{"id":127984,"date":"2021-09-15T15:52:32","date_gmt":"2021-09-15T10:22:32","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=127984"},"modified":"2021-10-13T15:48:03","modified_gmt":"2021-10-13T10:18:03","slug":"rd-sharma-class-10-solutions-chapter-11-exercise-11-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-11-exercise-11-2\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.2 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-127994\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-11-Exercise-11.2.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 11 Exercise 11.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-11-Exercise-11.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-11-Exercise-11.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 11 Exercise 11.2:\u00a0<\/strong>In this exercise, students can learn how to design a triangle that is comparable to a given triangle. All of the answers to the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> have been written by Kopykitab experts to dispel any concerns and provide the correct techniques for solving a question. Students can use the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-11-constructions\/\"><strong>RD Sharma Class 10 Solutions Chapter 11<\/strong><\/a> Exercise 11.2 PDF to aid them with the correct steps in solving questions.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 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\">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=\"ez-toc-toggle-icon-1\"><label for=\"item-69da25b85ae90\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" 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RD Sharma Class 10 Solutions Chapter 11 Exercise 11.2 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-11-Exercise-11.2.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-11-Exercise-11.2.pdf\">RD Sharma Class 10 Solutions Chapter 11 Exercise 11.2<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-11-exercise-112-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 11 Exercise 11.2- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>Construct a triangle of sides 4 cm, 5 cm, and 6 cm and then a triangle similar to it whose sides are (2\/3) of the corresponding sides of it.<br \/>Solution:<br \/>Steps of construction :<br \/>(i) Draw a line segment BC = 5 cm.<br \/>(ii) With centre B and radius 4 cm and with centre C and radius 6 cm, draw arcs intersecting each other at A.<br \/>(iii) Join AB and AC. Then ABC is the triangle.<br \/>(iv) Draw a ray BX making an acute angle with BC and cut off 3 equal parts making BB<sub>1<\/sub>\u00a0= B<sub>1<\/sub>B<sub>2<\/sub>= B<sub>2<\/sub>B<sub>3<\/sub>.<br \/>(v) Join B<sub>3<\/sub>C.<br \/>(vi) Draw B\u2019C\u2019 parallel to B<sub>3<\/sub>C and C\u2019A\u2019 parallel to CA then \u0394A\u2019BC\u2019 is the required triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1942\/44922310914_79414cb0bf_o.png\" alt=\"RD Sharma Class 10 Chapter 11 Constructions\" width=\"240\" height=\"291\" \/><\/p>\n<p>Question 2.<br \/>Construct a triangle similar to a given \u0394ABC such that each of its sides is (5\/7)<sup>th\u00a0<\/sup>of the corresponding sides of \u0394ABC. It is given that AB = 5 cm, BC = 7 cm and \u2220ABC = 50\u00b0.<br \/>Solution:<br \/>Steps of construction :<br \/>(i) Draw a line segment BC = 7 cm.<br \/>(ii) Draw a ray BX making an angle of 50\u00b0 and cut off BA = 5 cm.<br \/>(iii) Join AC. Then ABC is the triangle.<br \/>(iv) Draw a ray BY making an acute angle with BC and cut off 7 equal parts making BB, =B<sub>1<\/sub>B<sub>2<\/sub>=B<sub>2<\/sub>B<sub>3<\/sub>=B<sub>3<\/sub>B<sub>4<\/sub>=B<sub>4<\/sub>B<sub>s<\/sub>=B<sub>5<\/sub>B<sub>6<\/sub>=B<sub>6<\/sub>B<sub>7<br \/><\/sub>(v) Join B<sub>7<\/sub>\u00a0and C<br \/>(vi) Draw B<sub>5<\/sub>C\u2019 parallel to B<sub>7<\/sub>C and C\u2019A\u2019 parallel to CA.<br \/>Then \u0394A\u2019BC\u2019 is the required triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1978\/44922310744_6e33ab38a0_o.png\" alt=\"Constructions Class 10 RD Sharma \" width=\"262\" height=\"302\" \/><\/p>\n<p>Question 3.<br \/>Construct a triangle similar to a given \u2220ABC such that each of its sides is\u00a023rd of the corresponding sides of \u0394ABC. It is given that BC = 6 cm, \u2220B = 50\u00b0 and \u2220C = 60\u00b0.<br \/>Solution:<br \/>Steps of construction :<br \/>(i) Draw a line segment BC = 6 cm.<br \/>(ii) Draw a ray BX making an angle of 50\u00b0 and CY making 60\u00b0 with BC which intersect each other at A. Then ABC is the triangle.<br \/>(iii) From B, draw another ray BZ making an acute angle below BC and intersect 3 equal parts making BB<sub>1<\/sub>\u00a0=B<sub>1<\/sub>B<sub>2<\/sub>\u00a0= B<sub>2<\/sub>B<sub>2<\/sub><br \/>(iv) Join B<sub>3<\/sub>C.<br \/>(v) From B<sub>2<\/sub>, draw B<sub>2<\/sub>C\u2019 parallel to B<sub>3<\/sub>C and C\u2019A\u2019 parallel to CA.<br \/>Then \u0394A\u2019BC\u2019 is the required triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1903\/44922310684_0a6048e8d9_o.png\" alt=\"RD Sharma Class 10 Solutions Constructions \" width=\"257\" height=\"312\" \/><\/p>\n<p>Question 4.<br \/>Draw a \u0394ABC in which BC = 6 cm, AB = 4 cm and AC = 5 cm. Draw a triangle similar to \u0394ABC with its sides equal to\u00a034th of the corresponding sides of \u0394ABC.<br \/>Solution:<br \/>Steps of construction :<br \/>(i) Draw a line segment BC = 6 cm.<br \/>(ii) With centre B and radius 4 cm and with centre C and radius 5 cm, draw arcs intersecting each other at A.<br \/>(iii) Join AB and AC. Then ABC is the triangle,<br \/>(iv) Draw a ray BX making an acute angle with BC and cut off 4 equal parts making BB<sub>1<\/sub>= B<sub>1<\/sub>B<sub>2\u00a0<\/sub>= B<sub>2<\/sub>B<sub>3<\/sub>\u00a0= B<sub>3<\/sub>B<sub>4<\/sub>.<br \/>(v) Join B<sub>4<\/sub>\u00a0and C.<br \/>(vi) From B<sub>3<\/sub>C draw C<sub>3<\/sub>C\u2019 parallel to B<sub>4<\/sub>C and from C\u2019, draw C\u2019A\u2019 parallel to CA.<br \/>Then \u0394A\u2019BC\u2019 is the required triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1922\/45647274361_d483a93cdb_o.png\" alt=\"RD Sharma Class 10 Solutions Constructions Exercise 11.2\" width=\"242\" height=\"298\" \/><\/p>\n<p>Question 5.<br \/>Construct a triangle with sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides are 75 of the corresponding sides of the first triangle.<br \/>Solution:<br \/>Steps of construction :<br \/>(i) Draw a line segment BC = 5 cm.<br \/>(ii) With centre B and radius 6 cm and with centre C and radius 7 cm, draw arcs intersecting each other at A.<br \/>(iii) Join AB and AC. Then ABC is the triangle.<br \/>(iv) Draw a ray BX making an acute angle with BC and cut off 7 equal parts making BB<sub>1<\/sub>\u00a0= B<sub>1<\/sub>B<sub>2<\/sub>\u00a0= B<sub>2<\/sub>B<sub>3<\/sub>\u00a0= B<sub>3<\/sub>B<sub>4<\/sub>\u00a0= B<sub>4<\/sub>B<sub>5<\/sub>\u00a0= B<sub>5<\/sub>B<sub>6<\/sub>\u00a0= B<sub>6<\/sub>B<sub>7<\/sub>.<br \/>(v) Join B<sub>5<\/sub>\u00a0and C.<br \/>(vi) From B<sub>7<\/sub>, draw B<sub>7<\/sub>C\u2019 parallel to B<sub>5<\/sub>C and C\u2019A\u2019 parallel CA. Then \u0394A\u2019BC\u2019 is the required triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1937\/44922310594_ec7475bc55_o.png\" alt=\"RD Sharma Class 10 Solutions Chapter 11 Constructions \" width=\"244\" height=\"378\" \/><\/p>\n<p>Question 6.<br \/>Draw a right triangle ABC in which AC = AB = 4.5 cm and \u2220A = 90\u00b0. Draw a triangle similar to \u0394ABC with its sides equal to (54)th ot the corresponding sides of \u0394ABC.<br \/>Solution:<br \/>Steps of construction :<br \/>(i) Draw a line segment AB = 4.5 cm.<br \/>(ii) At A, draw a ray AX perpendicular to AB and cut off AC = AB = 4.5 cm.<br \/>(iii) Join BC. Then ABC is the triangle.<br \/>(iv) Draw a ray AY making an acute angle with AB and cut off 5 equal parts making AA<sub>1<\/sub>\u00a0= A<sub>1<\/sub>A<sub>2<\/sub>\u00a0= A<sub>2<\/sub>A<sub>3<\/sub>\u00a0=A<sub>3<\/sub>A<sub>4<\/sub>\u00a0= A<sub>4<\/sub>A<sub>5<\/sub><sub><br \/><\/sub>(v) Join A<sub>4<\/sub>\u00a0and B.<br \/>(vi) From 45, draw 45B\u2019 parallel to A<sub>4<\/sub>B and B\u2019C\u2019 parallel to BC.<br \/>Then \u0394AB\u2019C\u2019 is the required triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1966\/44733025025_635815ca39_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 11 Constructions \" width=\"281\" height=\"460\" \/><\/p>\n<p>Question 7.<br \/>Draw a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm. Then construct another triangle whose sides are\u00a053\u00a0times the corresponding sides of the given triangle. (C.B.S.E. 2008)<br \/>Solution:<br \/>Steps of construction :<br \/>(i) Draw a line segment BC = 5 cm.<br \/>(ii) At B, draw perpendicular BX and cut off BA = 4 cm.<br \/>(iii )join Ac, then ABC is the triangle<br \/>(iv) Draw a ray BY making an acute angle with BC, and cut off 5 equal parts making BB<sub>1<\/sub>\u00a0= B<sub>1<\/sub>B<sub>2<\/sub>\u00a0= B<sub>2<\/sub>B<sub>3<\/sub>\u00a0= B<sub>3<\/sub>B<sub>4<\/sub>\u00a0= B<sub>4<\/sub>B<sub>5<\/sub><br \/>(v) Join B<sub>3<\/sub>\u00a0and C.<br \/>(vi) From B<sub>5<\/sub>, draw B<sub>5<\/sub>C\u2019 parallel to B<sub>3<\/sub>C and C\u2019A\u2019 parallel to CA.<br \/>Then \u0394A\u2019BC\u2019 is the required triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1955\/44922310504_8850b5d485_o.png\" alt=\"Learncbse.In Class 10 Chapter 11 Constructions \" width=\"321\" height=\"416\" \/><\/p>\n<p>Question 8.<br \/>Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are\u00a032\u00a0times the corresponding sides of the isosceles triangle.<br \/>Solution:<br \/>Steps of construction :<br \/>(i) Draw a line segment BC = 8 cm and draw its perpendicular bisector DX and cut off DA = 4 cm.<br \/>(ii) Join AB and AC. Then ABC is the triangle.<br \/>(iii) Draw a ray DY making an acute angle with OA and cut off 3 equal parts making DD<sub>1<\/sub>\u00a0= D<sub>1<\/sub>D<sub>2<\/sub>\u00a0=D<sub>2<\/sub>D<sub>3<\/sub>\u00a0= D<sub>3<\/sub>D<sub>4<br \/><\/sub>(iv) Join D<sub>2<br \/><\/sub>(v) Draw D<sub>3<\/sub>A\u2019 parallel to D<sub>2<\/sub>A and A\u2019B\u2019 parallel to AB meeting BC at C\u2019 and B\u2019 respectively.<br \/>Then \u0394B\u2019A\u2019C\u2019 is the required triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1926\/45647274261_57a5f8348e_o.png\" alt=\"Class 10 RD Sharma Solutions Chapter 11 Constructions \" width=\"305\" height=\"253\" \/><\/p>\n<p>Question 9.<br \/>Draw a \u0394ABC with side BC = 6 cm, AB = 5 cm and \u2220ABC = 60\u00b0. Then construct a triangle whose sides are (34)th of the corresponding sides of the \u0394ABC.<br \/>Solution:<br \/>Steps of construction :<br \/>(i) Draw a line segment BC = 6 cm.<br \/>(ii) At B, draw a ray BX making an angle of 60\u00b0 with BC, and cut off BA = 5 cm.<br \/>(iii) Join AC. Then ABC is the triangle.<br \/>(iv) Draw a ray BY making an acute angle with BC and cut off 4 equal parts making BB<sub>1<\/sub>= B<sub>1<\/sub>B<sub>2<\/sub> B<sub>2<\/sub>B<sub>3<\/sub>=B<sub>3<\/sub>B<sub>4<\/sub>.<br \/>(v) Join B<sub>4<\/sub>\u00a0and C.<br \/>(vi) From B<sub>3<\/sub>, draw B<sub>3<\/sub>C\u2019 parallel to B<sub>4<\/sub>C and C\u2019A\u2019 parallel to CA.<br \/>Then \u0394A\u2019BC\u2019 is the required triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1944\/45647275781_cd3e0b2b1a_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 11 Constructions \" width=\"262\" height=\"341\" \/><\/p>\n<p>Question 10.<br \/>Construct a triangle similar to \u0394ABC in which AB = 4.6 cm, BC = 5.1 cm,\u2220A = 60\u00b0 with scale factor 4 : 5.<br \/>Solution:<br \/>Steps of construction :<br \/>(i) Draw a line segment AB = 4.6 cm.<br \/>(ii) At A, draw a ray AX making an angle of 60\u00b0.<br \/>(iii) With centre B and radius 5.1 cm draw an arc intersecting AX at C.<br \/>(iv) Join BC. Then ABC is the triangle.<br \/>(v) From A, draw a ray AX making an acute angle with AB and cut off 5 equal parts making AA<sub>1<\/sub>\u00a0= A<sub>1<\/sub>A<sub>2<\/sub>\u00a0= A<sub>2<\/sub>A<sub>3<\/sub>\u00a0= A<sub>3<\/sub>A<sub>4<\/sub>=A<sub>4<\/sub>A<sub>5<\/sub>.<br \/>(vi) Join A<sub>4<\/sub>\u00a0and B.<br \/>(vii) From A<sub>5<\/sub>, drawA<sub>5<\/sub>B\u2019 parallel to A<sub>4<\/sub>B and B\u2019C\u2019 parallel to BC.<br \/>Then \u0394C\u2019AB\u2019 is the required triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1928\/45647275711_4e39f71d83_o.png\" alt=\"RD Sharma Class 10 Solution Chapter 11 Constructions \" width=\"271\" height=\"376\" \/><\/p>\n<p>Question 11.<br \/>Construct a triangle similar to a given \u0394XYZ with its sides equal to\u00a0(32)\u00a0th of the corresponding sides of \u0394XYZ. Write the steps of construction. [CBSE 1995C]<br \/>Solution:<br \/>Steps of construction :<br \/>(i) Draw a triangle XYZ with some suitable data.<br \/>(ii) Draw a ray YL making an acute angle with XZ and cut off 5 equal parts making YY<sub>1<\/sub>= Y<sub>1<\/sub>Y<sub>2<\/sub>\u00a0= Y<sub>2<\/sub>Y<sub>3<\/sub>\u00a0= Y<sub>3<\/sub>Y<sub>4<\/sub>.<br \/>(iii) Join Y<sub>4<\/sub>\u00a0and Z.<br \/>(iv) From Y<sub>3<\/sub>, draw Y<sub>3<\/sub>Z\u2019 parallel to Y<sub>4<\/sub>Z and Z\u2019X\u2019 parallel to ZX.<br \/>Then \u0394X\u2019YZ\u2019 is the required triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1952\/45647275701_275349e1d4_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 11 Constructions \" width=\"239\" height=\"302\" \/><\/p>\n<p>Question 12.<br \/>Draw a right triangle in which sides (other than the hypotenuse) are of lengths 8 cm and 6 cm. Then construct another triangle whose sides are\u00a034\u00a0times the corresponding sides of the first triangle.<br \/>Solution:<br \/>(i) Draw right \u0394ABC right angle at B and BC = 8 cm and BA = 6 cm.<br \/>(ii) Draw a line BY making an a cut angle with BC and cut off 4 equal parts.<br \/>(iii) Join 4C and draw 3C\u2019 || 4C and C\u2019A\u2019 parallel to CA.<br \/>The BC\u2019A\u2019 is the required triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1926\/44922311814_373e6cb747_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Pdf Free Download Chapter 11 Constructions \" width=\"331\" height=\"422\" \/><\/p>\n<p>Question 13.<br \/>Construct a triangle with sides 5 cm, 5.5 cm, and 6.5 cm. Now construct another triangle, whose sides are 3\/5 times the corresponding sides of the given triangle. [CBSE 2014]<br \/>Solution:<br \/>Steps of construction:<br \/>(i) Draw a line segment BC = 5.5 cm.<br \/>(ii) With centre B and radius 5 cm and with centre C and radius 6.5 cm, draw arcs that intersect each other at A<br \/>(iii) Join BA and CA.<br \/>\u0394ABC is the given triangle.<br \/>(iv) At B, draw a ray BX making an acute angle, and cut off 5 equal parts from BX.<br \/>(v) Join C5 and draw 3D || 5C which meets BC at D.<br \/>From D, draw DE || CA which meets AB at E.<br \/>\u2234 \u0394EBD is the required triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1930\/45647275411_12ff170796_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 11 Constructions \" width=\"206\" height=\"352\" \/><\/p>\n<p>Question 14.<br \/>Construct a triangle PQR with side QR = 7 cm, PQ = 6 cm and \u2220PQR = 60\u00b0. Then construct another triangle whose sides are 3\/5 of the corresponding sides of \u0394PQR. [CBSE 2014]<br \/>Solution:<br \/>Steps of construction:<br \/>(i) Draw a line segment QR = 7 cm.<br \/>(ii) At Q draw a ray QX making an angle of 60\u00b0 and cut of PQ = 6 cm. Join PR.<br \/>(iii) Draw a ray QY making an acute angle and cut off 5 equal parts.<br \/>(iv) Join 5, R and through 3, draw 3, S parallel to 5, R which meet QR at S.<br \/>(v) Through S, draw ST || RP meeting PQ at T.<br \/>\u2234 \u0394QST is the required triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1972\/44922311704_cc25945c68_o.png\" alt=\"Class 10 RD Sharma Chapter 11 Constructions \" width=\"285\" height=\"405\" \/><\/p>\n<p>Question 15.<br \/>Draw a \u0394ABC in which base BC = 6 cm, AB = 5 cm and \u2220ABC = 60\u00b0. Then construct another triangle whose sides are\u00a034\u00a0of the corresponding sides of \u0394ABC. [CBSE 2017]<br \/>Solution:<br \/>Steps of construction:<\/p>\n<ol>\n<li>Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and \u2220ABC = 60\u00b0.<\/li>\n<li>\u00a0Draw a ray BX, which makes an acute angle \u2220CBX below the line BC.<\/li>\n<li>Locate four points B<sub>1<\/sub>, B<sub>2<\/sub>, B<sub>3<\/sub>and B<sub>4\u00a0<\/sub>on BX such that BB<sub>1<\/sub>\u00a0= B<sub>1<\/sub>B<sub>2<\/sub>=B<sub>2<\/sub>B<sub>3<\/sub>\u00a0= B<sub>3<\/sub>B<sub>4<\/sub>.<\/li>\n<li>Join B<sub>4<\/sub>C and draw a line through B<sub>3<\/sub>\u00a0parallel to B<sub>4<\/sub>C intersecting BC to C\u2019.<\/li>\n<li>Draw a line through C\u2019 parallel to the line CA to intersect BA at A\u2019.<\/li>\n<\/ol>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1914\/45647275251_2d58a86ef4_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 11 Constructions \" width=\"349\" height=\"620\" \/><\/p>\n<p>Question 16.<br \/>Draw a right triangle in which the sides (other than the hypotenuse) arc of lengths 4 cm and 3 cm. Now, construct another triangle whose sides are\u00a053\u00a0times the corresponding sides of the given triangle. [CBSE 2017]<br \/>Solution:<br \/>Steps of construction:<\/p>\n<ol>\n<li>Draw a right triangle ABC in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. \u2220B = 90\u00b0.<\/li>\n<li>Draw a line BX, which makes an acute angle \u2220CBX below the line BC.<\/li>\n<li>Locate 5 points B<sub>1<\/sub>, B<sub>2<\/sub>, B<sub>3<\/sub>, B<sub>4<\/sub>\u00a0and B<sub>5<\/sub>\u00a0on BX such that BB<sub>1<\/sub>\u00a0= B<sub>1<\/sub>B<sub>2<\/sub>=B<sub>2<\/sub>B<sub>3<\/sub>=B<sub>3<\/sub>B<sub>4<\/sub>=B<sub>4<\/sub>B<sub>5<\/sub>.<\/li>\n<li>Join B<sub>3<\/sub>\u00a0to C and draw a line through B<sub>5<\/sub>\u00a0parallel to B<sub>3<\/sub>C, intersecting the extended line segment BC at C\u2019.<\/li>\n<li>Draw a line through C\u2019 parallel to CA intersecting the extended line segment BA at A\u2019.<\/li>\n<\/ol>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1980\/45647275141_f37108b227_o.png\" alt=\"RD Sharma 10 Class Solutions Chapter 11 Constructions \" width=\"664\" height=\"602\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1912\/45647275021_21a198611a_o.png\" alt=\"RD Sharma Class 10 Textbook PDF Chapter 11 Constructions \" width=\"252\" height=\"177\" \/><\/p>\n<p>Question 17.<br \/>Construct a \u0394ABC in which AB = 5 cm, \u2220B = 60\u00b0, altitude CD = 3 cm. Construct a \u0394AQR similar to \u0394ABC such that the side of \u0394AQR is 1.5 times that of the corresponding sides of \u0394ACB.<br \/>Solution:<br \/>Steps of construction :<br \/>(i) Draw a line segment AB = 5 cm.<br \/>(ii) At A, draw a perpendicular and cut off AE = 3 cm.<br \/>(iii) From E, draw EF || AB.<br \/>(iv) From B, draw a ray making an angle of 60 meeting EF at C.<br \/>(v) Join CA. Then ABC is the triangle.<br \/>(vi) From A, draw a ray AX making an acute angle with AB and cut off 3 equal parts making A A<sub>1<\/sub>= A<sub>1<\/sub>A<sub>2<\/sub>\u00a0= A<sub>2<\/sub>A<sub>3<\/sub>.<br \/>(vii) Join A<sub>2<\/sub>\u00a0and B.<br \/>(viii) From A, draw A^B\u2019 parallel to A<sub>2<\/sub>B and B\u2019C\u2019 parallel toBC.<br \/>Then \u0394C\u2019AB\u2019 is the required triangle.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1915\/45596779952_45deda947c_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 11 Constructions \" width=\"310\" height=\"316\" \/><\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 11 Exercise 11.2. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-11-exercise-112\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 11 Exercise 11.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631700996444\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-the-rd-sharma-class-10-solutions-chapter-11-exercise-112-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 11 Exercise 11.2 available on the Kopykitab website?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, the PDFs of RD Sharma Class 10 Solutions Chapter 11 Exercise 11.2 are available. These solutions are created in a unique method by Kopykitab\u2019s expert faculty.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631701040047\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"does-rd-sharma-class-10-solutions-chapter-11-exercise-112-help-you-to-clear-board-exams\"><\/span>Does RD Sharma Class 10 Solutions Chapter 11 Exercise 11.2 help you to clear board exams?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, RD Sharma Class 10 Maths Solutions Chapter 11 Exercise 11.2 of Class 10 Maths is an essential chapter. These answers are centered on understanding several Math shortcuts and strategies for quick and easy calculations.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631701083597\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-rd-sharma-class-10-solutions-chapter-11-exercise-112-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 11 Exercise 11.2 Free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can get RD Sharma Class 10 Solutions Chapter 11 Exercise 11.2 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 11 Exercise 11.2:\u00a0In this exercise, students can learn how to design a triangle that is comparable to a given triangle. All of the answers to the RD Sharma Class 10 Solutions have been written by Kopykitab experts to dispel any concerns and provide the correct techniques for solving a &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.2 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-11-exercise-11-2\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 11 Constructions Exercise 11.2 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":127994,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/127984"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=127984"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/127984\/revisions"}],"predecessor-version":[{"id":138605,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/127984\/revisions\/138605"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/127994"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=127984"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=127984"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=127984"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}