Class 6

Students can practice TS SCERT Class 6 Maths Solutions Chapter 9 Introduction to Algebra Ex 9.3 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra Exercise 9.3

Question 1.
State which of the following are equations.
(i) x – 3 = 7
(ii) l + 5 > 9
(iii) p – 4 < 10
(iv) 5 + m = – 6
(v) 2s – 2 = 12
(vi) 3x + 5 > 13
(vii) 3x < 15
(viii) 2x – 5 = 3
(ix) 7y + 1 < 22
(x) – 3z + 6 = 12
(xi) 2x – 3y = 3
(xii) z = 4
Answer:
The following are the equations.
(i) x- 3 = 7
(iv) 5 + m = – 6
(v) 2s – 2 = 12
(viii) 2x – 5 = 3
(x) – 3z + 6 = 12
(xi) 2x – 3y = 3
(xii) z = 4

Question 2.
Write LHS and RHS of the following equations.
(i) x – 5 = 6
(ii) 4y = 12
(iii) 2z + 3 = 7
(iv) 3p = 24
(v) 4 = x – 2
(vi) 2a – 3 = – 5
Answer:
The value of expression to the left of the sign ‘=’ is called Left Hand Side (LHS) and that of expression to the right of the sign ‘=’ is called Right Hand Side
(RHS)
(i) x – 5 (LHS) 6(RHS)
(ii) 4y (LHS) 12 (RHS)
(iii) 2z + 3 (LHS) 7 (RHS)
(iv) 3p (LHS) 24 (RHS)
(v) 4 (LHS) x – 2 (RHS)
(vi) 2a – 3 (LHS) -5 (RHS)

Question 3.
Solve the following equations by Trial & Error method.
(i) x + 3 = 5
Answer:

We find that for x = 2, both LHS and RHS are equal. Therefore x = 2 is the solution of the equation.

(ii) y – 2 = 7
Answer:

We find that for y = 9, both LHS and RHS are equal. Therefore y = 9 is the solution of the equation.

(iii) a – 2 = 6
Answer:

We find that for a = 8, both LHS and RHS are equal. Therefore a = 8 is the solution of the equation.

(iv) 5y = 15
Answer:

We find that for y = 3, both LHS and RHS are equal. Therefore y = 3 is the solution of the equation.

(v) 6n = 30
Answer:

We find that for n = 5, both LHS and RHS are equal. Therefore n = 5 is the solution of the equation.

(vi) 3z = 27
Answer:

We find that for z = 9, both LHS and RHS are equal. Therefore z = 9 is the solution of the equation.

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 14 Understanding 3D and 2D Shapes Ex 14.2 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 14 Understanding 3D and 2D Shapes Exercise 14.2

Question 1.
Examine whether the following are polygons if not why ?

Answer:
i) Figure is kept open. So it is not a polygon.
ii) Figure is a closed one made up of 4 line segments. So it is a polygon.
iii) Figure is a circle. It is curved. It is not made up of line segments. So it is not a polygon.

Question 2.
Count the number of sides of the polygons given below and name them.

Answer:
Figure (i) has 5 sides. It is called a pentagon.
Figure (ii) has 8 sides. It is called an octagon.
Figure (iii) has 6 sides. It is called a hexagon.
Figure (iv) has 3 sides. It is called a triangle.

Question 3.
Identify the regular polygons among the figures given below :

Answer:
Figure (i) is a square. Its sides and angles are all equal. So it is a regular polygon.
Figure (iv) is a regular hexagon. It has equal sides and its angles are equal.
Figure (vi) is an equilateral triangle. Its sides are equal. Its angles are equal.

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 14 Understanding 3D and 2D Shapes Ex 14.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 14 Understanding 3D and 2D Shapes Exercise 14.1

Question 1.
A triangular pyramid has a triangle at its base. It is also known as a tetrahedron. Find the number of

i) No. of Faces :
ii) No. of Edges :
iii) No. of Vertices :
Answer:
A triangular pyramid (tetrahedron) has
i) No. of Faces : 4
ii) No. of Edges : 6
iii) No. of Vertices : 4

Question 2.
A square pyramid has a square at its base. Find the number of

i) No. of Faces :
ii) No. of Edges :
iii) No. of Vertices :
Answer:
A square pyramid has
i) No. of Faces 5
ii) No. of Edges : 8
iii) No. of Vertices : 5

Question 3.
Fill the table.

Answer:

Question 4.
A triapgular prism is often in the shape of a kaleidoscope. It has triangular faces.

i) No. of triangular Faces :
ii) No. of rectangular Faces :
iii) No. of Edges :
iv) No. of Vertices :
Answer:
i) No. of triangular Faces : 2
ii) No. of rectangular Faces : 3
iii) No. of Edges : 9
iv) No. of Vertices : 6

Students can practice TS SCERT Class 6 Maths Solutions Chapter 9 Introduction to Algebra Ex 9.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra Exercise 9.1

Question 1.
Find the rule which gives the number of matchsticks required to make the following matchsticks patterns.
(i) A pattern of letter ‘T’
Answer:
The rule which gives the number of sticks required to make the pattern T is 2n.

(ii) A pattern of letter ‘E’
Answer:
The rule which gives the number of sticks required to make the pattern E is 5n.

(iii) A pattern of letter ‘Z’
Answer:
The rule which gives the number of sticks required to make the pattern Z is 3n.

Question 2.
Make a rule between the number of blades required and the number of fans (say n) in a hall ?

Answer:
The number of blades that a fan has = 3
Number of fans = n (say)
∴ The number of blades for ‘n’ fans = 3 × n = 3n
The required rule is 3n.

Question 3.
Find a rule for the following patterns between number of shapes formed and number of matchsticks required.
(a)

Answer:
The first shape has 2 matchsticks, the second shape 4 and the third shape 6.
∴ The required rule 2s.
(‘s’ stands for number of shapes.)

(b)

Answer:
The first shape has 3 matchsticks, the second shape 6 and the third shape 9.
∴ The required rule is 3s.
(‘s’ stands for number of shapes.)

Question 4.
The cost of one pen is ₹ 7 then what is the rule for the cost of ‘n’ pens.
Answer:
Cost of one pen is Rs. 7.
Cost of ‘n’ pens = 7 × n = 7n
The rule for cost of n’ pens is 7n.

Question 5.
The cost of one bag is ₹ 90, what is the rule for the cost of ‘m’ bags ?
Answer:
The cost of one bag = Rs. 90.
Cost of ‘m’ bags = 90 × m = 90m
The rule for the cost of ‘m’ bags is 90m.

Question 6.
The rule for purchase of books is-that the cost of q books is ₹ 23; then find the price of one book ?
Answer:
Cost of ‘q’ books = Rs. 23
∴ Cost of one book = \(\frac{23}{q}\)
The rule for the price of one book is \(\frac{23}{q}\).

Question 7.
John says that he has two books less than Gayathri. Write the relationship using letter x.
Answer:
Let the number of books that Gayathri has be x.
Then the number of books that John has = (x – 2)
(∵ John says that he has two books less than Gayathri)
The required relationship is (x – 2).

Question 8.
Rekha has 3 books more than twice the books with Suresh. Write the relationship using letter y.
Answer:
Let the number of books with Suresh be ‘y’.
Twice the number of books with Suresh = 2 × y – 2y.
As per the problem, the number of ‘ books that Rekha has = (2y + 3)
The required relationship is 2y + 3.

Question 9.
A teacher distributes 6 pencils per student. Can you find how many pencils are needed for the given number of students (use V for the number of students).
Answer:
Let the number of students be ’z’.
The teacher gives 6 pencils to each student.
The number of pencils that are given to ‘z’ students = 6 × z = 6z
The rule that is required is 6z.

Question 10.
Complete each table to generate the given functional relationship.

Answer:

Question 11.
Observe the following pattern.

Count the number of line segments in each shape.
(i) How many line segments will the 9th shape contain ?
(ii) Write the rule for the above pattern.
Answer:
(i) Shape – 1 contains 3 line segments.
Shape – 2 contains 5 line segments.
Shape – 3 contains 7 line segments.
Shape – 4 contains 9 line segments.
Shape – 5 contains 11 line segments.
Shape – 6 contains 13 line segments.
Shape – 7 contains 15 line segments.
Shape – 8 contains 17 line segments.
Shape – 9 contains 19 line segments.
The number of line segments that 9 such shapes contains is 99.
(∵ 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)

(ii) The rule for the above pattern is 3 + 2(n – 1).

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 12 Symmetry InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 12 Symmetry InText Questions

Do This

Question 1.
Match each letter with its mirror image. The dotted line with every letter shows in the mirror.

Can you think of more such alphabets and words which will remain the same in their mirror image ?
Answer:

Yes. They are 0, X,H, I, which are same as mirror images.

Try This

Question 1.
Place a mirror along the dotted lines and draw their mirror images.

Do you observe any change ? Are angles in the images equal to the angles in the given figures ?
Answer:
We can observe the angles formed in the original figures and its mirror images are same.

Do This

Question 1.
In the figures given below find which are symmetric figures.

Can we find line of symmetry for every figure ?
Answer:
(i), (ii) figures are symmetric figures.
(iii) & (iv) are not symmetric figures.

Try This

Write the letters of English alphabet A to Z and find out which have
i) Vertical lines of symmetry
ii) Horizontal lines of symmetry
iii) No lines of symmetry
Answer:
i) Vertical lines of symmetry :

ii) Horizontal lines of symmetric

iii) No lines of symmetric

Try These

Question 1.
Draw any five objects which have a line of symmetry.
Answer:
The following figure having the line of symmetry.

Question 2.
Draw any five objects which are not symmetric.
Answer:

Think, Discuss And Write

Question 1.
If the paper is folded four times how many lines of symmetry can be formed with paper cutting ?
Answer:
If a paper is folded 4 times then 8 symmetric lines are formed.

Question 2.
To cut four similar figures side by side by folding the paper, how many folds are needed ?
Answer:
Two paper foldings are to be needed to form 4 similar figures.

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 5 Measures of Lines and Angles InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Exercise InText Questions

Think, Discuss and Write

Question 1.
How do we compare them ?
To compare them, we trace the line segments \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) on a tracing paper such that they are roughly aligned in the same direction. Do their end points coincide ?
Answer:
We can now say \(\overline{\mathrm{AB}}\) is longer than \(\overline{\mathrm{CD}}\). In the same way we can compare \(\overline{\mathrm{PQ}}\) with \(\overline{\mathrm{RS}}\). We can see \(\overline{\mathrm{PQ}}\) and \(\overline{\mathrm{RS}}\) are of equal length.
(Or)
We can compare the lengths of \(\overline{\mathrm{PQ}}\) and \(\overline{\mathrm{RS}}\) by a using a scale or a divider.
Now, we can say \(\overline{\mathrm{PQ}}\) > \(\overline{\mathrm{RS}}\)
∴ \(\overline{\mathrm{PQ}}\) = 3cm; \(\overline{\mathrm{RS}}\) = 25cm
∴ \(\overline{\mathrm{PQ}}\) > \(\overline{\mathrm{RS}}\)

Question 2.
What other errors can you find while measuring the length of line segment ?
For example, to find the length of a pencil, the eye Wrong Right Wrong should be correctly positioned as shown in the figure i.e., just vertically above the mark for both points. Other wise there may be an error due to angular viewing.

Answer:
(i) Eye error
(ii) Parallax error.

Try These

Question 1.
Take a post card and measure the length and breadth with ruler and divider. Do all post cards have the same dimensions ?
Answer:
By using ruler and a divider we can say that the dimensions of a post card are Length = 12 cm, Breadth = 7 cm
Same type of post cards are all having the same dimensions.

Question 2.
Select any three objects eraser, small pencil, etc. Trace their length on a paper. Measure the length of these line segments.
Answer:

The length of an eraser is 3 cm.
The length of a small pencil is 10 cms.

Try These

Question 1.
Use the right angle tester made of straw’s and identify the following angles.

Answer:
(i) Obtuse angle
(ii) Right angle
(iii) Acute angle
(iv) Obtuse angle

Question 2.
List out five daily life situations where you observe acute angles and obtuse angles
Answer:
Acute angles and obtuse angles are formed in following cases

1. The angle between the slant slopes of buffoon cap.
2. At the time of 2’o clock p.m. in a clock.
3. At the time of 10’o clock.
4. When the wrist is bonded.
5. When the knee is bonded.

Question 3.
Draw some angles of your choice. Test them by the “angle tester” and write which are acute and which are obtuse and which are right angles.
Answer:

Think, Discuss and Write

In the adjacent figure ∠AOB and ∠AOC are given. Which angle is clock-wise and which angle is anti clock-wise. Think and discuss with your friends.

Answer:
∠AOB is formed in a clock-wise direction.
∠AOC is formed in an anti clock-wise direction.

Try These

Question 1.
Which angle is greater ? Discuss with your friends.

Verify by measuring the angles using protractor. Is your estimation is correct ? Give reasons.
Answer:
The measure of ∠1 = 30° and that of ∠2 = 30°
∴ Both the angles ∠1 and ∠2 are of equal measure.

Question 2.
Which are acute angles ? Find and write their measures.

Answer:
(i) The measure of angle in figure (i) is 30°. So it is an acute angle,
(ii) The measure of angle in figure (ii) is 50°. So it is an acute angle.
(iii) The measure of angle in figure (iii) is 130°. It is not an acute angle. Because it is an obtuse angle.
(iv) The measure of angle in figure (iv) is 90°. So it is a right angle.
(v) The measure of angle in figure (v) is 80°. So it is an acute angle.

Question 3.
Which are obtuse angles?

Answer:
(i) The measure of angle in figure (i) is 110°. So it is an obtuse angle.
(ii) The measure of angle in figure (ii) is 90°. So it is a right angle.
(iii) The measure of angle in figure (iii) is 130°. So it is an obtuse angle.
(iv) The measure of angle in figure (iv) is 50°. So it is an acute angle.
(v) The measure of angle in figure (v) is 325°. So it is a reflex angle.

Question 4.
Draw any two acute and two obtuse angles of your choice.
Answer:
Acute angles:

AOB and POQ are acute angles.

Obtuse angles:

XOY and MON are obtuse angles.

Question 5.
Classify the following angles into acute, right, obtuse and straight angles.
40°, 140°, 90°, 210°, 44°, 215°, 345°, 125°
10°, 120°, 89°, 270°, 30°, 115°, 180°
Answer:
Acute angles : 40°, 44°, 10°, 89°, 30°
Right angle : 90°
Obtuse angles : 140°, 125°, 120°, 115°
Straight angle : 180°
Reflex angles : 210°, 215°, 345°, 270°

Think, Discuss and Write

Question 1.
If l ⊥ m, then can we say that m ⊥ l?
Answer:
If l ⊥ m then m ⊥ l.
Since both l, m are perpendicular to each other.

Question 2.
How many perpendicular lines can be drawn to a given line ?
Answer:
We can draw only one line which is perpendicular to the given line.

Question 3.
Which letters in English alphabet possess perpendicularity ?
Answer:
L, T possess the perpendicularity.

Try These

Question 1.
Draw two lines on a paper as shown in the figure.
Do they intersect each other ?
Can you call them parallel lines ? Give reason.

Answer:
The above lines are not intersecting. If, they are extended on either sides they will intersect at a point. They are called intersecting lines i.e., they are not parallel lines.

Question 2.
Make a pair of parallel lines. What is the angle formed between them ? Think, discuss with your friends and teacher.
Answer:
The angle between the parallel lines is 0° (zero). Since they do not intersect.

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 5 Measures of Lines and Angles Ex 5.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Exercise 5.1

Question 1.
Give any five examples of line segments observed in your classroom.
Eg: edge of black board.
Answer:
Edge of blackboard, edge of a book, edge of threshold, edge of window, height of the room, edge of table, edge of bench.

Question 2.
Why is it better to use a divider than a ruler, while comparing two line segments ?
Answer:
While finding the length of any object, the eye should be correctly positioned (i.e.) just vertically above the marks concerned. Otherwise there may be an error due to angular viewing. To avoid this problem a better way is to use a divider. In the same way, it would be better to use a divider while comparing two line segments.

Question 3.
Measure all the line segments in the figure given below and arrange them in the ascending order of their lengths.

Line segments \(\overline{\mathbf{A B}}, \overline{\mathbf{A C}}, \overline{\mathbf{A D}}, \overline{\mathbf{A E}}, \overline{\mathbf{B C}}, \overline{\mathbf{B D}}, \overline{\mathbf{B E}}, \overline{\mathbf{C D}}, \overline{\mathbf{C E}}, \overline{\mathbf{D E}}\)
Answer:
\(\overline{\mathrm{A}} \overline{\mathrm{B}}\) = 1 cm; \(\overline{\mathrm{A}} \overline{\mathrm{C}}\) = 2 cm; \(\overline{\mathrm{A}} \overline{\mathrm{D}}\) = 3 cm; \(\overline{\mathrm{A}} \overline{\mathrm{E}}\) = 4 cm
\(\overline{\mathrm{B}} \overline{\mathrm{C}}\) = 1 cm; \(\overline{\mathrm{B}} \overline{\mathrm{D}}\) = 2.0 cm; \(\overline{\mathrm{B}} \overline{\mathrm{E}}\) = 3.0 cm
\(\overline{\mathrm{C}} \overline{\mathrm{D}}\) = 1 cm; \(\overline{\mathrm{C}} \overline{\mathrm{E}}\) = 2.0 cm; \(\overline{\mathrm{D}} \overline{\mathrm{E}}\) = 1.0cm
Ascending order: \(\overline{\mathrm{AE}}, \overline{\mathrm{AD}}, \overline{\mathrm{BE}}, \overline{\mathrm{CE}}, \overline{\mathrm{BD}}, \overline{\mathrm{AC}}, \overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{DE}}\)

Question 4.
Mid point of \(\overline{\mathrm{AB}}\) is located by Swetha and Reshrna like this.

Which one do you feel correct ? Measure the lengths of \(\overline{\mathbf{A C}}, \overline{\mathbf{C B}}\) and verify.
Answer:
By observation, we can say that the mid point of \(\overline{\mathrm{AB}}\) located by Reshrna is correct. (∵ AC = CB)
AC = 1 cm; CB = 2 cm (Swetha)
AC = 1.5 cm; CB = 1.5 cm (Reshrna)

Question 5.
Each of these figures given along side has many line segments. For the almirah we have shown one line segment along the longer edge. Identify and mark all such line segments in these figures.

Answer:
(i) \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{DA}}\) and \(\overline{\mathrm{AC}}\) are the line segments.

(ii) \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{DA}}, \overline{\mathrm{AE}}, \overline{\mathrm{BF}}, \overline{\mathrm{CG}}, \overline{\mathrm{EF}}\) and \(\overline{\mathrm{FG}}\) are the line segments.

(iii) \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{BE}}, \overline{\mathrm{AF}}, \overline{\mathrm{FE}},\) and \(\overline{\mathrm{ED}}\) are the line segments.

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 13 Practical Geometry InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry InText Questions

Try These

Question 1.
Construct two circles with same radii in such a way that

i) the circles intersect at two points.
ii) touch each other at one point only.
Answer:
i) Circles intersect at two points :
ii) Touch each other at one point only :

Think, Discuss And Write

Question 1.
How would you check whether It Is perpendicular or not ? Note that it passes through P as required.
Answer:

The perpendicularity can be measured by protractor.
So, the perpendicular line always passes through P’.

Do This

Question 1.
Measure the lengths of \(\overline{\mathrm{A P}}\) and \(\overline{\mathrm{B P}}\) in both the constructions. Are they equal?
Answer:

\(\overline{\mathrm{A P}}\) = 1.7 cm
\(\overline{\mathrm{B P}}\) = 1.7 cm
yes. They are equal

Think, Discuss And Write

Question 1.
In the construction of perpendicular bisector in step 2. What would happen if we take the length of radius to be smaller than half the length of \(\overline{\mathrm{A B}}\)?
Answer:
Explanation:
1) Construct a line segment \(\overline{\mathrm{A B}}\) with a suitable radius.
2) To construct a perpendicular line to the given line, take the radius more than half of a given line segment \(\overline{\mathrm{A B}}\).
3) With the centres A, B draw two arcs as mentioned above.
4) The two arcs are intersected at a point. Name it P, Q.
5) Join these two points by a straight line. Name it as l. ∴ l ⊥ \(\overline{\mathrm{A B}}\).

If we does not taken the radius half of the given line segment, then the perpendicular bisector doesn’t formed. Since the arcs doesn’t interest each other.

Do This

Question 1.
Construct angles of 180°, 240°, 300°.
Answer:

Question 2.
Construct an angle of 45° by using compasses.
Answer:
Steps of construction:

1) Construct \(\overline{\mathrm{OA}}\) line segment with suitable radius.
2) Draw an arc on \(\overline{\mathrm{OA}}\) from the centre ‘O’, its cuts \(\overline{\mathrm{OA}}\) at ‘C’.
3) Draw two arcs with the centres O, C with equal radii the two arcs meet at ‘D’.
4) Draw another two arcs with the centres D, C they meet at point E. Join O, E.
5) With the centres D, G draw another two arcs, they meet at point F. Join O, F.
∴ ∠AOB = 45°
So we get the required angle.

Students can practice TS SCERT Class 6 Maths Solutions Chapter 9 Introduction to Algebra Ex 9.2 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra Exercise 9.2

Question 1.
Write the expressions for the following statements.
(i) q is multiplied by 5.
Answer:
q × 5 = 5q

(ii) y is divided by 4.
Answer:
\(\frac{y}{4}\)

(iii) One fourth of the product of numbers p and q.
Answer:
The product of p and q = p × q = pq
One fourth of the product = \(\frac{\mathrm{pq} \times 1}{4}\) = \(\frac{\mathrm{pq}}{4}\)

(iv) 5 is added to the three times z.
Answer:
3 times z = 3 × z = 3z
If 5 is added to 3z, then we get 3z + 5

(v) 9 times ‘n’ is added to ’10’.
Answer:
9 times ‘n’ = 9 × n = 9n;
If 10 is added 9n, then it becomes 9n + 10

(vi) 16 is subtracted from two times ‘y’.
Answer:
2 times ‘y’ = 2 × y = 2y;
If 16 is subtracted from 2y, it becomes (2y – 16)

(vii) ‘y’ is multiplied by 10 and then x is added to the product.
Answer:
y × 10 + x = 10y + x

Question 2.
Write two statements each for the following expressions.
(i) y – 11
Answer:
(a) In a class of ‘y’ students, 11 students failed in the examination. Find the number of passed students.
(b) The sum of two numbers is y. If one number is 11, then what is the other number ?

(ii) 10a
Answer:
(a) The cost of one pen is Rs. 10. What is the cost of ‘a’ such pens ?
(b) What is the value of 10 times ‘a’ ?

(iii) \(\frac{x}{5}\)
Answer:
(a) The cost of 5 books is .Rs. ’x’.
Then what is the cost of one book?
(b) What is the value of one fifth of x?

(iv) 3m + 11
Answer:
(a) Add 11 to 3 times of m.
(b) The length of a room is 11 more than 3 times of its breadth (say m).

(v) 2y – 5
Answer:
(a) Subtract 5 from 2 times of ‘y’.
(b) The age of A is y and the age of B is 5 less than 2 times of the age of A. What is the age of B ?

Question 3.
Peter has ‘p’ number of balls. Number of balls with David is 3 times the balls with Peter. Write this as an expression.
Answer:
Number of balls that Peter has = p
Number of balls with David is 3 times the balls with Peter.
∴ Number of balls with David is 3p.

Question 4.
Sita has 3 more notebooks than Githa. Find the number of books that Sita has ? Use any letter for the number of books that Githa has.
Answer:
Let the number of books that Githa be x.
Sita has 3 more books than Githa.
∴ The number of books that Sita has = x + 3

Question 5.
Cadets are marching in a parade. There are 5 cadets in each row. What is the rule for the number of cadets, for a given number of rows ? Use ‘n’ for the number of rows.
Answer:
The number of cadets in each row is 5. Let the number of rows be ‘n’.
Total number of cadets in ‘n’ rows = 5 × n = 5n
∴ The required rule is 5n.

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 14 Understanding 3D and 2D Shapes InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 14 Understanding 3D and 2D Shapes InText Questions

Try These

Question 1.
(i) What is the shape of the face of a cube ?
(ii) What is the shape of the face of a cuboid ?
Answer:
i) Square shape.
ii) Rectangular shape.

Question 2.
Ramesh has collected some boxes in his room. Pictures of these are given here. How many are cubes and how many are cuboids?

Answer:
Number of cubes = 3
Number of cuboids = 4

Question 3.
Ajith has made a cuboid by arranging cubes of 2 cms each. What is the length, breadth and height of the cuboid so formed ?

Answer:
Length of the first cuboid = 2 + 2 = 4 cm
breadth = 2 cm height = 2cm
Length of the second cuboid =2 + 2 + 2 = 6 cm
breadth = 2 cm height = 2 cm

Think, Discuss And Write

Question 1.
What is the difference between a cylinder and a cone with respect to the number of faces, vertices and edges? Discuss with your friends.
Answer:

Do This

Question 1.
Fill the table accordingly :

The cylinder, the cone and the sphere have no straight edges. What is the base of a cone ? Is it a circle ? The cylinder has two bases. What shape is the base ? Of course, a sphere has no face ! Think about it.
Answer:

The base of a cone is circle.
The shape of the base of a cylinder is circle.

Do This

Question 1.
Draw ten polygons with different shapes in your notebook.
Answer:

Question 2.
Use match-sticks or broom-sticks and form closed figures using :
i) Six sticks
ii) Five sticks
iii) Four sticks
iv) Three sticks
v) Two sticks
In which case was it not possible to form a polygon ? Why ?
Answer:

Observation : In the fifth case it is not possible to form a polygon.
Conclusion : We find that we could not form a polygon using two sticks.
Reason : A polygon must have at least three sides.

Try This

Question 1.
Find out the differences :

Measure the lengths of the sides and angles of (i) and (ii). What did you find ?
Answer:
i) Length of sides
AB = 3.2 cm
BC = 1. 2 cm
CD = 2 cm
ED – 2 6 cm
AE = 1. 6 cm
angles ∠A = 125°
∠B = 70°
∠C = 160°
∠D = 105°
∠E = 80°
∴ sides and angles are not equal.
The given figure is an irregular polygon (pentagon).
∴ Sides, angles are not equal.

ii) In the given figure AB = BC = CD = DE
= AE = 1.9 cm
Each angle = 108°
angles ∠A = ∠B = ∠C = ∠D = ∠E = 108°
∴ Sides, angles are equal.
The given figure is a regular pentagon.

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 12 Symmetry Ex 12.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 12 Symmetry Exercise 12.1

Question 1.
Check whether the given figures are symmetric or not ? Draw the line of symmetry as well.

Answer:

The dotted lines represents a line of symmetry.

Question 2.
Draw a line of symmetry for each of the figures, wherever possible.

Answer:

Question 3.
In the figure, l is the line of symmetry.
Complete the diagram to make it symmetric.

Answer:

Question 4.
Complete the figures such that the dotted line is the line of symmetry.

Answer: