{"id":4574,"date":"2023-02-16T05:33:12","date_gmt":"2023-02-16T05:33:12","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4574"},"modified":"2023-02-16T05:34:11","modified_gmt":"2023-02-16T05:34:11","slug":"ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-2\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 11 Constructions Ex 11.2"},"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; 11 (Constructions)\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; 11 Constructions <\/strong>Exercise 11.2 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 11: Constructions<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-11-constructions-ex-11-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 11.1<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 11.2<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Construct a triangle ABC in which BC = 7 cm, \u2220B = 75\u00b0 and AB+AC = 13 cm<\/strong>.<br \/>\n<strong>Answer &#8211;\u00a0 <\/strong>Construction Procedure<br \/>\n1. Draw a line segment of base BC = 7 cm<br \/>\n2. Measure and draw \u2220B = 75\u00b0 and draw the ray BX<br \/>\n3. Take a compass and measure AB + AC = 13 cm.<br \/>\n4. With B as the centre, draw an arc at the point be D<br \/>\n5. Join DC<br \/>\n6. Now, draw the perpendicular bisector of the line DC, and the intersection point is taken as A<br \/>\n7. Then, join AC<br \/>\n8. Therefore, ABC is the required triangle.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4622\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.2-Ans1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"350\" height=\"448\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.2-Ans1.png 350w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.2-Ans1-234x300.png 234w\" sizes=\"auto, (max-width: 350px) 100vw, 350px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Construct a triangle ABC in which BC = 8 cm, \u2220B = 45\u00b0 and AB\u2013AC = 3.5 cm<\/strong>.<br \/>\n<strong>Answer &#8211;\u00a0<\/strong>Construction Procedure<br \/>\n1. Draw a line segment of base BC = 8 cm<br \/>\n2. Measure and draw \u2220B = 45\u00b0 and draw the ray BX<br \/>\n3. Take a compass and measure AB-AC = 3.5 cm<br \/>\n4. With B as the centre, draw an arc at point D on the ray BX<br \/>\n5. Join DC<br \/>\n6. Now, draw the perpendicular bisector of the line CD, and the intersection point is taken as A<br \/>\n7. Then, join AC<br \/>\n8. Therefore, ABC is the required triangle.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4623\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.2-Ans2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"325\" height=\"319\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.2-Ans2.png 325w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.2-Ans2-300x294.png 300w\" sizes=\"auto, (max-width: 325px) 100vw, 325px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Construct a triangle PQR in which QR = 6 cm, \u2220Q = 60\u00b0 and PR\u2013PQ = 2cm.<br \/>\nAnswer &#8211;\u00a0<\/strong>Construction Procedure<br \/>\n1. Draw a line segment of base QR = 6 cm<br \/>\n2. Measure and draw \u2220Q = 60\u00b0, and let the ray be QX<br \/>\n3. Take a compass and measure PR \u2013 PQ = 2 cm.<br \/>\n4. Since PR \u2013 PQ is negative, QZ will be below the line QR.<br \/>\n5. With Q as the centre, draw an arc at the point Z on the ray QX<br \/>\n6. Join ZR<br \/>\n7. Now, draw the perpendicular bisector of the line DR and the intersection point is taken as P.<br \/>\n8. Then, join PR<br \/>\n9. Therefore, PQR is the required triangle.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4624\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.2-Ans3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"522\" height=\"448\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.2-Ans3.png 522w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.2-Ans3-300x257.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.2-Ans3-480x412.png 480w\" sizes=\"auto, (max-width: 522px) 100vw, 522px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. Construct a triangle XYZ in which \u2220Y = 30\u00b0, \u2220Z = 90\u00b0 and XY + YZ + ZX = 11 cm.<br \/>\nAnswer &#8211;\u00a0<\/strong>Construction Procedure<br \/>\n1. Draw a line segment AB which is equal to XY + YZ + ZX = 11 cm<strong>.<br \/>\n<\/strong>2. Make an angle \u2220LAB = 30\u00b0 from the point A.<br \/>\n3. Make an angle \u2220MBA = 90\u00b0 from the point B.<br \/>\n4. Bisect \u2220LAB and \u2220MBA at point X.<br \/>\n5. Now, take the perpendicular bisector of the lines as XA and XB and the intersection point as Y and Z, respectively.<br \/>\n6. Join XY and XZ<br \/>\n7. Therefore, XYZ is the required triangle<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4625\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.2-Ans4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"461\" height=\"322\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.2-Ans4.png 461w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.2-Ans4-300x210.png 300w\" sizes=\"auto, (max-width: 461px) 100vw, 461px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. Construct a right triangle whose base is 12 cm, and the sum of its hypotenuse and other side is 18 cm.<br \/>\nAnswer &#8211;\u00a0<\/strong>Construction Procedure:<br \/>\n1. Draw a line segment of base BC = 12 cm<br \/>\n2. Measure and draw \u2220B = 90\u00b0, and draw the ray BX<br \/>\n3. Take a compass and measure AB + AC = 18 cm<br \/>\n4. With B as the centre, draw an arc at the point D on the ray BX<br \/>\n5. Join DC<br \/>\n6. Now, draw the perpendicular bisector of the line CD, and the intersection point is taken as A.<br \/>\n7. Then, join AC<br \/>\n8. Therefore, ABC is the required triangle.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4626\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.2-Ans5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"449\" height=\"572\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.2-Ans5.png 449w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-11.2-Ans5-235x300.png 235w\" sizes=\"auto, (max-width: 449px) 100vw, 449px\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\">\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 11 (Constructions)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 11 Constructions Exercise 11.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 11: Constructions NCERT Solution Class 9 Maths Ex &#8211; 11.1 Exercise &#8211; 11.2 1. Construct [&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":[721,702,186,722,701,5],"class_list":["post-4574","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-11-constructions-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-11-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) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions Class 9 Maths Chapter 11 Constructions Ex 11.2 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 9 Maths\u00a0Chapter - 11 (Constructions)\u00a0The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter - 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