ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes

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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 viewing)



Solution 02:

By measuring the lengths of the given figure



(i) AB = CD

(ii) BC < AB

(iii) AC = BD

(iv) CD < BD

Solution 03:

Given that AC = 10 CM, AB = 6 CM and BC = 4 CM

By constructing line segment by the given data, the model drawn as below.



Point B lies in between A and C

Solution 04:



By measuring the Lengths of line segments in the above figure

AB = 3 CM

BC = 1.5 CM

(i) It can be observed that AC = AB + BC [i.e. 4.5 CM = 3 CM + 1.5 CM]

(ii) AC – BC = AB 4.5 CM – 1.5 CM = 3 CM by measurement fount that AB = 3 CM, so AC – BC = AB.

Solution 05:

By measuring the lengths of the given figure.



Given data

AB = 1.9 CM

BC = 0.7 CM

CD = 1.9 CM

AD = 4.5 CM

(i) 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.

(ii) 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.

Solution 06:

By measuring the lengths of the given triangle ABC as below



AB = 2.6 CM, AC = 3.8 CM and BC = 3.8 CM.

(i) AB + BC = 2.6 CM + 3.8 CM = 6.4 CM AC = 3.8 CM Hence, AB + BC > AC.

(ii) BC + AC = 3.8 CM + 3.8 CM = 7.6 CM AB = 2.6 CM Hence, AB > BC + AC.

(iii) AC + AB = 3.8 CM + 2.6 CM = 5.4 CM BC = 3.8 CM Hence, AC + AB > BC.

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.2

Solution 01:

(i) When the hour hand moves from 4 to 10 clockwise, fraction of revolution turned = ½ Number of right angles turned = 2.



(ii) When the hour hand moves from 2 to 5 clockwise, fraction of revolution turned = ¼. Number of right angles turned = 1.



(iii) When the hour hand moves from 7 to 10 clockwise, fraction of revolution turned = ¼. Number of right angles turned = 1.



(iv) When the hour hand moves from 8 to 5 clockwise, fraction of revolution turned = ¾. Number of right angles turned = 3.



(v) When the hour hand moves from 11 to 5 clockwise, fraction of revolution turned = ½ Number of right angles turned = 2.



(vi) When the hour hand moves from 6 to 3 clockwise, Fraction of revolution turned = ¾. Number of right angles turned = 3.

Solution 02:

(i) When the hour hand moves from 10 and makes half revolution, clockwise it will stop at 4.



(ii) When the hour hand moves from 4 and makes 1/4 revolution, clockwise it will stop at 7.



(iii) When the hour hand moves from 4 and makes 3/4 revolution, clockwise it will stop at 1.

Solution 03:

(i) When the hour hand moves from 6 and turns through 1 right angle, clockwise it will stop at 9.



(ii) When the hour hand moves from 8 and turns through 2 right angles, clockwise it will stop at 2.



(iii) When the hour hand moves from 10 and turns through 3 right angles, clockwise it will stop at 7.



(iv) When the hour hand moves from 7 and turns through 2 straight angles, clockwise it will stop at 7.

Solution 04:

(i) While turning from north to south Fraction of a revolution = ¾. Number of right angles = 3.



(ii) While turning from south to east Fraction of a revolution = 1/4. Number of right angles = 1.



(iii) While turning from east to west (clockwise). Fraction of a revolution = 1/2. Number of right angles = 2.

Solution 05:

(i) Straight angle – (c) Half of a revolution

(ii) Right angle – (d) One fourth of a revolution

(iii) Complete angle – (f) One complete revolution

(iv) Acute angle – (b) Less than one fourth of a revolution

(v) Obtuse angle – (e) Between ¼ and ½ of a revolution

(vi) Reflex angle – (a) More than half of a revolution

Solution 06:

(i) Acute angle

(ii) Obtuse angle

(iii) Right Angle

(iv) Straight angle

(v) Reflex angle

(vi) Reflex angle

(vii) Acute angle

(viii) Obtuse angle

Solution 07:

(i) Angle a and Angle c are acute, Angle b is obtuse

(ii) Angle x and Angle z are Obtuse, Angle y is acute

(iii) Angle p is obtuse, Angle q and Angle s are acute and Angle r is reflex.

Solution 08:





By measuring the protractor marked angles are as follows

(i) 62^o

(ii) 116^o

(iii) 121^o

Solution 09:





By measuring the protractor marked angles are as follows

(i) 315^o

(ii) 235^o

Solution 10:

In the clock the angle between every numeric is 30^o i.e. angle between 1 and 2 is 30^o, 2 and 3 is 30^o and 4 and 6 is 30^o x 2 = 60^o



Similarly,

(i) Angle between the hands of the clock – 60^o



(ii) Angle between the hands of the clock – 30^o



(iii) Angle between the hands of the clock – 150^o

Solution 011:



Smaller angle formed by the hour and minutes hands of a clock at 7’O clock is 150o [30o x 5 = 150o] (Type of Angle – Obtuse angle) as shown in the above model

Other Angle = 360^o – 150^o = 210^o (Type of Angle – Reflex angle)

Solution 12:

One is a 30^o – 60^o – 90^o set square; the other is a 45^o – 45^o -90^o set square. The angle of measure 90o is common between them.

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.3

Solution 01:

Two straight line are called perpendicular lines if they intersect at right angles.

In the given models (i), (iii) and (iv) are perpendicular lines.

Solution 02:

(i) Yes, CE = EG; E is the midpoint of CG

(ii) Yes, PF Line bisect segment BH – E is the midpoint of BH and Line P bisects line segment BH.

(iii) Line segment DF, Line segment BH

(iv) All are true (AC > FG, CD = GH and BC < EG)

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.4

Solution 01:



(i) Two sides are equal – Isosceles triangle

(ii) Three sides are different – Scalene triangle

(iii) Three sides are equal – Equilateral triangle

Solution 02:



(i) Angle is 90^o – Right angled triangle

(ii) Angle is more than 90^o – Obtuse angled triangle

(iii) Angle is less than 90^o – acute angled triangle

Solution 03:



(i) Angle is less than 90^o – acute angled triangle and two sides are in equal in length- Isosceles triangle.

(ii) Angle is 90^o – right angled triangle and three sides are in not equal in length- scalene triangle.

(iii) Angle is more than 90^o – Obtuse angled triangle and two sides are in equal in length- Isosceles triangle.

(iv) Angle is 90^o – right angled triangle and two sides are equal in length- Isosceles triangle.

(v) Angle is less than 90^o – acute angled triangle and three sides are in equal in length- Equilateral triangle.

(vi) Angle is more than 90^o – Obtuse angled triangle and three sides are in not equal in length- scalene triangle.

Solution 04:

(i) Three sides of equal length – (e) Equilateral

(ii) 2 Sides of length – (g) Isosceles

(iii) All sides of different length – (a) Scalene

(iv) 3 acute angles – (f) Acute angled

(v) 1 right angle – (d) Right Angled

(vi) 1 Obtuse angle- C) Obtuse Angled

(vii) 1 Right angle with two sides of equal length – (b)Right angled Isosceles

Solution 05:

(i) False

(ii) True

(iii) True

(iv) False

(v) False

(vi) False

(vii) True

(viii) False

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.5

Solution 01:

(i) True

(ii) True

(iii) True

(iv) True

(v) False

(vi) False

(vii) False

Solution 02:

(i) Not a polygon, because it is not a closed curve

(ii) Polygon, because it is a simple closed curve made up entirely of line segments

(iii) Not a polygon, because it is not a simple curve

(iv) Not a polygon, because it is not made up of entirely line segments.

Solution 03:

(i) Pentagon



(ii) Quadrilateral



(iii) Hexagon



(iv) Octagon

Solution 04:

ABCDE is a regular pentagon and diagonals as in the below figure.

Solution 05:



Let ABCDEF be regular hexagon then

(i) Triangle ABC is an Isosceles triangle.

(ii) Triangle CEF is a right angled triangle.

Solution 06:

ABCD is a regular quadrilateral – Square.

ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes Exercise 11.6

Solution 01:

(i) Cuboid

(ii) Cuboid

(iii) Cuboid

(iv) Cylinder

(v) Cube

(vi) Sphere

Solution 02:

(i) Cone



(ii) Sphere



(iii) Cube



(iv) Pyramid



(v) Cylinder



(vi) Cuboid

Solution 03:

(i) A cube has 6 square faces, 12 edges and 8 vertices.

(ii) A triangular prism has 2 triangular faces, 3 rectangular faces, 9 edges and 6 vertices.

(iii) A triangular pyramid has 4 faces, 6 edges and 4 vertices.