{"id":598595,"date":"2022-05-03T06:45:04","date_gmt":"2022-05-03T06:45:04","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=598595"},"modified":"2022-05-10T08:05:38","modified_gmt":"2022-05-10T08:05:38","slug":"selina-class-7-icse-solutions-mathematics-chapter-19-congruency-congruent-triangles","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-19-congruency-congruent-triangles\/","title":{"rendered":"Selina Class 7 ICSE Solutions Mathematics : Chapter 19-\u00a0Congruency: Congruent Triangles"},"content":{"rendered":"\n<p>Class 7: Maths Chapter 19 solutions. Complete Class 7 Maths Chapter 19 Notes.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-selina-class-7-icse-solutions-mathematics-chapter-19-congruency-congruent-triangles\">Selina Class 7 ICSE Solutions Mathematics : Chapter 19-&nbsp;Congruency: Congruent Triangles<\/h2>\n\n\n\n<p>Selina 7th Maths Chapter 19, Class 7 Maths Chapter 19 solutions<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Exercise 19 page: 213<\/strong><\/h3>\n\n\n\n<p><strong>1. State, whether the pairs of triangles given in the following figures are congruent or not:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-1.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 1\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 1\"\/><\/figure>\n\n\n\n<p><strong>(vii) \u2206ABC in which AB = 2 cm, BC = 3.5 cm and \u2220C = 80<sup>0<\/sup> and, \u2206DEF in which DE = 2cm, DF = 3.5 cm and \u2220D = 80<sup>0<\/sup>.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i) In the given figure, corresponding sides of the triangles are not equal.<\/strong><\/p>\n\n\n\n<p><strong>Therefore, the given triangles are not congruent.<\/strong><\/p>\n\n\n\n<p><strong>(ii) In the first triangle<\/strong><\/p>\n\n\n\n<p><strong>Third angle = 180<sup>0<\/sup> \u2013 (40<sup>0<\/sup> + 30<sup>0<\/sup>)<\/strong><\/p>\n\n\n\n<p><strong>By further calculation<\/strong><\/p>\n\n\n\n<p><strong>= 180<sup>0<\/sup> \u2013 70<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>So we get<\/strong><\/p>\n\n\n\n<p><strong>= 110<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>In the two triangles, the sides and included angle of one are equal to the corresponding sides and included angle.<\/strong><\/p>\n\n\n\n<p><strong>Therefore, the given triangles are congruent. (SAS axiom)<\/strong><\/p>\n\n\n\n<p><strong>(iii) In the given figure, corresponding two sides are equal and the included angles are not equal.<\/strong><\/p>\n\n\n\n<p><strong>Therefore, the given triangles are not congruent.<\/strong><\/p>\n\n\n\n<p><strong>(iv) In the given figure, the corresponding three sides are equal.<\/strong><\/p>\n\n\n\n<p><strong>Therefore, the given triangles are congruent. (SSS Axiom)<\/strong><\/p>\n\n\n\n<p><strong>(v) In the right triangles, one side and diagonal of one triangle are equal to the corresponding side and diagonal of the other.<\/strong><\/p>\n\n\n\n<p><strong>Therefore, the given triangles are congruent. (RHS Axiom)<\/strong><\/p>\n\n\n\n<p><strong>(vi) In the given figure, two sides and one angle of one triangle are equal to the corresponding sides and one angle of the other.<\/strong><\/p>\n\n\n\n<p><strong>Therefore, the given triangles are congruent. (SSA Axiom)<\/strong><\/p>\n\n\n\n<p><strong>(vii) In \u2206 ABC,<\/strong><\/p>\n\n\n\n<p><strong>It is given that<\/strong><\/p>\n\n\n\n<p><strong>AB = 2cm, BC = 3.5 cm, \u2220C = 80<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>In \u2206 DEF,<\/strong><\/p>\n\n\n\n<p><strong>It is given that<\/strong><\/p>\n\n\n\n<p><strong>DE = 2 cm, DF = 3.5 cm and \u2220D = 80<sup>0<\/sup><\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-2.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 2\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 2\"\/><\/figure>\n\n\n\n<p><strong>We get to know that two corresponding sides are equal but the included angles are not equal.<\/strong><\/p>\n\n\n\n<p><strong>Therefore, the triangles are not congruent.<\/strong><\/p>\n\n\n\n<p><strong>2. In the given figure, prove that:<\/strong><br><br><strong>\u2206 ABD \u2245&nbsp;\u2206 ACD<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-3.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 3\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 3\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>In \u2206 ABD and \u2206 ACD<\/strong><\/p>\n\n\n\n<p><strong>AD = AD is common<\/strong><\/p>\n\n\n\n<p><strong>It is given that<\/strong><\/p>\n\n\n\n<p><strong>AB = AC and BD = DC<\/strong><\/p>\n\n\n\n<p><strong>Here \u2206 ABD \u2245&nbsp;\u2206 ACD (SSS Axiom)<\/strong><\/p>\n\n\n\n<p><strong>Therefore, it is proved.<\/strong><\/p>\n\n\n\n<p><strong>3. Prove that:<\/strong><\/p>\n\n\n\n<p><strong>(i) \u2206 ABC \u2245&nbsp;\u2206 ADC<\/strong><\/p>\n\n\n\n<p><strong>(ii) \u2220B = \u2220D<\/strong><\/p>\n\n\n\n<p><strong>(iii) AC bisects angle DCB.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-4.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 4\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 4\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>In the figure<\/strong><\/p>\n\n\n\n<p><strong>AB = AD and CB = CD<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-5.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 5\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 5\"\/><\/figure>\n\n\n\n<p><strong>In \u2206 ABC and&nbsp;\u2206 ADC<\/strong><\/p>\n\n\n\n<p><strong>AC = AC is common<\/strong><\/p>\n\n\n\n<p><strong>It is given that<\/strong><\/p>\n\n\n\n<p><strong>AB = AD and CB = CD<\/strong><\/p>\n\n\n\n<p><strong>Here \u2206 ABC \u2245&nbsp;\u2206 ADC (SSS Axiom)<\/strong><\/p>\n\n\n\n<p><strong>\u2220B = \u2220D (c. p. c. t)<\/strong><\/p>\n\n\n\n<p><strong>So we get<\/strong><\/p>\n\n\n\n<p><strong>\u2220BCA = \u2220DCA<\/strong><\/p>\n\n\n\n<p><strong>Therefore, AC bisects \u2220DCB.<\/strong><\/p>\n\n\n\n<p><strong>4. Prove that:<\/strong><br><br><strong>(i) \u2206ABD&nbsp; \u2261 \u2206ACD<\/strong><br><br><strong>(ii) \u2220B = \u2220C<\/strong><br><br><strong>(iii) \u2220ADB = \u2220ADC<\/strong><br><br><strong>(iv) \u2220ADB = 90\u00b0<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-6.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 6\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 6\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>From the figure<\/strong><\/p>\n\n\n\n<p><strong>AD = AC and BD = CD<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-7.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 7\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 7\"\/><\/figure>\n\n\n\n<p><strong>In \u2206ABD&nbsp;and \u2206ACD<\/strong><\/p>\n\n\n\n<p><strong>AD = AD is common<\/strong><\/p>\n\n\n\n<p><strong>(i) \u2206ABD \u2261 \u2206ACD (SSS Axiom)<br><\/strong><\/p>\n\n\n\n<p><strong>(ii) \u2220B = \u2220C (c. p. c. t)<br><\/strong><\/p>\n\n\n\n<p><strong>(iii) \u2220ADB = \u2220ADC (c. p. c. t)<br><\/strong><\/p>\n\n\n\n<p><strong>(iv) We know that<\/strong><\/p>\n\n\n\n<p><strong>\u2220ADB + \u2220ADC = 180<sup>0<\/sup> is a linear pair<\/strong><\/p>\n\n\n\n<p><strong>Here \u2220ADB = \u2220ADC<\/strong><\/p>\n\n\n\n<p><strong>So we get<\/strong><\/p>\n\n\n\n<p><strong>\u2220ADB = 180<sup>0<\/sup>\/2<\/strong><\/p>\n\n\n\n<p><strong>\u2220ADB = 90\u00b0<\/strong><\/p>\n\n\n\n<p><strong>5. In the given figure, prove that:<\/strong><br><br><strong>(i) \u2206 ACB \u2245 \u2206 ECD<\/strong><br><br><strong>(ii) AB = ED<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-8.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 8\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 8\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i) In \u2206 ACB and \u2206 ECD<br>It is given that AC = CE and BC = CD<\/strong><\/p>\n\n\n\n<p><strong>\u2220ACB = \u2220DCE are vertically opposite angles<\/strong><\/p>\n\n\n\n<p><strong>Hence, \u2206 ACB \u2245 \u2206 ECD (SAS Axiom)<\/strong><\/p>\n\n\n\n<p><strong>(ii) Here AB = ED (c. p. c. t)<\/strong><\/p>\n\n\n\n<p><strong>Therefore, it is proved.<\/strong><\/p>\n\n\n\n<p><strong>6. Prove that<\/strong><\/p>\n\n\n\n<p><strong>(i) \u2206 ABC \u2245 \u2206 ADC<\/strong><br><br><strong>(ii) \u2220B = \u2220D<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-9.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 9\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 9\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i) In \u2206 ABC and \u2206 ADC<\/strong><\/p>\n\n\n\n<p>It is given that<\/p>\n\n\n\n<p>AB = DC and BC = AD<\/p>\n\n\n\n<p>AC = AC is common<\/p>\n\n\n\n<p>Hence, <strong>\u2206 ABC \u2245 \u2206 ADC (SSS Axiom)<\/strong><\/p>\n\n\n\n<p><strong>(ii) Here \u2220B = \u2220D (c. p. c. t)<\/strong><\/p>\n\n\n\n<p><strong>Therefore, it is proved.<\/strong><\/p>\n\n\n\n<p><strong>7. In the given figure, prove that:<\/strong><\/p>\n\n\n\n<p><strong>BD = BC.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-10.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 10\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 10\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>In right <strong>\u2206 ABD and \u2206 ABC<\/strong><\/p>\n\n\n\n<p><strong>AB = AB is common<\/strong><\/p>\n\n\n\n<p><strong>It is given that<\/strong><\/p>\n\n\n\n<p><strong>AD = AC<\/strong><\/p>\n\n\n\n<p><strong>Hence, \u2206 ABD \u2245 \u2206 ABC (RHS Axiom)<\/strong><\/p>\n\n\n\n<p><strong>Here BD = BC (c. p. c. t)<\/strong><\/p>\n\n\n\n<p><strong>Therefore, it is proved.<\/strong><\/p>\n\n\n\n<p><strong>8. In the given figure, \u22201 = \u22202 and AB = AC.<\/strong><\/p>\n\n\n\n<p><strong>Prove that:<br>(i) \u2220B = \u2220 C<br>(ii) BD = DC<br>(iii) AD is perpendicular to BC.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-11.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 11\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 11\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>In \u2206 ADB and \u2206 ADC<\/strong><\/p>\n\n\n\n<p><strong>It is given that<\/strong><\/p>\n\n\n\n<p><strong>AB = AC and \u22201 = \u22202<\/strong><\/p>\n\n\n\n<p><strong>AD = AD is common<\/strong><\/p>\n\n\n\n<p><strong>Hence, \u2206 ADB \u2245 \u2206 ADC (SAS Axiom)<\/strong><\/p>\n\n\n\n<p><strong>(i) \u2220B = \u2220C (c. p. c. t)<\/strong><\/p>\n\n\n\n<p><strong><br>(ii) BD = DC (c. p. c. t)<br>(iii) \u2220ADB = \u2220ADC (c. p. c. t)<\/strong><\/p>\n\n\n\n<p><strong>We know that<\/strong><\/p>\n\n\n\n<p><strong>\u2220ADB + \u2220ADC = 180<sup>0<\/sup> is a linear pair<\/strong><\/p>\n\n\n\n<p><strong>So we get<\/strong><\/p>\n\n\n\n<p><strong>\u2220ADB = \u2220ADC = 90<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>Here, AD is perpendicular to BC<\/strong><\/p>\n\n\n\n<p><strong>Therefore, it is proved.<\/strong><\/p>\n\n\n\n<p><strong>9. In the given figure, prove that:<br>(i) PQ = RS<br>(ii) PS = QR<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-12.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 12\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 12\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>In <strong>\u2206 PQR and \u2206 PSR<\/strong><\/p>\n\n\n\n<p><strong>PR = PR is common<\/strong><\/p>\n\n\n\n<p><strong>It is given that<\/strong><\/p>\n\n\n\n<p><strong>\u2220PRQ = \u2220RPS and \u2220PQR = \u2220PSR<\/strong><\/p>\n\n\n\n<p><strong>\u2206 PQR \u2245 \u2206 PSR (AAS Axiom)<\/strong><\/p>\n\n\n\n<p><strong>(i) PQ = RS (c. p. c. t)<\/strong><\/p>\n\n\n\n<p>(ii) QR = PS or PS = QR (c. p. c. t)<\/p>\n\n\n\n<p>Therefore, it is proved.<\/p>\n\n\n\n<p><strong>10. In the given figure, prove that:<\/strong><\/p>\n\n\n\n<p><strong>(i) \u2206 XYZ \u2245 \u2206 XPZ<br>(ii) YZ = PZ<br>(iii) \u2220YXZ = \u2220PXZ<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-13.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 13\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 13\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>In <strong>\u2206 XYZ and \u2206 XPZ<\/strong><\/p>\n\n\n\n<p><strong>It is given that<\/strong><\/p>\n\n\n\n<p><strong>XY = XP<\/strong><\/p>\n\n\n\n<p><strong>XZ = XZ is common<\/strong><\/p>\n\n\n\n<p><strong>(i) \u2206 XYZ \u2245 \u2206 XPZ (RHS Axiom)<\/strong><\/p>\n\n\n\n<p>(ii) <strong>YZ = PZ (c. p. c. t)<\/strong><\/p>\n\n\n\n<p><strong><br>(iii) \u2220YXZ = \u2220PXZ (c. p. c. t)<\/strong><\/p>\n\n\n\n<p><strong>Therefore, it is proved.<\/strong><\/p>\n\n\n\n<p><strong>11. In the given figure, prove that:<\/strong><br><br><strong>(i) \u2206 ABC \u2245 \u2206 DCB<\/strong><br><br><strong>(ii) AC = DB<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-14.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 14\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 14\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>In <strong>\u2206 ABC and \u2206 DCB<\/strong><\/p>\n\n\n\n<p><strong>CB = CB is common<\/strong><\/p>\n\n\n\n<p><strong>\u2220ABC = \u2220BCD = 90<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>It is given that<\/strong><\/p>\n\n\n\n<p><strong>AB = CD<\/strong><\/p>\n\n\n\n<p><strong>(i) \u2206 ABC \u2245 \u2206 DCB (SAS Axiom)<\/strong><\/p>\n\n\n\n<p><strong><br>(ii) AC = DB (c. p. c. t)<\/strong><\/p>\n\n\n\n<p><strong>Therefore, it is proved.<\/strong><\/p>\n\n\n\n<p><strong>12. In the given figure, prove that:<\/strong><br><br><strong>(i) \u2206 AOD \u2245 \u2206 BOC<\/strong><br><br><strong>(ii) AD = BC<\/strong><br><br><strong>(iii) \u2220ADB = \u2220ACB<\/strong><br><br><strong>(iv) \u2206ADB \u2245 \u2206BCA<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-15.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 15\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 15\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>In <strong>\u2206 AOD and \u2206 BOC<\/strong><\/p>\n\n\n\n<p><strong>It is given that<\/strong><\/p>\n\n\n\n<p><strong>OA = OB and OD = OC<\/strong><\/p>\n\n\n\n<p><strong>\u2220AOD = \u2220BOC are vertically opposite angles<\/strong><\/p>\n\n\n\n<p><strong>(i) \u2206 AOD \u2245 \u2206 BOC (SAS Axiom)<\/strong><\/p>\n\n\n\n<p><strong><br>(ii) AD = BC (c. p. c. t)<\/strong><\/p>\n\n\n\n<p><strong><br>(iii) \u2220ADB = \u2220ACB (c. p. c. t)<\/strong><\/p>\n\n\n\n<p><strong><br>(iv) \u2206ADB \u2245 \u2206BCA<\/strong><\/p>\n\n\n\n<p><strong>It is given that<\/strong><\/p>\n\n\n\n<p><strong>\u2206ADB = \u2206BCA<\/strong><\/p>\n\n\n\n<p>AB = AB is common<\/p>\n\n\n\n<p>Here <strong>\u2206 AOB \u2245 \u2206 BCA<\/strong><\/p>\n\n\n\n<p><strong>Therefore, it is proved.<\/strong><\/p>\n\n\n\n<p><strong>13. ABC is an equilateral triangle, AD and BE are perpendiculars to BC and AC respectively. Prove that:<br>(i) AD = BE<br>(ii)BD = CE<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-16.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 16\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 16\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>In <strong>\u2206 ABC<\/strong><\/p>\n\n\n\n<p><strong>AB = BC = CA<\/strong><\/p>\n\n\n\n<p><strong>We know that<\/strong><\/p>\n\n\n\n<p><strong>AD is perpendicular to BC and BE is perpendicular to AC<\/strong><\/p>\n\n\n\n<p>In <strong>\u2206 ADC and \u2206 BEC<\/strong><\/p>\n\n\n\n<p><strong>\u2220ADC = \u2220BEC = 90<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>\u2220ACD = \u2220BCE is common<\/strong><\/p>\n\n\n\n<p><strong>AC = BC are the sides of an equilateral triangle<\/strong><\/p>\n\n\n\n<p><strong>\u2206 ADC \u2245 \u2206 BEC (AAS Axiom)<\/strong><\/p>\n\n\n\n<p><strong>(i) AD = BE (c. p. c. t)<\/strong><\/p>\n\n\n\n<p>(ii) BD = CE (c. p. c. t)<\/p>\n\n\n\n<p>Therefore, it is proved.<\/p>\n\n\n\n<p><strong>14. Use the informations given in the following figure to prove triangles ABD and CBD are congruent.<\/strong><br><br><strong>Also, find the values of x and y.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-17.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 17\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 17\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>In the figure<\/p>\n\n\n\n<p>AB = BC and AD = DC<\/p>\n\n\n\n<p><strong>\u2220ABD = 50<sup>0<\/sup>, \u2220ADB = y \u2013 7<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>\u2220CBD = x + 5<sup>0<\/sup>, \u2220CDB = 38<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p>In <strong>\u2206 ABD and \u2206 CBD<\/strong><\/p>\n\n\n\n<p><strong>BD = BD is common<\/strong><\/p>\n\n\n\n<p><strong>It is given that<\/strong><\/p>\n\n\n\n<p><strong>AB = BC and AD = CD<\/strong><\/p>\n\n\n\n<p><strong>Here \u2206 ABD \u2245 \u2206 CBD (SSS Axiom)<\/strong><\/p>\n\n\n\n<p><strong>\u2220ABD = \u2220CBD<\/strong><\/p>\n\n\n\n<p><strong>So we get<\/strong><\/p>\n\n\n\n<p><strong>50 = x + 5<\/strong><\/p>\n\n\n\n<p><strong>x = 50 \u2013 5 = 45<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>\u2220ADB = \u2220CDB<\/strong><\/p>\n\n\n\n<p><strong>y \u2013 7 = 38<\/strong><\/p>\n\n\n\n<p><strong>y = 38 + 7 = 45<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>Therefore, x = 45<sup>0<\/sup> and y = 45<sup>0<\/sup>.<\/strong><\/p>\n\n\n\n<p><strong>15. The given figure shows a triangle ABC in which AD is perpendicular to side BC and BD = CD. Prove that:<br>(i) \u2206 ABD \u2245 \u2206 ACD<br>(ii) AB = AC<br>(iii) \u2220B = \u2220C<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2022\/05\/selina-solutions-concise-maths-class-7-chapter-19-image-18.png\" alt=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 18\" title=\"Selina Solutions Concise Maths Class 7 Chapter 19 Image 18\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>(i) In \u2206 ABC<\/strong><\/p>\n\n\n\n<p>AD is perpendicular to BC<\/p>\n\n\n\n<p>BD = CD<\/p>\n\n\n\n<p><strong>In \u2206 ABD and \u2206 ACD<\/strong><\/p>\n\n\n\n<p><strong>AD = AD is common<\/strong><\/p>\n\n\n\n<p><strong>\u2220ADB = \u2220ADC = 90<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>It is given that<\/strong><\/p>\n\n\n\n<p><strong>BD = CD<\/strong><\/p>\n\n\n\n<p><strong>\u2206 ABD \u2245 \u2206 CAD (SAS Axiom)<\/strong><\/p>\n\n\n\n<p><strong>(ii) AB = AC (c. p. c. t)<\/strong><\/p>\n\n\n\n<p><strong>(iii) \u2220B = \u2220C as \u2206 ADB \u2245 \u2206 ADC<\/strong><\/p>\n\n\n\n<p><strong>Therefore, it is proved.<\/strong><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-download-pdf\"><strong>Download PDF<\/strong><\/h2>\n\n\n\n<p>Selina Class 7 ICSE Solutions Mathematics : Chapter 19-&nbsp;Congruency: Congruent Triangles<\/p>\n\n\n\n<p><a href=\"https:\/\/www.indcareer.com\/docs\/8b9a9f37-95ed-474c-b8e1-417a75229aa0\" target=\"_blank\" rel=\"noreferrer noopener\"><strong>Download PDF<\/strong>: Selina Class 7 ICSE Solutions Mathematics : Chapter 19-\u00a0Congruency: Congruent Triangles PDF<\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Chapterwise Selina Publishers&nbsp;ICSE Solutions for Class 7&nbsp;Mathematics :<\/strong><\/h2>\n\n\n\n<ul class=\"wp-block-list\"><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-1-integers\/\">Chapter 1- Integers<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-2-rational-numbers\/\">Chapter 2- Rational Numbers<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-3-fraction-including-problems\/\">Chapter 3- Fraction (Including Problems)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-4-decimal-fractions-decimals\/\">Chapter 4- Decimal Fractions (Decimals)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-5-exponents-including-laws-of-exponents\/\">Chapter 5- Exponents (Including Laws of Exponents)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-6-ratio-and-proportion-including-sharing-in-a-ratio\/\">Chapter 6- Ratio and Proportion (Including Sharing in a Ratio)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-7-unitary-method-including-time-and-work\/\">Chapter 7- Unitary Method (Including Time and Work)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-8-percent-and-percentage\/\">Chapter 8- Percent and Percentage<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-9-profit-loss-and-discount\/\">Chapter 9- Profit, Loss and Discount<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-10-simple-interest\/\">Chapter 10- Simple Interest<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-11-fundamental-concepts-including-fundamental-operations\/\">Chapter 11- Fundamental Concepts (Including Fundamental Operations)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-12-simple-linear-equations-including-word-problems\/\">Chapter 12- Simple Linear Equations (Including Word Problems)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-13-set-concepts\/\">Chapter 13- Set Concepts<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-14-lines-and-angles-including-construction-of-angles\/\">Chapter 14- Lines and Angles (Including Construction of Angles)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-15-triangles\/\">Chapter 15- Triangles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-16-pythagoras-theorem\/\">Chapter 16- Pythagoras Theorem<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-17-symmetry-including-reflection-and-rotation\/\">Chapter 17- Symmetry (Including Reflection and Rotation)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-18-recognition-of-solids-representing-3-d-in-2-d\/\">Chapter 18- Recognition of Solids (Representing 3-D in 2-D)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-19-congruency-congruent-triangles\/\">Chapter 19- Congruency: Congruent Triangles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-20-mensuration-perimeter-and-area-of-plane-figures\/\">Chapter 20- Mensuration (Perimeter and Area of Plane Figures)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-21-data-handling\/\">Chapter 21- Data Handling<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-22-probability\/\">Chapter 22- Probability<\/a><\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">About Selina Publishers&nbsp;ICSE<\/h2>\n\n\n\n<p>Selina Publishers has been serving the students since 1976 and is one of the quality ICSE school textbooks publication houses. Mathematics and Science books for classes 6-10 form the core of our business, apart from certain English and Hindi literature as well as a few primary books. All these books are based upon the syllabus published by the Council for the I.C.S.E. Examinations, New Delhi. The textbooks are composed by a panel of subject experts and vetted by teachers practising in ICSE schools all over the country. Continuous efforts are made in complying with the standards and ensuring lucidity and clarity in content, which makes them stand tall in the industry.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Class 7: Maths Chapter 19 solutions. Complete Class 7 Maths Chapter 19 Notes. Selina Class 7 ICSE Solutions Mathematics : Chapter 19-&nbsp;Congruency: Congruent Triangles Selina 7th Maths Chapter 19, Class 7 Maths Chapter 19 solutions Exercise 19 page: 213 1. State, whether the pairs of triangles given in the following figures are congruent or not: [&hellip;]<\/p>\n","protected":false},"author":302,"featured_media":598597,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,907],"tags":[2261],"boards":[],"class_list":["post-598595","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-7","tag-icse-solutions","entry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.0 (Yoast SEO v27.1.1) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>Selina Solutions for Class 7, maths Chapter 19 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Selina Class 7 ICSE Solutions Mathematics : Chapter 19-\u00a0Congruency: Congruent Triangles | Browse all Class 7 maths - IndCareer Schools\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-19-congruency-congruent-triangles\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Selina Class 7 ICSE Solutions Mathematics : Chapter 19-\u00a0Congruency: Congruent Triangles\" \/>\n<meta property=\"og:description\" content=\"Class 7: Maths Chapter 19 solutions. 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