ML Aggarwal Solutions for Class 6 Maths Chapter 11 Understanding Symmetrical Shapes are available in PDF. ML Aggarwal solutions assist students in developing problem-solving and higher-order thinking skills. It aims to provide students with different types of problems and their solutions in an easy-to-understand manner. Students who wish to score good marks in Maths, practise ML Aggarwal Solutions for Class 6 Maths Chapter 11.
Chapter 11 – Understanding Symmetrical Shapes consists of problems based on the concept of symmetrical shapes. The images which are divided into identical halves are known as symmetrical. 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 will learn the use of divider and ruler to measure the length.
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:
The divider gives up accurate measurement, whereas the ruler may cause difficulties in reading length. Therefore, it is better to use a divider while measuring the length of a line segment.
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
Here, point B lies between the other two, i.e., A and C.
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 – 1.5 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, AB + CD = 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
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