{"id":63812,"date":"2021-09-04T17:09:30","date_gmt":"2021-09-04T11:39:30","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=63812"},"modified":"2023-12-27T10:41:19","modified_gmt":"2023-12-27T05:11:19","slug":"rd-sharma-solutions-class-9-maths-chapter-17-constructions","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-17-constructions\/","title":{"rendered":"RD Sharma Solutions Class 9 Maths Chapter 17 &#8211; Constructions (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-124578\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-17-Constructions.png\" alt=\"RD Sharma Solutions Class 9 Maths Chapter 17\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-17-Constructions.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-17-Constructions-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 9 Maths Chapter 17 Constructions:<\/strong> <span style=\"font-weight: 400;\">Each question is solved using figures for easy as well as better understanding. <\/span><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a> Chapter 17 <span style=\"font-weight: 400;\">builds a good foundation for students. <\/span><span style=\"font-weight: 400;\">They can secure excellent marks in the exams for answering more difficult questions accurately by practicing these solutions. All answers to the questions of this chapter constructions are accurate &amp; reliable. <\/span><\/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-69eb0680e7a56\" 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\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-17-constructions\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-17\" title=\"From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 17?\">From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 17?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-17-constructions\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-17\" title=\"How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 17?\">How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 17?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-17-constructions\/#can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-17-pdf-offline\" title=\"Can I access the RD Sharma Solutions for Class 9 Maths Chapter 17\u00a0PDF offline?\">Can I access the RD Sharma Solutions for Class 9 Maths Chapter 17\u00a0PDF offline?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-9-maths-chapter-17-pdf\"><\/span><strong>Download RD Sharma Solutions Class 9 Maths Chapter 17 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/17-1-2.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 17<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<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\/17-1-2.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"exercise-wise-rd-sharma-solutions-class-9-maths-chapter-17-constructions\"><\/span><strong>Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 17 Constructions<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-17-class-9-maths-exercise-17-1-solutions\/\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">RD Sharma class 9 chapter 17 exercise 17a<\/span><\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-17-class-9-maths-exercise-17-2-solutions\/\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">RD Sharma class 9 chapter 17 exercise 17b<\/span><\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-17-class-9-maths-exercise-17-3-solutions\/\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">RD Sharma class 9 chapter 17 exercise 17c<\/span><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-solutions-class-9-maths-chapter-17-constructions\"><\/span><strong>Access answers of R<\/strong><strong>D Sharma Solutions Class 9 Maths Chapter 17 Constructions<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-17-constructions-ex-171\"><\/span>RD Sharma Solutions Class 9 Chapter 17 Constructions Ex 17.1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>Find the area of a triangle whose sides are respectively 150 cm, 120 cm and 200 cm.<br \/>Solution:<br \/>Sides of triangle are 120 cm, 150 cm, 200 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1920\/44732409055_2e73462253_o.png\" alt=\"RD Sharma Class 9 Chapter 17 Constructions\" width=\"345\" height=\"610\" \/><\/p>\n<p>Question 2.<br \/>Find the area of a triangle whose sides are 9 cm, 12 cm and 15 cm.<br \/>Solution:<br \/>Sides of a triangle are 9 cpi, 12 cm, 15 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1943\/44732408805_76d4ce0fc0_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 17 Constructions\" width=\"250\" height=\"189\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1926\/44732408665_05e4015196_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 17 Constructions\" width=\"343\" height=\"192\" \/><\/p>\n<p>Question 3.<br \/>Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.<br \/>Solution:<br \/>Perimeter of a triangle = 42 cm<br \/>Two sides are 18 cm and 10 cm<br \/>Third side = 42 \u2013 (18 + 10)<br \/>= 42 \u2013 28 = 14 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1926\/44732408485_4d5b114fde_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 17 Constructions\" width=\"302\" height=\"268\" \/><\/p>\n<p>Question 4.<br \/>In a \u2206ABC, AB = 15 cm, BC = 13 cm and AC = 14 cm. Find the area of \u2206ABC and hence its altitude on AC.<br \/>Solution:<br \/>Sides of triangle ABC are AB = 15 cm, BC = 13 cm, AC = 14 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1970\/44732408245_24cd5b293e_o.png\" alt=\"Constructions Class 9 RD Sharma Solutions\" width=\"259\" height=\"222\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1967\/44732408005_d8778cf782_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 17 Constructions\" width=\"332\" height=\"369\" \/><\/p>\n<p>Question 5.<br \/>The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle. [NCERT]<br \/>Solution:<br \/>Perimeter of a triangle = 540 m<br \/>Ratio in sides = 25 : 17 : 12<br \/>Sum of ratios = 25 + 17 + 12 = 54<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1932\/44732407695_f48bae4dbc_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 17 Constructions\" width=\"356\" height=\"452\" \/><\/p>\n<p>Question 6.<br \/>The perimeter of a triangle is 300 m. If its sides are in the ratio 3:5:7. Find the area of the triangle. [NCERT]<br \/>Solution:<br \/>Perimeter of a triangle = 300 m<br \/>Ratio in the sides = 3 : 5 : 7<br \/>\u2234 Sum of ratios = 3 + 5 + 7= 15<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1937\/30705645607_f2ac5fcdb4_o.png\" alt=\"Class 9 Maths Chapter 17 Constructions RD Sharma Solutions\" width=\"352\" height=\"431\" \/><\/p>\n<p>Question 7.<br \/>The perimeter of a triangular field is 240 dm. If two of its sides are 78 dm and 50 dm, find the length of the perpendicular on the side of length 50 dm from the opposite vertex.<br \/>Solution:<br \/>Perimeter of a triangular field = 240 dm<br \/>Two sides are 78 dm and 50 dm<br \/>\u2234 Third side = 240 \u2013 (78 + 50)<br \/>= 240 \u2013 128 = 112 dm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1934\/30705645377_e00d340e5f_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 17 Constructions\" width=\"314\" height=\"172\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1976\/30705645197_ec2fc86a38_o.png\" alt=\"RD Sharma Class 9 Book Chapter 17 Constructions\" width=\"357\" height=\"256\" \/><\/p>\n<p>Question 8.<br \/>A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest of its altitudes.<br \/>Solution:<br \/>Sides of a triangle are 35 cm, 54 cm, 61 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1953\/45596105882_980b1295c3_o.png\" alt=\"Constructions With Solutions PDF RD Sharma Class 9 Solutions\" width=\"349\" height=\"474\" \/><\/p>\n<p>Question 9.<br \/>The lengths of the sides of a triangle are in the ratio 3:4:5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side.<br \/>Solution:<br \/>Ratio in the sides of a triangle = 3:4:5<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1940\/30705644937_ff259887cb_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 17 Constructions\" width=\"350\" height=\"623\" \/><\/p>\n<p>Question 10.<br \/>The perimeter of an isosceles triangle is 42 cm and its base is (3\/2) times each of the equal sides. Find the length of each side of the triangle, area of the triangle and the height of the triangle.<br \/>Solution:<br \/>Perimeter of an isosceles triangle = 42 cm<br \/>Base =\u00a0<span id=\"MathJax-Element-1-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-1\" class=\"math\"><span id=\"MathJax-Span-2\" class=\"mrow\"><span id=\"MathJax-Span-3\" class=\"mfrac\"><span id=\"MathJax-Span-4\" class=\"mn\">3<\/span><span id=\"MathJax-Span-5\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0of its one of equal sides<br \/>Let each equal side = x, then 3<br \/>Base =\u00a0<span id=\"MathJax-Element-2-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-6\" class=\"math\"><span id=\"MathJax-Span-7\" class=\"mrow\"><span id=\"MathJax-Span-8\" class=\"mfrac\"><span id=\"MathJax-Span-9\" class=\"mn\">3<\/span><span id=\"MathJax-Span-10\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1918\/45596105482_de33c54755_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 17 Constructions\" width=\"263\" height=\"43\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1905\/44921634784_75c378625b_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 17 Constructions\" width=\"314\" height=\"591\" \/><\/p>\n<p>Question 11.<br \/>Find the area of the shaded region in figure.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1953\/45596105332_dc6abbddd0_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 17 Constructions\" width=\"248\" height=\"168\" \/><br \/>Solution:<br \/>In \u2206ABC, AC = 52 cm, BC = 48 cm<br \/>and in right \u2206ADC, \u2220D = 90\u00b0<br \/>AD = 12 cm, BD = 16 cm<br \/>\u2234 AB\u00b2=AD\u00b2 + BD\u00b2 (Pythagoras Theorem)<br \/>(12)\u00b2 + (16)\u00b2 = 144 + 256 = 400 = (20)\u00b2<br \/>\u2234 AB = 20 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1920\/44921634234_163e835d24_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 17 Constructions\" width=\"351\" height=\"422\" \/><\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-solution-chapter-17-constructions-ex-172\"><\/span>RD Sharma Class 9 Solution Chapter 17 Constructions Ex 17.2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>Find the area of a quadrilateral ABCD in which AB = 3cm, BC = 4cm, CD = 4cm, DA = 5cm and AC = 5cm (NCERT)<br \/>Solution:<br \/>In the quadrilateral, AC is the diagonal which divides the figure into two triangles<br \/>Now in \u2206ABC, AB = 3 cm, BC = 4cm, AC = 5cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1960\/45596104842_0f5aab3178_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 17 Constructions\" width=\"333\" height=\"336\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1945\/44921633964_3cb0949d3a_o.png\" alt=\"Constructions Class 9 RD Sharma Solutions\" width=\"346\" height=\"276\" \/><\/p>\n<p>Question 2.<br \/>The sides of a quadrangular field taken in order are 26 m, 27 m, 7 m and 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.<br \/>Solution:<br \/>In quad. ABCD, AB = 26 m, BC = 27 m CD = 7m, DA = 24 m, \u2220CDA = 90\u00b0<br \/>Join AC,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1943\/43828366370_70d143d4be_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 17 Constructions\" width=\"278\" height=\"530\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1946\/30705644157_da5ef43f6f_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 17 Constructions\" width=\"342\" height=\"266\" \/><\/p>\n<p>Question 3.<br \/>The sides of a quadrilateral taken in order are 5, 12, 14 and 15 metres respectively, and the angle contained by the first two sides is a right angle. Find its area.<br \/>Solution:<br \/>In quad. ABCD,<br \/>AB = 5m, BC = 12 m, CD = 14m,<br \/>DA = 15 m and \u2220ABC = 90\u00b0<br \/>Join AC,<br \/>Now in right \u2206ABC,<br \/>AC\u00b2 = AB\u00b2 + BC\u00b2 = (5)\u00b2 + (12)\u00b2<br \/>= 25 + 144 = 169 = (13)\u00b2<br \/>\u2234 AC = 13 m<br \/>Now area of right \u2206ABC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1955\/43828365970_1018d00e72_o.png\" alt=\"Class 9 Maths Chapter 17 Constructions RD Sharma Solutions\" width=\"320\" height=\"385\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1964\/30705643997_341eabb122_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 17 Constructions\" width=\"333\" height=\"287\" \/><\/p>\n<p>Question 4.<br \/>A park, in shape of a quadrilateral ABCD, has \u2220C = 90\u00b0, AB = 9m, BC = 12m, CD = 5m and AD = 8m. How much area does it occupy? (NCERT)<br \/>Solution:<br \/>In quadrilateral ABCD,<br \/>AB = 9m, BC = 12m, CD = 5m and<br \/>DA = 8m, \u2220C = 90\u00b0<br \/>Join BD,<br \/>Now in right \u2206BCD,<br \/>BD\u00b2 = BC\u00b2+ CD\u00b2 = (12)\u00b2 + (5)\u00b2<br \/>= 144 + 25 = 169 = (13)\u00b2<br \/>\u2234 BD = 13m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1957\/30705643837_345172862b_o.png\" alt=\"RD Sharma Class 9 Book Chapter 17 Constructions\" width=\"350\" height=\"408\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1908\/30705643677_76f795b391_o.png\" alt=\"Constructions With Solutions PDF RD Sharma Class 9 Solutions\" width=\"337\" height=\"181\" \/><\/p>\n<p>Question 5.<br \/>Find the area of a rhombus whose perimeter is 80m and one of whose diagonal is 24m.<br \/>Solution:<br \/>Perimeter of rhombus ABCD = 80 m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1929\/43828365270_6c624651d0_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 17 Constructions\" width=\"316\" height=\"669\" \/><\/p>\n<p>Question 6.<br \/>A rhombus sheet whose perimeter = 32 m and whose one diagonal is 10 m long, is painted on both sides at the rate of \u20b95 per m\u00b2. Find the cost of painting.<br \/>Solution:<br \/>Perimeter of the rhombus shaped sheet = 32 m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1916\/30705643427_e7364a386b_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 17 Constructions\" width=\"234\" height=\"182\" \/><br \/>\u2234 Length of each side =\u00a0<span id=\"MathJax-Element-3-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-11\" class=\"math\"><span id=\"MathJax-Span-12\" class=\"mrow\"><span id=\"MathJax-Span-13\" class=\"mfrac\"><span id=\"MathJax-Span-14\" class=\"mn\">32<\/span><span id=\"MathJax-Span-15\" class=\"mn\">4<\/span><\/span><\/span><\/span><\/span>\u00a0= 8m<br \/>and length of one diagonal AC = 10 m<br \/>In \u2206ABC, sides are 8m, 8m, 10m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1948\/44921632564_cd8dd38f1f_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 17 Constructions\" width=\"394\" height=\"456\" \/><\/p>\n<p>Question 7.<br \/>Find the area of a quadrilateral ABCD in which AD = 24 cm, \u2220BAD = 90\u00b0 and BCD forms an equilateral triangle whose each side is equal to 26 cm. (Take\u00a0<span id=\"MathJax-Element-4-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-16\" class=\"math\"><span id=\"MathJax-Span-17\" class=\"mrow\"><span id=\"MathJax-Span-18\" class=\"msqrt\"><span id=\"MathJax-Span-19\" class=\"mrow\"><span id=\"MathJax-Span-20\" class=\"mn\">3<\/span><\/span>\u2013\u221a<\/span><\/span><\/span><\/span>\u00a0= 1.73 )<br \/>Solution:<br \/>In quadrilateral ABCD, AD = 24cm, \u2220BAD = 90\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1932\/30705643247_814625710f_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 17 Constructions\" width=\"273\" height=\"288\" \/><br \/>BCD is an equilateral triangle with side 26cm<br \/>In right \u2206ABD,<br \/>BD\u00b2 = AB\u00b2+ AD\u00b2<br \/>(26)\u00b2 = AB\u00b2 + (24)\u00b2<br \/>\u21d2 676 = AB\u00b2 + 576<br \/>AB\u00b2 = 676 \u2013 576 = 100 = (10)\u00b2<br \/>\u2234 AB = 10cm<br \/>Now area of right \u2206ABD,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1903\/30705643137_2e283317bb_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 17 Constructions\" width=\"344\" height=\"268\" \/><\/p>\n<p>Question 8.<br \/>Find the area of a quadrilateral ABCD in which AB = 42cm, BC = 21cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.<br \/>Solution:<br \/>In quadrilateral ABCD,<br \/>AB = 42 cm, BC = 21 cm, CD = 29cm DA = 34 cm, BD = 20 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1908\/44921631804_bc2bddda92_o.png\" alt=\"RD Sharma Class 9 Chapter 17 Constructions\" width=\"316\" height=\"585\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1938\/30705642947_2af0486d3a_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 17 Constructions\" width=\"294\" height=\"287\" \/><\/p>\n<p>Question 9.<br \/>The adjacent sides of a parallelogram ABCD measures 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of the parallelogram.<br \/>Solution:<br \/>In ||gm ABCD,<br \/>AB = 34cm, BC = 20 cm<br \/>and AC = 42 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1941\/44921631524_34fd3f3661_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 17 Constructions\" width=\"235\" height=\"181\" \/><br \/>\u2235 The diagonal of a parallelogram divides into two triangles equal in area,<br \/>Now area of \u2206ABC,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1950\/30705642797_66edd0254c_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 17 Constructions\" width=\"363\" height=\"282\" \/><\/p>\n<p>Question 10.<br \/>Find the area of the blades of the magnetic compass shown in figure. (Take\u00a0<span id=\"MathJax-Element-5-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-21\" class=\"math\"><span id=\"MathJax-Span-22\" class=\"mrow\"><span id=\"MathJax-Span-23\" class=\"msqrt\"><span id=\"MathJax-Span-24\" class=\"mrow\"><span id=\"MathJax-Span-25\" class=\"mn\">11<\/span><\/span>\u2212\u2212\u221a<\/span><\/span><\/span><\/span>\u00a0= 3.32).<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1980\/44921631044_e86c0f4799_o.png\" alt=\"Constructions Class 9 RD Sharma Solutions\" width=\"255\" height=\"223\" \/><br \/>Solution:<br \/>ABCD is a rhombus with each side 5cm and one diagonal 1cm<br \/>Diagonal BD divides into two equal triangles Now area of \u2206ABD,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1980\/30705642587_8119e4bd85_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 17 Constructions\" width=\"328\" height=\"529\" \/><\/p>\n<p>Question 11.<br \/>A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.<br \/>Solution:<br \/>Area of a triangle with same base and area of a 11gm with equal sides of triangle are 13, 14, 15 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1942\/44921630694_3bef52bd86_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 17 Constructions\" width=\"305\" height=\"162\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1958\/44921630504_f9031048da_o.png\" alt=\"Class 9 Maths Chapter 17 Constructions RD Sharma Solutions\" width=\"330\" height=\"389\" \/><\/p>\n<p>Question 12.<br \/>Two parallel sides of a trapezium are 60cm and 77 cm and other sides are 25 cm and 26 cm. Find the area of the trapezium.<br \/>Solution:<br \/>In trapezium ABCD, AB || DC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1922\/30705642197_03240f928b_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 17 Constructions\" width=\"270\" height=\"151\" \/><br \/>AB = 77cm, BC = 26 cm, CD 60cm DA = 25 cm<br \/>Through, C, draw CE || DA meeting AB at E<br \/>\u2234 AE = CD = 60 cm and EB = 77 \u2013 60 = 17 cm,<br \/>CE = DA = 25 cm<br \/>Now area of \u2206BCE, with sides 17 cm, 26 cm, 25 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1914\/44921630074_1cac36d77b_o.png\" alt=\"RD Sharma Class 9 Book Chapter 17 Constructions\" width=\"316\" height=\"131\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1968\/30705642057_f09ec62fbc_o.png\" alt=\"Constructions With Solutions PDF RD Sharma Class 9 Solutions\" width=\"347\" height=\"427\" \/><\/p>\n<p>Question 13.<br \/>Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9cm, CD = 12cm, \u2220ACB = 90\u00b0 and AC = 15cm.<br \/>Solution:<br \/>In right \u0394ABC, \u2220ACB = 90\u00b0<br \/>AB\u00b2 = AC\u00b2 + BC\u00b2<br \/>(17)\u00b2 = (15)\u00b2+ BC\u00b2 = 289 = 225 + BC\u00b2<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1965\/30705641837_19d128a8bb_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 17 Constructions\" width=\"310\" height=\"340\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1957\/44921629414_a08dde7d01_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 17 Constructions\" width=\"329\" height=\"382\" \/><\/p>\n<p>Question 14.<br \/>A hand fan is made by stitching 10 equal size triangular strips of two different types of paper as shown figure. The dimensions of equal strips are 25 cm, 25 cm and 14 cm. Find the area of each types of paper needed to make the hand fan.<br \/>Solution:<br \/>In the figure, a hand fan has 5 isosceles and triangle. With sides 25 cm, 25 cm and 14 cm each.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1923\/30705641567_e5340954a6_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 17 Constructions\" width=\"334\" height=\"353\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1941\/44921628764_9b92d752cc_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 17 Constructions\" width=\"346\" height=\"86\" \/><\/p>\n<h3><span class=\"ez-toc-section\" id=\"class-9-rd-sharma-solutions-chapter-17-constructions-vsaqs\"><\/span>Class 9 RD Sharma Solutions Chapter 17 Constructions VSAQS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>Find the area of a triangle whose base and altitude are 5 cm and 4 cm respectively.<br \/>Solution:<br \/>In \u2206ABC,<br \/>Base BC = 5cm<br \/>Altitude AD = 4cm<br \/><img class=\"alignnone\" src=\"https:\/\/farm2.staticflickr.com\/1962\/44732411265_19fa274ced_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 17 Constructions vsaqs\" width=\"291\" height=\"281\" \/><\/p>\n<p>Question 2.<br \/>Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm respectively.<br \/>Solution:<br \/>Sides of triangle are 3 cm, 4cm and 5cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1958\/44732411135_d252bc2b4c_o.png\" alt=\"Class 9 Maths Chapter 17 Constructions RD Sharma Solutions\" width=\"321\" height=\"305\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1903\/44732410915_15bbaca4f9_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 17 Constructions\" width=\"271\" height=\"39\" \/><\/p>\n<p>Question 3.<br \/>Find the area of an isosceles triangle having the base x cm and one side y cm.<br \/>Solution:<br \/>In isosceles \u2206ABC,<br \/>AB = AC = y cm<br \/>BC = x cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1950\/45596109542_0374536e05_o.png\" alt=\"RD Sharma Class 9 Book Chapter 17 Constructions\" width=\"342\" height=\"510\" \/><\/p>\n<p>Question 4.<br \/>Find the area of an equilateral triangle having each side 4 cm.<br \/>Solution:<br \/>Each side of equilateral triangle (a) = 4cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1917\/44732410735_038002a474_o.png\" alt=\"Constructions With Solutions PDF RD Sharma Class 9 Solutions\" width=\"302\" height=\"156\" \/><\/p>\n<p>Question 5.<br \/>Find the area of an equilateral triangle having each side x cm.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1945\/44732410445_339ae1cc05_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 17 Constructions\" width=\"345\" height=\"154\" \/><\/p>\n<p>Question 6.<br \/>The perimeter of a triangular field is 144 m and the ratio of the sides is 3 : 4 : 5. Find the area of the field.<br \/>Solution:<br \/>Perimeter of the field = 144 m<br \/>Ratio in the sides = 3:4:5<br \/>Sum of ratios = 3 + 4 + 5 = 12<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1903\/45596109022_a6ce722fb0_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 17 Constructions\" width=\"314\" height=\"434\" \/><\/p>\n<p>Question 7.<br \/>Find the area of an equilateral triangle having altitude h cm.<br \/>Solution:<br \/>Altitude of an equilateral triangle = h<br \/>Let side of equilateral triangle = x<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1968\/45596108702_3baa0e7ba3_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 17 Constructions\" width=\"255\" height=\"123\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1970\/45596108562_d3d7d4d842_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 17 Constructions\" width=\"279\" height=\"131\" \/><\/p>\n<p>Question 8.<br \/>Let \u2206 be the area of a triangle. Find the area of a triangle whose each side is twice the side of the given triangle.<br \/>Solution:<br \/>Let a, b, c be the sides of the original triangle<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1979\/44732409795_b7294bdd01_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 17 Constructions\" width=\"350\" height=\"369\" \/><br \/>Hence area of new triangle = 4 x area of original triangle.<\/p>\n<p>Question 9.<br \/>If each side of a triangle is doubled, then find percentage increase in its area.<br \/>Solution:<br \/>Sides of original triangle be a, b, c<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1945\/44732409585_51aef4013a_o.png\" alt=\"RD Sharma Class 9 Chapter 17 Constructions\" width=\"346\" height=\"220\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1970\/44732409385_3829763867_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 17 Constructions\" width=\"379\" height=\"244\" \/><\/p>\n<p>Question 10.<br \/>If each side of an equilateral triangle is tripled then what is the percentage increase in the area of the triangle?<br \/>Solution:<br \/>Let the sides of the original triangle be a, b, c and area \u2206, then<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1929\/45596108102_83261d16e0_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 17 Constructions\" width=\"350\" height=\"436\" \/><\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-17-constructions-mcqs\"><\/span>RD Sharma Solutions Class 9 Chapter 17 Constructions MCQS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Mark the correct alternative in each of the following:<br \/>Question 1.<br \/>The sides of a triangle are 16 cm, 30 cm, 34 cm. Its area is<br \/>(a) 225 cm\u00b2<br \/>(b) 225<span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-26\" class=\"math\"><span id=\"MathJax-Span-27\" class=\"mrow\"><span id=\"MathJax-Span-28\" class=\"msqrt\"><span id=\"MathJax-Span-29\" class=\"mrow\"><span id=\"MathJax-Span-30\" class=\"mn\">3<\/span><\/span>\u2013\u221a<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<br \/>(c) 225<span id=\"MathJax-Element-7-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-31\" class=\"math\"><span id=\"MathJax-Span-32\" class=\"mrow\"><span id=\"MathJax-Span-33\" class=\"msqrt\"><span id=\"MathJax-Span-34\" class=\"mrow\"><span id=\"MathJax-Span-35\" class=\"mn\">2<\/span><\/span>\u2013\u221a<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<br \/>(d) 240 cm\u00b2<br \/>Solution:<br \/>Sides of triangle and 16 cm, 30 cm, 34 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1918\/30705654197_b26e42336e_o.png\" alt=\"Class 9 Maths Chapter 17 Constructions RD Sharma Solutions\" width=\"351\" height=\"231\" \/><\/p>\n<p>Question 2.<br \/>The base of an isosceles right triangle is 30 cm. Its area is<br \/>(a) 225 cm\u00b2<br \/>(b) 225<span id=\"MathJax-Element-8-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-36\" class=\"math\"><span id=\"MathJax-Span-37\" class=\"mrow\"><span id=\"MathJax-Span-38\" class=\"msqrt\"><span id=\"MathJax-Span-39\" class=\"mrow\"><span id=\"MathJax-Span-40\" class=\"mn\">3<\/span><\/span>\u2013\u221a<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<br \/>(c) 225<span id=\"MathJax-Element-9-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-41\" class=\"math\"><span id=\"MathJax-Span-42\" class=\"mrow\"><span id=\"MathJax-Span-43\" class=\"msqrt\"><span id=\"MathJax-Span-44\" class=\"mrow\"><span id=\"MathJax-Span-45\" class=\"mn\">2<\/span><\/span>\u2013\u221a<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<br \/>(d) 450 cm\u00b2<br \/>Solution:<br \/>Base of isosceles triangle \u2206ABC = 30cm<br \/>Let each of equal sides = x<br \/>Then AB = AC = x<br \/>Now in right \u2206ABC,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1921\/44732413965_315f781554_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 17 Constructions\" width=\"343\" height=\"461\" \/><\/p>\n<p>Question 3.<br \/>The sides of a triangle are 7cm, 9cm and 14cm. Its area is<br \/>(a) 12<span id=\"MathJax-Element-10-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-46\" class=\"math\"><span id=\"MathJax-Span-47\" class=\"mrow\"><span id=\"MathJax-Span-48\" class=\"msqrt\"><span id=\"MathJax-Span-49\" class=\"mrow\"><span id=\"MathJax-Span-50\" class=\"mn\">5<\/span><\/span>\u2013\u221a<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<br \/>(b) 12<span id=\"MathJax-Element-11-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-51\" class=\"math\"><span id=\"MathJax-Span-52\" class=\"mrow\"><span id=\"MathJax-Span-53\" class=\"msqrt\"><span id=\"MathJax-Span-54\" class=\"mrow\"><span id=\"MathJax-Span-55\" class=\"mn\">3<\/span><\/span>\u2013\u221a<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<br \/>(c) 24\u00a0<span id=\"MathJax-Element-12-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-56\" class=\"math\"><span id=\"MathJax-Span-57\" class=\"mrow\"><span id=\"MathJax-Span-58\" class=\"msqrt\"><span id=\"MathJax-Span-59\" class=\"mrow\"><span id=\"MathJax-Span-60\" class=\"mn\">5<\/span><\/span>\u2013\u221a<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<br \/>(d) 63 cm\u00b2<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1979\/30705653967_41db44e0a5_o.png\" alt=\"RD Sharma Class 9 Book Chapter 17 Constructions\" width=\"353\" height=\"261\" \/><\/p>\n<p>Question 4.<br \/>The sides of a triangular field are 325 m, 300 m and 125 m. Its area is<br \/>(a) 18750 m\u00b2<br \/>(b) 37500 m\u00b2<br \/>(c) 97500 m\u00b2<br \/>(d) 48750 m\u00b2<br \/>Solution:<br \/>Sides of a triangular field are 325m, 300m, 125m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1952\/44732413705_c86f23148a_o.png\" alt=\"Constructions With Solutions PDF RD Sharma Class 9 Solutions\" width=\"344\" height=\"330\" \/><\/p>\n<p>Question 5.<br \/>The sides of a triangle are 50 cm, 78 cm and 112 cm. The smallest altitude is<br \/>(a) 20 cm<br \/>(b) 30 cm<br \/>(c) 40 cm<br \/>(d) 50 cm<br \/>Solution:<br \/>The sides of a triangle are 50 cm, 78 cm, 112cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1923\/30705653647_f306603ba1_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 17 Constructions\" width=\"338\" height=\"407\" \/><\/p>\n<p>Question 6.<br \/>The sides of a triangle are 11m, 60m and 61m. Altitude to the smallest side is<br \/>(a) 11m<br \/>(b) 66 m<br \/>(c) 50 m<br \/>(d) 60 m<br \/>Solution:<br \/>Sides of a triangle are 11m, 60m and 61m<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1939\/44732413375_089150068e_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 17 Constructions\" width=\"349\" height=\"280\" \/><\/p>\n<p>Question 7.<br \/>The sides of a triangle are 11 cm, 15 cm and 16 cm. The altitude to the largest side is<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1960\/30705653217_128308869b_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 17 Constructions\" width=\"297\" height=\"60\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1905\/44732413195_d1ec67f318_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 17 Constructions\" width=\"270\" height=\"53\" \/><br \/>Solution:<br \/>Sides of a triangle are 11 cm, 15 cm, 16 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1914\/45596112282_680543b4b6_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 17 Constructions\" width=\"344\" height=\"392\" \/><\/p>\n<p>Question 8.<br \/>The base and hypotenuse of a right triangle are respectively 5cm and 13cm long. Its area is<br \/>(a) 25 cm\u00b2<br \/>(b) 28 cm\u00b2<br \/>(c) 30 cm\u00b2<br \/>(d) 40 cm\u00b2<br \/>Solution:<br \/>In a right triangle base = 5 cm<br \/>base hypotenuse = 13 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1958\/44732413075_986669209c_o.png\" alt=\"RD Sharma Class 9 Chapter 17 Constructions\" width=\"350\" height=\"266\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1926\/45596111892_8443d6f68f_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 17 Constructions\" width=\"327\" height=\"107\" \/><\/p>\n<p>Question 9.<br \/>The length of each side of an equilateral triangle of area 4\u00a0<span id=\"MathJax-Element-13-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-61\" class=\"math\"><span id=\"MathJax-Span-62\" class=\"mrow\"><span id=\"MathJax-Span-63\" class=\"msqrt\"><span id=\"MathJax-Span-64\" class=\"mrow\"><span id=\"MathJax-Span-65\" class=\"mn\">3<\/span><\/span>\u2013\u221a<\/span><\/span><\/span><\/span>\u00a0cm\u00b2, is<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1979\/44732412825_9ff276f681_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 17 Constructions\" width=\"270\" height=\"122\" \/><br \/>Solution:<br \/>Area of an equilateral triangle = 4<span id=\"MathJax-Element-14-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-66\" class=\"math\"><span id=\"MathJax-Span-67\" class=\"mrow\"><span id=\"MathJax-Span-68\" class=\"msqrt\"><span id=\"MathJax-Span-69\" class=\"mrow\"><span id=\"MathJax-Span-70\" class=\"mn\">3<\/span><\/span>\u2013\u221a<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<br \/>Let each side be = a<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1906\/44732412735_f159d9137d_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 17 Constructions\" width=\"352\" height=\"176\" \/><\/p>\n<p>Question 10.<br \/>If the area of an isosceles right triangle is 8cm, what is the perimeter of the triangle.<br \/>(a) 8 +\u00a0<span id=\"MathJax-Element-15-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-71\" class=\"math\"><span id=\"MathJax-Span-72\" class=\"mrow\"><span id=\"MathJax-Span-73\" class=\"msqrt\"><span id=\"MathJax-Span-74\" class=\"mrow\"><span id=\"MathJax-Span-75\" class=\"mn\">2<\/span><\/span>\u2013\u221a<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<br \/>(b) 8 + 4<span id=\"MathJax-Element-16-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-76\" class=\"math\"><span id=\"MathJax-Span-77\" class=\"mrow\"><span id=\"MathJax-Span-78\" class=\"msqrt\"><span id=\"MathJax-Span-79\" class=\"mrow\"><span id=\"MathJax-Span-80\" class=\"mn\">2<\/span><\/span>\u2013\u221a<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<br \/>(c) 4 + 8<span id=\"MathJax-Element-17-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-81\" class=\"math\"><span id=\"MathJax-Span-82\" class=\"mrow\"><span id=\"MathJax-Span-83\" class=\"msqrt\"><span id=\"MathJax-Span-84\" class=\"mrow\"><span id=\"MathJax-Span-85\" class=\"mn\">2<\/span><\/span>\u2013\u221a<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<br \/>(b) 12<span id=\"MathJax-Element-18-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-86\" class=\"math\"><span id=\"MathJax-Span-87\" class=\"mrow\"><span id=\"MathJax-Span-88\" class=\"msqrt\"><span id=\"MathJax-Span-89\" class=\"mrow\"><span id=\"MathJax-Span-90\" class=\"mn\">2<\/span><\/span>\u2013\u221a<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<br \/>Solution:<br \/>Let base = x<br \/>ABC an isosceles right triangle, which has 2 sides same<br \/>\u21d2 Height = x<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1926\/45596111632_def71f5ef3_o.png\" alt=\"Constructions Class 9 RD Sharma Solutions\" width=\"197\" height=\"224\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1946\/44732412545_a6f9410c42_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 17 Constructions\" width=\"355\" height=\"514\" \/><\/p>\n<p>Question 11.<br \/>The length of the sides of \u2206ABC are consecutive integers. If \u2206ABC has the same perimeter as an equilateral triangle with a side of length 9cm, what is the length of the shortest side of \u2206ABC?<br \/>(a) 4<br \/>(b) 6<br \/>(c) 8<br \/>(d) 10<br \/>Solution:<br \/>Side of an equilateral triangle = 9 cm<br \/>Its perimeter = 3 x 9 = 27 cm<br \/>Now perimeter of \u2206ABC = 27 cm<br \/>and let its sides be x, x + 1, x +2<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1935\/45596111382_89b5350ae8_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 17 Constructions\" width=\"353\" height=\"132\" \/><\/p>\n<p>Question 12.<br \/>In the figure, the ratio of AD to DC is 3 to 2. If the area of \u2206ABC is 40cm2, what is the area of \u2206BDC?<br \/>(a) 16 cm\u00b2<br \/>(b) 24 cm\u00b2<br \/>(c) 30 cm\u00b2<br \/>(d) 36 cm\u00b2<br \/>Solution:<br \/>Ratio in AD : DC = 3:2<br \/>and area \u2206ABC = 40 cm\u00b2<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1911\/44732412335_31f1fdf453_o.png\" alt=\"Class 9 Maths Chapter 17 Constructions RD Sharma Solutions\" width=\"351\" height=\"682\" \/><\/p>\n<p>Question 13.<br \/>If the length of a median of an equilateral triangle is x cm, then its area is<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1958\/44732412125_fe28aab7f7_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 17 Constructions\" width=\"240\" height=\"127\" \/><br \/>Solution:<br \/>\u2235 The median of an equilateral triangle is the perpendicular to the base also,<br \/>\u2234 Let side of the triangle = a<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1912\/45596111042_bc643cc7dc_o.png\" alt=\"RD Sharma Class 9 Book Chapter 17 Constructions\" width=\"351\" height=\"439\" \/><\/p>\n<p>Question 14.<br \/>If every side of a triangle is doubled, then increase in the area of the triangle is<br \/>(a) 100<span id=\"MathJax-Element-19-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-91\" class=\"math\"><span id=\"MathJax-Span-92\" class=\"mrow\"><span id=\"MathJax-Span-93\" class=\"msqrt\"><span id=\"MathJax-Span-94\" class=\"mrow\"><span id=\"MathJax-Span-95\" class=\"mn\">2<\/span><\/span>\u2013\u221a<\/span><\/span><\/span><\/span>\u00a0%<br \/>(b) 200%<br \/>(c) 300%<br \/>(d) 400%<br \/>Solution:<br \/>Let the sides of the original triangle be a, b, c<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1975\/44732412025_7016bd61f0_o.png\" alt=\"Constructions With Solutions PDF RD Sharma Class 9 Solutions\" width=\"341\" height=\"312\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1922\/44732411865_53a0470ee5_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 17 Constructions\" width=\"344\" height=\"139\" \/><\/p>\n<p>Question 15.<br \/>A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 1272 cm, then area of the triangle is<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1968\/44732411775_0fd1c3d8e1_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 17 Constructions\" width=\"313\" height=\"69\" \/><br \/>Solution:<br \/>A square and an equilateral triangle have equal perimeter<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1932\/44732411675_2dd1dfcb53_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 17 Constructions\" width=\"358\" height=\"322\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"detailed-exercise-wise-explanation-with-listing-of-important-topics-in-the-exercise\"><\/span><strong>Detailed Exercise-wise Explanation with Listing of Important Topics in the Exercise<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\"><strong>RD Sharma class 9 chapter 17 exercise 17a<\/strong>: This exercise includes concepts related to construction &amp; also include problems based on it. In RD Sharma class 9 chapter 17 exercise 17a, the students will study how to draw a perpendicular bisector for a given line segment. These study material assist the students to obtain higher marks in the exam. The solutions for all questions are provided as per the CBSE syllabus.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\"><strong>RD Sharma class 9 chapter 17 exercise 17b:<\/strong> This exercise includes topics on the construction of the bisector of a given angle. Each question is prescribed as per the syllabus and guidelines of CBSE. RD Sharma class 9 chapter 17 exercise 17b enables the students to learn how to construct different angles by using ruler and compass only. The students can definitely acquire better marks in the exams by solving questions from RD Sharma textbook.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\"><strong>RD Sharma class 9 chapter 17 exercise 17c:<\/strong> This exercise includes word problems on constructions. The students can consider these detailed solutions to understand the in-depth topics of constructions of different kinds of triangles. By practicing the textbook questions with the help of RD Sharma solutions, a student&#8217;s problem-solving abilities are also enhanced.<\/span><\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"important-topics-from-class-9-maths-chapter-17-constructions\"><\/span>Important Topics from Class 9 Maths Chapter 17 Constructions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><span style=\"font-weight: 400;\">The students will study some important concepts on RD Sharma class 9 solutions chapter 17 constructions which are listed below:<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Introduction of Construction<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Construction of Bisector of a Line Segment<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Construction of the Bisector of a Given Angle<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Constructions of Triangles\u00a0<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">RD Sharma <a href=\"http:\/\/cbse.nic.in\" target=\"_blank\" rel=\"noopener noreferrer\">CBSE<\/a> class 9 solutions chapter 17 constructions are explained in detail &amp; in an easy language to foster quick knowledge among students. If you have any doubts regarding the Class 9 Maths exam, ask in the comments.<\/span><\/p>\n<h2><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-maths-chapter-17-constructions\"><\/span>RD Sharma Solutions Class 9 Maths Chapter 17 Constructions<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-1630756361375\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-17\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 17?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630756387769\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-17\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 17?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630756406065\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-17-pdf-offline\"><\/span>Can I access the RD Sharma Solutions for Class 9 Maths Chapter 17\u00a0PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline as well.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 9 Maths Chapter 17 Constructions: Each question is solved using figures for easy as well as better understanding. RD Sharma Solutions Class 9 Maths Chapter 17 builds a good foundation for students. They can secure excellent marks in the exams for answering more difficult questions accurately by practicing these solutions. All &#8230; <a title=\"RD Sharma Solutions Class 9 Maths Chapter 17 &#8211; Constructions (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-17-constructions\/\" aria-label=\"More on RD Sharma Solutions Class 9 Maths Chapter 17 &#8211; Constructions (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":124578,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2985,73411,73410],"tags":[3081,3037,3086],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63812"}],"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\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=63812"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63812\/revisions"}],"predecessor-version":[{"id":527214,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63812\/revisions\/527214"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/124578"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=63812"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=63812"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=63812"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}