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

ML Aggarwal Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes are available in pdf format. ML Aggarwal problems assist students in developing problem-solving and higher-order thinking skills. It aims to provide students with a variety of quality problems in different formats. Students who wish to score good marks in Maths, practise ML Aggarwal Solutions for Class 6 Maths Chapter 11.

Chapter 11 – Understanding Symmetrical Shapes, something is symmetrical when it is the same on both sides. A shape has symmetry if a central dividing line can be drawn on it. This chapter contains different concepts of symmetrical shapes and also students learn the use of divider and ruler to measure the length.

## Download the PDF of ML Aggarwal Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes    ### Access answers to ML Aggarwal Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes

1. Why is it better to use a divider and a ruler than a ruler only, while measuring the length of a line segment?
Solution:-

It is better to use a divider than a ruler, because the thickness of the ruler may cause difficulties in reading off her length. However, divider gives up accurate measurement.

2. In the given figure, compare the line segments with the help of a divider and fill in the blanks by using the symbol >, = or <:
(i) AB ………. CD
(ii) BC ………. AB
(iii) AC ………. BD
(iv) CD ………. BD Solution:-

(i) From the figure,

AB = CD … [in parallelograms opposites are equal]
(ii) BC < AB

(iii) AC > BD

(iv) CD < BD

3. If A, B and C are collinear points such that AB = 6 cm, BC = 4 cm and AC = 10 cm, which one of them lies between the other two?
Solution:-

From the question it is given that,

A, B and C are collinear points

AB = 6 cm

BC = 4 cm

AC = 10 cm 4. In the given figure, verify the following by measurement:
(i) AB + BC = AC
(ii) AC – BC = AB Solution:-

From the figure,

By using the divider and ruler we measured the given figure,

So,

AB = 3 cm, BC = 1.5 cm and AC = 4.5 cm

(i) AB + BC = AC

3 cm + 1.5 cm = 4.5 cm

4.5 cm = 4.5 cm

(ii) AC – BC = AB

4.5 cm – 5 cm = 3 cm

3 cm = 3 cm

5. In the given figure, verify by measurement that:
(i) AC + BD = AD + BC
(ii) AB + CD = AD – BC Solution:-

From the figure,

By using the divider and ruler we measured the given figure,

So,

AB = 1.8 cm, BC = 0.8 cm, BD = 2.7 cm CD = 1.9 cm, AC = 2.6 cm and AD = 4.5 cm

(i) AC + BD = AD + BC

Consider Left hand side (LHS) = AC + BD

= 2.6 cm + 2.7 cm

= 5.3 cm

Now, Right hand side (RHS) = AD + BC

= 4.5 cm + 0.8 cm

= 5.3 cm

By comparing LHS and RHS

LHS = RHS

5.3 cm = 5.3 cm

Therefore, AC + BD = AD + BC

(ii) AB + CD = AD – BC

Consider Left hand side (LHS) = AB + CD

= 1.8 cm + 1.9 cm

= 3.7 cm

Now, Right hand side (RHS) = AD – BC

= 4.5 cm – 0.8 cm

= 3.7 cm

By comparing LHS and RHS

LHS = RHS

3.7 cm = 3.7 cm

Therefore, AC + BD = AD + BC

6. In the given figure, measure the lengths of the sides of the triangle ABC and verify:
(i) AB + BC > AC
(ii) BC + AC > AB
(iii) AC + AB > BC Solution:-

From the figure,

By using the divider and ruler we measured the given figure,

So,

AB = 2.6 cm, BC = 3.9 cm, AC = 4 cm

(i) AB + BC > AC

2.6 + 3.9 cm > 4 cm

6.5 cm > 4 cm

(ii) BC + AC > AB

3.9 + 4 cm > 2.6 cm

7.9 cm > 2.6 cm

(iii) AC + AB > BC

4 cm + 2.6 cm > 3.9 cm

6.6 cm > 3.9 cm