{"id":6051,"date":"2023-08-16T08:19:08","date_gmt":"2023-08-16T08:19:08","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6051"},"modified":"2023-08-16T08:19:08","modified_gmt":"2023-08-16T08:19:08","slug":"ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-6-triangles-ex-6-3\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 6 Triangles Ex 6.3"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"font-family: Georgia, Palatino;\"><strong><span style=\"color: #000000;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><\/span><br \/>\n<span style=\"font-family: Georgia, Palatino;\"><strong><span style=\"color: #000000;\">Chapter &#8211; 6 (Triangles)\u00a0<\/span><\/strong><\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 6 Triangles <\/strong>Exercise 6.3 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-6-2\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 6.2<\/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;\"><span style=\"font-family: Georgia, Palatino;\"><strong><span style=\"color: #000000;\">Exercise &#8211; 6.3<\/span><\/strong><\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. State which pairs of triangles in the figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6342\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Que1.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"631\" height=\"566\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Que1.png 631w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Que1-300x269.png 300w\" sizes=\"auto, (max-width: 631px) 100vw, 631px\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;<\/strong><\/span><\/p>\n<div class=\"separator\" style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) In \u00a0\u0394ABC and \u0394PQR, we have<br \/>\n<\/strong>\u2220A =\u00a0\u2220P = 60\u00b0 (Given)<\/span><br \/>\n<span style=\"font-family: Georgia, Palatino;\"><span style=\"color: #000000;\">\u2220B = \u2220Q = 80\u00b0 (Given)<br \/>\n\u2220C = \u2220R = 40\u00b0 (Given)<br \/>\n\u2234 \u0394ABC ~ \u0394PQR (AAA similarity criterion)<\/span><span style=\"color: #000000;\"><strong>(ii) In \u00a0\u0394ABC and \u0394PQR, we have<br \/>\n<\/strong>AB\/QR = BC\/RP = CA\/PQ<\/span><\/span><br \/>\n<span style=\"font-family: Georgia, Palatino;\"><span style=\"color: #000000;\">\u2234 \u00a0\u0394ABC ~ \u0394QRP (SSS similarity criterion)<\/span><span style=\"color: #000000;\"><strong>(iii) In \u0394LMP and \u0394DEF, we have<br \/>\n<\/strong>LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6<\/span><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">MP\/DE = 2\/4 = 1\/2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">PL\/DF = 3\/6 = 1\/2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">LM\/EF= 2.7\/5 = 27\/50<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Here, MP\/DE = PL\/DF \u2260 LM\/EF<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, \u0394LMP and \u0394DEF are not similar.<\/span><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iv) In \u0394MNL and \u0394QPR, we have<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">MN\/QP = ML\/QR = 1\/2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220M = \u2220Q = 70\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 \u0394MNL ~ \u0394QPR (SAS similarity criterion)<\/span><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(v) In \u0394ABC and \u0394DEF, we have<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AB = 2.5, BC = 3, \u2220A = 80\u00b0, EF = 6, DF = 5, \u2220F = 80\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Here, AB\/DF = 2.5\/5 = 1\/2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">And, BC\/EF = 3\/6 = 1\/2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 \u2220B \u2260 \u2220F<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, \u0394ABC and \u0394DEF are not similar.<\/span><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(vi) In \u0394DEF,we have<\/strong><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220D\u00a0+\u00a0\u2220E\u00a0+\u00a0\u2220F = 180\u00b0 (sum of angles of a triangle)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 70\u00b0\u00a0+ 80\u00b0\u00a0+ \u2220F = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 \u2220F = 180\u00b0 &#8211; 70\u00b0 &#8211; 80\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 \u2220F = 30\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">In PQR, we have<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220P\u00a0+\u00a0\u2220Q\u00a0+\u00a0\u2220R = 180 (Sum of angles of \u0394)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 \u2220P\u00a0+ 80\u00b0\u00a0+ 30\u00b0 = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 \u2220P = 180\u00b0 &#8211; 80\u00b0 -30\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 \u2220P = 70\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">In \u0394DEF and \u0394PQR, we have<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220D = \u2220P = 70\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220F = \u2220Q = 80\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220F = \u2220R = 30\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, \u0394DEF ~ \u0394PQR (AAA similarity criterion)<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>2.\u00a0\u00a0In figure 6.35, \u0394ODC ~ \u0394OBA, \u2220 BOC = 125\u00b0 and \u2220 CDO = 70\u00b0. Find \u2220 DOC, \u2220 DCO and \u2220 OAB.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6344\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Que2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"245\" height=\"170\" \/><\/strong><\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong style=\"text-align: justify;\">Solution &#8211;<br \/>\n<\/strong><span style=\"text-align: justify;\">DOB is a straight line.<br \/>\n<\/span><span style=\"text-align: justify;\">Therefore, \u2220DOC + \u2220 COB = 180\u00b0<br \/>\n<\/span><span style=\"text-align: justify;\">\u21d2 \u2220DOC = 180\u00b0 &#8211; 125\u00b0 <\/span><span style=\"text-align: justify;\">= 55\u00b0<\/span><\/span><\/p>\n<\/div>\n<div style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">In \u0394DOC, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220DCO + \u2220 CDO + \u2220 DOC = 180\u00b0 (Sum of the measures of the angles of a triangle is 180\u00ba.)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 \u2220DCO + 70\u00ba + 55\u00ba = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 \u2220DCO = 55\u00b0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">It is given that \u0394ODC ~ \u0394OBA.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 \u2220OAB = \u2220OCD [Corresponding angles are equal in similar triangles.]<br \/>\n\u21d2 \u2220 OAB = 55\u00b0<br \/>\n\u2234 \u2220OAB = \u2220OCD [Corresponding angles are equal in similar triangles.]<br \/>\n<\/span><\/div>\n<div style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 \u2220OAB = 55\u00b0<\/span><\/div>\n<div style=\"text-align: justify;\"><\/div>\n<div style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO\/OC = OB\/OD<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6345\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Ans3.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"130\" \/><br \/>\n<\/strong>In \u0394DOC and \u0394BOA,<br \/>\n\u2220CDO = \u2220ABO [Alternate interior angles as AB || CD]<br \/>\n\u2220DCO = \u2220BAO [Alternate interior angles as AB || CD]<br \/>\n\u2220DOC = \u2220BOA [Vertically opposite angles]<br \/>\n\u2234 \u0394DOC ~ \u0394BOA [AAA similarity criterion]<br \/>\n\u2234 DO\/BO = OC\/OA \u00a0[ Corresponding sides are proportional]<\/span><\/div>\n<div style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 OA\/OC = OB\/OD<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, proved.<\/span><\/div>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4.\u00a0In the fig.6.36, QR\/QS = QT\/PR and\u00a0\u22201 =\u00a0\u22202. Show that\u00a0\u0394PQS ~\u00a0\u0394TQR.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6346\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Que4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"316\" height=\"194\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Que4.png 316w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Que4-300x184.png 300w\" sizes=\"auto, (max-width: 316px) 100vw, 316px\" \/><br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>In \u0394PQR, \u2220PQR = \u2220PRQ<br \/>\n\u2234 PQ = PR\u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (i)<br \/>\nGiven,<br \/>\nQR\/QS = QT\/PR<br \/>\nUsing\u00a0(i), we get<br \/>\nQR\/QS = QT\/QP\u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (ii)<br \/>\nIn \u0394PQS and \u0394TQR,<br \/>\nQR\/QS = QT\/QP\u00a0 \u00a0[using (ii)]<br \/>\n\u2220Q = \u2220Q<br \/>\n\u2234 \u0394PQS ~ \u0394TQR [SAS similarity criterion]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5. S and T are point on sides PR and QR of \u0394PQR such that \u2220P = \u2220RTS. Show that \u0394RPQ ~ \u0394RTS.<br \/>\n<\/strong><strong>Solution &#8211;<\/strong> Given, S and T are point on sides PR and QR of \u0394PQR\u00a0 and \u2220P = \u2220RTS.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6347\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Ans5.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"185\" height=\"189\" \/><br \/>\nIn \u0394RPQ and \u0394RTS,<br \/>\n\u2220RTS = \u2220QPS (Given)<br \/>\n\u2220R = \u2220R (Common angle)<br \/>\n\u2234 \u0394RPQ ~ \u0394RTS (AA similarity criterion)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>6. In the figure, if \u0394ABE \u2245 \u0394ACD, show that \u0394ADE ~ \u0394ABC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6348\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Que6.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"164\" height=\"200\" \/><br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Given, \u0394ABE \u2245 \u0394ACD.<br \/>\n\u2234 AB = AC [By CPCT]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (i)<br \/>\nand,<br \/>\nAD = AE [By CPCT]\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (ii)<br \/>\nIn \u0394ADE and \u0394ABC, dividing eq. (ii) by eq. (i),<br \/>\nAD\/AB = AE\/AC<br \/>\n\u2220A = \u2220A [Common angle]<br \/>\n\u2234 \u0394ADE ~ \u0394ABC [SAS similarity criterion]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>7.\u00a0In the figure, altitudes AD and CE of \u0394ABC intersect each other at the point P. Show that:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6349\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Que7.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"310\" height=\"239\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Que7.png 310w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Que7-300x231.png 300w\" sizes=\"auto, (max-width: 310px) 100vw, 310px\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">(i) \u0394AEP ~ \u0394CDP<br \/>\n(ii) \u0394ABD ~ \u0394CBE<br \/>\n(iii) \u0394AEP ~ \u0394ADB<br \/>\n(iv) \u0394PDC ~ \u0394BEC<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong>Given, altitudes AD and CE of \u0394ABC intersect each other at the point P.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) In \u0394AEP and \u0394CDP,<br \/>\n<\/strong>\u2220AEP = \u2220CDP (90\u00b0 each)<br \/>\n\u2220APE = \u2220CPD (Vertically opposite angles)<br \/>\nHence, by AA similarity criterion,<br \/>\n\u0394AEP ~ \u0394CDP<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) In \u0394ABD and \u0394CBE,<br \/>\n<\/strong>\u2220ADB = \u2220CEB ( 90\u00b0 each)<br \/>\n\u2220ABD = \u2220CBE (Common Angles)<br \/>\nHence, by AA similarity criterion,<br \/>\n\u0394ABD ~ \u0394CBE<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii)\u00a0In \u0394AEP and \u0394ADB,<br \/>\n<\/strong>\u2220AEP = \u2220ADB (90\u00b0 each)<br \/>\n\u2220PAE = \u2220DAB (Common Angles)<br \/>\nHence, by AA similarity criterion,<br \/>\n\u0394AEP ~ \u0394ADB<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iv) In \u0394PDC and \u0394BEC,<br \/>\n<\/strong>\u2220PDC = \u2220BEC (90\u00b0 each)<br \/>\n\u2220PCD = \u2220BCE (Common angles)<br \/>\nHence, by AA similarity criterion,<br \/>\n\u0394PDC ~ \u0394BEC<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that \u0394ABE ~ \u0394CFB.<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below,<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6350\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Ans8.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"166\" \/><br \/>\nIn \u0394ABE and \u0394CFB, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220A = \u2220C (Opposite angles of a parallelogram) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220AEB = \u2220CBF (Alternate interior angles as AE || BC) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 \u0394ABE ~ \u0394CFB (By AA similarity criterion)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>9. In the figure, ABC and AMP are two right triangles, right angled at B and M, respectively, prove that:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6351\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Que9.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"200\" height=\"181\" \/><br \/>\n<\/strong>(i) \u0394ABC ~ \u0394AMP<br \/>\n(ii) CA\/PA = BC\/MP<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong>Given, ABC and AMP are two right triangles, right angled at B and M, respectively.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) In \u0394ABC and \u0394AMP,<\/strong><br \/>\nwe have,<br \/>\n\u2220CAB = \u2220MAP (common angles)<br \/>\n\u2220ABC = \u2220AMP = 90\u00b0 (each 90\u00b0)<br \/>\n\u2234 \u0394ABC ~ \u0394AMP (AA similarity criterion)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii)<\/strong> As, \u0394ABC ~ \u0394AMP (AA similarity criterion)<br \/>\nIf two triangles are similar then the corresponding sides are always equal,<br \/>\nHence, CA\/PA = BC\/MP<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>10. CD and GH are respectively the bisectors of \u2220ACB and \u2220EGF such that D and H lie on sides AB and FE of \u0394ABC and \u0394EFG respectively. If \u0394ABC ~ \u0394FEG, Show that:<br \/>\n<\/strong>(i) CD\/GH = AC\/FG<br \/>\n(ii) \u0394DCB\u00a0~\u00a0\u0394HGE<br \/>\n(iii) \u0394DCA ~ \u0394HGF <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong>Given, CD and GH are respectively the bisectors of \u2220ACB and \u2220EGF such that D and H lie on sides AB and FE of \u0394ABC and \u0394EFG, respectively.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6352\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Ans10.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"320\" height=\"137\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Ans10.png 320w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Ans10-300x128.png 300w\" sizes=\"auto, (max-width: 320px) 100vw, 320px\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(i) From the given condition,<br \/>\n<\/strong>\u0394ABC ~ \u0394FEG.<br \/>\n\u2234 \u2220A = \u2220F, \u2220B = \u2220E, and \u2220ACB = \u2220FGE<br \/>\nSince, \u2220ACB = \u2220FGE<br \/>\n\u2234 \u2220ACD = \u2220FGH (Angle bisector)<br \/>\nand,<br \/>\n\u2220DCB = \u2220HGE (Angle bisector)<br \/>\nIn \u0394ACD and \u0394FGH,<br \/>\n\u2220A = \u2220F<br \/>\n\u2220ACD = \u2220FGH<br \/>\n\u2234 \u0394ACD ~ \u0394FGH (AA similarity criterion)<br \/>\n\u21d2CD\/GH = AC\/FG<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(ii) In \u0394DCB and \u0394HGE,<br \/>\n<\/strong>\u2220DCB = \u2220HGE (Already proved)<br \/>\n\u2220B = \u2220E (Already proved)<br \/>\n\u2234 \u0394DCB ~ \u0394HGE (AA similarity criterion)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>(iii) In \u0394DCA and \u0394HGF,<br \/>\n<\/strong>\u2220ACD = \u2220FGH (Already proved)<br \/>\n\u2220A = \u2220F (Already proved)<br \/>\n\u2234 \u0394DCA ~ \u0394HGF (AA similarity criterion)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD \u22a5 BC and EF \u22a5 AC, prove that \u0394ABD ~ \u0394ECF.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6353\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Que11.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"206\" height=\"200\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong>Given, ABC is an isosceles triangle.<br \/>\n\u2234 AB = AC<br \/>\n\u2220ABD = \u2220ECF<br \/>\nIn \u0394ABD and \u0394ECF,<br \/>\n\u2220ADB = \u2220EFC (Each 90\u00b0)<br \/>\n\u2220BAD = \u2220CEF (Already proved)<br \/>\n\u2234 \u0394ABD ~ \u0394ECF (using AA similarity criterion)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>12.\u00a0Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of \u0394PQR (see Fig 6.41). Show that \u0394ABC ~ \u0394PQR.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6354\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Que12.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"320\" height=\"166\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Que12.png 320w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Que12-300x156.png 300w\" sizes=\"auto, (max-width: 320px) 100vw, 320px\" \/><br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Given, \u0394ABC and \u0394PQR, AB, BC and median AD of \u0394ABC are proportional to sides PQ, QR and median PM of \u0394PQR<br \/>\ni.e. AB\/PQ = BC\/QR = AD\/PM<br \/>\n<strong>We have to prove :<\/strong> \u0394ABC ~ \u0394PQR<br \/>\nAs we know here,<br \/>\nAB\/PQ = BC\/QR = AD\/PM<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{AB}{PQ}&amp;space;=&amp;space;\\frac{\\frac{1}{2}BC}{\\frac{1}{2}QR}&amp;space;=&amp;space;\\frac{AD}{PM}\" alt=\"\\frac{AB}{PQ} = \\frac{\\frac{1}{2}BC}{\\frac{1}{2}QR} = \\frac{AD}{PM}\" align=\"absmiddle\" \/> &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (i)<br \/>\n\u21d2AB\/PQ = BC\/QR = AD\/PM (D is the midpoint of BC. M is the midpoint of QR)<br \/>\n\u21d2 \u0394ABD ~ \u0394PQM [SSS similarity criterion]<br \/>\n\u2234 \u2220ABD = \u2220PQM [Corresponding angles of two similar triangles are equal]<br \/>\n\u21d2 \u2220ABC = \u2220PQR<br \/>\nIn \u0394ABC and \u0394PQR<br \/>\nAB\/PQ = BC\/QR\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (ii)<strong><br \/>\n<\/strong>\u2220ABC = \u2220PQR\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (iii)<br \/>\nFrom equation\u00a0<strong>(ii)<\/strong>\u00a0and\u00a0<strong>(iii)<\/strong>, we get,<br \/>\n\u0394ABC ~ \u0394PQR [SAS similarity criterion]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>13. D is a point on the side BC of a triangle ABC such that \u2220ADC = \u2220BAC. Show that CA<sup>2<\/sup>\u00a0= CB.CD<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Given, D is a point on the side BC of a triangle ABC such that \u2220ADC = \u2220BAC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6355\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Ans13.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"253\" height=\"149\" \/><br \/>\nIn \u0394ADC and \u0394BAC,<br \/>\n\u2220ADC = \u2220BAC (Already given)<br \/>\n\u2220ACD = \u2220BCA (Common angles)<br \/>\n\u2234 \u0394ADC ~ \u0394BAC (AA similarity criterion)<br \/>\nWe know that corresponding sides of similar triangles are in proportion.<br \/>\n\u2234 CA\/CB = CD\/CA<br \/>\n\u21d2\u00a0CA<sup>2<\/sup> = CB \u00d7 CD.<br \/>\nHence, proved.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that \u0394ABC ~ \u0394PQR.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong><strong>Given :<\/strong> Two triangles \u0394ABC and \u0394PQR in which AD and PM are medians such that AB\/PQ = AC\/PR = AD\/PM<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>To Prove :<\/strong> \u0394ABC ~ \u0394PQR<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Construction :<\/strong> Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6356\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Ans14.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"320\" height=\"189\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Ans14.png 320w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Ans14-300x177.png 300w\" sizes=\"auto, (max-width: 320px) 100vw, 320px\" \/><br \/>\n<strong>Proof :<\/strong> In \u0394ABD and \u0394CDE, we have<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AD = DE \u00a0[By Construction]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">BD = DC [\u2234 AP is the median]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">and, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220ADB = \u2220CDE [Vertically opp. angles] <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 \u0394ABD\u00a0\u2245\u00a0\u0394CDE [By SAS criterion of congruence]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 AB = CE [CPCT]\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212; (i)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Also, in \u0394PQM and \u0394MNR, we have<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">PM = MN [By Construction]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">QM = MR [\u2234 PM is the median]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">and, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220PMQ = \u2220NMR [Vertically opposite angles] <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 \u0394PQM = \u0394MNR [By SAS criterion of congruence] <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 PQ = RN [CPCT] &#8212;&#8212;&#8212;&#8212; (ii)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Now, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AB\/PQ = AC\/PR = AD\/PM <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 CE\/RN = AC\/PR = AD\/PM\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[From <b>(i)<\/b>\u00a0and\u00a0<b>(ii)<\/b>]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 CE\/RN = AC\/PR = 2AD\/2PM<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 CE\/RN = AC\/PR = AE\/PN [\u2234 2AD = AE and 2PM = PN]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 \u0394ACE ~ \u0394PRN [By SSS similarity criterion]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, \u22202 = \u22204<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Similarly, \u22201 = \u22203<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 \u22201\u00a0+\u00a0\u22202\u00a0=\u00a0\u22203\u00a0+\u00a0\u22204<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2 \u2220A = \u2220P\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212; (iii)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Now, In \u0394ABC and \u0394PQR, we have<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">AB\/PQ = AC\/PR (Given)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220A = \u2220P [From\u00a0(iii)]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 \u0394ABC ~ \u0394PQR [By SAS similarity criterion]<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>Given, Length of the vertical pole = 6m<br \/>\nShadow of the pole = 4 m<br \/>\nLet Height of tower =\u00a0<em>h<\/em> m<br \/>\nLength of shadow of the tower = 28 m<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6357\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Ans15.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"308\" height=\"251\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Ans15.png 308w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Ans15-300x244.png 300w\" sizes=\"auto, (max-width: 308px) 100vw, 308px\" \/><br \/>\n<\/span><\/p>\n<div class=\"separator\" style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">In \u0394ABC and \u0394DEF,<\/span><\/div>\n<div class=\"separator\" style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220C = \u2220E (angular elevation of sum)<\/span><\/div>\n<div class=\"separator\" style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2220B = \u2220F = 90\u00b0<\/span><\/div>\n<div class=\"separator\" style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 \u0394ABC ~ \u0394DEF (By AA similarity criterion)<\/span><\/div>\n<div class=\"separator\" style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 AB\/DF = BC\/EF (If two triangles are similar corresponding sides are proportional)<\/span><\/div>\n<div class=\"separator\" style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u2234 6\/<i>h<\/i>\u00a0= 4\/28<\/span><\/div>\n<div class=\"separator\" style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2\u00a0<i>h<\/i> = 6 \u00d7 28\/4<\/span><\/div>\n<div class=\"separator\" style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2\u00a0<i>h<\/i>\u00a0=\u00a06 \u00d7 7<\/span><\/div>\n<div class=\"separator\" style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">\u21d2\u00a0<i>h\u00a0<\/i>= 42 m<\/span><\/div>\n<div class=\"separator\" style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, the height of the tower is 42 m.<\/span><\/div>\n<div class=\"separator\" style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>16. If AD and PM are medians of triangles ABC and PQR, respectively where \u0394ABC ~ \u0394PQR prove that AB\/PQ = AD\/PM.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>Given, \u0394ABC ~ \u0394PQR<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6358\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Ans16.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"320\" height=\"134\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Ans16.png 320w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex6.3-Ans16-300x126.png 300w\" sizes=\"auto, (max-width: 320px) 100vw, 320px\" \/><br \/>\nWe know that the corresponding sides of similar triangles are in proportion.<br \/>\n\u2234 AB\/PQ = AC\/PR = BC\/QR\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (i)<br \/>\nAlso, \u2220A = \u2220P, \u2220B = \u2220Q, \u2220C = \u2220R\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (ii)<br \/>\nSince AD and PM are medians, they will divide their opposite sides.<br \/>\n\u2234 BD = BC\/2 and QM = QR\/2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (iii)<br \/>\nFrom equations\u00a0<strong>(i)<\/strong>\u00a0and\u00a0<strong>(iii)<\/strong>, we get<br \/>\nAB\/PQ = BD\/QM\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (iv)<br \/>\nIn \u0394ABD and \u0394PQM,<br \/>\nFrom equation (ii), we have<br \/>\n\u2220B = \u2220Q<br \/>\nFrom equation\u00a0<strong>(iv), we have,<br \/>\n<\/strong>AB\/PQ = BD\/QM<br \/>\n\u2234 \u0394ABD ~ \u0394PQM (SAS similarity criterion)<br \/>\n\u21d2 AB\/PQ = BD\/QM = AD\/PM<\/span><\/div>\n<p>&nbsp;<\/p>\n<p><span style=\"color: #0000ff; font-family: Georgia, Palatino;\"><b><i><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\">Go Back To Chapters<\/a><\/i><\/b><\/span><\/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 6.3 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":[1045,1079,1082,1044,1049,1048],"class_list":["post-6051","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-6-3-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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