{"id":5024,"date":"2016-05-19T11:49:00","date_gmt":"2016-05-19T06:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-9-maths-exercise-11-2\/"},"modified":"2018-06-18T17:51:46","modified_gmt":"2018-06-18T12:21:46","slug":"ncert-solutions-for-class-9-maths-exercise-11-2","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-11-2\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Exercise 11.2"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-toggle-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-11-2\/#NCERT_Solutions_for_Class_9_Mathematics_Constructions\" >NCERT Solutions for Class 9 Mathematics Constructions<\/a><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-11-2\/#1_Construct_a_triangle_ABC_in_which_BC_7_cm_B_and_AB_AC_13_cm\" >1. Construct a triangle ABC in which BC = 7 cm, B =  and AB + AC = 13 cm.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-11-2\/#2_Construct_a_triangle_ABC_in_which_BC_8_cm_B_and_AB_-AC_35_cm\" >2. Construct a triangle ABC in which BC = 8 cm, B =  and AB -AC = 3.5 cm.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-11-2\/#3_Construct_a_triangle_PQR_in_which_QR_6_cm_Q_and_PR_-PQ_2_cm\" >3. Construct a triangle PQR in which QR = 6 cm, Q =  and PR -PQ = 2 cm.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-11-2\/#4_Construct_a_triangle_XYZ_in_which_Y_Z_and_XY_YZ_ZX_11_cm\" >4. Construct a triangle XYZ in which Y = Z =  and XY + YZ + ZX = 11 cm.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-11-2\/#5_Construct_a_right_triangle_whose_base_is_12_cm_and_sum_of_its_hypotenuse_and_other_side_is_18_cm\" >5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.<\/a><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-11-2\/#NCERT_Solutions_for_Class_9_Maths_Exercise_112\" >NCERT Solutions for Class 9 Maths Exercise 11.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-11-2\/#CBSE_app_for_Class_9\" >CBSE app for Class 9<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 9 Maths Exercise 11.2 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.<\/p>\n<p style=\"text-align: center;\"><strong>NCERT solutions for Class 9 Maths\u00a0<\/strong><strong>Constructions\u00a0<\/strong><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-09-mathematics-constructions\/1245\/ncert-solutions\/5\/\">Download as PDF<\/a><\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/09_Class_Maths_Book.jpg\" alt=\"NCERT Solutions for Class 9 Maths Exercise 11.2\" width=\"181\" height=\"233\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_9_Mathematics_Constructions\"><\/span>NCERT Solutions for Class 9 Mathematics Constructions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"1_Construct_a_triangle_ABC_in_which_BC_7_cm_B_and_AB_AC_13_cm\"><\/span><strong>1. Construct a triangle ABC in which BC = 7 cm, <\/strong><img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/><strong>B = <\/strong><img decoding=\"async\" style=\"height: 19px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image002.png\" \/><strong> and AB + AC = 13 cm.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. Given<\/strong> : Base BC = 7 cm, <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>B = <img decoding=\"async\" style=\"height: 19px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image002.png\" \/> and sum of two sides AB + AC = 13 cm.<\/p>\n<p style=\"text-align: justify;\"><strong>To construct<\/strong> : A triangle ABC.<\/p>\n<p style=\"text-align: justify;\"><strong>Steps of construction<\/strong>:<\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 478\" style=\"height: 176px; width: 163px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image003.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(a)<\/strong> Draw a ray BX and cut off a line segment BC = 7 cm from it.<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong> At B, construct <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>YBX = <img decoding=\"async\" style=\"height: 19px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image002.png\" \/>.<\/p>\n<p style=\"text-align: justify;\"><strong>(c)<\/strong> With B as centre and radius = 13 cm (<img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image004.png\" \/> AB + AC = 13 cm) draw an arc to meet BY at D.<\/p>\n<p style=\"text-align: justify;\"><strong>(d) <\/strong>Join CD.<\/p>\n<p style=\"text-align: justify;\"><strong>(e)<\/strong> Draw perpendicular bisector PQ of CD intersecting BD at A.<\/p>\n<p style=\"text-align: justify;\"><strong>(f)<\/strong> Join AC.<\/p>\n<p style=\"text-align: justify;\">Then ABC is required triangle.<\/p>\n<p style=\"text-align: justify;\"><strong>Justification<\/strong>:<\/p>\n<p style=\"text-align: justify;\">A lies on perpendicular bisector of CD.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image005.png\" \/> AC = AD<\/p>\n<p style=\"text-align: justify;\">And AB = BD -AD <img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image006.png\" \/> AB = BD -AC<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image006.png\" \/> AB + AC = BD = 13 cm<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 11.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2_Construct_a_triangle_ABC_in_which_BC_8_cm_B_and_AB_-AC_35_cm\"><\/span><strong>2. Construct a triangle ABC in which BC = 8 cm, <\/strong><img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/><strong>B = <\/strong><img decoding=\"async\" style=\"height: 21px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image007.png\" \/><strong> and AB -AC = 3.5 cm.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. Given<\/strong>: Base BC = 8 cm, One Base angle <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>B = <img decoding=\"async\" style=\"height: 21px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image007.png\" \/> and AB -AC = 3.5 cm<\/p>\n<p style=\"text-align: justify;\"><strong>To construct<\/strong>: A triangle ABC.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 481\" style=\"height: 214px; width: 193px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image008.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>Steps of construction<\/strong>:<\/p>\n<p style=\"text-align: justify;\"><strong>(a)<\/strong> Draw a ray BX and cut off a line segment BC = 8 cm from it.<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong> Cut <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>YBC = <img decoding=\"async\" style=\"height: 21px; width: 28px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image009.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(c)<\/strong> Cut off a line segment BD = 3.5 cm<\/p>\n<p style=\"text-align: justify;\">(<img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image004.png\" \/> AB -AC = 3.5 cm) from BY.<\/p>\n<p style=\"text-align: justify;\"><strong>(d)<\/strong> Join CD.<\/p>\n<p style=\"text-align: justify;\"><strong>(e)<\/strong> Draw perpendicular bisector PQ of CD intersecting BY at a point A.<\/p>\n<p style=\"text-align: justify;\"><strong>(f)<\/strong> Join AC.<\/p>\n<p style=\"text-align: justify;\">Then ABC is the required triangle.<\/p>\n<p style=\"text-align: justify;\"><strong>Justification<\/strong>:<\/p>\n<p style=\"text-align: justify;\">A lies on the perpendicular bisector of CD.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image005.png\" \/> AD = AC<\/p>\n<p style=\"text-align: justify;\">Now BD = AB -AD<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image006.png\" \/> BD = AB -AC = 3.5 cm<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 11.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"3_Construct_a_triangle_PQR_in_which_QR_6_cm_Q_and_PR_-PQ_2_cm\"><\/span><strong>3. Construct a triangle PQR in which QR = 6 cm, <\/strong><img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/><strong>Q = <\/strong><img decoding=\"async\" style=\"height: 21px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image010.png\" \/><strong> and PR -PQ = 2 cm.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. Given<\/strong>: Base QR = 6 cm, one base angle <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>Q = <img decoding=\"async\" style=\"height: 21px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image010.png\" \/> and PR -PQ = 2 cm.<\/p>\n<p style=\"text-align: justify;\"><strong>To construct<\/strong>: A triangle PQR.<\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 484\" style=\"height: 186px; width: 140px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image011.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Steps of construction<\/strong>:<\/p>\n<p style=\"text-align: justify;\"><strong>(a)<\/strong> Draw a ray QX and cut off a line segment QR = 6 cm from it.<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong> Construct a ray QY making an angle of <img decoding=\"async\" style=\"height: 21px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image010.png\" \/> with QR and produce YQ to form a line YQY\u2019.<\/p>\n<p style=\"text-align: justify;\"><strong>(c)<\/strong> Cut off a line segment QO = 2 cm (<img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image004.png\" \/> PR -PQ = 2 cm) from QY\u2019.<\/p>\n<p style=\"text-align: justify;\"><strong>(d)<\/strong> Join OR.<\/p>\n<p style=\"text-align: justify;\"><strong>(e)<\/strong> Draw perpendicular bisector MN of OR.<\/p>\n<p style=\"text-align: justify;\"><strong>(f)<\/strong> Join PR.<\/p>\n<p style=\"text-align: justify;\">Then PQR is the required triangle.<\/p>\n<p style=\"text-align: justify;\">Justification:<\/p>\n<p style=\"text-align: justify;\">P lies on perpendicular bisector of OR.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image005.png\" \/> PO = PR<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image006.png\" \/> PQ + QO = PR<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image006.png\" \/> QO = PR -PQ = 2 cm<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 11.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"4_Construct_a_triangle_XYZ_in_which_Y_Z_and_XY_YZ_ZX_11_cm\"><\/span><strong>4. Construct a triangle XYZ in which <\/strong><img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/><strong>Y = <\/strong><img decoding=\"async\" style=\"height: 24px; width: 44px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image012.png\" \/><strong>Z = <\/strong><img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image013.png\" \/><strong> and XY + YZ + ZX = 11 cm.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. Given<\/strong>: Base angles <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>Y = <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image014.png\" \/> and <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>Z = <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image013.png\" \/> and XY + YZ + ZX = 11 cm.<\/p>\n<p style=\"text-align: justify;\"><strong>To construct<\/strong>: <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image015.png\" \/>XYZ<\/p>\n<p style=\"text-align: justify;\"><strong>Steps of construction<\/strong>:<\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 487\" style=\"height: 205px; width: 265px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image016.jpg\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(a)<\/strong> Draw a line segment PQ = 11 cm.<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong> Draw <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>KPQ = <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image014.png\" \/> and <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>LQP = <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image013.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(c)<\/strong> Bisect the <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>KPQ and <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>LQP. Let these intersect at point X.<\/p>\n<p style=\"text-align: justify;\"><strong>(d)<\/strong> Draw perpendicular bisectors, MN of PX and RS of XQ.<\/p>\n<p style=\"text-align: justify;\"><strong>(e)<\/strong> Let MN intersects PQ at Y and RS intersects PQ at Z.<\/p>\n<p style=\"text-align: justify;\"><strong>(f)<\/strong> Join XY and XZ.<\/p>\n<p style=\"text-align: justify;\">Then XYZ is the required triangle.<\/p>\n<p style=\"text-align: justify;\"><strong>Justification<\/strong>:<\/p>\n<p style=\"text-align: justify;\">Y lies on perpendicular bisector MN of PX.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image005.png\" \/> PY = XY and similarly QZ = XZ<\/p>\n<p style=\"text-align: justify;\">This gives XY + YZ + XZ = PY + YZ + QZ = PQ = 11 cm<\/p>\n<p style=\"text-align: justify;\">Again <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>YXP = <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>XPY [Since XY = PY]\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image006.png\" \/> <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>XYZ = <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>YXP + <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>XPY = 2<img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>XPY = <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>KPQ<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image006.png\" \/> <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>XYZ = <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image014.png\" \/><\/p>\n<p style=\"text-align: justify;\">Similarly, <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>XZY = <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>LQP<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image006.png\" \/> <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image001.png\" \/>XZY = <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image013.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 11.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"5_Construct_a_right_triangle_whose_base_is_12_cm_and_sum_of_its_hypotenuse_and_other_side_is_18_cm\"><\/span><strong>5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. Given<\/strong>: Base BC = 12 cm and AB + AC = 18 cm.<\/p>\n<p style=\"text-align: justify;\"><strong>To construct<\/strong>: A right angled triangle ABC right angled at B.<\/p>\n<p style=\"text-align: justify;\"><strong>Steps of construction<\/strong>:<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 490\" style=\"height: 169px; width: 179px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image017.jpg\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(a)<\/strong> Draw a ray BX and cut off a line segment BC = 12 cm from it.<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong> Draw an angle XBY = <img decoding=\"async\" style=\"height: 21px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image018.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(c)<\/strong> From BY cut off a line segment BD = 18 cm.<\/p>\n<p style=\"text-align: justify;\"><strong>(d) <\/strong>Join CD.<\/p>\n<p style=\"text-align: justify;\"><strong>(e)<\/strong> Draw the perpendicular bisector of CD intersecting BD at A.<\/p>\n<p style=\"text-align: justify;\"><strong>(f)<\/strong> Join AC.<\/p>\n<p style=\"text-align: justify;\">Then ABC is the required right angled triangle.<\/p>\n<p style=\"text-align: justify;\"><strong>Justification<\/strong>:<\/p>\n<p style=\"text-align: justify;\">A lies on the perpendicular bisector of CD.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image005.png\" \/> AC = AD<\/p>\n<p style=\"text-align: justify;\">And then AB = BD -AD<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image006.png\" \/> AB = BD -AC<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch11\/Ex11.2\/image006.png\" \/> AB + AC = BD = 18 cm.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_9_Maths_Exercise_112\"><\/span>NCERT Solutions for Class 9 Maths Exercise 11.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>NCERT Solutions for Class 9 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 9 Maths includes text book solutions from Mathematics Book. NCERT Solutions for CBSE Class 9 Maths have total 15 chapters. 9 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 9 solutions PDF and Maths ncert class 9 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_app_for_Class_9\"><\/span>CBSE app for Class 9<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>To download NCERT Solutions for Class 9 Maths, Computer Science, Home Science,Hindi ,English, Social Science do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through\u00a0<a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\"><strong>the best app for CBSE students<\/strong><\/a>\u00a0and myCBSEguide website.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions for Class 9 Maths Exercise 11.2 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter &#8230; <a title=\"NCERT Solutions for Class 9 Maths Exercise 11.2\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-11-2\/\" aria-label=\"More on NCERT Solutions for Class 9 Maths Exercise 11.2\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":-1,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1376,281],"tags":[283,1344,321,1485,216],"class_list":["post-5024","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematics-cbse-class-09","category-ncert-solutions","tag-cbse-study-material","tag-class-9","tag-mathematics","tag-maths","tag-ncert-solutions"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>NCERT Solutions for Class 9 Maths Exercise 11.2 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for Class 9 Maths Exercise 11.2 in PDF format for free download. 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