{"id":4354,"date":"2023-02-09T06:58:28","date_gmt":"2023-02-09T06:58:28","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4354"},"modified":"2023-02-09T07:00:23","modified_gmt":"2023-02-09T07:00:23","slug":"ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-2\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 7 Triangles Ex 7.2"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 9 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 7 (Triangles)\u00a0<\/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 9 Mathematics\u00a0<strong>Chapter &#8211; 7 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;\"><strong>Chapter 7: Triangles<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.1<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.3<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-4\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.4<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-5\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.5<\/a><\/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;\"><strong>1. In an isosceles triangle ABC, with AB = AC, the bisectors of \u2220B and \u2220C intersect each other at O. Join A to O. Show that:<br \/>\n<\/strong><strong>(i)<\/strong> OB = OC<br \/>\n<strong>(ii)<\/strong> AO bisects \u2220A<br \/>\n<strong>Answer &#8211;\u00a0 <\/strong><strong>Given:\u00a0<\/strong>AB = AC and the bisectors of \u2220B and \u2220C intersect each other at O.<br \/>\nLet&#8217;s construct a diagram according to the given question.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4396\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Q1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"254\" height=\"170\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Q1.png 835w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Q1-300x200.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Q1-768x513.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Q1-720x480.png 720w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Q1-480x321.png 480w\" sizes=\"auto, (max-width: 254px) 100vw, 254px\" \/><br \/>\n<strong>(i)<\/strong> <strong>OB = OC<br \/>\n<\/strong>It is given that in triangle ABC, <\/span><br \/>\n<span style=\"color: #000000;\">AB = AC (given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ACB = \u2220ABC (Angles opposite to equal sides of an isosceles triangle are equal)<\/span><br \/>\n<span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u2220ACB = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u2220ABC<br \/>\n\u21d2 \u2220OCB = \u2220OBC (Since OB and OC are the\u00a0angle bisectors\u00a0of\u00a0\u2220ABC and \u2220ACB)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 OB = OC (Sides opposite to equal angles of an isosceles triangle are<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) AO bisects \u2220A<br \/>\n<\/strong>In \u0394OAB and \u0394OAC,<\/span><br \/>\n<span style=\"color: #000000;\">AO = AO (Common)<\/span><br \/>\n<span style=\"color: #000000;\">AB = AC (Given)<\/span><br \/>\n<span style=\"color: #000000;\">OB = OC (Proved above)<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, <\/span><br \/>\n<span style=\"color: #000000;\">\u0394OAB \u2245 \u0394OAC (By\u00a0SSS\u00a0congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">Also, we can use an alternative approach as shown below, <\/span><br \/>\n<span style=\"color: #000000;\">\u2220OBA = \u2220OCA (OB and OC bisects angle \u2220B and \u2220C) <\/span><br \/>\n<span style=\"color: #000000;\">AB = AC (Given)<\/span><br \/>\n<span style=\"color: #000000;\">OB = OC (Proved above)<\/span><br \/>\n<span style=\"color: #000000;\">\u0394OAB \u2245 \u0394OAC (By SAS\u00a0congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 \u2220BAO = \u2220CAO (CPCT)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AO bisects \u2220A or AO is the angle bisector of \u2220A.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. In \u0394ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that \u0394ABC is an isosceles triangle in which AB = AC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4397\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Q2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"195\" height=\"206\" \/><br \/>\n<\/strong><strong>Answer &#8211;\u00a0 \u00a0<\/strong><strong>Given:<\/strong> AD is the perpendicular bisector of BC means \u2220ADB = \u2220ADC = 90\u00b0 and BD = DC<\/span><br \/>\n<span style=\"color: #000000;\"><strong>To Prove:<\/strong> \u0394ABC is an\u00a0isosceles triangle\u00a0in which AB = AC.<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394ADC and \u0394ADB,<\/span><br \/>\n<span style=\"color: #000000;\">AD = AD (Common)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ADC = \u2220ADB (Each 90\u00b0)<\/span><br \/>\n<span style=\"color: #000000;\">CD = BD (AD is the\u00a0perpendicular bisector\u00a0of BC)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394ADC \u2245 \u0394ADB (By\u00a0SAS\u00a0congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AB = AC (By\u00a0CPCT)<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, ABC is an isosceles triangle in which AB = AC.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB, respectively (see Fig. 7.31). Show that these altitudes are equal.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4398\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Q3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"239\" height=\"209\" \/><br \/>\n<\/strong><strong>Answer &#8211;\u00a0 <\/strong><strong>Given:\u00a0<\/strong>\u0394ABC is an isosceles triangle<\/span><br \/>\n<span style=\"color: #000000;\"><strong>To prove:<\/strong> BE = CF<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394AEB and \u0394AFC,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AEB =\u00a0\u2220AFC (Each 90\u00b0 as BE and CF are\u00a0altitudes)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220A = \u2220A (Common angle)<\/span><br \/>\n<span style=\"color: #000000;\">AB = AC (Given \u0394ABC is an isosceles triangle)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394AEB \u2245 \u0394AFC (By\u00a0AAS\u00a0congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 BE = CF (By\u00a0CPCT)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that<\/strong><br \/>\n<strong>(i)<\/strong> \u0394ABE \u2245 \u0394ACF<br \/>\n<strong>(ii)<\/strong> AB = AC, i.e. ABC is an isosceles triangle.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4399\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Q4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"213\" height=\"188\" \/><br \/>\n<strong>Answer &#8211;\u00a0 <\/strong><strong>Given:<\/strong> BE = CF<br \/>\n<strong>To prove:<\/strong> \u00a0<strong>(i) \u0394ABE \u2245 \u0394ACF<br \/>\n<\/strong>In \u0394ABE and \u0394ACF,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AEB =\u00a0\u2220AFC (Each 90\u00b0)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220A = \u2220A (Common angle)<\/span><br \/>\n<span style=\"color: #000000;\">BE = CF (Given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394ABE \u2245 \u0394ACF (By\u00a0AAS\u00a0congruence rule)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>To prove:<\/strong> \u00a0<strong>(ii) AB = AC, i.e. ABC is an isosceles triangle.<br \/>\n<\/strong>We have\u00a0proved above that \u0394ABE \u2245 \u0394ACF<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AB = AC (By CPCT)<\/span><br \/>\n<span style=\"color: #000000;\">Hence,\u00a0\u0394ABC is an\u00a0isosceles triangle.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that \u2220ABD = \u2220ACD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4400\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Q5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"196\" height=\"269\" \/><br \/>\n<\/strong><strong>Answer &#8211;\u00a0 <\/strong><strong>Given:\u00a0<\/strong>ABC and DBC are\u00a0isosceles triangles<\/span><br \/>\n<span style=\"color: #000000;\"><strong>To Prove:\u00a0<\/strong>\u2220ABD = \u2220ACD<\/span><br \/>\n<span style=\"color: #000000;\">Let&#8217;s join point A and point B.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4401\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Ans5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"200\" height=\"230\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Ans5.png 314w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Ans5-261x300.png 261w\" sizes=\"auto, (max-width: 200px) 100vw, 200px\" \/><\/span><br \/>\n<span style=\"color: #000000;\">In \u25b3DAB\u00a0and \u25b3DAC,<\/span><br \/>\n<span style=\"color: #000000;\">AB = AC (Given)<\/span><br \/>\n<span style=\"color: #000000;\">BD = CD (Given)<\/span><br \/>\n<span style=\"color: #000000;\">AD = AD (Common side)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u25b3 ABD \u2245 \u25b3 ACD (By\u00a0SSS\u00a0congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220ABD = \u2220ACD (By\u00a0CPCT)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. \u0394ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that \u2220BCD is a right angle.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4402\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Q6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"158\" height=\"222\" \/><br \/>\n<\/strong><strong>Answer &#8211;\u00a0 <\/strong><strong>Given :<\/strong> AB = AC and AD = AB<br \/>\n<strong>To Prove: <\/strong>\u2220BCD is a right angle.<br \/>\nIn\u00a0isosceles triangle\u00a0ABC,<\/span><br \/>\n<span style=\"color: #000000;\">AB = AC (Given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220ACB = \u2220ABC (Angles opposite to equal sides of a triangle are\u00a0equal)<\/span><br \/>\n<span style=\"color: #000000;\">Let \u2220ACB = \u2220ABC be x.\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8211; (1)<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394ACD,<\/span><br \/>\n<span style=\"color: #000000;\">AC = AD (Since, AB = AD)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220ADC = \u2220ACD (Angles opposite to equal sides of a triangle are equal) <\/span><br \/>\n<span style=\"color: #000000;\">Let \u2220ADC = \u2220ACD be y.\u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8211; (2)<\/span><br \/>\n<span style=\"color: #000000;\">Thus,\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BCD =\u00a0\u2220ACB +\u00a0\u2220ACD = x + y &#8212;&#8212;&#8212;&#8211; (3)<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394BCD,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ABC + \u2220BCD + \u2220ADC = 180\u00b0 (Angle sum property of a triangle)<\/span><br \/>\n<span style=\"color: #000000;\">Substituting the values we get,<\/span><br \/>\n<span style=\"color: #000000;\">x + (x + y) + y = 180\u00b0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [From equation (1), (2) and (3)]<\/span><br \/>\n<span style=\"color: #000000;\">2 (x + y) = 180\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">2(\u2220BCD) = 180\u00b0 [From equation(3)]<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220BCD = 90\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. ABC is a right-angled triangle in which \u2220A = 90\u00b0 and AB = AC. Find \u2220B and \u2220C.<br \/>\n<\/strong><strong>Answer &#8211;<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4403\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Q7.jpeg\" alt=\"NCERT Class 9 Solutions Maths\" width=\"142\" height=\"176\" \/><br \/>\n<\/strong><strong>Given:\u00a0<\/strong>\u2220A = 90\u00b0 and AB = AC<br \/>\n\u2234 \u2220C = \u2220B (Angles opposite to equal sides are also equal)<\/span><br \/>\n<span style=\"color: #000000;\">Let \u2220B = \u2220C = x <\/span><br \/>\n<span style=\"color: #000000;\">In \u0394ABC, <\/span><br \/>\n<span style=\"color: #000000;\">\u2220A + \u2220B + \u2220C = 180\u00b0 (Angle sum property of a triangle)<\/span><br \/>\n<span style=\"color: #000000;\">90\u00b0+ x + x = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">90\u00b0+ 2x = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">2x\u00a0= 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">x = 45\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220B = \u2220C = 45\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. Show that the angles of an equilateral triangle are 60\u00b0 each.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0 <\/strong>Let ABC be an equilateral triangle, as shown below:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4404\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Q8.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"255\" height=\"187\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Q8.png 718w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Q8-300x221.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.2-Q8-480x353.png 480w\" sizes=\"auto, (max-width: 255px) 100vw, 255px\" \/><br \/>\nAB = BC = AC <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220C = \u2220A\u00a0=\u00a0\u2220B (Angles opposite to equal sides of a triangle are equal)<\/span><br \/>\n<span style=\"color: #000000;\">Let\u00a0\u2220A = \u2220B = \u2220C be x.\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">In \u25b3 ABC,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220A + \u2220B + \u2220C = 180\u00b0 (Angle sum property of a triangle)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x + x + x\u00a0= 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 3x\u00a0= 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x = 60\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220A = \u2220B = \u2220C = 60\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Hence, in an equilateral triangle, all interior angles are of measure 60\u00b0.<\/span><\/p>\n<p style=\"text-align: justify;\">\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 7 (Triangles)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 7 Triangles Exercise 7.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 7: Triangles NCERT Solution Class 9 Maths Ex &#8211; 7.1 NCERT Solution Class 9 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":[698],"tags":[713,702,186,714,701,5],"class_list":["post-4354","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-7-triangles-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-7-in-english","tag-ncert-solutions-class-9-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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