{"id":503,"date":"2023-10-18T09:17:16","date_gmt":"2023-10-18T03:47:16","guid":{"rendered":"https:\/\/tnboardsolutions.com\/?p=503"},"modified":"2023-11-10T14:41:53","modified_gmt":"2023-11-10T09:11:53","slug":"samacheer-kalvi-9th-maths-guide-chapter-4-ex-4-6","status":"publish","type":"post","link":"https:\/\/tnboardsolutions.com\/samacheer-kalvi-9th-maths-guide-chapter-4-ex-4-6\/","title":{"rendered":"Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.6"},"content":{"rendered":"

Students can download Maths Chapter 4 Geometry Ex 4.6 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide<\/a> Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.<\/p>\n

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.6<\/h2>\n

Question 1.
\nDraw a triangle ABC, where AB = 8 cm, BC = 6 cm and \u2220B = 70\u00b0 and locate its circumcentre and draw the circumcircle.
\nSolution:
\n\"Samacheer
\nSteps for construction:
\nStep 1: Draw the \u0394ABC with the given measures.
\nStep 2: Construct the perpendicular bisector of (AB and BC) any two sides and let them, meet at S which is the circumcenter.
\nStep 3: With S as centre and SA = SB = SC as radius draw the circumcircle to passes through A, B and C.
\nCircum radius = 4.3 cm.<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nConstruct the right triangle PQR whose perpendicular sides are 4.5 cm and 6 cm. Also locate its circumcentre and draw circumcircle.
\nSolution:
\n\"Samacheer
\nSteps for construction:
\nStep 1: Draw the \u0394PQR with the given measures.
\nStep 2: Construct the perpendicular bisector of (PQ and PR) any two sides and let them meet at S which is the circumcenter.
\nStep 3: With S as centre and SP = SQ = SR as radius draw the circumcircle to passes through P, Q and R.
\nCircum radius = 3.8 cm.<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nConstruct \u0394ABC with AB = 5 cm \u2220B = 100\u00b0 and BC = 6 cm. Also locate its circumcentre draw circumcircle.
\nSolution:
\n\"Samacheer
\nSteps for construction:
\nStep 1: Draw the \u0394ABC with the given measures.
\nStep 2: Construct the perpendicular bisector of any two sides (AB and BC) and let them meet at S which is the circumcenter.
\nStep 3: With S as centre and SA = SB = SC as radius draw the circumcircle to passes through A, B and C.
\nCircum radius = 4.3 cm.<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nConstruct an isosceles triangle PQR where PQ = PR and \u2220Q = 50\u00b0, QR = 7cm. Also draw its circumcircle.
\nSolution:
\n\"Samacheer
\nGiven PQ = PR
\n\u2234 \u2220R = 50\u00b0 (opposite angles are equal)
\nSteps for construction:
\nStep 1: Draw the \u0394ABC with the given measures.
\nStep 2: Construct the perpendicular bisector of any two sides (QR and PR) and let them meet at S. S is the circumcenter of \u0394PQR.
\nStep 3: With S as centre SP = SQ = SR as radius. Draw the circumcircle.
\nCircum radius = 3.5 cm.<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nDraw an equilateral triangle of side 6.5 cm and locate its incentre. Also draw the incircle.
\nSolution:
\n\"Samacheer
\nSteps for construction:
\nStep 1: Draw the \u0394ABC with the each side measure 6.5 cm.
\nStep 2: Construct the angles bisectors of any two angles (A and B) and let them meet at I. Then I is the incentre of \u0394ABC.
\nStep 3: Draw perpendicular from I to any one of the side (AB) to meet AB at D.
\nStep 4: With I as centre and ID as radius draw the circle. This circle touches all the sides of the triangle internally.
\nStep 5: Measure of In-radius = 1.5 cm.<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\nDraw a right triangle whose hypotenuse is 10 cm and one of the legs is 8 cm. Locate its incentre and also draw the incircle
\nSolution:
\n\"Samacheer
\nSteps for construction:
\nStep 1: Draw the \u0394ABC with AB = 8cm, BC = 10 cm and \u2220A = 90\u00b0.
\nStep 2: Construct the angle bisectors of any two angles (\u2220B and \u2220C) and let them meet at I. Then I is the incentre of \u0394ABC.
\nStep 3: Draw perpendicular from I to any one of the side to meet AB at D.
\nStep 4: With I as centre and ID as radius draw the circle. This circle touches all the sides of the triangle internally.
\nStep 5: Measure of In-radius = 1.8 cm.<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nDraw \u0394ABC given AB = 9 cm, \u2220CAB = 115\u00b0 and \u2220ABC = 40\u00b0. Locate its incentre and also draw the incircle. (Note: You can check from the above examples that the incentre of any triangle is always in its interior).
\nSolution:
\n\"Samacheer
\nStep 1: Draw the \u0394ABC with AB = 9cm, \u2220B = 40\u00b0 and \u2220A = 115\u00b0.
\nStep 2: Construct the angle bisectors of any two angles (A and B) and let them meet at I. Then I is the incentre of \u0394ABC.
\nStep 3: Draw perpendicular from I to any one of the side (AB) to meet AB at D.
\nStep 4: With I as centre and ID as radius draw the circle. This circle touches all the sides of the triangle internally.
\nStep 5: Measure of In-radius = 2.7 cm.<\/p>\n

\"Samacheer<\/p>\n

Question 8.
\nConstruct \u0394ABC in which AB = BC = 6cm and \u2220B = 80\u00b0. Locate its incentre and draw the incircle.
\nSolution:
\n\"Samacheer
\nSteps for construction:
\nStep 1: Draw the \u0394ABC with AB = 6 cm, \u2220B = 80\u00b0 and BC = 6 cm.
\nStep 2: Construct the angle bisectors of any two angles (A and B) and let them meet at I. Then I is the incentre of \u0394ABC.
\nStep 3: Draw perpendicular from I to any one of the side (AB) to meet AB at D.
\nStep 4: With I as centre and ID as radius draw the circle. This circle touches all the sides of the triangle internally.
\nStep 5: Measure of In-radius = 1.9 cm.<\/p>\n

\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can download Maths Chapter 4 Geometry Ex 4.6 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams. Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.6 Question …<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts\/503"}],"collection":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/comments?post=503"}],"version-history":[{"count":1,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts\/503\/revisions"}],"predecessor-version":[{"id":43490,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts\/503\/revisions\/43490"}],"wp:attachment":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/media?parent=503"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/categories?post=503"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/tags?post=503"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}