{"id":1132,"date":"2024-02-12T09:30:27","date_gmt":"2024-02-12T04:00:27","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=1132"},"modified":"2024-02-12T17:09:05","modified_gmt":"2024-02-12T11:39:05","slug":"samacheer-kalvi-6th-maths-term-1-chapter-4-ex-4-4","status":"publish","type":"post","link":"https:\/\/samacheerguru.com\/samacheer-kalvi-6th-maths-term-1-chapter-4-ex-4-4\/","title":{"rendered":"Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.4"},"content":{"rendered":"<p>You can Download <a href=\"https:\/\/samacheerguru.com\/samacheer-kalvi-6th-maths-book-solutions\/\">Samacheer Kalvi 6th Maths Book Solutions<\/a> Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.<br \/>\n<img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Ex 4.4\" width=\"224\" height=\"19\" \/><\/p>\n<h2>Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.4<\/h2>\n<p><span style=\"color: #0000ff;\">Miscellaneous Practice Problems<\/span><\/p>\n<p>Question 1.<br \/>\nFind the type of lines marked in thick lines (Parallel, intersecting or perpendicular)<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22167\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q1.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q1\" width=\"528\" height=\"435\" srcset=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q1.png 528w, https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q1-300x247.png 300w\" sizes=\"(max-width: 528px) 100vw, 528px\" \/><br \/>\nSolution:<br \/>\n(i) Parallel lines<br \/>\n(ii) Parallel lines<br \/>\n(iii) Parallel lines and Perpendicular lines<br \/>\n(iv) Intersecting lines<\/p>\n<p>Question 2.<br \/>\nFind the parallel and intersecting line segments in the picture given below.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22168\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q2.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q2\" width=\"226\" height=\"225\" srcset=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q2.png 226w, https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q2-150x150.png 150w\" sizes=\"(max-width: 226px) 100vw, 226px\" \/><br \/>\nSolution:<br \/>\n(a) Parallel line segments<\/p>\n<ul>\n<li>\\(\\overline{\\mathrm{YZ}} \\text { and } \\overline{\\mathrm{DE}}\\)<\/li>\n<li>\\(\\overline{\\mathrm{EA}} \\text { and } \\overline{\\mathrm{ZV}}\\)<\/li>\n<li>\\(\\overline{\\mathrm{VW}} \\text { and } \\overline{\\mathrm{AB}}\\)<\/li>\n<li>\\(\\overline{\\mathrm{WX}} \\text { and } \\overline{\\mathrm{BC}}\\)<\/li>\n<li>\\(\\overline{\\mathrm{YX}} \\text { and } \\overline{\\mathrm{DC}}\\)<\/li>\n<li>\\(\\overline{\\mathrm{YD}} \\text { and } \\overline{\\mathrm{XC}}\\)<\/li>\n<li>\\(\\overline{\\mathrm{XC}} \\text { and } \\overline{\\mathrm{WB}}\\)<\/li>\n<li>\\(\\overline{\\mathrm{WB}} \\text { and } \\overline{\\mathrm{VA}}\\)<\/li>\n<li>\\(\\overline{\\mathrm{VA}} \\text { and } \\overline{\\mathrm{ZE}}\\)<\/li>\n<li>\\(\\overline{\\mathrm{ZE}} \\text { and } \\overline{\\mathrm{YD}}\\)<\/li>\n<\/ul>\n<p>(b) Intersecting line segments<\/p>\n<ul>\n<li>DE and ZV<\/li>\n<li>WX and DC<\/li>\n<\/ul>\n<p>Question 3.<br \/>\nName the following angles as shown in the figure.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22170\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q3.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q3\" width=\"193\" height=\"209\" \/><br \/>\nSolution:<br \/>\n(i) \u22201 = \u2220DBC or \u2220CBD<br \/>\n(ii) \u22202 = \u2220DBE or \u2220EBD<br \/>\n(iii) \u22203 = \u2220ABE or \u2220EBA<br \/>\n(iv) \u22201 + \u22202 = \u2220EBC or \u2220CBE<br \/>\n(v) \u22202 + \u22203 = \u2220ABD or \u2220DHA<br \/>\n(vi) \u22201 + \u22202 + \u22203 = \u2220ABC or \u2220B or \u2220CBA<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Ex 4.4\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 4.<br \/>\nMeasure the angles of the given figures using a protractor and identify the type of angle as acute, obtuse, right or straight.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22171\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q4.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q4\" width=\"605\" height=\"147\" srcset=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q4.png 605w, https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q4-300x73.png 300w\" sizes=\"(max-width: 605px) 100vw, 605px\" \/><br \/>\nSolution:<br \/>\n(i) 90\u00b0 &#8211; Right Angle<br \/>\n(ii) 45\u00b0 &#8211; Acute Angle<br \/>\n(iii) 180\u00b0 &#8211; Straight Angle<br \/>\n(iv) 105\u00b0 &#8211; Obtuse Angle<\/p>\n<p>Question 5.<br \/>\nDraw the following angles using the protractor.<br \/>\n(i) 45\u00b0<br \/>\n(ii) 120\u00b0<br \/>\n(iii) 65\u00b0<br \/>\n(iv) 135\u00b0<br \/>\n(v) 0\u00b0<br \/>\n(vi) 180\u00b0<br \/>\n(vii) 38\u00b0<br \/>\n(viii) 90\u00b0<br \/>\nSolution:<br \/>\n(i) 45\u00b0<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22173\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.2.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q5.2\" width=\"452\" height=\"191\" srcset=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.2.png 452w, https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.2-300x127.png 300w\" sizes=\"(max-width: 452px) 100vw, 452px\" \/><br \/>\nConstruction:<br \/>\n1. Drawn the base ray PQ.<br \/>\n2. Placed the centre of the protractor at the vertex P. Lined up the ray \\(\\overrightarrow{\\mathrm{PQ}}\\) with the 0\u00b0 line. Then drawn and labelled a pointed (R) at the 45\u00b0 mark on the inner scale (a) anticlockwise and (b) outer scale (clockwise)<br \/>\n3. Removed the protractor and drawn at \\(\\overrightarrow{\\mathrm{PR}}\\) to complete the angle<br \/>\nNow \u2220P = \u2220QPR &#8211; \u2220RPQ = 45\u00b0.<\/p>\n<p>(ii) 120\u00b0<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22174\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.1.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q5.1\" width=\"515\" height=\"157\" srcset=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.1.png 515w, https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.1-300x91.png 300w\" sizes=\"(max-width: 515px) 100vw, 515px\" \/><br \/>\nConstruction:<br \/>\n1. Placed the centre of the protractor at the vertex X. Lined up the ray \\(\\overline{\\mathrm{XY}}\\) with the 0\u00b0 Line. Then draw and label a point Z at 120\u00b0 mark on the (a) inner scale (anti-clockwise) and (b) outer scale (clockwise).<br \/>\n2. Removed the protractor and draw \\(\\overline{\\mathrm{XZ}}\\) to complete the angle.<br \/>\nNow, \u2220X = \u2220ZXY = \u2220YXZ = 120\u00b0.<\/p>\n<p>(iii) 65\u00b0<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22173\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.2.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q5.2\" width=\"452\" height=\"191\" srcset=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.2.png 452w, https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.2-300x127.png 300w\" sizes=\"(max-width: 452px) 100vw, 452px\" \/><br \/>\nConstruction:<br \/>\n1. Placed the centre of the protractor at the vertex A. Line up the ray \\(\\overrightarrow{\\mathrm{AB}}\\) with the 0\u00b0 line. Then draw and label a point C at the 65\u00b0 mark on the (a) inner scale (anti-clockwise) (b) outer scale (clockwise).<br \/>\n2. Removed the protractor and draw \\(\\overrightarrow{\\mathrm{AB}}\\) to complete the angle.<br \/>\nNow \u2220A = \u2220BAC = \u2220CAB = 65\u00b0.<\/p>\n<p>(iv) 135\u00b0<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22175\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.3.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q5.3\" width=\"509\" height=\"156\" srcset=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.3.png 509w, https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.3-300x92.png 300w\" sizes=\"(max-width: 509px) 100vw, 509px\" \/><br \/>\nConstruction:<br \/>\n1. Placed the centre of the protractor at the Vertex A. Lined up the ray \\(\\overrightarrow{\\mathrm{EG}}\\) with the 0\u00b0 line. Then draw and label a point F at the 135\u00b0 mark on the (a) inner scale (anti-clockwise) and (b) outer scale (clockwise)<br \/>\n2. Removed the protractor and draw \\(\\overrightarrow{\\mathrm{EF}}\\) to complete the angle.<br \/>\nNow \u2220E = \u2220FEG = \u2220GEF = 135\u00b0.<\/p>\n<p>(v) 0\u00b0<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22177\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.4.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q5.4\" width=\"502\" height=\"69\" srcset=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.4.png 502w, https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.4-300x41.png 300w\" sizes=\"(max-width: 502px) 100vw, 502px\" \/><br \/>\nConstruction:<br \/>\n1. Placed the centre of the protractor at the vertex G. Lined up the ray \\(\\overrightarrow{\\mathrm{GH}}\\) with the 0\u00b0 line. Then draw and label a point I at the 0\u00b0 mark on the<br \/>\n(a) inner scale (anti-clockwise)<br \/>\n(b) outer scale (clockwise)<br \/>\n2. Removed the protractor and seen \\(\\overrightarrow{\\mathrm{GI}}\\) lies exactly on \\(\\overrightarrow{\\mathrm{GH}}\\)<br \/>\nNow \u2220G = \u2220HGI = \u2220IGH = 0\u00b0, which is a zero angle.<\/p>\n<p>(vi) 180\u00b0<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22189\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.5.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q5.5\" width=\"498\" height=\"101\" srcset=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.5.png 498w, https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.5-300x61.png 300w\" sizes=\"(max-width: 498px) 100vw, 498px\" \/><br \/>\nConstruction:<br \/>\n1. Placed the centre of the protractor at the vertex I. Lined up the ray \\(\\overrightarrow{\\mathrm{IJ}}\\) with the 0\u00b0 line. Then draw and labelled a point K at the 180\u00b0 mark on the (a) inner scale (anticlockwise) (b) outer scale (clockwise)<br \/>\n2. Removed the protractor and draw \\(\\overrightarrow{\\mathrm{IK}}\\) to complete the angle.<br \/>\nNow \u2220I = \u2220JHK = \u2220KIJ = 180\u00b0, which is a straight Angle.<\/p>\n<p>(vii) 38\u00b0<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22197\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.6.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q5.6\" width=\"351\" height=\"150\" srcset=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.6.png 351w, https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.6-300x128.png 300w\" sizes=\"(max-width: 351px) 100vw, 351px\" \/><br \/>\nConstruction:<br \/>\n1. Placed the centre of the protractor at the vertex L. Lined up the ray \\(\\overrightarrow{\\mathrm{LM}}\\) with the 0\u00b0 line. Then draw and label a point N at 38\u00b0 mark on the (a) inner scale (anticlockwise) and (b)*huter scale (clockwise).<br \/>\n2. Removed the protractor and draw \\(\\overrightarrow{\\mathrm{LN}}\\) to complete the angle.<br \/>\nNow \u2220L = \u2220MLN = \u2220NLM = 38\u00b0.<\/p>\n<p>(viii) 90\u00b0<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22198\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.7.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q5.7\" width=\"425\" height=\"170\" srcset=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.7.png 425w, https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q5.7-300x120.png 300w\" sizes=\"(max-width: 425px) 100vw, 425px\" \/><br \/>\nConstruction:<br \/>\n1. Placed the centre of the protractor at the vertex \u2018O\u2019. Lined up the ray \\(\\overrightarrow{\\mathrm{OP}}\\) with the 0\u00b0 line. Then draw and label a point Q at 90\u00b0 mark on the (a) inner scale (anticlockwise) and (b) outer scale (clockwise)<br \/>\n2. Removed the protractor and draw \\(\\overrightarrow{\\mathrm{OQ}}\\) to complete the angle.<br \/>\nNow \u2220O = \u2220POQ = \u2220QOP = 90\u00b0.<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Ex 4.4\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 6.<br \/>\nFrom the figures given below, classify the following pairs of angles into complementary and non-complementary.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22199\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q6.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q6\" width=\"462\" height=\"358\" srcset=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q6.png 462w, https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q6-300x232.png 300w\" sizes=\"(max-width: 462px) 100vw, 462px\" \/><br \/>\nSolution:<br \/>\nWe know that the two angles are complementary if they add up to 90\u00b0.<br \/>\nTherefore (a) (i) is complementary.<br \/>\nIn (v) \u2220ABC and \u2220CBD are complementary<br \/>\n(b) (ii), (iii), (iv) and (v) are non-complementary<\/p>\n<p>Question 7.<br \/>\nFrom the figures given below, classify the following pairs of angles into supplementary and non-supplementary.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22200\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q7.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q7\" width=\"627\" height=\"220\" srcset=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q7.png 627w, https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q7-300x105.png 300w\" sizes=\"(max-width: 627px) 100vw, 627px\" \/><br \/>\nSolution:<br \/>\nIf two angles add up to 180\u00b0, then they are supplementary angles.<br \/>\n(a) In (ii) \u2220AOB and \u2220BOD are supplementary. In (iv) the pair is supplementary<br \/>\n(b) (i) and (iii) are not supplementary.<\/p>\n<p>Question 8.<br \/>\nFrom the figure.<br \/>\n(i) name a pair of complementary angles<br \/>\n(ii) name a pair of supplementary angles<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22201\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q8.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q8\" width=\"265\" height=\"192\" \/><br \/>\nSolution:<br \/>\n(i) \u2220FAE and \u2220DAE are complementary<br \/>\n(ii) \u2220FAD and \u2220DAC are supplementary<\/p>\n<p>Question 9.<br \/>\nFind the complementary angle of<br \/>\n(i) 30\u00b0<br \/>\n(ii) 26\u00b0<br \/>\n(iii) 85\u00b0<br \/>\n(iv) 0\u00b0<br \/>\nSolution:<br \/>\nWhen we have an angle, how far we need to go to reach the right angle is called the complementary angle.<br \/>\n(i) Complementary angle of 30\u00b0 is 90\u00b0 &#8211; 30\u00b0 = 60\u00b0<br \/>\n(ii) Complementary angle of 26\u00b0 is 90\u00b0 &#8211; 26\u00b0 = 64\u00b0<br \/>\n(iii) Complementary angle of 85\u00b0 is 90\u00b0 &#8211; 85\u00b0 = 5\u00b0<br \/>\n(iv) Complementary angle of 0\u00b0 is 90\u00b0 &#8211; 0\u00b0 = 90\u00b0<br \/>\n(v) Complementary angle of 90\u00b0 is 90\u00b0 &#8211; 90\u00b0 = 0\u00b0<\/p>\n<p>Question 10.<br \/>\nFind the supplementary angle of<br \/>\n(i) 70\u00b0<br \/>\n(ii) 35\u00b0<br \/>\n(iii) 165\u00b0<br \/>\n(iv) 90\u00b0<br \/>\n(v) 0\u00b0<br \/>\n(vi) 180\u00b0<br \/>\n(vii) 95\u00b0<br \/>\nSolution:<br \/>\nHow far we should go in the same direction to reach the straight angle (180\u00b0) is called supplementary angle.<br \/>\n(i) Supplementary angle of 70\u00b0 = 180\u00b0 &#8211; 70\u00b0 = 110\u00b0<br \/>\n(ii) Supplementary angle of 35\u00b0 is 180\u00b0 &#8211; 35\u00b0 = 145\u00b0<br \/>\n(iii) Supplementary angle of 165\u00b0 is 180\u00b0 &#8211; 165\u00b0 = 15\u00b0<br \/>\n(iv) Supplementary angle of 90\u00b0 is 180\u00b0 &#8211; 90\u00b0 = 90\u00b0<br \/>\n(v) Supplementary angle of 0\u00b0 is 180\u00b0 &#8211; 0\u00b0 = 180\u00b0<br \/>\n(vi) Supplementary angle of 180\u00b0 is 180\u00b0 &#8211; 180\u00b0 = 0\u00b0<br \/>\n(vii) Supplementary angle of 95\u00b0 is 180\u00b0 &#8211; 95\u00b0 = 85\u00b0<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Ex 4.4\" width=\"224\" height=\"19\" \/><\/p>\n<p><span style=\"color: #0000ff;\">Challenging Problems<\/span><\/p>\n<p>Question 11.<br \/>\nThink and write and object having.<br \/>\n(i) Parallel Lines<br \/>\n1. _____________<br \/>\n2. _____________<br \/>\n3. _____________<br \/>\n(ii) Perpendicular lines<br \/>\n1. _____________<br \/>\n2. _____________<br \/>\n3. _____________<br \/>\n(iii) Intersecting lines<br \/>\n1. _____________<br \/>\n2. _____________<br \/>\n3. _____________<br \/>\nSolution:<br \/>\n(i) 1. Opposite edges of a Table.<br \/>\n2. Path traced by the wheels of a car on a straight road<br \/>\n3. Opposite edges of a black board<br \/>\n(ii) 1. Adjacent edges of a Table.<br \/>\n2. Hands of the block when it shows 3.30<br \/>\n3. Strokes of the letter \u2018L\u2019<br \/>\n(iii) 1. Sides of a triangle<br \/>\n2. Strokes of letter \u2018V\u2019<br \/>\n3. Hands of a scissors<\/p>\n<p>Question 12.<br \/>\nWhich angle is equal to twice of its complement?<br \/>\nSolution:<br \/>\nWe know that the sum of complementary angles 90\u00b0<br \/>\nGiven Angle = 2 \u00d7 Complementary angle<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22203\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q12.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q12\" width=\"591\" height=\"184\" srcset=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q12.png 591w, https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q12-300x93.png 300w\" sizes=\"(max-width: 591px) 100vw, 591px\" \/><br \/>\nBy trial and error, we find that Angle = 2 \u00d7 Complement for 60\u00b0<br \/>\nThe required angle = 60\u00b0<br \/>\nAnother method:<br \/>\nLet the angle be x given<br \/>\nx = 2 (90 &#8211; x)<br \/>\n\u21d2 x = 180 &#8211; 2x<br \/>\n\u21d2 x + 2x = 180<br \/>\n\u21d2 3x = 180<br \/>\n\u21d2 x = 60\u00b0<\/p>\n<p>Question 13.<br \/>\nWhich angle is equal to two-thirds of its supplement.<br \/>\nSolution:<br \/>\nSupplementary angles sum upto 180\u00b0<br \/>\nGiven Angle = \\(\\frac{2}{3}\\) \u00d7 Supplement.<br \/>\nForming the Table.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22204\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q13.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q13\" width=\"589\" height=\"278\" srcset=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q13.png 589w, https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q13-300x142.png 300w\" sizes=\"(max-width: 589px) 100vw, 589px\" \/><br \/>\nBy trial and error, we find that angle = \\(\\frac{2}{3}\\) \u00d7 supplement for 72\u00b0.<br \/>\nThe required angle 72\u00b0.<\/p>\n<p>Question 14.<br \/>\nGiven two angles are supplementary and one angle is 20\u00b0 more than other. Find the two angles.<br \/>\nSolution:<br \/>\nGiven two angles are supplementary i.e. their sum = 180\u00b0.<br \/>\nLet the angle be x<br \/>\nThen another angle = x + 20 (given)<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-22205\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q14.png\" alt=\"Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q14\" width=\"529\" height=\"124\" srcset=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q14.png 529w, https:\/\/samacheerguru.com\/wp-content\/uploads\/2020\/06\/Samacheer-Kalvi-6th-Maths-Term-1-Chapter-4-Geometry-Ex-4.4-Q14-300x70.png 300w\" sizes=\"(max-width: 529px) 100vw, 529px\" \/><br \/>\nThe two angles are 80\u00b0 and 100\u00b0.<\/p>\n<p>Question 15.<br \/>\nTwo complementary angles are in the ratio 7 : 2. Find the angles.<br \/>\nSolution:<br \/>\nLet the angles be 7x, 2x<br \/>\nAccording to the problem,<br \/>\n7x + 2x = 90<br \/>\n9x = 90<br \/>\nx = \\(\\frac{90}{9}\\)<br \/>\nx = 10<br \/>\n7x = 7 \u00d7 10<br \/>\n= 70<br \/>\n2x = 2 \u00d7 10<br \/>\n= 20<br \/>\n\u2234 Two angles are 70\u00b0 and 20\u00b0<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/samacheerguru.com\/wp-content\/uploads\/2019\/12\/SamacheerKalvi.Guru_.png\" alt=\"Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Ex 4.4\" width=\"224\" height=\"19\" \/><\/p>\n<p>Question 16.<br \/>\nTwo supplementary angles are in ratio 5 : 4. Find the angles.<br \/>\nSolution:<br \/>\nTotal of two supplementary angles = 180\u00b0<br \/>\nGiven they are in the ratio 5 : 4<br \/>\nDividing total angles to 5 + 4 = 9 equal parts.<br \/>\nOne angle \\(=\\frac{5}{9} \\times 180=100^{\\circ}\\)<br \/>\nAnother angle \\(=\\frac{4}{9} \\times 180=80^{\\circ}\\)<br \/>\nTwo angles are 100\u00b0 and 80\u00b0.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.4 Miscellaneous Practice Problems Question 1. Find the type of lines marked in thick lines [&hellip;]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[6],"tags":[],"blocksy_meta":"","jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/1132"}],"collection":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/comments?post=1132"}],"version-history":[{"count":1,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/1132\/revisions"}],"predecessor-version":[{"id":55700,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/1132\/revisions\/55700"}],"wp:attachment":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/media?parent=1132"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/categories?post=1132"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/tags?post=1132"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}