{"id":31427,"date":"2018-08-28T04:55:26","date_gmt":"2018-08-28T04:55:26","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=31427"},"modified":"2021-02-12T12:52:52","modified_gmt":"2021-02-12T07:22:52","slug":"ml-aggarwal-icse-solutions-for-class-6-maths-chapter-11","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-icse-solutions-for-class-6-maths-chapter-11\/","title":{"rendered":"ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes"},"content":{"rendered":"

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes<\/strong><\/span><\/h2>\n

ML Aggarwal Solutions<\/a>ICSE Solutions<\/a>Selina ICSE Solutions<\/a><\/p>\n

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.1<\/strong><\/span><\/h3>\n

Solution 01:<\/strong><\/span>
\nIt gives accurate measurement and avoids error due to thickness of ruler or positioning of eye (due to angular viewing)
\n\"ML
\nSolution 02:<\/span><\/strong>
\nBy measuring the lengths of the given figure
\n\"ML
\n(i)<\/strong> AB = CD
\n(ii)<\/strong> BC < AB
\n(iii)<\/strong> AC = BD
\n(iv)<\/strong> CD < BD<\/p>\n

Solution 03:<\/span><\/strong>
\nGiven that AC = 10 CM, AB = 6 CM and BC = 4 CM
\nBy constructing line segment by the given data, the model drawn as below.
\n\"ML
\nPoint B lies in between A and C<\/p>\n

Solution 04:<\/strong><\/span>
\n\"ML
\nBy measuring the Lengths of line segments in the above figure
\nAB = 3 CM
\nBC = 1.5 CM
\n(i)<\/strong> It can be observed that AC = AB + BC [i.e. 4.5 CM = 3 CM + 1.5 CM]
\n(ii)<\/strong> AC \u2013 BC = AB 4.5 CM \u2013 1.5 CM = 3 CM by measurement fount that AB = 3 CM, so AC \u2013 BC = AB.<\/p>\n

Solution 05:<\/strong><\/span>
\nBy measuring the lengths of the given figure.
\n\"ML
\nGiven data<\/strong>
\nAB = 1.9 CM
\nBC = 0.7 CM
\nCD = 1.9 CM
\nAD = 4.5 CM
\n(i)<\/strong> AC + BD = 2.6 CM + 2.6 CM = 5.2 CM AD + BC = 4.5 CM + 0.7 CM = 5.2 CM Hence, AC + BD = AD +BC.
\n(ii)<\/strong> AB + CD = 1.9 CM + 1.9 CM = 3.8 CM AD – BC = 4.5 CM – 0.7 CM = 3.8 CM Hence, AC + BD = AD +BC.<\/p>\n

Solution 06:<\/span><\/strong>
\nBy measuring the lengths of the given triangle ABC as below
\n\"ML
\nAB = 2.6 CM, AC = 3.8 CM and BC = 3.8 CM.
\n(i)<\/strong> AB + BC = 2.6 CM + 3.8 CM = 6.4 CM AC = 3.8 CM Hence, AB + BC > AC.
\n(ii)<\/strong> BC + AC = 3.8 CM + 3.8 CM = 7.6 CM AB = 2.6 CM Hence, AB > BC + AC.
\n(iii)<\/strong> AC + AB = 3.8 CM + 2.6 CM = 5.4 CM BC = 3.8 CM Hence, AC + AB > BC.<\/p>\n

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.2<\/strong><\/span><\/h3>\n

Solution 01:<\/strong><\/span>
\n(i)<\/strong> When the hour hand moves from 4 to 10 clockwise, fraction of revolution turned = \u00bd Number of right angles turned = 2.
\n\"ML
\n(ii)<\/strong> When the hour hand moves from 2 to 5 clockwise, fraction of revolution turned = \u00bc. Number of right angles turned = 1.
\n\"ML
\n(iii)<\/strong> When the hour hand moves from 7 to 10 clockwise, fraction of revolution turned = \u00bc. Number of right angles turned = 1.
\n\"ML
\n(iv)<\/strong> When the hour hand moves from 8 to 5 clockwise, fraction of revolution turned = \u00be. Number of right angles turned = 3.
\n\"ML
\n(v)<\/strong> When the hour hand moves from 11 to 5 clockwise, fraction of revolution turned = \u00bd Number of right angles turned = 2.
\n\"ML
\n(vi)<\/strong> When the hour hand moves from 6 to 3 clockwise, Fraction of revolution turned = \u00be. Number of right angles turned = 3.<\/p>\n

Solution 02:<\/strong><\/span>
\n(i)<\/strong> When the hour hand moves from 10 and makes half revolution, clockwise it will stop at 4.
\n\"ML
\n(ii)<\/strong> When the hour hand moves from 4 and makes 1\/4 revolution, clockwise it will stop at 7.
\n\"ML
\n(iii)<\/strong> When the hour hand moves from 4 and makes 3\/4 revolution, clockwise it will stop at 1.
\n\"ML<\/p>\n

Solution 03:<\/strong><\/span>
\n(i)<\/strong> When the hour hand moves from 6 and turns through 1 right angle, clockwise it will stop at 9.
\n\"ML
\n(ii)<\/strong> When the hour hand moves from 8 and turns through 2 right angles, clockwise it will stop at 2.
\n\"ML
\n(iii)<\/strong> When the hour hand moves from 10 and turns through 3 right angles, clockwise it will stop at 7.
\n\"ML
\n(iv)<\/strong> When the hour hand moves from 7 and turns through 2 straight angles, clockwise it will stop at 7.
\n\"ML<\/p>\n

Solution 04:<\/strong><\/span>
\n(i)<\/strong> While turning from north to south Fraction of a revolution = \u00be. Number of right angles = 3.
\n\"ML
\n(ii)<\/strong> While turning from south to east Fraction of a revolution = 1\/4. Number of right angles = 1.
\n\"ML
\n(iii)<\/strong> While turning from east to west (clockwise). Fraction of a revolution = 1\/2. Number of right angles = 2.
\n\"ML<\/p>\n

Solution 05:<\/strong><\/span>
\n(i)<\/strong> Straight angle \u2013 (c) Half of a revolution
\n(ii)<\/strong> Right angle \u2013 (d) One fourth of a revolution
\n(iii)<\/strong> Complete angle \u2013 (f) One complete revolution
\n(iv)<\/strong> Acute angle \u2013 (b) Less than one fourth of a revolution
\n(v)<\/strong> Obtuse angle \u2013 (e) Between \u00bc and \u00bd of a revolution
\n(vi)<\/strong> Reflex angle \u2013 (a) More than half of a revolution<\/p>\n

Solution 06:<\/strong><\/span>
\n(i)<\/strong> Acute angle
\n(ii)<\/strong> Obtuse angle
\n(iii)<\/strong> Right Angle
\n(iv)<\/strong> Straight angle
\n(v)<\/strong> Reflex angle
\n(vi)<\/strong> Reflex angle
\n(vii)<\/strong> Acute angle
\n(viii)<\/strong> Obtuse angle<\/p>\n

Solution 07:<\/strong><\/span>
\n(i)<\/strong> Angle a and Angle c are acute, Angle b is obtuse
\n(ii)<\/strong> Angle x and Angle z are Obtuse, Angle y is acute
\n(iii)<\/strong> Angle p is obtuse, Angle q and Angle s are acute and Angle r is reflex.
\n\"ML<\/p>\n

Solution 08:<\/strong><\/span>
\n\"ML
\n\"ML
\nBy measuring the protractor marked angles are as follows
\n(i)<\/strong> 62o<\/sup>
\n(ii)<\/strong> 116o<\/sup>
\n(iii)<\/strong> 121o<\/sup><\/p>\n

Solution 09:<\/strong><\/span>
\n\"ML
\n\"ML
\nBy measuring the protractor marked angles are as follows
\n(i)<\/strong> 315o<\/sup>
\n(ii)<\/strong> 235o<\/sup><\/p>\n

Solution 10:<\/strong><\/span>
\nIn the clock the angle between every numeric is 30o<\/sup> i.e. angle between 1 and 2 is 30o<\/sup>, 2 and 3 is 30o<\/sup> and 4 and 6 is 30o<\/sup> x 2 = 60o<\/sup>
\n\"ML
\nSimilarly,
\n(i)<\/strong> Angle between the hands of the clock – 60o<\/sup>
\n\"ML
\n(ii)<\/strong> Angle between the hands of the clock – 30o<\/sup>
\n\"ML
\n(iii)<\/strong> Angle between the hands of the clock – 150o<\/sup>
\n\"ML<\/p>\n

Solution 011:<\/strong><\/span>
\n\"ML
\nSmaller angle formed by the hour and minutes hands of a clock at 7\u2019O clock is 150o [30o x 5 = 150o] (Type of Angle \u2013 Obtuse angle) as shown in the above model
\nOther Angle = 360o<\/sup> \u2013 150o<\/sup> = 210o<\/sup> (Type of Angle \u2013 Reflex angle)<\/p>\n

Solution 12:<\/strong><\/span>
\nOne is a 30o<\/sup> \u2013 60o<\/sup> \u2013 90o<\/sup> set square; the other is a 45o<\/sup> \u2013 45o<\/sup> -90o<\/sup> set square. The angle of measure 90o is common between them.<\/p>\n

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.3<\/strong><\/span><\/h3>\n

Solution 01:<\/strong><\/span>
\nTwo straight line are called perpendicular lines if they intersect at right angles.
\nIn the given models (i)<\/strong>, (iii)<\/strong> and (iv)<\/strong> are perpendicular lines.<\/p>\n

Solution 02:<\/strong><\/span>
\n\"ML (i)<\/strong> Yes, CE = EG; E is the midpoint of CG
\n(ii)<\/strong> Yes, PF Line bisect segment BH \u2013 E is the midpoint of BH and Line P bisects line segment BH.
\n(iii)<\/strong> Line segment DF, Line segment BH
\n(iv)<\/strong> All are true (AC > FG, CD = GH and BC < EG)<\/p>\n

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.4<\/strong><\/span><\/h3>\n

Solution 01:<\/strong><\/span>
\n\"ML
\n(i)<\/strong> Two sides are equal \u2013 Isosceles triangle
\n(ii)<\/strong> Three sides are different \u2013 Scalene triangle
\n(iii)<\/strong> Three sides are equal \u2013 Equilateral triangle<\/p>\n

Solution 02:<\/strong><\/span>
\n\"ML
\n(i)<\/strong> Angle is 90o<\/sup> \u2013 Right angled triangle
\n(ii)<\/strong> Angle is more than 90o<\/sup> \u2013 Obtuse angled triangle
\n(iii)<\/strong> Angle is less than 90o<\/sup> \u2013 acute angled triangle<\/p>\n

Solution 03:<\/strong><\/span>
\n\"ML
\n(i)<\/strong> Angle is less than 90o<\/sup> \u2013 acute angled triangle and two sides are in equal in length- Isosceles triangle.
\n(ii)<\/strong> Angle is 90o<\/sup> \u2013 right angled triangle and three sides are in not equal in length- scalene triangle.
\n(iii)<\/strong> Angle is more than 90o<\/sup> \u2013 Obtuse angled triangle and two sides are in equal in length- Isosceles triangle.
\n(iv)<\/strong> Angle is 90o<\/sup> \u2013 right angled triangle and two sides are equal in length- Isosceles triangle.
\n(v)<\/strong> Angle is less than 90o<\/sup> \u2013 acute angled triangle and three sides are in equal in length- Equilateral triangle.
\n(vi)<\/strong> Angle is more than 90o<\/sup> \u2013 Obtuse angled triangle and three sides are in not equal in length- scalene triangle.<\/p>\n

Solution 04:<\/strong><\/span>
\n(i)<\/strong> Three sides of equal length \u2013 (e) Equilateral
\n(ii)<\/strong> 2 Sides of length \u2013 (g) Isosceles
\n(iii)<\/strong> All sides of different length \u2013 (a) Scalene
\n(iv)<\/strong> 3 acute angles \u2013 (f) Acute angled
\n(v)<\/strong> 1 right angle \u2013 (d) Right Angled
\n(vi)<\/strong> 1 Obtuse angle- C) Obtuse Angled
\n(vii)<\/strong> 1 Right angle with two sides of equal length – (b)Right angled Isosceles<\/p>\n

Solution 05:<\/strong><\/span>
\n(i)<\/strong> False
\n(ii)<\/strong> True
\n(iii)<\/strong> True
\n(iv)<\/strong> False
\n(v)<\/strong> False
\n(vi)<\/strong> False
\n(vii)<\/strong> True
\n(viii)<\/strong> False<\/p>\n

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.5<\/strong><\/span><\/h3>\n

Solution 01:<\/strong><\/span>
\n(i)<\/strong> True
\n(ii)<\/strong> True
\n(iii)<\/strong> True
\n(iv)<\/strong> True
\n(v)<\/strong> False
\n(vi)<\/strong> False
\n(vii)<\/strong> False
\nSolution 02:<\/strong><\/span>
\n(i)<\/strong> Not a polygon, because it is not a closed curve
\n(ii)<\/strong> Polygon, because it is a simple closed curve made up entirely of line segments
\n(iii)<\/strong> Not a polygon, because it is not a simple curve
\n(iv)<\/strong> Not a polygon, because it is not made up of entirely line segments.<\/p>\n

Solution 03:<\/strong><\/span>
\n(i)<\/strong> Pentagon
\n\"ML
\n(ii)<\/strong> Quadrilateral
\n\"ML
\n(iii)<\/strong> Hexagon
\n\"ML
\n(iv)<\/strong> Octagon
\n\"ML<\/p>\n

Solution 04:<\/strong><\/span>
\nABCDE is a regular pentagon and diagonals as in the below figure.
\n\"ML<\/p>\n

Solution 05:<\/strong><\/span>
\n\"ML
\nLet ABCDEF be regular hexagon then
\n(i)<\/strong> Triangle ABC is an Isosceles triangle.
\n(ii)<\/strong> Triangle CEF is a right angled triangle.<\/p>\n

Solution 06:<\/strong><\/span>
\nABCD is a regular quadrilateral \u2013 Square.
\n\"ML<\/p>\n

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.6<\/strong><\/span><\/h3>\n

Solution 01:<\/strong><\/span>
\n(i)<\/strong> Cuboid
\n(ii)<\/strong> Cuboid
\n(iii)<\/strong> Cuboid
\n(iv)<\/strong> Cylinder
\n(v)<\/strong> Cube
\n(vi)<\/strong> Sphere<\/p>\n

Solution 02:<\/strong><\/span>
\n(i)<\/strong> Cone
\n\"ML
\n(ii)<\/strong> Sphere
\n\"ML
\n(iii)<\/strong> Cube
\n\"ML
\n(iv)<\/strong> Pyramid
\n\"ML
\n(v)<\/strong> Cylinder
\n\"ML
\n(vi)<\/strong> Cuboid
\n\"ML<\/p>\n

Solution 03:<\/strong><\/span>
\n(i)<\/strong> A cube has 6 square faces, 12 edges and 8 vertices.
\n(ii)<\/strong> A triangular prism has 2 triangular faces, 3 rectangular faces, 9 edges and 6 vertices.
\n(iii)<\/strong> A triangular pyramid has 4 faces, 6 edges and 4 vertices.<\/p>\n

\"ML<\/p>\n

 <\/p>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes ML Aggarwal SolutionsICSE SolutionsSelina ICSE Solutions ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.1 Solution 01: It gives accurate measurement and avoids error due to thickness of ruler or positioning of eye (due to angular … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":31428,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[22095,38526,38525,38529,38528,38522,38527,38524,38523,22091],"yoast_head":"\nML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes<\/title>\n<meta name=\"description\" content=\"ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, 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