{"id":29217,"date":"2018-07-18T06:12:47","date_gmt":"2018-07-18T06:12:47","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=29217"},"modified":"2020-12-01T16:47:58","modified_gmt":"2020-12-01T11:17:58","slug":"rs-aggarwal-solutions-class-10-chapter-13-constructions","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/rs-aggarwal-solutions-class-10-chapter-13-constructions\/","title":{"rendered":"RS Aggarwal Solutions Class 10 Chapter 13 Constructions"},"content":{"rendered":"<h2>RS Aggarwal Solutions Class 10 Chapter 13 Constructions<\/h2>\n<p>These Solutions are part of <a href=\"https:\/\/cbselibrary.com\/rs-aggarwal-class-10-solutions\/\">RS Aggarwal Solutions Class 10<\/a>. Here we have given RS Aggarwal Solutions Class 10 Chapter 13 Constructions<\/p>\n<h3 style=\"text-align: center;\"><span style=\"color: #0000ff;\"><strong>Exercise 13A<\/strong><\/span><\/h3>\n<p><span style=\"color: #993366;\"><strong>Question 1:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-116998\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-1.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13a 1.1\" width=\"244\" height=\"274\" \/><br \/>\n<strong>Steps of construction:<\/strong><br \/>\nStep 1 : Draw a line segment AB = 6.5 cm<br \/>\nStep 2: Draw a ray AX making \u2220 BAX.<br \/>\nStep 3: Along AX mark (4+7) = 11 points<br \/>\nA<sub>1<\/sub>, A<sub>2<\/sub>, A<sub>3<\/sub>, A<sub>4<\/sub>, A<sub>5<\/sub>, A<sub>6<\/sub>, A<sub>7<\/sub>, A<sub>8<\/sub>, A<sub>9<\/sub>, A<sub>10<\/sub>, A<sub>11<\/sub>, such that<br \/>\nAA<sub>1<\/sub>\u00a0= A<sub>1<\/sub>A<sub>2<\/sub><br \/>\nStep 4: Join A<sub>11<\/sub>\u00a0and B.<br \/>\nStep 5: Through A<sub>4<\/sub>\u00a0draw a line parallel to A<sub>11<\/sub>\u00a0B meeting AB at C.<br \/>\nTherefore, C is the point on AB, which divides AB in the ratio 4 : 7<br \/>\nOn measuring,<br \/>\nAC = 2.4 cm<br \/>\nCB = 4.1 cm<br \/>\n<b>Read More:<\/b><\/p>\n<ul>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/cbselibrary.com\/construction-of-triangles-with-compass-and-protractor\/\"><span style=\"font-weight: 400;\">Construction of an Equilateral Triangle<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/cbselibrary.com\/construction-of-similar-triangles\/\"><span style=\"font-weight: 400;\">Construction Of Similar Triangle As Per Given Scale Factor<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/cbselibrary.com\/construction-of-line-segment\/\"><span style=\"font-weight: 400;\">Construction Of A Line Segment<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/cbselibrary.com\/construction-of-bisector-of-an-angle\/\"><span style=\"font-weight: 400;\">Construction Of The Bisector Of A Given Angle<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/cbselibrary.com\/construction-of-perpendicular-bisector-of-a-line-segment\/\"><span style=\"font-weight: 400;\">Construction Of Perpendicular Bisector Of A Line Segment<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/cbselibrary.com\/construction-of-angles-using-compass-ruler\/\"><span style=\"font-weight: 400;\">Construction Of An Angle Using Compass And Ruler<\/span><\/a><\/li>\n<\/ul>\n<p><span style=\"color: #993366;\"><strong>Question 2:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-116999\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-2.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13a 2.1\" width=\"201\" height=\"194\" \/><br \/>\n<strong>Steps of Construction:<\/strong><br \/>\nStep 1 : Draw a line segment PQ = 5.8 cm<br \/>\nStep 2: Draw a ray PX making an acute angle QPX.<br \/>\nStep 3: Along PX mark (5 + 3) = 8 points<br \/>\nA<sub>1<\/sub>, A<sub>2<\/sub>, A<sub>3<\/sub>, A<sub>4<\/sub>, A<sub>5<\/sub>, A<sub>6<\/sub>, A<sub>7\u00a0<\/sub>and A<sub>8<\/sub>\u00a0such that<br \/>\nPA<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>\u00a0= A<sub>5<\/sub>A6 = A<sub>6<\/sub>A7 = A<sub>7<\/sub>A<sub>8<\/sub><br \/>\nStep 4: Join A<sub>8<\/sub>Q.<br \/>\nStep 5: From A<sub>5<\/sub> draw A<sub>5<\/sub>C || A<sub>8<\/sub>Q meeting PQ at C.<br \/>\nC is the point on PQ, which divides PQ in the ratio 5 : 3<br \/>\nOn measurement,<br \/>\nPC = 3.6 cm, CQ = 2.2 cm<\/p>\n<p><iframe loading=\"lazy\" title=\"How to Construct a Triangle Similar to a Given Triangle | Geometric Constructions | Letstute\" width=\"1200\" height=\"675\" src=\"https:\/\/www.youtube.com\/embed\/vNCBBlfO0DI?start=535&#038;feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p><strong>More Resources<\/strong><\/p>\n<ul>\n<li><a title=\"RS Aggarwal Class 10 Solutions\" href=\"https:\/\/cbselibrary.com\/course\/rs-aggarwal-class-10-solutions\/\">RS Aggarwal Solutions Class 10<\/a><\/li>\n<li><a title=\"RD Sharma Class 10 Solutions\" href=\"http:\/\/www.learncbse.in\/rd-sharma-class-10-solutions\/\">RD Sharma Class 10 Solutions<\/a><\/li>\n<li><a title=\"NCERT Solutions Home Page\" href=\"http:\/\/www.learncbse.in\/ncert-solutions-2\/\">NCERT Solutions<\/a><\/li>\n<\/ul>\n<p><span style=\"color: #993366;\"><strong>Question 3:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117000\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-3.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13a 3.1\" width=\"282\" height=\"358\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-3.1.jpg 282w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-3.1-236x300.jpg 236w\" sizes=\"(max-width: 282px) 100vw, 282px\" \/><br \/>\n<strong>Steps of construction:<\/strong><br \/>\nStep 1: Draw a line segment BC = 6 cm<br \/>\nStep 2: With B as centre and radius equal to 5 cm draw an arc.<br \/>\nStep 3: With C as centre and radius equal to 7 cm draw another arc cutting the previous arc at A.<br \/>\nStep 4: Join AB and AC. Thus, \u2206ABC is obtained.<br \/>\nStep 5: Below BC draw another line BX.<br \/>\nStep 6: Mark 7 points B<sub>1<\/sub>B<sub>2<\/sub>B<sub>3<\/sub>B<sub>4<\/sub>B<sub>5<\/sub>B<sub>6<\/sub>B<sub>7<\/sub>\u00a0such that<br \/>\nBB<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 \/>\nStep 7: Join \u00a0B<sub>7<\/sub>C.<br \/>\nStep 8: from B<sub>5<\/sub>, draw B<sub>5<\/sub>D || B<sub>7<\/sub>C.<br \/>\nStep 9: Draw a line DE through D parallel to CA.<br \/>\nHence \u2206 BDE is the required triangle.<\/p>\n<p><span style=\"color: #993366;\"><strong>Question 4:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117001\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-4.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13a 4.1\" width=\"322\" height=\"299\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-4.1.jpg 322w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-4.1-300x279.jpg 300w\" sizes=\"(max-width: 322px) 100vw, 322px\" \/><br \/>\n<strong>Steps of construction:<\/strong><br \/>\nStep 1: Draw a line segment QR = 6 cm<br \/>\nStep 2: At Q, draw an angle RQA of 60\u25e6.<br \/>\nStep 3: From QA cut off a segment QP = 5 cm.<br \/>\nJoin PR. \u2206PQR is the given triangle.<br \/>\nStep 4: Below QR draw another line QX.<br \/>\nStep 5: Along QX cut &#8211; off equal distances Q<sub>1<\/sub>Q<sub>2<\/sub>Q<sub>3<\/sub>Q<sub>4<\/sub>Q<sub>5<\/sub><br \/>\nQQ<sub>1<\/sub>\u00a0= Q<sub>1<\/sub>Q2<sub>\u00a0<\/sub>= Q<sub>2<\/sub>Q<sub>3<\/sub>\u00a0= Q<span style=\"font-size: 13.3333px; line-height: 20px;\">3<\/span>Q<sub>4<\/sub>\u00a0= Q<sub>4<\/sub>Q<sub>5<\/sub><br \/>\nStep 6: Join Q<sub>5<\/sub>R.<br \/>\nStep 7: Through Q<sub>3<\/sub>\u00a0draw Q<sub>3<\/sub>S || Q<sub>5<\/sub>R.<br \/>\nStep 8: Through S, draw ST || PR.<br \/>\n\u2206 TQS is the required triangle.<\/p>\n<p><span style=\"color: #993366;\"><strong>Question 5:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117002\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-5.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13a 5.1\" width=\"311\" height=\"274\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-5.1.jpg 311w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-5.1-300x264.jpg 300w\" sizes=\"(max-width: 311px) 100vw, 311px\" \/><br \/>\n<strong>Steps of construction:<\/strong><br \/>\nStep 1: Draw a line segment BC = 6 cm<br \/>\nStep 2: Draw a right bisector PQ of BC meeting it at M.<br \/>\nStep 3: From QP cut &#8211; off a distance MA = 4 cm<br \/>\nStep 4: Join AB, AC.<br \/>\n\u2206 ABC is the given triangle.<br \/>\nStep 5: Below BC, draw a line BX.<br \/>\nStep 6: Along BX, cut &#8211; off 3 equal distances such that<br \/>\nBR<sub>1<\/sub> = R<sub>1<\/sub>R<sub>2<\/sub><sub>\u00a0<\/sub>= R<sub>2<\/sub>R<sub>3<\/sub><sub>\u00a0<\/sub><br \/>\nStep 7: Join R<sub>2<\/sub>C.<br \/>\nStep 8: Through R<sub>3<\/sub>\u00a0draw a line\u00a0R<sub>3<\/sub>C<sub>1<\/sub> ||\u00a0R<sub>2<\/sub>C.<br \/>\nStep 9 : Through C<sub>1<\/sub>\u00a0draw line C<sub>1<\/sub>A<sub>1\u00a0<\/sub>|| CA<sub>\u00a0.<\/sub><br \/>\n\u2206 A<sub>1<\/sub>BC<sub>1<\/sub> is the required triangle.<\/p>\n<p><span style=\"color: #993366;\"><strong>Question 6:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117003\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-6.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13a 6.1\" width=\"287\" height=\"285\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-6.1.jpg 287w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-6.1-150x150.jpg 150w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-6.1-100x100.jpg 100w\" sizes=\"(max-width: 287px) 100vw, 287px\" \/><br \/>\n<strong>Steps of Construction:<\/strong><br \/>\nStep 1: Draw a line segment BC = 5.4 cm<br \/>\nStep 2. At B, draw \u2220 CBM = 45\u00b0<br \/>\nStep 3: Now \u2220 A = 105\u00b0, \u2220 B = 45\u00b0, \u2220 C = 180\u00b0 &#8211; (105\u00b0+ 45\u00b0) = 30\u00b0<br \/>\nAt C draw \u2220 BCA = 30\u00b0.<br \/>\n\u2206 ABC is the given triangle.<br \/>\nStep 4: Draw a line BX below BC.<br \/>\nStep 5: Cut-off equal distances such that \u00a0BR<sub>1<\/sub> = R<sub>1<\/sub>R<sub>2<\/sub><sub>\u00a0<\/sub>= R<sub>2<\/sub>R<sub>3\u00a0<\/sub>=\u00a0R<sub>3<\/sub>R<sub>4<\/sub><br \/>\nStep 6: Join\u00a0R<sub>3<\/sub>C.<br \/>\nStep 7: Through R<sub>4<\/sub>, draw a line\u00a0R<sub>4<\/sub>C<sub>1<\/sub> ||\u00a0R<sub>3<\/sub>C.<br \/>\nStep 8: Through C<sub>1<\/sub>\u00a0draw a line C<sub>1<\/sub>A<sub>1<\/sub>\u00a0parallel to CA.<br \/>\n\u2206 A<sub>1<\/sub>BC<sub>1\u00a0<\/sub>is the required triangle.<\/p>\n<p><span style=\"color: #993366;\"><strong>Question 7:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117004\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-7.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13a 7.1\" width=\"446\" height=\"469\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-7.1.jpg 446w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-7.1-285x300.jpg 285w\" sizes=\"(max-width: 446px) 100vw, 446px\" \/><\/p>\n<p><strong>Steps of Construction:<\/strong><br \/>\nStep 1: Draw a line segment BC = 4 cm<br \/>\nStep 2: Draw a right- angle CBM at B.<br \/>\nStep 3: Cut-off BA = 3cm from BM.<br \/>\nStep 4: Join AC.<br \/>\n\u0394ABC is the given triangle.<br \/>\nStep 5: Below BC draw a line BX.<br \/>\nStep 6: Along BX, cut-off 7 equal distances such that<br \/>\nBR<sub>1<\/sub> = R<sub>1<\/sub>R<sub>2<\/sub><sub>\u00a0<\/sub>= R<sub>2<\/sub>R<sub>3\u00a0<\/sub>=\u00a0R<sub>3<\/sub>R<sub>4<\/sub>\u00a0= R<sub>4<\/sub>R<sub>5<\/sub>\u00a0= R<sub>5<\/sub>R<sub>6<\/sub>\u00a0= R<sub>6<\/sub>R<sub>7<\/sub><br \/>\nStep 7: Join\u00a0R<sub>5<\/sub>C.<br \/>\nStep 8: Through R<sub>7<\/sub>\u00a0draw a line parallel to R<sub>5<\/sub>C\u00a0cutting BC produced at\u00a0C<sub>1<\/sub><br \/>\nStep 9: Through C<sub>1<\/sub>\u00a0draw a line parallel to CA cutting BA at\u00a0A<sub>1<\/sub><br \/>\n\u2206 A<sub>1<\/sub>BC<sub>1<\/sub> is the required triangle.<\/p>\n<p><span style=\"color: #993366;\"><strong>Question 8:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117005\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-8.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13a 8.1\" width=\"249\" height=\"370\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-8.1.jpg 249w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-8.1-202x300.jpg 202w\" sizes=\"(max-width: 249px) 100vw, 249px\" \/><br \/>\n<strong>Steps of Construction:<\/strong><br \/>\nStep 1: draw a line segment BC = 5 cm<br \/>\nStep 2: With B as centre and radius 7cm an arc is drawn.<br \/>\nStep 3: With C as centre and radius 6 cm another arc is drawn intersecting the previous arc at A.<br \/>\nStep 4: Join AB and AC.<br \/>\nStep 5: \u2206 ABC is the given triangle.<br \/>\nStep 6: Draw a line BX below BC.<br \/>\nStep 7: Cut- off equal distances from DX such that<br \/>\nBB<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 \/>\nStep 8: join\u00a0B<sub>5<\/sub>C.<br \/>\nStep 9: Draw a line through B<sub>7<\/sub>\u00a0parallel to B<sub>5<\/sub>C\u00a0cutting BC produced at C&#8217;.<br \/>\nStep 10: Through C&#8217; draw a line parallel to CA, cutting BA produced at A&#8217;.<br \/>\nStep 11: \u2206 A&#8217;BC&#8217; is the required triangle.<\/p>\n<p><span style=\"color: #993366;\"><strong>Question 9:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117006\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-9.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13a 9.1\" width=\"277\" height=\"301\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-9.1.jpg 277w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-9.1-276x300.jpg 276w\" sizes=\"(max-width: 277px) 100vw, 277px\" \/><br \/>\n<strong>Steps of construction:<\/strong><br \/>\nStep 1: Draw a line segment AB = 6.5 cm<br \/>\nStep 2: With B as centre and some radius draw an arc cutting AB at D.<br \/>\nStep 3: With centre D and same radius draw another arc cutting previous arc at E. \u2220 ABE = 60\u00b0<br \/>\nStep 4: Join BE and produce it to a point X.<br \/>\nStep 5: With centre B and radius 5.5 cm draw an arc intersecting BX at C.<br \/>\nStep 6: Join AC.<br \/>\n\u2206 ABC is the required triangle.<br \/>\nStep 7: Draw a line AP below AB.<br \/>\nStep 8: Cut- off 3 equal distances such that<br \/>\nAA<sub>1<\/sub>\u00a0= A<sub>1<\/sub>A<sub>2<\/sub>\u00a0= A<sub>2<\/sub>A<sub>3<\/sub><br \/>\nStep 9: Join\u00a0BA<sub>2<\/sub><br \/>\nStep 10: Draw A<sub>3<\/sub>B&#8217;\u00a0through A<sub>3<\/sub>\u00a0parallel to\u00a0A<sub>3<\/sub>B.<br \/>\nStep 11: Draw a line parallel to BC through B&#8217; intersecting AY at C&#8217;.<br \/>\n\u2206 AB&#8217;C&#8217; is the required triangle.<\/p>\n<p><span style=\"color: #993366;\"><strong>Question 10:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117007\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-10.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13a 10.1\" width=\"203\" height=\"184\" \/><br \/>\n<strong>Steps of construction:<\/strong><br \/>\nStep 1: Draw a line segment BC = 6.5 cm<br \/>\nStep 2: Draw an angle of 60\u00b0 at B so that \u2220 XBC = 60\u00b0.<br \/>\nStep 3: With centre B and radius 4.5cm, draw an arc intersecting XB at A.<br \/>\nStep 4: Join AC.<br \/>\n\u2206 ABC is the required triangle.<br \/>\nStep 5: Draw a line BY below BC.<br \/>\nStep 6: Cut- off 4 equal distances from BY.<br \/>\nSuch that\u00a0BB<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><br \/>\nStep 7: Join\u00a0CB<sub>4<\/sub><br \/>\nStep 8: draw B<sub>3<\/sub>C&#8217;\u00a0parallel to\u00a0CB<sub>4<\/sub><br \/>\nStep 9: Draw C&#8217;A&#8217;\u00a0parallel to CA through C&#8217; intersecting BA produced at A&#8217;.<br \/>\n\u2206 A&#8217;BC&#8217; is the required similar triangle.<\/p>\n<p><span style=\"color: #993366;\"><strong>Question 11:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117008\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13a-11.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13a 11.1\" width=\"255\" height=\"297\" \/><br \/>\n<strong>Steps of Construction:<\/strong><br \/>\nStep 1: Draw a line segment BC = 9 cm<br \/>\nStep 2: with centre B and radius more than 1\/2 BC, draw arcs on both sides of BC.<br \/>\nStep 3: With centre C and same radius draw other arcs on both sides of BC intersecting previous arcs at P and Q.<br \/>\nStep 4: join PQ and produce it to a point X. PQ meets BC at M.<br \/>\nStep 5: With centre M and radius 5 cm, draw an arc intersecting MX at A.<br \/>\nStep 6: Join AB and AC.<br \/>\n\u2206 ABC is the required triangle.<br \/>\nStep 7: Draw a line BY below BC.<br \/>\nStep 8: Cut off 4 equal distances from BY so that<br \/>\nBB<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><br \/>\nStep 9: Join CB<sub>4<\/sub><br \/>\nStep 10: Draw C&#8217;B<sub>3<\/sub> parallel to CB<sub>4<\/sub><br \/>\nStep 11: Draw C&#8217;A&#8217; parallel to CA, through C&#8217; intersecting BA at A&#8217;.<br \/>\n\u2206 A&#8217;BC&#8217; is the required similar triangle.<\/p>\n<h3 style=\"text-align: center;\"><span style=\"color: #0000ff;\"><strong>Exercise 13B<\/strong><\/span><\/h3>\n<p><iframe loading=\"lazy\" title=\"Construction of tangents to a circle from a point outside a circle | Math | Letstute\" width=\"1200\" height=\"675\" src=\"https:\/\/www.youtube.com\/embed\/HEdUc-q9_es?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p><span style=\"color: #993366;\"><strong>Question 1:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117009\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13b-1.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13b 1.1\" width=\"193\" height=\"238\" \/><br \/>\n<strong>Steps of construction:<\/strong><br \/>\nStep 1: Draw a circle of radius 5 cm with centre O.<br \/>\nStep 2: A point P at a distance of 8cm from O is taken.<br \/>\nStep 3: A right bisector of OP meeting OP at M is drawn.<br \/>\nStep 4: With centre M radius OM a circle is drawn intersecting the previous circle at T<sub>1<\/sub>\u00a0and\u00a0T<sub>2<\/sub><br \/>\nStep 5: Join PT<sub>1<\/sub> and PT<sub>2<\/sub><br \/>\nPT<sub>1<\/sub> and PT<sub>2<\/sub> and the required tangents. Measuring PT<sub>1<\/sub> and PT<sub>2<\/sub><br \/>\nWe find, PT<sub>1<\/sub>= PT<sub>2<\/sub>\u00a0=6.2 cm<\/p>\n<p><span style=\"color: #993366;\"><strong>Question 2:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117010\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13b-2.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13b 2.1\" width=\"200\" height=\"235\" \/><br \/>\n<strong>Steps of construction:<\/strong><br \/>\nStep 1: Two concentric circles with centre O and radii 4 cm and 6 cm are drawn.<br \/>\nStep 2: A point P is taken on outer circle and O, P are joined.<br \/>\nStep 3: A right bisector of OP is drawn bisecting OP at M.<br \/>\nStep 4: With centre M and radius OM a circle is drawn cutting the inner circle at\u00a0T<sub>1<\/sub>\u00a0and\u00a0T<sub>2<\/sub><br \/>\nStep 5: Join PT<sub>1<\/sub> and PT<sub>2<\/sub><br \/>\nPT<sub>1<\/sub> and PT<sub>2\u00a0<\/sub>are the required tangents. Further\u00a0PT<sub>1<\/sub>= PT<sub>2<\/sub>\u00a0=4.8 cm<\/p>\n<p><span style=\"color: #993366;\"><strong>Question 3:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117011\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13b-3.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13b 3.1\" width=\"353\" height=\"220\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13b-3.1.jpg 353w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13b-3.1-300x187.jpg 300w\" sizes=\"(max-width: 353px) 100vw, 353px\" \/><br \/>\n<strong>Steps of construction:<\/strong><br \/>\nStep 1: Draw a circle with centre O and radius 3.5 cm<br \/>\nStep 2: the diameter P<sub>1<\/sub>P<sub>2<\/sub>\u00a0is extended to the points A and B such that AO = OB = 7 cm<br \/>\nStep 3: With centre P<sub>1<\/sub>\u00a0and radius 3.5 cm draw a circle cutting the first circle at\u00a0T<sub>1<\/sub>\u00a0and\u00a0T<sub>2<\/sub><br \/>\nStep 4: join\u00a0AT<sub>1<\/sub> and AT<sub>2<\/sub><br \/>\nStep 5: With centre P<sub>2<\/sub>\u00a0and radius 3.5 cm draw another circle cutting the first circle at\u00a0T<sub>3<\/sub>\u00a0and\u00a0T<sub>4<\/sub><br \/>\nStep 6: Join BT<sub>3<\/sub>\u00a0and BT<sub>4<\/sub>\u00a0. Thus AT<sub>1<\/sub>, AT<sub>2<\/sub>\u00a0and BT<sub>3<\/sub>, BT<sub>4<\/sub>\u00a0are the required tangents to the given circle from A and B.<\/p>\n<p><span style=\"color: #993366;\"><strong>Question 4:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117012\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13b-4.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13b 4.1\" width=\"362\" height=\"249\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13b-4.1.jpg 362w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13b-4.1-300x206.jpg 300w\" sizes=\"(max-width: 362px) 100vw, 362px\" \/><br \/>\n<strong>Steps of construction:<\/strong><br \/>\n(i) A circle of radius 4.2 cm at centre O is drawn.<br \/>\n(ii) A diameter AB is drawn.<br \/>\n(iii) With OB as base, an angle BOC of 45\u00b0 is drawn.<br \/>\n(iv) At A, a line perpendicular to OA is drawn.<br \/>\n(v) At C, a line perpendicular to OC is drawn.<br \/>\n(vi) These lines intersect each other at P.<br \/>\nPA and PC are the required tangents.<\/p>\n<p><span style=\"color: #993366;\"><strong>Question 5:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117013\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13b-5.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13b 5.1\" width=\"345\" height=\"261\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13b-5.1.jpg 345w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13b-5.1-300x227.jpg 300w\" sizes=\"(max-width: 345px) 100vw, 345px\" \/><br \/>\n<strong>Steps of construction:<\/strong><br \/>\n(i) A line segment AB = 8,5 cm is drawn.<br \/>\n(ii) Draw a right bisector of AB which meets AB at M.<br \/>\n(iii) With M as centre AM as radius a circle is drawn intersecting the given circles at\u00a0T<sub>1<\/sub>,\u00a0T<sub>2<\/sub>, T<sub>3<\/sub>\u00a0and\u00a0T<sub>4<\/sub><br \/>\n(iv) Join AT<sub>3<\/sub>, AT<sub>4\u00a0<\/sub>and\u00a0BT<sub>1<\/sub>, BT<sub>2<\/sub>.<br \/>\nThus AT<sub>3<\/sub>, AT<sub>4<\/sub>, BT<sub>1<\/sub>, BT<sub>2<\/sub>\u00a0are the required tangents.<\/p>\n<p><span style=\"color: #993366;\"><strong>Question 6:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117014\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13b-6.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13b 6.1\" width=\"275\" height=\"215\" \/><br \/>\n<strong>Steps of construction:<\/strong><br \/>\n(i) Draw a line segment AB = 7cm<br \/>\n(ii) Taking A as centre and radius 3 cm, a circle is drawn.<br \/>\n(iii) With centre B and radius 2.5 cm, another circle is drawn.<br \/>\n(iv) With centre A and radius more than 1\/2 AB, arcs are drawn on both sides of AB.<br \/>\n(v) With centre B and the same radius, [as in step (iv)] arcs are drawn on both sides of AB intersecting previous arcs at P and Q.<br \/>\n(vi) Join PQ which meets AB at M.<br \/>\n(vii) With centre M and radius AM, a circle is drawn which intersects circle with centre A at T<sub>1<\/sub>\u00a0and\u00a0T<sub>2<\/sub>\u00a0and the circle with centre B at\u00a0T<sub>3<\/sub>\u00a0and\u00a0T<sub>4<\/sub><br \/>\n(viii) Join AT<sub>3<\/sub>, AT<sub>4<\/sub>, BT<sub>1\u00a0<\/sub>and BT<sub>2<\/sub><br \/>\nThus, AT<sub>3<\/sub>, AT<sub>4<\/sub>, BT<sub>1<\/sub>, BT<sub>2\u00a0<\/sub>are the required tangents.<\/p>\n<p><span style=\"color: #993366;\"><strong>Question 7:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117015\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13b-7.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13b 7.1\" width=\"134\" height=\"205\" \/><br \/>\n<strong>Steps of construction:<\/strong><br \/>\n(i) A circle of radius 3 cm with centre O is drawn.<br \/>\n(ii) A radius OC is drawn making an angle of 60\u00b0 with the diameter AB.<br \/>\n(iii) At C, \u2220OCP = 90\u00b0 is drawn.<br \/>\nCP is required tangent.<\/p>\n<p><span style=\"color: #993366;\"><strong>Question 8:<\/strong><\/span><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-117016\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-10-Chapter-13-Constructions-13b-8.1.jpg\" alt=\"RS Aggarwal Solutions Class 10 Chapter 13 Constructions 13b 8.1\" width=\"174\" height=\"251\" \/><br \/>\n<strong>Steps of construction :<\/strong><br \/>\n(i) Draw a circle of radius 4 cm with centre O.<br \/>\n(ii) With diameter AB, a line OC is drawn making an angle of 30\u00b0 i.e., \u2220BOC = 30\u00b0<br \/>\n(iii) At C a perpendicular to OC is drawn meeting OB at P.<br \/>\nPC is the required tangent.<\/p>\n<p>Hope given <a href=\"https:\/\/cbselibrary.com\/rs-aggarwal-solutions-class-10-chapter-13-constructions\/\">RS Aggarwal Solutions Class 10 Chapter 13 Constructions<\/a> are helpful to complete your math homework.<\/p>\n<p>If you have any doubts, please comment below. <a href=\"https:\/\/cbselibrary.com\">A Plus Topper<\/a> try to provide online math tutoring for you.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>RS Aggarwal Solutions Class 10 Chapter 13 Constructions These Solutions are part of RS Aggarwal Solutions Class 10. 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CBSE Library","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/cbselibrary.com\/rs-aggarwal-solutions-class-10-chapter-13-constructions\/","og_locale":"en_US","og_type":"article","og_title":"RS Aggarwal Solutions Class 10 Chapter 13 Constructions","og_description":"RS Aggarwal Solutions Class 10 Chapter 13 Constructions These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 13 Constructions Exercise 13A Question 1: Steps of construction: Step 1 : Draw a line segment AB = 6.5 cm Step 2: Draw a ray AX making ... 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