{"id":19476,"date":"2018-01-30T11:46:46","date_gmt":"2018-01-30T11:46:46","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=19476"},"modified":"2020-11-25T14:43:10","modified_gmt":"2020-11-25T09:13:10","slug":"frank-icse-solutions-class-9-maths-constructions-quadrilaterals","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/frank-icse-solutions-class-9-maths-constructions-quadrilaterals\/","title":{"rendered":"Frank ICSE Solutions for Class 9 Maths – Constructions of Quadrilaterals"},"content":{"rendered":"
Ex No: 20.1<\/strong><\/span> Solution 2:<\/strong><\/span> Solution 3:<\/strong><\/span> Solution 4:<\/strong><\/span> Solution 5:<\/strong><\/span> Solution 6:<\/strong><\/span> Solution 7:<\/strong><\/span> Solution 8:<\/strong><\/span> Solution 9:<\/strong><\/span> Solution 10:<\/strong><\/span> Solution 11:<\/strong><\/span> Solution 12:<\/strong><\/span> Solution 13:<\/strong><\/span> Solution 14(a):<\/strong><\/span> Solution 14(b):<\/strong><\/span> Solution 15:<\/strong><\/span> Solution 16:<\/strong><\/span> Solution 17:<\/strong><\/span> Solution 18(a):<\/strong><\/span> Solution 18(b):<\/strong><\/span> Solution 19:<\/strong><\/span> Solution 20:<\/strong><\/span> Maths<\/a>Physics<\/a>Chemistry<\/a>Biology<\/a><\/p>\n","protected":false},"excerpt":{"rendered":" Frank ICSE Solutions for Class 9 Maths – Constructions of Quadrilaterals Ex No: 20.1 Solution 1: Solution 2: Solution 3: Solution 4: Solution 5: Solution 6: Solution 7: Solution 8: Solution 9: Solution 10: Solution 11: Solution 12: Solution 13: Solution 14(a): Steps of construction: Draw AD = 3.2 cm Draw \u2220XAD = 90\u00b0. With […]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[6442,5],"tags":[],"yoast_head":"\n
\nSolution 1:<\/strong><\/span>
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\nSteps of construction:<\/strong>
\nDraw AD = 3.2 cm
\nDraw \u2220XAD = 90\u00b0.
\nWith D as centre and radius BD = 5.5 cm, draw an arc to cut AX at point B.
\nJoin BD.
\nWith B as centre and radius 3.2 cm draw an arc and with D as centre and radius = AB, draw another arc to cut the previous arc at C.
\nJoin BC and CD.
\nThus, ABCD is the required rectangle.
\nCD = 4.5 cm
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\nSteps of construction:<\/strong>
\nDraw BC = 6.2 cm
\nThrough B, draw BP such that \u2220B = 90\u00b0
\nFrom BP, cut BA = 5 cm
\nWith A and C as centres and radii 6.2 cm and 5 cm respectively, draw arcs cutting each other at D.
\nJoin AD and CD.
\nThus, ABCD is the required triangle.
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\nSince area of rectangle = 21 cm2<\/sup>
\nAnd, length = 4.2 cm
\nBreadth = Area \u00f7 Length = 21 \u00f7 4.2 = 5 cm
\nSteps of construction:<\/strong>
\nDraw BC = 5 cm
\nThrough B, draw BP such that \u2220B = 90\u00b0
\nFrom BP, cut BA = 4.2 cm
\nWith A and C as centres and radii 5 cm and 4.2 cm respectively, draw arcs cutting each other at D.
\nJoin AD and CD.
\nThus, ABCD is the required triangle.
\n<\/p>\n
\nSince area of rectangle = 33.8 cm2<\/sup>
\nAnd, breadth = 6.5 cm
\nLength = Area \u00f7 Breadth = 33.8 \u00f7 6.5 = 5.2 cm
\nSteps of construction:<\/strong>
\nDraw BC = 6.5 cm
\nThrough B, draw BP such that \u2220B = 90\u00b0
\nFrom BP, cut BA = 5.2 cm
\nWith A and C as centres and radii 6.5 cm and 5.2 cm respectively, draw arcs cutting each other at D.
\nJoin AD and CD.
\nThus, ABCD is the required triangle.
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