{"id":599611,"date":"2022-05-06T04:43:27","date_gmt":"2022-05-06T04:43:27","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=599611"},"modified":"2022-05-10T05:30:12","modified_gmt":"2022-05-10T05:30:12","slug":"selina-class-6-icse-solutions-mathematics-chapter-26-triangles","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-26-triangles\/","title":{"rendered":"Selina Class 6 ICSE Solutions Mathematics : Chapter 26-\u00a0Triangles"},"content":{"rendered":"\n<p>Class 6: Maths Chapter 26 solutions. Complete Class 6 Maths Chapter 26 Notes.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-selina-class-6-icse-solutions-mathematics-chapter-26-triangles\">Selina Class 6 ICSE Solutions Mathematics : Chapter 26-&nbsp;Triangles<\/h2>\n\n\n\n<p>Selina 6th Maths Chapter 26, Class 6 Maths Chapter 26 solutions<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Exercise 26(A)<\/h3>\n\n\n\n<p><strong>1. In each of the following, find the marked unknown angles:<\/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-mathematics-class-6-chapter-26-1.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 1\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 1\"\/><\/figure>\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-mathematics-class-6-chapter-26-2.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 2\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 2\"\/><\/figure>\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-mathematics-class-6-chapter-26-3.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 3\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 3\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) We know that,<\/p>\n\n\n\n<p>Sum of all angles of triangle = 180<sup>0<\/sup><\/p>\n\n\n\n<p>Therefore,<\/p>\n\n\n\n<p>70<sup>0<\/sup>&nbsp;+ 72<sup>0<\/sup>&nbsp;+ z = 180<sup>0<\/sup><\/p>\n\n\n\n<p>142<sup>0<\/sup>&nbsp;+ z = 180<sup>0<\/sup><\/p>\n\n\n\n<p>z = 180<sup>0<\/sup>&nbsp;\u2013 142<sup>0<\/sup><\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>z = 38<sup>0<\/sup><\/p>\n\n\n\n<p>(ii) We know that,<\/p>\n\n\n\n<p>Sum of all angles of a triangle = 180<sup>0<\/sup><\/p>\n\n\n\n<p>First triangle<\/p>\n\n\n\n<p>50<sup>0<\/sup>&nbsp;+ 80<sup>0<\/sup>&nbsp;+ b = 180<sup>0<\/sup><\/p>\n\n\n\n<p>130<sup>0<\/sup>&nbsp;+ b = 180<sup>0<\/sup><\/p>\n\n\n\n<p>b = 180<sup>0<\/sup>&nbsp;\u2013 130<sup>0<\/sup><\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>b = 50<sup>0<\/sup><\/p>\n\n\n\n<p>Second triangle<\/p>\n\n\n\n<p>40<sup>0<\/sup>&nbsp;+ 45<sup>0<\/sup>&nbsp;+ a = 180<sup>0<\/sup><\/p>\n\n\n\n<p>85<sup>0<\/sup>&nbsp;+ a = 180<sup>0<\/sup><\/p>\n\n\n\n<p>a = 180<sup>0<\/sup>&nbsp;\u2013 85<sup>0<\/sup><\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>a = 95<sup>0<\/sup><\/p>\n\n\n\n<p>(iii) 60<sup>0<\/sup>&nbsp;+ 45<sup>0<\/sup>&nbsp;+ 20<sup>0<\/sup>&nbsp;+ x = 180<sup>0<\/sup><\/p>\n\n\n\n<p>125<sup>0<\/sup>&nbsp;+ x = 180<sup>0<\/sup><\/p>\n\n\n\n<p>x = 180<sup>0<\/sup>&nbsp;\u2013 125<sup>0<\/sup><\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>x = 55<sup>0<\/sup><\/p>\n\n\n\n<p><strong>2. Can a triangle together have the following angles?<\/strong><\/p>\n\n\n\n<p><strong>(i) 55<sup>0<\/sup>, 55<sup>0<\/sup>&nbsp;and 80<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>(ii) 33<sup>0<\/sup>, 74<sup>0<\/sup>&nbsp;and 73<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>(iii) 85<sup>0<\/sup>, 95<sup>0<\/sup>&nbsp;and 22<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Sum of all angles of a triangle = 180<sup>0<\/sup><\/p>\n\n\n\n<p>Here,<\/p>\n\n\n\n<p>55<sup>0<\/sup>&nbsp;+ 55<sup>0<\/sup>&nbsp;+ 80<sup>0<\/sup>&nbsp;= 180<sup>0<\/sup><\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>190<sup>0<\/sup>&nbsp;\u2260 180<sup>0<\/sup><\/p>\n\n\n\n<p>Therefore, it cannot form a triangle<\/p>\n\n\n\n<p>(ii) 33<sup>0<\/sup>&nbsp;+ 74<sup>0<\/sup>&nbsp;+ 73<sup>0<\/sup>&nbsp;= 180<sup>0<\/sup><\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>180<sup>0<\/sup>&nbsp;= 180<sup>0<\/sup><\/p>\n\n\n\n<p>Therefore, it form a triangle<\/p>\n\n\n\n<p>(iii) 85<sup>0<\/sup>&nbsp;+ 95<sup>0<\/sup>&nbsp;+ 22<sup>0<\/sup>&nbsp;= 180<sup>0<\/sup><\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>202<sup>0<\/sup>&nbsp;\u2260 180<sup>0<\/sup><\/p>\n\n\n\n<p>Therefore, it cannot form a triangle<\/p>\n\n\n\n<p><strong>3. Find x, if the angles of a triangle are:<\/strong><\/p>\n\n\n\n<p><strong>(i) x<sup>0<\/sup>, x<sup>0<\/sup>, x<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>(ii) x<sup>0<\/sup>, 2x<sup>0<\/sup>, 2x<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>(iii) 2x<sup>0<\/sup>, 4x<sup>0<\/sup>, 6x<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We know that,<\/p>\n\n\n\n<p>The sum of all the angles in a triangle is 180<sup>0<\/sup><\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>x<sup>0&nbsp;<\/sup>+ x<sup>0<\/sup>&nbsp;+ x<sup>0<\/sup>&nbsp;= 180<sup>0<\/sup><\/p>\n\n\n\n<p>3x = 180<sup>0<\/sup><\/p>\n\n\n\n<p>x = 180<sup>0<\/sup>&nbsp;\/ 3<\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>x = 60<sup>0<\/sup><\/p>\n\n\n\n<p>The value of x = 60<sup>0<\/sup><\/p>\n\n\n\n<p>(ii) We know that,<\/p>\n\n\n\n<p>The sum of all the angles in a triangle is 180<sup>0<\/sup><\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>x + 2x + 2x = 180<sup>0<\/sup><\/p>\n\n\n\n<p>5x = 180<sup>0<\/sup><\/p>\n\n\n\n<p>x = 180<sup>0<\/sup>&nbsp;\/ 5<\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>x = 36<sup>0<\/sup><\/p>\n\n\n\n<p>Therefore, the value of x = 36<sup>0<\/sup><\/p>\n\n\n\n<p>(iii) We know that,<\/p>\n\n\n\n<p>The sum of all the angles in a triangle is 180<sup>0<\/sup><\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>2x + 4x + 6x = 180<sup>0<\/sup><\/p>\n\n\n\n<p>12x = 180<sup>0<\/sup><\/p>\n\n\n\n<p>x = 180<sup>0<\/sup>&nbsp;\/ 12<\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>x = 15<sup>0<\/sup><\/p>\n\n\n\n<p>Therefore, the value of x = 15<sup>0<\/sup><\/p>\n\n\n\n<p><strong>4. One angle of a right-angled triangle is 70<sup>0<\/sup>. Find the other acute angle.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We know that,<\/p>\n\n\n\n<p>Sum of all the angles in a triangle = 180<sup>0<\/sup><\/p>\n\n\n\n<p>Let us consider the acute angle as x<\/p>\n\n\n\n<p>Hence,<\/p>\n\n\n\n<p>x + 90<sup>0<\/sup>&nbsp;+ 70<sup>0<\/sup>&nbsp;= 180<sup>0<\/sup><\/p>\n\n\n\n<p>x + 160<sup>0<\/sup>&nbsp;= 180<sup>0<\/sup><\/p>\n\n\n\n<p>x = 180<sup>0<\/sup>&nbsp;\u2013 160<sup>0<\/sup><\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>x = 20<sup>0<\/sup><\/p>\n\n\n\n<p>Therefore, the acute angle is 20<sup>0<\/sup><\/p>\n\n\n\n<p><strong>5. In \u25b3ABC, \u2220A = \u2220B = 62<sup>0<\/sup>; find \u2220C<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given<\/p>\n\n\n\n<p>\u2220A = \u2220B = 62<sup>0<\/sup><\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>\u2220A + \u2220B + \u2220C = 180<sup>0<\/sup><\/p>\n\n\n\n<p>62<sup>0<\/sup>&nbsp;+ 62<sup>0<\/sup>&nbsp;+ \u2220C = 180<sup>0<\/sup><\/p>\n\n\n\n<p>124<sup>0<\/sup>&nbsp;+ \u2220C = 180<sup>0<\/sup><\/p>\n\n\n\n<p>\u2220C = 180<sup>0<\/sup>&nbsp;\u2013 124<sup>0<\/sup><\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>\u2220C = 56<sup>0<\/sup><\/p>\n\n\n\n<p>Hence, \u2220C = 56<sup>0<\/sup><\/p>\n\n\n\n<p><strong>6. In \u25b3ABC, \u2220B = \u2220C and \u2220A = 100<sup>0<\/sup>; find \u2220B.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given<\/p>\n\n\n\n<p>\u2220B = \u2220C<\/p>\n\n\n\n<p>We know that,<\/p>\n\n\n\n<p>Sum of all the angles in a triangle is 180<sup>0<\/sup><\/p>\n\n\n\n<p>\u2220A + \u2220B + \u2220C = 180<sup>0<\/sup><\/p>\n\n\n\n<p>100<sup>0<\/sup>&nbsp;+ \u2220B + \u2220B = 180<sup>0<\/sup><\/p>\n\n\n\n<p>100<sup>0<\/sup>&nbsp;+ 2\u2220B = 180<sup>0<\/sup><\/p>\n\n\n\n<p>2\u2220B = 180<sup>0<\/sup>&nbsp;\u2013 100<sup>0<\/sup><\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>2\u2220B = 80<sup>0<\/sup><\/p>\n\n\n\n<p>\u2220B = 80<sup>0<\/sup>&nbsp;\/ 2<\/p>\n\n\n\n<p>\u2220B = 40<sup>0<\/sup><\/p>\n\n\n\n<p>Therefore, \u2220B + \u2220C = 40<sup>0<\/sup><\/p>\n\n\n\n<p><strong>7. Find, giving reasons, the unknown marked angles, in each triangle drawn below:<\/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-mathematics-class-6-chapter-26-4.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 4\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 4\"\/><\/figure>\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-mathematics-class-6-chapter-26-5.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 5\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 5\"\/><\/figure>\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-mathematics-class-6-chapter-26-6.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 6\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 6\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We know that,<\/p>\n\n\n\n<p>Exterior angle of a triangle is always equal to the sum of its two interior opposite angles (property)<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>(i) 110<sup>0<\/sup>&nbsp;= x + 30<sup>0<\/sup>&nbsp;[By property]<\/p>\n\n\n\n<p>x = 110<sup>0<\/sup>&nbsp;\u2013 30<sup>0<\/sup><\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>x = 80<sup>0<\/sup><\/p>\n\n\n\n<p>(ii) x + 115<sup>0<\/sup>&nbsp;= 180<sup>0<\/sup>&nbsp;[By linear property of angles]<\/p>\n\n\n\n<p>x = 180<sup>0<\/sup>&nbsp;\u2013 115<sup>0<\/sup><\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>x = 65<sup>0<\/sup><\/p>\n\n\n\n<p>By exterior angle property<\/p>\n\n\n\n<p>x + y = 115<sup>0<\/sup><\/p>\n\n\n\n<p>65<sup>0<\/sup>&nbsp;+ y = 115<sup>0<\/sup><\/p>\n\n\n\n<p>y = 115<sup>0<\/sup>&nbsp;\u2013 65<sup>0<\/sup><\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>y = 50<sup>0<\/sup><\/p>\n\n\n\n<p>Therefore the value of angle x is 65<sup>0<\/sup>&nbsp;and y is 50<sup>0<\/sup><\/p>\n\n\n\n<p>(iii) By exterior angle property,<\/p>\n\n\n\n<p>110<sup>0<\/sup>&nbsp;= 2x + 3x<\/p>\n\n\n\n<p>5x = 110<sup>0<\/sup><\/p>\n\n\n\n<p>x = 110<sup>0<\/sup>&nbsp;\/ 5<\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>x = 22<sup>0<\/sup><\/p>\n\n\n\n<p>Hence,<\/p>\n\n\n\n<p>The value of 2x = 2 \u00d7 22<\/p>\n\n\n\n<p>= 44<sup>0<\/sup><\/p>\n\n\n\n<p>The value of 3x = 3 \u00d7 22<\/p>\n\n\n\n<p>= 66<sup>0<\/sup><\/p>\n\n\n\n<p><strong>8. Classify the following triangles according to angle:<\/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-mathematics-class-6-chapter-26-7.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 7\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 7\"\/><\/figure>\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-mathematics-class-6-chapter-26-8.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 8\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 8\"\/><\/figure>\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-mathematics-class-6-chapter-26-9.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 9\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 9\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Since, one of the angle of a triangle is 120<sup>0<\/sup>.<\/p>\n\n\n\n<p>Therefore, it is obtuse angled triangle<\/p>\n\n\n\n<p>(ii) Since, all the angles of a triangle is less than 90<sup>0<\/sup><\/p>\n\n\n\n<p>Therefore, it is acute angled triangle<\/p>\n\n\n\n<p>(iii) Since \u2220MNL = 90<sup>0<\/sup>&nbsp;and<\/p>\n\n\n\n<p>Sum of two acute angle s,<\/p>\n\n\n\n<p>\u2220M + \u2220N = 30<sup>0<\/sup>&nbsp;+ 60<sup>0<\/sup><\/p>\n\n\n\n<p>= 90<sup>0<\/sup><\/p>\n\n\n\n<p>Therefore, it is right angled triangle<\/p>\n\n\n\n<p><strong>9. Classify the following triangles according to sides:<\/strong><\/p>\n\n\n\n<p><strong>(i)<\/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-mathematics-class-6-chapter-26-10.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 10\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 10\"\/><\/figure>\n\n\n\n<p><strong>(ii)<\/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-mathematics-class-6-chapter-26-11.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 11\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 11\"\/><\/figure>\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-mathematics-class-6-chapter-26-12.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 12\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 12\"\/><\/figure>\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-mathematics-class-6-chapter-26-13.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 13\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 13\"\/><\/figure>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) In the given triangle, we find two sides are equal.<\/p>\n\n\n\n<p>Therefore, it is isosceles triangle<\/p>\n\n\n\n<p>(ii) In the given triangle, all the three sides are unequal.<\/p>\n\n\n\n<p>Therefore, it is scalene triangle<\/p>\n\n\n\n<p>(iii) In the given triangle, all the three sides are unequal.<\/p>\n\n\n\n<p>Therefore, it is scalene triangle<\/p>\n\n\n\n<p>(iv) In the given triangle, all the three sides are equal.<\/p>\n\n\n\n<p>Therefore, it is equilateral triangle<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Exercise 26(B)<\/strong><\/h3>\n\n\n\n<p><strong>1. Construct triangle ABC, when:<\/strong><\/p>\n\n\n\n<p><strong>AB = 6 cm, BC = 8 cm and AC = 4 cm<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/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-mathematics-class-6-chapter-26-14.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 14\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26\"\/><\/figure>\n\n\n\n<p>Given<\/p>\n\n\n\n<p>AB = 6 cm<\/p>\n\n\n\n<p>BC = 8 cm<\/p>\n\n\n\n<p>AC = 4 cm<\/p>\n\n\n\n<p>Now,<\/p>\n\n\n\n<p>Steps of Construction:<\/p>\n\n\n\n<p>(i) Draw a line AB of length 6 cm<\/p>\n\n\n\n<p>(ii) Using compasses, take B as centre, and draw an arc of 8 cm radius<\/p>\n\n\n\n<p>(iii) Again, taking A as centre, draw another arc of 4 cm radius, which cuts the previous arc at point C<\/p>\n\n\n\n<p>(iv) Now join AC and BC<\/p>\n\n\n\n<p>The obtained triangle ABC is the required triangle.<\/p>\n\n\n\n<p><strong>2. AB = 3.5 cm, AC = 4.8 cm and BC = 5.2 cm<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/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-mathematics-class-6-chapter-26-15.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 15\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26\"\/><\/figure>\n\n\n\n<p>Given<\/p>\n\n\n\n<p>AB = 3.5 cm<\/p>\n\n\n\n<p>AC = 4.8 cm<\/p>\n\n\n\n<p>BC = 5.2 cm<\/p>\n\n\n\n<p>Steps of Construction:<\/p>\n\n\n\n<p>(i) Draw a line AB of length 3.5 cm<\/p>\n\n\n\n<p>(ii) With the help of compasses, taking B as centre, draw an arc of 5.2 cm radius<\/p>\n\n\n\n<p>(iii) Again with A as centre, draw an arc of 4.8 radius<\/p>\n\n\n\n<p>(iv) Now, join AC and BC<\/p>\n\n\n\n<p><strong>3. AB = BC = 5 cm and AC = 3 cm. Measure angles A and C. Is \u2220A = \u2220C?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/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-mathematics-class-6-chapter-26-16.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 16\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26\"\/><\/figure>\n\n\n\n<p>Given<\/p>\n\n\n\n<p>AB = BC = 5 cm<\/p>\n\n\n\n<p>AC = 3 cm<\/p>\n\n\n\n<p>Steps of Construction:<\/p>\n\n\n\n<p>(i) Draw a line AB of length 5 cm<\/p>\n\n\n\n<p>(ii) Using compasses take B as centre and draw an arc of 5 cm radius<\/p>\n\n\n\n<p>(iii) Now, taking A as centre, draw another arc of 3 cm radius, which cuts the previous arc at point C<\/p>\n\n\n\n<p>(iv) Now, join AC and BC<\/p>\n\n\n\n<p>Measuring angles A and C, we get<br>\u2220A = 72.5o and \u2220C = 72.5o<br>Hence, yes \u2220A = \u2220C.<\/p>\n\n\n\n<p><strong>4. AB = BC = CA = 4.5 cm. Measure all the angles of the triangle. Are they equal?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/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-mathematics-class-6-chapter-26-17.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 17\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26\"\/><\/figure>\n\n\n\n<p>Given<\/p>\n\n\n\n<p>AB = BC = CA = 4.5 cm<\/p>\n\n\n\n<p>Steps of Construction:<\/p>\n\n\n\n<p>(i) Draw a line AB of length 4.5 cm<\/p>\n\n\n\n<p>(ii) Using compasses and taking BC as centre, draw an arc of 4.5 cm radius<\/p>\n\n\n\n<p>(iii) Again taking AC as centre, draw another arc of 4.5 cm radius, which cuts the previous arc at point C<\/p>\n\n\n\n<p>(iv) Now, join AC and BC<\/p>\n\n\n\n<p>(v) All the angles in ABC i.e \u2220A = \u2220B = \u2220C = 60<sup>0<\/sup><\/p>\n\n\n\n<p>Since AB = BC = CA = 4.5 cm and all the angles are equal. Hence, it is an equilateral triangle<\/p>\n\n\n\n<p><strong>5. AB = 3 cm, BC = 7 cm and \u2220B = 90<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/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-mathematics-class-6-chapter-26-18.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 18\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26\"\/><\/figure>\n\n\n\n<p>Given<\/p>\n\n\n\n<p>AB = 3 cm,<\/p>\n\n\n\n<p>BC = 7 cm and<\/p>\n\n\n\n<p>\u2220B = 90<sup>0<\/sup><\/p>\n\n\n\n<p>Steps of Construction:<\/p>\n\n\n\n<p>(i) Draw a line segment AB of length 3 cm<\/p>\n\n\n\n<p>(ii) Using compasses, construct \u2220ABC = 90<sup>0<\/sup><\/p>\n\n\n\n<p>(iii) Taking B as centre, draw an arc of 7 cm length and mark as point C i.e BC = 7 cm<\/p>\n\n\n\n<p>(iv) Now, join A and C<\/p>\n\n\n\n<p>(v) The obtained \u25b3ABC, is the required triangle<\/p>\n\n\n\n<p><strong>6. AC = 4.5 cm, BC = 6 cm and \u2220C = 60<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/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-mathematics-class-6-chapter-26-19.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 19\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26\"\/><\/figure>\n\n\n\n<p>Given<\/p>\n\n\n\n<p>AC = 4.5 cm<\/p>\n\n\n\n<p>BC = 6 cm<\/p>\n\n\n\n<p>\u2220C = 60<sup>0<\/sup><\/p>\n\n\n\n<p>Steps of Construction:<\/p>\n\n\n\n<p>(i) Draw a line AC of length 4.5 cm<\/p>\n\n\n\n<p>(ii) Using compasses, construct \u2220ACB = 60<sup>0<\/sup><\/p>\n\n\n\n<p>(iii) Draw an arc of 6 cm radius and mark it as B such that BC = 6 cm<\/p>\n\n\n\n<p>(iv) Now, join B and A<\/p>\n\n\n\n<p><strong>7. AC = 6 cm, \u2220A = 60<sup>0<\/sup>&nbsp;and \u2220C = 45<sup>0<\/sup>. Measure AB and BC.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/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-mathematics-class-6-chapter-26-20.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 20\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26\"\/><\/figure>\n\n\n\n<p>Given<\/p>\n\n\n\n<p>AC = 6 cm<\/p>\n\n\n\n<p>\u2220A = 60<sup>0<\/sup><\/p>\n\n\n\n<p>\u2220C = 45<sup>0<\/sup><\/p>\n\n\n\n<p>Steps of Construction:<\/p>\n\n\n\n<p>(i) Draw a line segment AC of length 6 cm<\/p>\n\n\n\n<p>(ii) With the help of Compass, construct \u2220A = 60<sup>0<\/sup><\/p>\n\n\n\n<p>(iii) Again, using compass, construct \u2220C = 45<sup>0<\/sup><\/p>\n\n\n\n<p>(iv) AD and CE intersect each other at point B<\/p>\n\n\n\n<p>(v) Now, the obtained \u25b3ABC is the required triangle<\/p>\n\n\n\n<p>(vi) Measure the side AB and BC with the help of a scale<\/p>\n\n\n\n<p>(vii) We get, AB = 4.4 cm and BC = 5.4 cm<\/p>\n\n\n\n<p><strong>8. AB = 5.4 cm, \u2220A = 30<sup>0<\/sup>&nbsp;and \u2220B = 90<sup>0<\/sup>. Measure \u2220C and side BC.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/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-mathematics-class-6-chapter-26-21.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 21\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 21\"\/><\/figure>\n\n\n\n<p>Given<\/p>\n\n\n\n<p>AB = 5.4 cm<\/p>\n\n\n\n<p>\u2220A = 30<sup>0<\/sup>&nbsp;and<\/p>\n\n\n\n<p>\u2220B = 90<sup>0<\/sup><\/p>\n\n\n\n<p>Steps of Construction:<\/p>\n\n\n\n<p>(i) Draw a line segment AB of length 5.4 cm<\/p>\n\n\n\n<p>(ii) With the help of compass, construct \u2220A = 30<sup>0<\/sup><\/p>\n\n\n\n<p>(iii) Similarly, construct \u2220B = 90<sup>0<\/sup><\/p>\n\n\n\n<p>(iv) AD and BE intersect each other at point C<\/p>\n\n\n\n<p>(v) Hence, the obtained \u25b3ABC is the required triangle<\/p>\n\n\n\n<p>(vi) On measuring we get, \u2220C = 60 and side BC = 3.1 cm approximately<\/p>\n\n\n\n<p><strong>9. AB = 7 cm, \u2220B = 120<sup>0<\/sup>&nbsp;and \u2220A = 30<sup>0<\/sup>. Measure AC and BC.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/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-mathematics-class-6-chapter-26-22.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 22\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 22\"\/><\/figure>\n\n\n\n<p>Given<\/p>\n\n\n\n<p>AB = 7 cm<\/p>\n\n\n\n<p>\u2220B = 120<sup>0<\/sup><\/p>\n\n\n\n<p>\u2220A = 30<sup>0<\/sup><\/p>\n\n\n\n<p>Steps of Construction:<\/p>\n\n\n\n<p>(i) Draw a line segment AB of length 7 cm<\/p>\n\n\n\n<p>(ii) With the help of a compass, construct \u2220A = 30<sup>0<\/sup><\/p>\n\n\n\n<p>(iii) AE and BD intersect each other at point C<\/p>\n\n\n\n<p>(iv) Hence, the obtained \u25b3ABC is the required triangle<\/p>\n\n\n\n<p>(v) On measuring the lengths, we get AC = 12 cm and BC = 7 cm respectively<\/p>\n\n\n\n<p><strong>10. BC = 3 cm, AC = 4 cm and AB = 5 cm. Measure angle ACB. Give a special name to this triangle<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/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-mathematics-class-6-chapter-26-23.png\" alt=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 23\" title=\"Selina Solutions Concise Mathematics Class 6 Chapter 26 - 23\"\/><\/figure>\n\n\n\n<p>Given<\/p>\n\n\n\n<p>BC = 3 cm<\/p>\n\n\n\n<p>AC = 4 cm and<\/p>\n\n\n\n<p>AB = 5 cm<\/p>\n\n\n\n<p>Steps of Construction:<\/p>\n\n\n\n<p>(i) Draw a line segment AB of length 5 cm<\/p>\n\n\n\n<p>(ii) From B, using compass cut an arc of 3 cm radius<\/p>\n\n\n\n<p>(iii) Similarly, from A again with the help of compass, cut an arc of 4 cm bisecting the previous arc formed from point B<\/p>\n\n\n\n<p>(iv) Now, join point C with A and B<\/p>\n\n\n\n<p>(v) The obtained triangle is the required \u25b3ABC<\/p>\n\n\n\n<p>(vi) On measuring \u2220ACB, we get \u2220ACB = 90<sup>0<\/sup>.<\/p>\n\n\n\n<p>Therefore, the obtained triangle ABC is a right angled triangle<\/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 6 ICSE Solutions Mathematics : Chapter 26-&nbsp;Triangles<\/p>\n\n\n\n<p><a href=\"https:\/\/www.indcareer.com\/docs\/2a99bcb6-7d5c-4aed-9880-212b225cc73e\" target=\"_blank\" rel=\"noreferrer noopener\"><strong>Download PDF<\/strong>: Selina Class 6 ICSE Solutions Mathematics : Chapter 26-\u00a0Triangles PDF<\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Chapterwise Selina Publishers&nbsp;ICSE Solutions for Class 6&nbsp;Mathematics :<\/strong><\/h2>\n\n\n\n<ul class=\"wp-block-list\"><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-1-number-system\/\">Chapter 1- Number System<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-2-estimation\/\">Chapter 2- Estimation<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-3-numbers-in-indian-and-international-systems\/\">Chapter 3- Numbers In Indian And International Systems<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-4-place-value\/\">Chapter 4- Place Value<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-5-natural-numbers-and-whole-numbers\/\">Chapter 5- Natural Numbers And Whole Numbers<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-6-negative-numbers-and-integers\/\">Chapter 6- Negative Numbers And Integers<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-7-number-line\/\">Chapter 7- Number Line<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-8-hcf-and-lcm\/\">Chapter 8- HCF And LCM<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-9-playing-with-numbers\/\">Chapter 9- Playing With Numbers<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-10-sets\/\">Chapter 10- Sets<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-11-ratio\/\">Chapter 11- Ratio<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-12-proportion\/\">Chapter 12- Proportion<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-13-unitary-method\/\">Chapter 13- Unitary Method<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-14-fractions\/\">Chapter 14- Fractions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-15-decimal-fractions\/\">Chapter 15- Decimal Fractions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-16-percent-percentage\/\">Chapter 16- Percent (Percentage)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-17-idea-of-speed-distance-and-time\/\">Chapter 17- Idea of Speed, Distance and Time<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-18-fundamental-concepts\/\">Chapter 18- Fundamental Concepts<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-19-fundamental-operations\/\">Chapter 19- Fundamental Operations<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-20-substitution\/\">Chapter 20- Substitution<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-21-framing-algebraic-expressions-including-evaluation\/\">Chapter 21- Framing Algebraic Expressions (Including Evaluation)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-22-simple-linear-equations\/\">Chapter 22- Simple (Linear) Equations<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-23-fundamental-concepts\/\">Chapter 23- Fundamental Concepts<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-24-angles\/\">Chapter 24- Angles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-25-properties-of-angles-and-lines\/\">Chapter 25- Properties of Angles and Lines<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-26-triangles\/\">Chapter 26- Triangles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-27-quadrilateral\/\">Chapter 27- Quadrilateral<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-28-polygons\/\">Chapter 28- Polygons<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-29-the-circle\/\">Chapter 29- The Circle<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-30-revision-exercise-symmetry\/\">Chapter 30- Revision Exercise Symmetry<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-31-recognition-of-solids\/\">Chapter 31- Recognition of Solids<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-32-perimeter-and-area-of-plane-figures\/\">Chapter 32- Perimeter and Area of Plane Figures<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-33-data-handling\/\">Chapter 33- Data Handling<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-34-mean-and-median\/\">Chapter 34- Mean and Median<\/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 6: Maths Chapter 26 solutions. Complete Class 6 Maths Chapter 26 Notes. Selina Class 6 ICSE Solutions Mathematics : Chapter 26-&nbsp;Triangles Selina 6th Maths Chapter 26, Class 6 Maths Chapter 26 solutions Exercise 26(A) 1. In each of the following, find the marked unknown angles: Solution: (i) We know that, Sum of all angles [&hellip;]<\/p>\n","protected":false},"author":302,"featured_media":599614,"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":[1411,876],"tags":[2261],"boards":[],"class_list":["post-599611","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-6","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 6, maths Chapter 26 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Selina Class 6 ICSE Solutions Mathematics : Chapter 26-\u00a0Triangles | Browse all Class 6 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-6-icse-solutions-mathematics-chapter-26-triangles\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Selina Class 6 ICSE Solutions Mathematics : Chapter 26-\u00a0Triangles\" \/>\n<meta property=\"og:description\" content=\"Class 6: Maths Chapter 26 solutions. 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