**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.

img src=”https://farm2.staticflickr.com/1891/43429106585_c8b5b41836_o.png” width=”355″ height=”318″ alt=”ML Aggarwal ICSE Solutions for Class 6 Maths Chapter 11 Understanding Elementary Shapes Ex 11.5 Solution 06″>

**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.

Abu Imran says

Thanks

For the Answer

Abu Imran says

Thanks

For the Answer of the Maths book 😄😄