{"id":1995,"date":"2022-11-05T05:02:26","date_gmt":"2022-11-05T05:02:26","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=1995"},"modified":"2022-11-05T05:02:26","modified_gmt":"2022-11-05T05:02:26","slug":"ncert-solutions-class-7-maths-chapter-7-congruence-of-triangles-ex-7-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-7-congruence-of-triangles-ex-7-2\/","title":{"rendered":"NCERT Solutions Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 7 Mathematics\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 7 (Congruence of Triangles)<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 7 Mathematics <strong>Chapter &#8211; 7 Congruence of Triangles <\/strong>Exercise 7.2 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Chapter : 7 Congruence of Triangles<\/strong><\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-7-maths-chapter-7-congruence-of-triangles-ex-7-1\" target=\"_blank\" rel=\"noopener\"><span style=\"font-family: georgia, palatino, serif;\">NCERT Solution Class 7 Maths Exercise &#8211; 7.1<\/span><\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 7.2\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>1. Which congruence criterion do you use in the following?<br \/>\n<\/strong><strong>(a) Given: AC = DF<br \/>\n<\/strong><strong>AB = DE<br \/>\n<\/strong><strong>BC = EF<br \/>\n<\/strong><strong>So, \u0394ABC \u2245 \u0394DEF<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1999\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q1a.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"362\" height=\"126\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q1a.png 362w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q1a-300x104.png 300w\" sizes=\"auto, (max-width: 362px) 100vw, 362px\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>By SSS congruence property &#8211;\u00a0<\/strong> Two triangles are congruent if the three sides of one triangle are respectively equal to the three sides of the other triangle.<br \/>\n\u0394ABC \u2245 \u0394DEF<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(b) Given: ZX = RP<br \/>\n<\/strong><strong>RQ = ZY<br \/>\n<\/strong><strong>\u2220PRQ = \u2220XZY<br \/>\n<\/strong><strong>So, \u0394PQR \u2245 \u0394XYZ<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2000\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q1b.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"453\" height=\"118\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q1b.png 453w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q1b-300x78.png 300w\" sizes=\"auto, (max-width: 453px) 100vw, 453px\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<\/strong><br \/>\n<strong>By SAS congruence property &#8211;<\/strong> Two triangles are congruent if the two sides and the included angle of one are respectively equal to the two sides and the included angle of the other.<br \/>\n\u0394ACB \u2245 \u0394DEF<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(c) Given: \u2220MLN = \u2220FGH<br \/>\n<\/strong><strong>\u2220NML = \u2220GFH<br \/>\n<\/strong><strong>\u2220ML = \u2220FG<br \/>\n<\/strong><strong>So, \u0394LMN \u2245 \u0394GFH<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2001\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q1c.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"455\" height=\"149\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q1c.png 455w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q1c-300x98.png 300w\" sizes=\"auto, (max-width: 455px) 100vw, 455px\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>By ASA congruence property &#8211;<\/strong> Two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.<br \/>\n\u0394LMN \u2245 \u0394GFH<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(d) Given: EB = DB<br \/>\n<\/strong><strong>AE = BC<br \/>\n<\/strong><strong>\u2220A = \u2220C = 90<sup>o<br \/>\n<\/sup><\/strong><strong>So, \u0394ABE \u2245 \u0394ACD<br \/>\n<\/strong><strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2002\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q1d.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"224\" height=\"160\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>By RHS congruence property &#8211;<\/strong> Two right triangles are congruent if the hypotenuse and one side of the first triangle are respectively equal to the hypotenuse and one side of the second.<br \/>\n\u0394ABE \u2245 \u0394ACD<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>2. You want to show that \u0394ART \u2245 \u0394PEN,<br \/>\n<\/strong><strong>(a) If you have to use SSS criterion, then you need to show<br \/>\n<\/strong><strong>(i) AR = (ii) RT = (iii) AT =<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2003\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q2a.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"261\" height=\"133\" \/> <img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2004\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q2b.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"261\" height=\"133\" \/><\/strong><\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(b) If it is given that \u2220T = \u2220N and you are to use SAS criterion, you need to have<br \/>\n<\/strong><strong>(i) RT = and <\/strong><strong>(ii) PN =<\/strong><\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(c) If it is given that AT = PN and you are to use ASA criterion, you need to have<br \/>\n<\/strong><strong>(i) ZA\u00a0 <\/strong><strong>(ii) ZT<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(a)<\/strong> For SSS criterion, we need<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(i) AR = PE<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(ii) RT = EN<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(iii) AT = PN<\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(b)<\/strong> For SAS criterion, we need<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(i) RT = EN and<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(ii) PN = AT<\/span><\/p>\n<p><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(c)<\/strong> For ASA criterion, we need<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(i) \u2220A = \u2220P<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">(ii) \u2220T = \u2220N<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>3. You have to show that \u0394AMP \u2245 \u0394AMQ. <\/strong><strong>In the following proof, supply the missing reasons.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-2005 alignleft\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q3.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"242\" height=\"180\" \/>\u00a0<\/strong><\/span><\/p>\n<div class=\"table-responsive\" style=\"text-align: justify;\">\n<table class=\"table table-bordered\" style=\"width: 50%;\" border=\"1\">\n<tbody>\n<tr>\n<td><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Steps<\/strong><\/span><\/td>\n<td><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Reasons<\/strong><\/span><\/td>\n<\/tr>\n<tr>\n<td><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(i) PM = QM<\/span><\/td>\n<td><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(i) _______<\/span><\/td>\n<\/tr>\n<tr>\n<td><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(ii) \u2220PMA = \u2220QMA<\/span><\/td>\n<td><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(ii) _______<\/span><\/td>\n<\/tr>\n<tr>\n<td><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(iii) AM = AM<\/span><\/td>\n<td><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(iii) _______<\/span><\/td>\n<\/tr>\n<tr>\n<td><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(iv) \u0394AMP \u2245 \u0394AMQ<\/span><\/td>\n<td><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(iv) _______<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<\/strong><\/span><\/p>\n<div class=\"table-responsive\" style=\"text-align: justify;\">\n<table class=\"table table-bordered\" style=\"width: 60%;\">\n<tbody>\n<tr style=\"height: 24px;\">\n<td style=\"height: 24px; width: 33.0505%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Steps<\/strong><\/span><\/td>\n<td style=\"height: 24px; width: 65.9798%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Reasons<\/strong><\/span><\/td>\n<\/tr>\n<tr style=\"height: 24px;\">\n<td style=\"height: 24px; width: 33.0505%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(i) PM = QM<\/span><\/td>\n<td style=\"height: 24px; width: 65.9798%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(i) From the given figure<\/span><\/td>\n<\/tr>\n<tr style=\"height: 48px;\">\n<td style=\"height: 48px; width: 33.0505%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(ii) \u2220PMA = \u2220QMA<\/span><\/td>\n<td style=\"height: 48px; width: 65.9798%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(ii) From the given figure<\/span><\/td>\n<\/tr>\n<tr style=\"height: 48px;\">\n<td style=\"height: 48px; width: 33.0505%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(iii) AM = AM<\/span><\/td>\n<td style=\"height: 48px; width: 65.9798%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(iii) Common side for the both triangles<\/span><\/td>\n<\/tr>\n<tr style=\"height: 48px;\">\n<td style=\"height: 48px; width: 33.0505%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(iv) \u0394AMP \u2245 \u0394AMQ<\/span><\/td>\n<td style=\"height: 48px; width: 65.9798%;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">(iv) By SAS congruence property\u00a0<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>4. In \u0394ABC, \u2220A = 30<sup>o<\/sup>, \u2220B = 40<sup>o<\/sup>\u00a0and \u2220C = 110<sup>o<br \/>\n<\/sup><\/strong><strong>In \u0394PQR, \u2220P = 30<sup>o<\/sup>, \u2220Q = 40<sup>o<\/sup>\u00a0and \u2220R = 110<sup>o<br \/>\n<\/sup><\/strong><strong>A student says that \u0394ABC \u2245 \u0394PQR by AAA congruence criterion. Is he justified? Why or Why not?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2006\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Ans4.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"306\" height=\"149\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Ans4.png 306w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Ans4-300x146.png 300w\" sizes=\"auto, (max-width: 306px) 100vw, 306px\" \/><br \/>\nNo, because the two triangles with equal corresponding angles need not be congruent. In such a correspondence, one of them can be enlarged copy of the other.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>5. In the figure, the two triangles are congruent. The corresponding parts are marked. We can write \u0394RAT \u2245 ?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2007\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q5.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"266\" height=\"240\" \/><br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>From the given figure,<br \/>\nWe may observe that,<br \/>\n\u2220TRA = \u2220OWN<br \/>\n\u2220TAR = \u2220NOW<br \/>\n\u2220ATR = \u2220ONW<br \/>\nHence, \u0394RAT \u2245 \u0394WON<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>6. Complete the congruence statement:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2008\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q6.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"536\" height=\"182\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q6.png 536w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q6-300x102.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q6-480x163.png 480w\" sizes=\"auto, (max-width: 536px) 100vw, 536px\" \/><br \/>\n<\/strong><strong>\u0394BCA\u00a0 \u2245\u00a0 ?\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u0394QRS \u2245 ?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>First consider the \u0394BCA and \u0394BTA<br \/>\n<\/strong>From the figure, it is given that,<br \/>\nBT = BC<br \/>\nThen,<br \/>\nBA is common side for the \u0394BCA and \u0394BTA<br \/>\nHence, \u0394BCA \u2245 \u0394BTA<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Consider the \u0394QRS and \u0394TPQ<\/strong><br \/>\nFrom the figure, it is given that<br \/>\nPT = QR<br \/>\nTQ = QS<br \/>\nPQ = RS<br \/>\nHence, \u0394QRS \u2245 \u0394TPQ<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>7. In a squared sheet, draw two triangles of equal areas such that<br \/>\n<\/strong><strong>(i) The triangles are congruent.<br \/>\n<\/strong><strong>(ii) The triangles are not congruent.<br \/>\n<\/strong><strong>What can you say about their perimeters?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong>(i) On the given square sheet, we have draw two congruent triangles i.e.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">\u2206ABC = \u2206DEF<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2009\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Ans7i.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"207\" height=\"211\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">such that<\/span><br \/>\n<span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" style=\"color: #000000; font-family: georgia, palatino, serif;\" tabindex=\"0\"><span id=\"MathJax-Span-76\" class=\"math\"><span id=\"MathJax-Span-77\" class=\"mrow\"><span id=\"MathJax-Span-84\" class=\"mo\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\overline{AB}=\\overline{DE},&amp;space;\\overline{BC}=\\overline{EF},&amp;space;\\overline{AC}=\\overline{DF}\" alt=\"\\overline{AB}=\\overline{DE}, \\overline{BC}=\\overline{EF}, \\overline{AC}=\\overline{DF}\" align=\"absmiddle\" \/><\/span><\/span><\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">On adding, we get<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\overline{AB}+\\overline{BC}+\\overline{CA}\" alt=\"\\overline{AB}+\\overline{BC}+\\overline{CA}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\overline{DE}+\\overline{EF}+\\overline{FD}\" alt=\"\\overline{DE}+\\overline{EF}+\\overline{FD}\" align=\"absmiddle\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">i.e. perimeters of \u2206ABC = Perimeter of \u2206DEF<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(ii)\u00a0<\/strong>On the other square sheet, we have drawn two triangles ABC and PQR which are not congruent.<br \/>\nSuch that<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\overline{AB}&amp;space;\\neq&amp;space;\\overline{PQ}\" alt=\"\\overline{AB} \\neq \\overline{PQ}\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\overline{BC}&amp;space;\\neq&amp;space;\\overline{QR}\" alt=\"\\overline{BC} \\neq \\overline{QR}\" align=\"absmiddle\" \/> and <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\overline{CA}&amp;space;\\neq&amp;space;\\overline{RP}\" alt=\"\\overline{CA} \\neq \\overline{RP}\" align=\"absmiddle\" \/><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2010\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Ans7ii.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"214\" height=\"125\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Adding both sides, we get<\/span><br \/>\n<span id=\"MathJax-Element-8-Frame\" class=\"MathJax\" style=\"color: #000000; font-family: georgia, palatino, serif;\" tabindex=\"0\"><span id=\"MathJax-Span-162\" class=\"math\"><span id=\"MathJax-Span-163\" class=\"mrow\"><span id=\"MathJax-Span-164\" class=\"munderover\"><span id=\"MathJax-Span-165\" class=\"texatom\"><span id=\"MathJax-Span-166\" class=\"mrow\"><span id=\"MathJax-Span-167\" class=\"mi\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\overline{AB}+\\overline{BC}+\\overline{CA}\" alt=\"\\overline{AB}+\\overline{BC}+\\overline{CA}\" align=\"absmiddle\" \/> <\/span><\/span><\/span><\/span><span id=\"MathJax-Span-184\" class=\"mo\">\u2260 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\overline{PQ}&amp;space;+\\overline{QR}&amp;space;+\\overline{RP}\" alt=\"\\overline{PQ} +\\overline{QR} +\\overline{RP}\" align=\"absmiddle\" \/><\/span><\/span><\/span><\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">i.e., perimeter of \u2206ABC \u2260 the perimeter of \u2206PQR.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>8. Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong>Let us draw triangles LMN and FGH.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2011\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q8.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"460\" height=\"135\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q8.png 460w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q8-300x88.png 300w\" sizes=\"auto, (max-width: 460px) 100vw, 460px\" \/><br \/>\nIn the above figure, all angles of two triangles are equal. But, out of three sides only two sides are equal.<br \/>\nHence, \u0394LMN is not congruent to \u0394FGH.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>9. If \u0394ABC and \u0394PQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2012\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q9.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"345\" height=\"200\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q9.png 345w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q9-300x174.png 300w\" sizes=\"auto, (max-width: 345px) 100vw, 345px\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong>By observing the given figure, we can say that<br \/>\n\u2220ABC = \u2220PQR<br \/>\n\u2220BCA = \u2220PRQ<br \/>\nThe other additional pair of corresponding part is BC = QR<br \/>\n\u2234 \u0394ABC \u2245 \u0394PQR<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>10. Explain, why \u0394ABC \u2245 \u0394FED<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2013\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q10.png\" alt=\"NCERT Class 7 Maths Solution\" width=\"327\" height=\"135\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q10.png 327w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/10\/NCERT-Solutions-Class-7-Maths-Ex-7.1-Q10-300x124.png 300w\" sizes=\"auto, (max-width: 327px) 100vw, 327px\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solution &#8211;<br \/>\n<\/strong>From the figure, it is given that,<br \/>\n\u2220ABC = \u2220DEF = 90<sup>o<br \/>\n<\/sup>\u2220BAC = \u2220DFE<br \/>\nBC = DE<br \/>\nBy ASA congruence property, two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.<br \/>\n\u0394ABC \u2245 \u0394FED<\/span><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 7 Mathematics\u00a0 Chapter &#8211; 7 (Congruence of Triangles) The NCERT Solutions in English Language for Class 7 Mathematics Chapter &#8211; 7 Congruence of Triangles Exercise 7.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 7 Congruence of Triangles NCERT Solution Class 7 Maths [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[219],"tags":[298,299,283,5],"class_list":["post-1995","post","type-post","status-publish","format-standard","hentry","category-class-7-maths","tag-ncert-class-7-maths-chapter-7-congruence-of-triangles-in-english","tag-ncert-solutions-class-7-maths-chapter-7-in-english","tag-ncert-solutions-class-7-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.4) - 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