Srinivas

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 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 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:

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 7 Fractions and Decimals InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals InText Questions

Try These

Question 1.
How will you represent the following pictorially ?
(i) \(\frac{3}{4}\)
Answer:
\(\frac{3}{4}\)

(ii) \(\frac{2}{8}\)
Answer:
\(\frac{2}{8}\)

(iii) \(\frac{1}{3}\)
Answer:
\(\frac{1}{3}\)

(iv) \(\frac{5}{8}\)
Answer:
\(\frac{5}{8}\)

Question 2.
Write the fraction representing the shaded portion.

Answer:
(i) \(\frac{1}{3}\)
(ii) \(\frac{1}{4}\)
(iii) \(\frac{2}{6}\)

Do This

Question 1.
Write 5 proper fractions and draw them pictorially.
Answer:

Question 2.
Rani says that shaded portion in the given figure represents \(\frac{1}{4}\). Do you agree with her? Give reason to support your answer.

Answer:
Yes. The reason is out of fourportions only one portion is shaded.

Do This

Question 1.
Write improper fractions represented by the following pictures.

Answer:

Answer:

Answer:

Question 2.
Represent the following fractions pictorially :
\(\frac{7}{4}, \frac{5}{3}, \frac{7}{6}\)
Answer:

Do This

Question 1.
Write the following as mixed fractions
\(\frac{7}{2}, \frac{8}{5}, \frac{9}{4}, \frac{13}{5}, \frac{17}{3}\)
Answer:

Try These

Question 1.
Write the numerator and denominators of the following fractional numbers:
\(\frac{1}{3}, \frac{2}{5}, \frac{7}{2}, \frac{19}{3}, \frac{7}{29}, \frac{11}{13}, \frac{1}{7}, \frac{8}{3}\)
Answer:

Question 2.
Sort the following fractions into the category of proper and improper fractions. Also write improper fraction as mixed fractions:
\(\frac{1}{3}, \frac{2}{5}, \frac{7}{2}, \frac{19}{3}, \frac{7}{29}, \frac{11}{13}, \frac{1}{7}, \frac{8}{3}\)
Answer:
Improper fractions:

Proper fractions:
\(\frac{1}{3}, \frac{2}{7}, \frac{3}{5}, \frac{1}{9}\)

Do This

Question 1.
Show that following on number lines.
(i) \(\frac{7}{6}\)
Answer:

(ii) \(\frac{5}{2}\)
Answer:

(iii) \(\frac{7}{5}\)
Answer:

(iv) \(\frac{9}{6}\)
Answer:

Question 2.
Consider the following numbers. Which of these would lie on the number line
i) before 1
Answer:
Fractions before 1 are \(\frac{1}{3}, \frac{7}{9}, \frac{6}{11}\)

ii) between 1 and 2
Answer:
Fractions between 1 and 2 are \(\frac{7}{5}, \frac{9}{5}\)

Try These

Question 1.
Write 5 fractional numbers that are in the standard form.
Answer:
Standard form of 5 fractional numbers
= \(\frac{1}{2}, \frac{2}{3}, \frac{4}{5}, \frac{5}{6}, \frac{8}{9}\)

Question 2.
Write 5 fractional numbers that are not in standard form.
Answer:
\(\frac{10}{6}, \frac{144}{100}, \frac{51}{17}, \frac{48}{16}, \frac{36}{44}\)

Question 3.
Convert the following fractions into their standard form.
i) \(\frac{7}{28}\)
ii) \(\frac{15}{90}\)
iii) \(\frac{11}{33}\)
iv) \(\frac{39}{13}\)
Answer:

Think, Discuss And Write

Question 1.
Rafi says “there can be no equivalent fractions that are also like fractions”. Do you agree with him ? Explain your answer and justify.
Answer:
Equivalent fractions = \(\frac{1}{2}, \frac{2}{4}, \frac{8}{16}, \frac{32}{64}\)
Like fraction = \(\frac{3}{2}, \frac{5}{2}, \frac{7}{2}, \frac{8}{2}, \frac{9}{2}\)
Equivalent fractions are like fractions also. So, I agree with him.

Do This

Question 1.
Identify the biggest and the smallest in these group of fractional numbers.
(i) \(\frac{1}{7}, \frac{3}{7}, \frac{2}{7}, \frac{5}{7}\)
Answer:
Biggest fraction = \(\frac{5}{7}\)
Smallest fraction = \(\frac{1}{7}\)

(ii) \(\frac{1}{9}, \frac{13}{9}, \frac{11}{9}, \frac{5}{9}\)
Answer:
Biggest fraction = \(\frac{13}{9}\)
Smallest fraction = \(\frac{1}{9}\)

(iii) \(\frac{1}{3}, \frac{5}{3}, \frac{17}{3}, \frac{9}{3}\)
Answer:
Biggest fraction = \(\frac{17}{3}\)
Smallest fraction = \(\frac{1}{3}\)

Question 2.
Which of these is the smaller fraction ?
(i) \(\frac{2}{5}, \frac{3}{7}\)
Answer:

(ii) \(\frac{7}{8}, \frac{5}{4}\)
Answer:

(iii) \(\frac{3}{11}, \frac{1}{2}\)
Answer:

(iv) \(\frac{5}{6}, \frac{2}{3}\)
Answer:

Question 3.
Write the following fractional number in ascending order.
(i) \(\frac{1}{7}, \frac{13}{7}, \frac{11}{7}, \frac{5}{7}, \frac{15}{7}\)
Answer:
Ascending order:
\(\frac{1}{7}<\frac{5}{7}<\frac{11}{7}<\frac{13}{7}<\frac{15}{7}\)

(ii) \(\frac{2}{3}, \frac{5}{6}, \frac{3}{9}, \frac{24}{18}\)
Answer:

(iii) \(\frac{2}{3}, \frac{1}{2}, \frac{5}{6}, \frac{7}{12}\)
Answer:

(iv) \(\frac{1}{5}, \frac{1}{2}, \frac{1}{8}, \frac{1}{3}, \frac{1}{12}\)
Answer:

Do this

Question 1.
Write the following in descending order.
i) \(\frac{1}{9}, \frac{13}{9}, \frac{11}{9}, \frac{15}{9}, \frac{3}{9}\)
Answer:

ii) \(\frac{1}{6}, \frac{2}{3}, \frac{3}{9}, \frac{5}{6}\)
Answer:

iii) \(\frac{1}{5}, \frac{9}{5}, \frac{3}{5}, \frac{6}{5}\)
Answer:
Descending Order = \(\frac{9}{5}>\frac{6}{5}>\frac{3}{5}>\frac{1}{5}\)

iv) \(\frac{1}{4}, \frac{1}{2}, \frac{1}{8}, \frac{3}{4}\)
Answer:
Descending Order = \(\frac{3}{4}>\frac{1}{2}>\frac{1}{4}>\frac{1}{8}\)

Question 8.
Simplify:
(i) \(\frac{1}{18}+\frac{1}{18}\)
Answer:
\(\frac{1}{18}+\frac{1}{18}=\frac{1+1}{18}=\frac{2}{18}=\frac{1}{9}\)

(ii) \(\frac{8}{15}+\frac{3}{15}\)
Answer:
\(\frac{8}{15}+\frac{3}{15}=\frac{8+3}{15}=\frac{11}{15}\)

(iii) \(\frac{7}{7}-\frac{5}{7}\)
Answer:
\(\frac{7}{7}-\frac{5}{7}=\frac{7-5}{7}=\frac{2}{7}\)

(iv) \(\frac{1}{22}+\frac{21}{22}\)
Answer:
\(\frac{1}{22}+\frac{21}{22}=\frac{1+21}{22}=\frac{22}{22}\) = 1

(v) \(\frac{12}{15}-\frac{7}{15}\)
Answer:
\(\frac{12}{15}-\frac{7}{15}=\frac{12-7}{15}=\frac{5}{15}=\frac{1}{3}\)

(vi) \(\frac{5}{8}+\frac{3}{8}\)
Answer:
\(\frac{5}{8}+\frac{3}{8}=\frac{5+3}{8}=\frac{8}{8}\) = 1

(vii) 1 – \(\frac{2}{3}\left(1=\frac{3}{3}\right)\)
Answer:
1 – \(\frac{5}{8}+\frac{3}{8}=\frac{5+3}{8}=\frac{8}{8}\) [∵ 1 = \(\frac{3}{3}\)]

(viii) \(\frac{1}{4}+\frac{0}{4}\)
Answer:
\(\frac{1}{4}+\frac{0}{4}=\frac{1+0}{4}=\frac{1}{4}\)

(ix) 3 – \(\frac{12}{5}\)
Answer:
3 – \(\frac{12}{5}=\frac{3 \times 5}{1 \times 5}-\frac{12}{5}=\frac{15}{5}-\frac{12}{5}=\frac{15-12}{5}=\frac{3}{5}\)

Question 9.
Fill in the missing fractions.
i) \(\frac{7}{10}\) – ___ = \(\frac{3}{10}\)
Answer:

ii) __ – \(\frac{3}{21}=\frac{5}{21}\)
Answer:

iii) ___ – \(\frac{3}{3}=\frac{3}{6}\)
Answer:

iv) ___ – \(\frac{5}{27}=\frac{12}{27}\)
Answer:

Question 10.
Narendra painted \(\frac{2}{3}\) area of the wall in his room. His brother Ritesh helped and painted \(\frac{1}{3}\) area of the wall. How much did they paint together ?
Answer:
Area of the wall painted by Narendra = \(\frac{2}{3}\)
Area of the wall painted by Ritesh = \(\frac{1}{3}\)
Area of the wall painted by both
Narendra and Ritesh = \(\frac{2}{3}+\frac{1}{3}\)
= \(\frac{2+1}{3}=\frac{3}{3}\) = 1
Narendra and his brother Ritesh painted the complete wall.

Question 11.
Neha was given \(\frac{5}{7}\) of a basket of bananas. What fraction of bananas was left in the basket?
Answer:
The part of a basket of bananas given to Neha = \(\frac{5}{7}\)
The part of bananas left in the basket
= 1 – \(\frac{5}{7}=\frac{1 \times 7}{1 \times 7}-\frac{5}{7}=\frac{7}{7}-\frac{5}{7}=\frac{7-5}{7}=\frac{2}{7}\)

Question 12.
A piece of rod \(\frac{7}{8}\) metre long is broken into two pieces. One piece was \(\frac{1}{4}\) metre long. How long is the other piece?
Answer:
Length of a piece of rod = \(\frac{7}{8}\) metre
Length of one broken piece of rod = \(\frac{1}{4}\) metre.
Length of the other piece = \(\frac{7}{8}-\frac{1}{4}\)
= \(\frac{7}{8}-\frac{1 \times 2}{4 \times 2}=\frac{7}{8}-\frac{2}{8}=\frac{7-2}{8}=\frac{5}{8}\)m
∴ \(\frac{5}{8}\)m long is the other piece.

Question 13.
Renu takes 2\(\frac{1}{5}\) minutes to walk 5 around the school ground. Snigdha takes \(\frac{7}{4}\) minutes to do the same. Who takes less time and by what fraction?
Answer:
Time taken by Renu to walk around the school ground = 2\(\frac{1}{5}\) minutes
= \(\frac{11}{5}\) minutes
Time taken by Snigdha to walk around the school ground = \(\frac{7}{4}\) minutes
To find the person who takes less time to do the same, we write the fractions \(\frac{11}{5}\) and \(\frac{7}{4}\) having the same denominators.
\(\frac{11}{5} \times \frac{4}{4}=\frac{44}{20} ; \frac{7}{4} \times \frac{5}{5}=\frac{35}{20}\)
We know that \(\frac{35}{20}<\frac{44}{20}\)
Therefore, Snigdha takes \(\frac{9}{20}\) minutes less time to walk around the school ground.
(∵ \(\frac{44}{20}-\frac{35}{20}=\frac{44-35}{20}=\frac{9}{20}\))

Try These

Question 1.
(i) Write fractions for the following decimal and also find how many tenth parts are there in each :
0.4, 0.2, 0.8, 1.6, 5.4, 555.3, 0.9
Answer:
0.4 = \(\frac{4}{10}\)
0.2 = \(\frac{2}{10}\)
0.8 = \(\frac{8}{10}\)
4 tenth parts 2 tenth parts 8 tenth parts

1.6 = 1\(\frac{6}{10}\) = 1 + \(\frac{6}{10}\) 6 tenth parts
5.4 = 5\(\frac{4}{10}\) = 5 + \(\frac{4}{10}\) 4 tenth parts
555.3 = 553 + \(\frac{3}{10}\) = 3 tenth parts
0.9 = \(\frac{9}{10}\) = 9 tenth parts

(ii) Complete the following table.

Answer:

(iii) Complete the following table.

Answer:

(iv) Measure the length of these line segments and fill it in the table given below.

Answer:

Do This

Question 1.
Fill in the blanks.
(i) 325 paise = _________ rupees _________ paise = ₹ _________
(ii) 570 paise = _________ rupees _________ paise = ₹ _________
(iii) 2050 paise = _________ rupees _________ paise = ₹ _________
Answer:
(i) 325 paise = 3 rupees 25 paise = ₹ 3.25
(ii) 570 paise = 5 rupees 70 paise = ₹ 5.70
(iii) 2050 paise = 20 rupees 50 paise = ₹ 20.50

Do This

Question 1.
Find: i) 0.39 + 0.26
ii) 0.8 + 0.07
iii) 1.45 + 1.90
iv) 3.44 + 1.58
Answer:

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 7 Fractions and Decimals Ex 7.5 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Exercise 7.5

Question 1.
Sonu went to a shop. He wanted to buy a chiki and a toffee. One chiki costs ₹ 0.75 and a toffee costs ₹ 0.50. If he buys one each of them how much he has to pay to the shopkeeper. Sonu’s mother gave him ₹ 2. How much he will get in return? Suppose if his mother gave him ₹ 5, then how much will the shopkeeper return ?
Answer:
Cost of one chiki is ₹ 0.75
Cost of one toffee is₹ 0.50
Cost of one chiki and one toffee
= ₹ 0.75 + ₹ 0.50
= \(\frac{75}{100}+\frac{50}{100}=\frac{125}{100}\) = ₹ 1.25
The amount to be paid to the shop-keeper by Sonu = ₹ 1.25
Amount given to Sonu by his mother
= ₹ 2.00
The amount Sonu will get in return
= ₹ 2.00 – ₹ 1.25 = ₹ 0.75
If Rs. 5 is given by his mother, the amount returned by shopkeeper.
= ₹ 5.00 – ₹ 1.25 = ₹ 3.75
₹ 3.75 will the shopkeeper return.

Question 2.
Add the following decimal fractions,
i) 25.11 + 3.80
ii) 14.01 + 1.1 + 1.98
iii) 9.85 + 0.61
iv) 2.3 + 18.94
v) 2.57+ 3.75
Answer:
i) 25.11 + 3.80

∴ 25.11 + 3.80 = 28.91

ii) 14.01 + 1.1 + 1.98

∴ 14.01 + 1.1 + 1.98 = 17.09

iii) 9.85 + 0.61

∴ 9.85 + 0.61 = 10.46

iv) 2.3 + 18.94

∴ 2.3 + 18.94 = 21.24

v) 2.57 + 3.75

∴ 2.57 + 3.75 = 6.32

Question 3.
Abhishek travelled 5 km. 28 m. by bus, 2 km. 265 m. by car and the rest 1 km. 30 m. on foot. How much distance did he travel in all ?
Answer:
Abhishek travelled a distance by bus-= 5 km. 28m
Abhishek travelled a distance by car = 2 km. 265 m
Abhishek travelled a distance by foot = 1 km. 30 m
∴ Total distance did he travel in all = 5 km.
28 m + 2 km. 265 m + 1 km.30 m = 8 km. 323 m.

Question 4.
Mrs.Vykuntam bought 6.25 m of dress material for her elder daughter and 5.75 m for the younger one. How much dress material did she buy for her daughters ?
Answer:
Dress material for elder daughter
= 6.25 m
Dress material for younger one
= 5.75 m
∴ Total dress material bought for two daughters
= 6.25 m + 5.75 m
= 12 m.

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 7 Fractions and Decimals Ex 7.4 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Exercise 7.4

Question 1.
Fill in the blanks.
i) The fractional form of 0.8 is
ii) The integral part of 15.9 is
iii) The digit in the tenths place of 171.9 is
iv) The place value of 8 in 9.8 is
v) The point between the integral part and the decimal part of the decimal number is called
Answer:
i) \(\frac{8}{10}\)
ii) 15
iii) 9
iv) tenth (or) \(\frac{8}{10}\)
v) decimal point

Question 2.
Write the decimal for each of the following.
i) One hundred twenty five and four tenths
ii) Twenty and two tenths
iii) Eight and six tenths.
Answer:
i) 125.4
ii) 20.2
iii) 8.6

Question 3.
Write the following fractions in the decimal form using the decimal point.
(i) \(\frac{16}{100}\)
Answer:
\(\frac{16}{100}\) = 0.16

(ii) \(\frac{278}{1000}\)
Answer:
\(\frac{278}{1000}\) = 0.278

(iii) \(\frac{6}{100}\)
Answer:
\(\frac{6}{100}\) = 0.06

(iv) \(\frac{369}{100}\)
Answer:
\(\frac{369}{100}\) = 3.69

(v) \(\frac{16}{1000}\)
Answer:
\(\frac{16}{1000}\) = 0.016

(vi) \(\frac{345}{10}\)
Answer:
\(\frac{345}{10}\) = 34.5

(vii) \(\frac{907}{100}\)
Answer:
\(\frac{907}{100}\) = 9.07

Question 4.
Write the place value of each underlined digit. ‘
i) 34.26
ii) 8.88
iii) 0.91
iv) 0.50
v) 3.03
vi) 6.74
Answer:
i) The underlined digit is in ones place. So its place value is 4 x 1 = 4
ii) The underlined digit is in hundredths place. So its place value is \(\frac{8}{100}\)
iii) The underlined digit is in tenths place. So its place value is \(\frac{9}{10}\).
iv) The underlined digit is in tenths place. So its place value is \(\frac{5}{10}\).
v) The underlined digit is in hundredths place. So its place value is \(\frac{3}{100}\).
vi) The underlined digit is in tenths place. So its place value is \(\frac{7}{10}\).

Question 5.
Find greater in the following.
(i) 0 0.2 (or) 0.4
(ii) 70.08 (or) 70.7
(iii) 6.6 (or) 6.58
(iv) 7.4 (or) 7.35
(v) 0.76 (or) 0.8
Answer:
(i) 0.2 or 0.4 2
0.2 = \(\frac{2}{10}\) and 0.4 = \(\frac{4}{10}\)
\(\frac{2}{10}\) means 2 parts out of 10 parts.
\(\frac{4}{10}\) means 4 parts out of 10 parts.
∴ 4 > 2
(i.e) 0.4 > 0.2

(ii) 70.08 (or) 70.7
Whole part of both is 70.
0.08 = \(\frac{8}{100}\); 0.7 = \(\frac{7}{10}=\frac{7 \times 10}{10 \times 10}=\frac{70}{100}\)
0.08 means 8 parts out of 100 parts
0.7 means 70 parts out of 100 parts
∴ 70 > 8
(i.e.,) 70.7 > 70.08

(iii) 6.6 or 6.58
Whole part of both is 6.
0.6 = \(\frac{6}{10}=\frac{6 \times 10}{10 \times 10}=\frac{60}{100}\); 0.58 = \(\frac{58}{100}\)
0.6 means 60 parts out of 100 parts.
0.58 means 58 parts out of 100 parts.
∴ 60 > 58
(i.e.,) 6.6 > 6.58

(iv) 7.4 or 7.35
Whole part of both is 7.
0.4 = \(\frac{4}{10}=\frac{4 \times 10}{10 \times 10}=\frac{40}{100}\); 0.35 = \(\frac{35}{100}\)
0.4 means 40 parts out of 100 parts.
0.35 means 35 parts out of 100 parts.
∴ 40 > 35
(i.e.,) 7.4 > 7.35

(v) 0.76 or 0.8
Whole part of both is zero.
0.76 = \(\frac{76}{100}\); 0.8 = \(\frac{8}{10}=\frac{8 \times 10}{10 \times 10}=\frac{80}{100}\)
0.76 means 76 parts out of 100 parts.
0.8 means 80 parts out of 100 parts.
∴ 80 > 76
(i.e.,) 0.8 > 0.76

Question 6.
Rewrite in ascending order.
i) 0.04, 1.04, 0.14, 1.14
ii) 9.09, 0.99, 1.1, 7
Answer:
i) 0.04,1.04,0.14,1.14
By observation, we can say that
0.04 < 0.14 < 1.04 < 1.14 (ascending order)

ii) 9.09, 0.99, 1.1, 7
By observation, we can say that
0.99 < 1.1 < 7 < 9.09 (ascending order)

Question 7.
Rewrite in descending order
i) 8.6, 8.59, 8.09, 8.8
ii) 6.8, 8.66, 8.06, 8.68
Answer:
i) 8.6,8.59,8.09, 8.8
By observation, we can say that
8.8 > 8.6 > 8.59 > 8.09 (descending order)

ii) 6.8, 8.66, 8.06, 8.68
By observation, we can say that
8.68 > 8.66 > 8.06 > 6.8 (descending order)

Students can practice TS SCERT Class 6 Maths Solutions Chapter 10 Perimeter and Area Ex 10.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 10 Perimeter and Area Exercise 10.1

Question 1.
Find the perimeter of each of the following shapes :

Answer:
(i) Shape – 1 :
Given that AB = 40 cm ; BC = 50 cm
CD = 35 cm ; DE = 60 cm
and EA = 45 cm
Perimeter of shape – 1 = AB + BC + CD + DE + EA
= (40 + 50 + 35 + 60 + 45) cm
= 230 cm

(ii) Shape – 2 : Given that
AB = 8 cm; i.e., AB = HG = 8 cm;
GF = 2 cm; i.e., GF = HI = 2 cm;
ED = 5 cm; i.e., ED = JK = 5 cm;
IJ = 3 cm; i.e., IJ = LK = 3 cm;
AL = 3 cm; i.e., AL = BC = 3 cm
CD = EF = 3 cm
Perimeter of shape – 2 = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA
= (8 + 3 + 3 + 5 + 3 + 2 + 8 + 2 + 3 + 5 + 3+ 3)cm
= 48 cm

(iii) Shape – 3 : Given that
AB = 6 cm ; BC = 2 cm ; CD = 2 cm ;
DE = 2 cm ; EF = 2 cm ; FG = 2 cm ;
GH = 2 cm ; and HA = 6 cm
Perimeter of shape – 3
= AB + BC + CD + DE + EF + FG + GH + HA
= (6 + 2 + 2 + 2 + 2 + 2 + 2 + 6) cm
= 24 cm

(iv) Shape – 4 : Given that
AL = 4 cm (i.e.) AL = BC = 4 cm
AB = 2 cm (i.e.) AB = HG = 2 cm
JK = 2 cm (i.e.) JK = ED = 2 cm
GF = 4 cm (i.e.) GF = HI = 4 cm
KL = 4 cm (i.e.) KL = IJ = 4 cm
EF = 4 cm (i.e.) EF = CD = 4 cm
Perimeter of shape – 4
= AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA
= 2 + 4 + 4 + 2 + 4 + 4 + 2 + 4 + 4 + 2 + 4 + 4
= 40 cm

Question 2.
Find the perimeter of each of the following figures.

What would be cost of putting a wire around each of these shapes given that 1 cm wire costs ₹ 15 ?
Answer:
(i)

Perimeter = AB + BC + CA
= (30 + 55 + 35) cm
= 120 cm
Cost of putting a wire around the given shape = 120 × Rs.15 = Rs. 1800

(ii)

ABCD is a rectangle so opposite sides are equal.
AB = CD = 40 cm
BC = AD = 20 cm
Perimeter = AB + BC + CD + DA
= 40 + 20 + 40 + 20
= 120 cm
Cost of putting a wire around the given shape at Rs.15 per 1 cm = 120 × Rs.15
= Rs. 1800

(iii)

ABCD is a square, so all the sides are equal.
∴ AB = BC = CD = DA = 30 cm
Perimeter = AB + BC + CD + DA
= 30 + 30 + 30 + 30
= 120 cm
Cost of putting a wire around the given shape = 120 × Rs.15
= Rs. 1800

(iv)

Given that the length of one side = 24 cm
ABCDEF is a regular hexagon (i.e.) it contains 6 equal sides.
Perimeter = 6 × side
= 6 × 24cm = 144 cm
Cost of putting a wire around the given shape = 144 × Rs.15 = Rs. 2160

Question 3.
How many different rectangles can you make with a 24 cm long string with integral sides and what are the sides of those rectangles in cm ?
Answer:
The length of string = 24 cm
Perimeter of the rectangle = 2(l + b) =24 cm
l + b = \(\frac{24}{2}\)cm = 12 cm
We can make different rectangles with 24 cm. long string as follows.

Question 4.
A flower bed is in the shape of a square with a side 3.5 m. Each side is to be fenced with 4 rows of ropes. Find the cost of rope required at ₹ 15 per meter.
Answer:
The shape of a flower bed is a square.
The length of the side of the square
= 3.5 m
Length of 4 rows of ropes on each side = 4 × 3.5 m = 14 m
Length of the rope on 4 sides
= 14 m × 4 = 56 m
Cost of rope required at Rs. 15 per meter = Rs. 56 × 15 = Rs. 840

Question 5.
A piece of wire is 60 cm long. What will be the length of each side if the string is used to form :
(i) an equilateral triangle
(ii) a square
(iii) a regular hexagon
(iv) a regular pentagon
Answer:
(i) Length of the wire = 60 cm
The perimeter of an equilateral triangle with side x cm = 3 × x = 3x
By problem,
3x = 60 cm
∴ x = \(\frac{60}{3}\) cm = 20 cm
Length of each side of the equilateral triangle = 20 cm

(ii) Length of the wire = 60 cm
The perimeter of the square with side ‘x’ cm = 4 × x = 4x
By problem,
4x = 60 cm
x= \(\frac{60}{4}\) cm = 15 cm
Length of each side of the square =15 cm

(iii) Length of the wire = 60 cm
The perimeter of the regular hexagon with side x’ cm = 6 × x = 6x
By problem,
6x = 60 cm
∴ x = \(\frac{60}{5}\) cm = 10 cm 6
Length of each side of the regular hexagon = 10 cm

(iv) Length of the wire = 60 cm
The perimeter of the regular pentagon with side ‘x’ cm = 5 × x = 5x
By problem,
5x = 60 cm
x = \(\frac{60}{5}\) cm = 12 cm
Length of each side of the regular pentagon = 12 cm.

Question 6.
Bunty and Bubly go for jogging every morning. Bunty goes around a square park of side 80 m. Bubly goes around a rectangular park with length 00 m and breadth 60 m. If they both fake 3 rounds, who covers more distance and by how much ?
Answer:
Bunty goes around a square park.
The length of the side of square park
= 80 m.
The perimeter of the square park
= 4 × 80 = 320 m.
Distance covered by Bunty in 3 rounds = 320m × 3 = 960 m.
Bubly goes around a rectangular park.
The length and breadth of the park are 90 m and 60 m respectively.
The perimeter of the rectangular park = 2 (length + breadth)
= 2[90 + 60]
= 2 × 150
= 300 m
Distance covered by Bubly in 3 rounds = 300m × 3 = 900 m.
Bunty covers greater distance by 60 m. (∵ 960 – 900 = 60 m)

Question 7.
The length of a rectangle is twice of its breadth. If its perimeter is 48 cm, find the dimensions of the rectangle.
Answer:
The perimeter of the rectangle = 48 cm Let the breadth of the rectangle be x cm The length of the rectangle .
= 2 × x = 2x cm Perimeter of the rectangle
= 2 (length + breadth)
= 2 (2x + x)cm
= 2 × 3x = 6x
By problem,
6x = 48
∴ x = \(\frac{48}{6}\) = 8
Breadth of the rectangle = 8 cm
Length of the rectangle = 2 × 8 = 16 cm

Question 8.
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is the length of third side ?
Answer:
Two sides of a triangle are 12 cm and 14 cm.
Length of third side = x cm .
Perimeter of the triangle
= sum of the three sides = 12 + 14 + x = (26 + x) cm
By problem,
(26 + x) cm = 36 cm
∴ x = 36 – 26 = 10 cm
Length of third side of the triangle
= 10 cm.

Question 9.
Find the perimeter of each of the following shapes:
(i) A triangle of sides 3 cm., 4 cm. and 5 cm.
(ii) An equilateral triangle of side 9 cm.
(iii) An isosceles triangle with equal sides 8 cm each and third side of 6 cm.
Answer:
(i) The sides of a triangle are 3 cm, 4 cm, 5 cm
Perimeter of the given triangle
= sum of the three sides = (3 + 4 + 5)cm
= 12 cm

(ii) Side of the equilateral triangle = 9 cm
Perimeter of the equilateral triangle = (9 + 9 + 9)cm
= 27 cm (∵ All the 3 sides are equal)

(iii) The length of one of the equal sides of the isosceles triangle is 8 cm.
Length of the third side is 6 cm.
Perimeter of the given triangle
= (8 + 8 + 6) cm
= 22 cm

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 8 Data Handling InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 8 Data Handling InText Questions

Try This

Question 1.
Give two examples of data in numerical figures.
Answer:
In a class of 36 students 15 like mangoes, 12 like apples, 9 like bananas.

Total number of students = 36

Question 2.
Give two examples of data in words.
Answer:
In a village of 100 families 45 read Eenadu, 35 read Sakshi, 20 read Andra Jyothi news papers.

Think, Discuss And Write

Question 1.
In what way Is the bar graph better than the pictograph?
Answer:

• Pictographs are difficult and time consuming to construct. But bargraphs easy to draw.
• In pictographs, each pictorial symbol represents a fixcd number of units. So, it becomes a problem to represent and read any fraction of that unit. For example suppose each pictorial symbol represents 50 units. Then 231 units are represented by 4 full pictures and a proportional fraction of the fifth. This proportional representation introduces error and is quite different to guess the correct value.
• But each bar represents only one value. The length or height of the bar indicates the value of the item. So, we represent correct value with bargraphs.
• Thats why hargraph is better than pictograph.

Students can practice TS SCERT Class 6 Maths Solutions Chapter 10 Perimeter and Area Ex 10.2 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 10 Perimeter and Area Exercise 10.2

Question 1.
Find the areas of the following figures by counting squares.
(i) 50 cm and 20 cm
(ii) 65 in and 45 in
(iii) 25 cm and 16 cm
(iv) 7 kin and 19 km
Answer:
(i) Length of the rectangle = 50 cm
Breadth of the rectangle = 20 cm
∴ Area of the rectangle
= length × breadth
= 50 cm × 20 cm
= 1000 sq. cm (cm²)

(ii) Length of the rectangle = 65 m
Breadth of the rectangle = 45 m
∴ Area of the rectangle
= length × breadth
= 65 cm × 45 cm
= 2925 sq. m (m²)

(iii) Length of the rectangle = 25 cm
Breadth of the rectangle = 16 cm
∴ Area of the rectangle
= length × breadth
= 25 cm × 16 cm
= 400 sq. cm (cm²)

(iv) Length of the rectangle = 7 km
Breadth of the rectangle =19 km
∴ Area of the rectangle
= length × breadth
= 7 cm × 19 cm
= 133 sq. km (km²)

Question 2.
Find the area of squares with the given sides:
(i) 26 m
(ii) 17 km
(iii) 52 cm
(iv) 8 cm
Answer:
(i) Side of the square = 26 m
Area of the square = side × side
= 26m × 26m
= 676 m²

(ii) Side of the square =17 km
Area of the square = side × side
= 17 cm × 17cm
= 289 km²

(iii) Side of the square = 52 cm
Area of the square = side × side
= 52 cm × 52cm
= 2704 cm²

iv) Side of the square = 8 cm
Area of the square = side × side
= 8 cm × 8 cm
= 64 cm²

Question 3.
The area of rectangular frame is 1,125 sq. cm. If its width is 25 cm, what is its length ?
Answer:
Area of rectangular frame =1125 sq.cm
Width of the frame = 25 cm
Area of rectangular frame
= length × width

Question 4.
The length of a rectangular field is 60 m and the breadth is half of its length. Find the area of the field.
Answer:
The length of a rectangular field = 60 m By problem,
Area of the rectangular field = \(\frac{60 \times 1}{2}\) = 30 m
= length × breadth
= 60 m × 30 m
= 1800 m²

Question 5.
A square sheet of paper has a perimeter of 40 cm. What is the length of its side ? Also find the area of the square sheet ?
Answer:
Let the side of the square sheet of paper = x cm.
Perimeter of the sheet of paper = 4 × side
= 4 × x
= 4x cm
By problem,
4x = 40
∴ x = \(\frac{40}{4}\) = 10 cm
Side of the square sheet of paper = 10 cm
Area of the sheet of paper = side × side
= 10 cm × 10 cm
= 100 cm²
Area of sheet of paper = 100 cm²

Question 6.
The area of rectangular plot is 2400 square meters and it’s length is 1\(\frac{1}{2}\) times to it’s breadth.What is the perimeter?
Answer:
Area of rectangular plot = 2400 m²
Length is 1\(\frac{1}{2}\) to its breadth i.e, their ratio of its sides = 3:2
Let the length of the plot be 3x m.
Breadth of the plot = 2x m
Area of the plot = length × breadth
= 3x m × 2x m
= 6x²
By problem, 6x² = 2400
x² = \(\frac{2400}{6}\) = 400
x² = 400 = 20 × 20
∴ x = 20
Length of the plot = 3x = 3 × 20 = 60 m
Breadth of the plot = 2x = 2 × 20 = 40 m
Perimeter of the plot
= 2(length + breadth)
= 2(60 + 40) m
= 2 × 100
= 200 m

Question 7.
The length and breadth of a room are 6 m and 4 m respectively. How many square meters of carpet is required to completely cover the floor of the room ? If the carpet costs ₹ 240 a square meter, what will be the total cost of the carpet for completely covering the floor ?
Answer:
Length of the room = 6 m
Breadth of the room = 4 m
Area of the floor of the room
= length × breadth
= 6m × 4m
= 24m²
The carpet required to completely cover the floor of the room = 24 m²
The cost of 1 sq. m of the carpet
= Rs. 240
∴ The cost of 24 sq. m of the carpet
= Rs. 240 × 24 = Rs. 5760

Question 8.
Two fields have the same perimeter. One is a square of side 72 m and another is a rectangle of length 80 m. Which field has the greater area and by how much ?
Answer:

Area of square = l × b
= 72 m × 72 m = 5184 sq.m
Length of a rectangle l = 80 m
Perimeter of a rectangle 2(l + b) = 288 m
= 2 × 80 m + 2b
= 288 m
= 160 m + 2b
= 288 m
⇒ 2b = 288 – 160 = 128 m
b = \(\frac{128}{2}\) = 64 m
Area of rectangle = 80 m × 64 m
= 5120 sq.m
∴ The area of square field is greater than that of rectangular field.
5184 – 5120
= 64 sq.m

Question 9.
The area of a square is 49 sq. dm. A rectangle has the same perimeter as the square. If the; length of the rectangle is 9.3 cm., What is its breadth ? Also find which has greater area ?
Answer:
Area of the square = 49 sq. cm
side × side = 7 cm × 7 cm
Side of the square = 7 cm
The perimeter of the square = 4 × side = 4 × 7 cm
= 28 cm
By problem,
The perimeter of the rectangle = 28 cm
2(l + b) = 28 cm 28
(l + b) = \(\frac{28}{2}\) = 14 cm
9.3 + b = 14
∴ b= 14 – 9.3 = 4.7 cm
Area of the rectangle = length × breadth
= 9.3 cm × 4.7 cm = 43.71 cm²
∴ The square has greater area when compared to the rectangle.

Question 10.
Rahul owns a rectangular field of length 400 m and breadth 200 m. His friend Ramu owns a square field of length 300 m. Find the cost of fencing the two fields at ₹ 150 per meter. If one tree can be planted in an area of 10 sq. m. who can plant more trees in his field ? How many more trees can he plant ?
Answer:
Length of the rectangular field of Rahul
= 400 m
Breadth of the rectangular field of Rahul
= 200 m
Perimeter of rectangular field of Rahul
= 2 [length + breadth]
= 2 [400 + 200]m = 2 × 600 m = 1200 m
Side of the square field of Ramu
= 300 m
Perimeter of square field of Ramu = 4 × side
= 4 × 300 m = 1200 m
Total length of the wire to be fenced = 1200 + 1200 = 2400 m
Cost of fencing 1 meter = Rs. 150

Cost of fencing 2400 meters
= Rs. 2400 × 150
= Rs. 3,60,000

Cost of fencing Ramu’s field
= Rs. 1200 × 150
= Rs. 1,80,000

Area of Rahul’s field = length × breadth
= 400 m × 200 m
= 80,000 m²

A tree occupies an area of 10 m²
Number of trees that can planted in 80,000
Rahul’s field = \(\frac{80,000}{10}\) = 8000
Area of Ramu’s field = side × side
= 300 × 300
= 90,000 m²

A tree occupies an area of 10 m²
Number of trees that can be planted in
Ramu’s field = \(\frac{90,000}{10}\) = 9000
Ramu can plant 1000 trees
(= 9000 – 8000) more.

Question 11.
The length of a rectangular floor is 20 m., more than its breadth. If the perimeter of the floor is 280 m, what is its length ?
Answer:
Let the breadth of the rectangular field = x m
∴ The length of the rectangular field = (x + 20) m

Perimeter of the rectangular field
= 2(length + breadth)
= 2(x + 20 + x) m
= 2(2x + 20)m
= 4x + 40 m
By problem,
4x + 40 = 280 m
4x = 280 – 40
= 240 m
x = \(\frac{240}{2}\) = 60 m

The breadth of the rectangular field
= 60 m
The length of the rectangular field
= x + 20 m
= 60 + 20 m
= 80 m
Its length is 80 m.

Question 12.
A rectangular plot of land is 240 m. by 200 m. The cost of fencing per meter is ₹ 30. What is the cost of fencing the entire field ?
Answer:
The dimensions of the rectangular plot of land are 240 m and 200 m.
Perimeter of the plot of land
= 2 [240 + 200]
= 2[440]m
= 2 × 440 m
= 880 m
Cost of fencing 1 meter is Rs. 30.
Cost of fencing 880 metres is x
Rs. 880 × 30 = Rs. 26400

Question 13.
The side of a square field is 120 meters. The cost of preparing a grass lawn is ₹ 35 per square meter. How much will it cost, if the entire field is converted into a lawn ?
Answer:
The side of the square field = 120 m
Area of the square field = side × side
= 120 m × 120 m
= 14,400 m²
Cost of preparing a sq. m of grass lawn is Rs. 35.
Cost of converting the entire field (14400 m²) into a lawn
= Rs. 14,400 × 35
= Rs. 5,04,000

Question 14.
What will happen to the area of rectangle, if
(i) its length and breadfii are doubled?
(ii) its length is doubled and breadth is tripled ?
Answer:
(i) Let the length of a rectangle be x m and the breadth be y m.
∴ Area of the rectangle = x × y = xy m²
The length and breadth of it are doubled.
Then the length will be 2x (= x + x) and breadth 2y (= y + y).
The area of the rectangle will be
2x × 2y = 4xy m² = 4 (xy m²)
∴ The area of the rectangle increases by 4 times.

(ii) The length after it is doubled will be
x + x = 2x
The breadth after it is tripled will be y + y + y = 3y
The area of the rectangle will be 2x × 3y
= 6 xy m²
= 6 × (xy m²)
∴ The area of the rectangle increases by 6 times.

Question 15.
What will happen to the area of square if its side is:
(i) doubled
(ii) halved
Answer:
(i) Let the side of a square be x metres.
Area of the square = side × side
= x × x
= x² sq. metres
If the side is doubled, it becomes x + x
= 2x metres

Area of the square = side x side
= 2x m × 2x m
= 4x² sq. metres
= 4 x (x² sq. metres)
The area increases by 4 times,

(ii) If the side of the square is halved, it
Area of the square = side × side
= \(\frac{1}{2}\) x m x \(\frac{1}{2}\) x m
= \(\frac{1}{4}\) x² sq. metres
= \(\frac{1}{4}\) × (x² sq. metres)
∴ The area of the square becomes \(\frac{1}{4}\) of the original area.