportionality theorem, we get,<br \/>\nAM\/MB = AL\/LC\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (i)<b><br \/>\n<\/b>Similarly, LN || CD<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 AN\/AD = AL\/LC\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (ii)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">From\u00a0(i)\u00a0and\u00a0(ii), we get<br \/>\nAM\/MB = AN\/AD<br \/>\nHence, proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4. In the figure, DE||AC and DF||AE. Prove that BF\/FE = BE\/EC<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6332\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.2-Que4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"151\" \/><br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>In \u0394ABC, DE || AC (Given)<br \/>\n\u2234 BD\/DA = BE\/EC\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (i)\u00a0[By using Basic Proportionality Theorem]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">In \u00a0\u0394ABC, DF || AE (Given)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 BD\/DA = BF\/FE\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (ii) [By using Basic Proportionality Theorem]<br \/>\nFrom equation (i)\u00a0and\u00a0(ii), we get<br \/>\nBE\/EC = BF\/FE<br \/>\nHence, proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5. In the figure, DE||OQ and DF||OR, show that EF||QR.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6333\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.2-Que5.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"233\" height=\"179\" \/><br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>In \u0394PQO, DE || OQ (Given) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 PD\/DO = PE\/EQ &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (i)\u00a0[By using Basic Proportionality Theorem]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">In \u0394PQO, DE || OQ (Given) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 PD\/DO = PF\/FR &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (ii) [By using Basic Proportionality Theorem]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">From equation\u00a0(i)\u00a0and\u00a0(ii), we get<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">PE\/EQ = PF\/FR<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">In \u0394PQR, EF\u00a0||\u00a0QR. [By converse of Basic Proportionality Theorem]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>6. In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6334\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.2-Que6.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"247\" height=\"187\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong>In \u0394OPQ, AB || PQ (Given)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 OA\/AP = OB\/BQ &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (i)\u00a0[By using Basic Proportionality Theorem]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">In \u0394OPR, AC || PR (Given)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 OA\/AP = OC\/CR &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (ii) [By using Basic Proportionality Theorem]<\/span><\/p>\n<div class=\"separator\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">From equation\u00a0(i)\u00a0and\u00a0(ii), we get<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">OB\/BQ = OC\/CR<\/span><\/div>\n<div class=\"separator\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">In \u0394OQR, BC ||\u00a0QR. [By converse of Basic Proportionality Theorem].<\/span><\/div>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>7. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX). <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6335\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.2-Ans7.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"185\" height=\"200\" \/><br \/>\nGiven :<\/strong> \u0394ABC in which D is the mid point of AB such that AD=DB.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>To Prove :<\/strong> E is the mid point of AC.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Proof :<\/strong> D is the mid-point of AB.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 AD = DB<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 AD\/BD = 1 &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (i)<br \/>\nIn \u0394ABC, DE || BC,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, AD\/DB = AE\/EC [By using Basic Proportionality Theorem]<br \/>\n\u21d2 1 = AE\/EC [From equation (i)]<br \/>\n\u2234 AE = EC<br \/>\nHence, E is the mid point of AC.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>8. Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<\/strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6335\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.2-Ans7.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"185\" height=\"200\" \/><br \/>\n<strong>Given :<\/strong> \u0394ABC in which D and E are the mid points of AB and AC respectively such that AD=BD and AE=EC.<br \/>\n<strong>To Prove :<\/strong> DE || BC<br \/>\n<strong>Proof :<\/strong> D is the mid point of AB (Given)<br \/>\n\u2234 AD = DB<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 AD\/BD = 1 &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (i)<br \/>\nAlso, E is the mid-point of AC (Given)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 AE = EC<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 AE\/EC = 1 [From equation (i)] &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (ii)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">From equation (i) and (ii), we get<br \/>\nAD\/BD = AE\/EC<br \/>\nHence, DE || BC [By converse of Basic Proportionality Theorem]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO\/BO = CO\/DO.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6336\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.2-Ans9.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"230\" height=\"151\" \/><br \/>\n<strong>Given :<\/strong> ABCD is a trapezium in which AB || DC in which diagonals AC and BD intersect each other at O.<br \/>\n<strong>To Prove :<\/strong> AO\/BO = CO\/DO<br \/>\n<strong>Construction :<\/strong> Through O, draw EO || DC || AB<br \/>\n<strong>Proof :<\/strong> In \u0394ADC, we have<br \/>\nOE || DC (By Construction)<br \/>\n\u2234 AE\/ED = AO\/CO &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (i) [By using Basic Proportionality Theorem]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">In \u0394ABD, we have<br \/>\nOE || AB (By Construction)<br \/>\n\u2234 DE\/EA = DO\/BO &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (ii) [By using Basic Proportionality Theorem]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">From equation <b>(i)<\/b>\u00a0and\u00a0<b>(ii)<\/b>, we get<br \/>\nAO\/CO = BO\/DO<br \/>\n\u21d2 \u00a0AO\/BO = CO\/DO<br \/>\nHence, proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that\u00a0AO\/BO = CO\/DO.\u00a0Show that ABCD is a trapezium.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6337\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.2-Ans10.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"145\" \/><br \/>\n<\/strong><strong>Given :<\/strong> Quadrilateral ABCD in which diagonals AC and BD intersects each other at O such that AO\/BO = CO\/DO.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>To Prove :<\/strong> ABCD is a trapezium<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Construction :<\/strong> Through O, draw line EO, where EO || AB, which meets AD at E.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Proof :<\/strong> In \u0394DAB, we have<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">EO || AB<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 DE\/EA = DO\/OB &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (i) [By using Basic Proportionality Theorem]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Also, \u00a0AO\/BO = CO\/DO (Given)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 AO\/CO = BO\/DO<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 CO\/AO = BO\/DO<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 DO\/OB = CO\/AO &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (ii) <b>\u00a0<\/b><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">From equation\u00a0(i)\u00a0and\u00a0(ii), we get<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">DE\/EA = CO\/AO <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, By using converse of Basic Proportionality Theorem, EO || DC also EO || AB <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 AB || DC. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, quadrilateral ABCD is a trapezium with AB || CD.<\/span><\/p>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\"><span style=\"color: #0000ff;\"><b><i>Go Back To Chapters<\/i><\/b><\/span><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 6 (Triangles)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 6 Triangles Exercise 6.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 6 Triangles NCERT Class 10 Maths Solution Ex &#8211; 6.1 NCERT Class 10 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1079,1081,1044,1049,1048],"class_list":["post-6050","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-6-triangles-solutions","tag-ncert-class-10-mathematics-exercise-6-2-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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