Class 6

Students can Download Chapter 5 Understanding Elementary Shapes Ex 5.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.2

Question 1.
What fraction of a clock wise revolution does the hour hand of a clock turn through, When it goes from
Solution:
We may observe that in 1 complete clock Wise revolution the hour hand will rotate by 360’
a) 3 to 9
When the hour hand goes from 3 to 9 clock wise, it will rotate by 2 right angles or 180°

b) 4 to 7
When the hour hand goes from 4 to 7 clock wise, it Will rotate by 1 right angle or 90°

c) 7 to 10
When the hour goes from 7 to 10 clock wise it will rotate by 1 right angle or 90°

d) 12 to 9
When the hour hand goes from 12 to 9 clock wise it rotate by 3 right angles or 270°

e) 1 to 10
When the hour hand goes from 1 to 10 clock wise it rotate by 3 right angles or 170°

f) 6 to 3
When the hour hand goes from 6 to 3 clock wise, it will rotate by 3 right angles or 270°

Question 2.
Where will the hand of a clock stop if it-
In 1 complete clock wise revolution the hand of a clock will rotate by 360°
Solution:
a) Starts at 12 and makes \(\frac{1}{2}\) of a revolution, clockwise?
If the hand of the clock start 12 and makes \(\frac{1}{2}\) of a revolution, clockwise, it will rotate by 180° and hence, it will stop at 6

b) Starts at 12 and makes \(\frac{1}{2}\) of a revolution, clockwise?
If the hand of the clock starts at 2 and makes \(\frac{1}{2}\) of a revolution clock wise, then it will rotate by 180° and hence, it will stop at 8

c) Starts at 5 and makes \(\frac{1}{4}\) of a revolution, clockwise?
If the hand of the clock starts at 5 and makes of a revolution clockwise than it will rotate by 90° and hence it will stop at 8

d) Starts 5 makes \(\frac{3}{4}\) of a revolution Clock wise?
Solution:
If the hand of the clock starts at 5 and makes of a revolution clockwise, than it will rotate by 270° and hence. it will stop at 2.

Question 3.
Which direction will you face if you start facing

a) East and make \(\frac{1}{2}\) of a revolution clockwise ?
If we start facing east and make \(\frac{1}{2}\) of a revolution clockwise, than we will face the west direction.

b) East and make \(1 \frac{1}{2}\) of a revolution clockwise?
If we starts facing east and make \(1 \frac{1}{2}\) of a revolution clock wise, than we will face the West direction

c) West and makes at \(\frac{3}{4}\) of a revolution anti-clock wise ?
If we starts facing west and make of a revolution anti – clockwise, than we face the north direction.

d) South and make one full revolution (should we specify clockwise or anti – clockwise for this last question? Why not?
There is no need to specify clockwise or anti-clockwise in the last question as turning by one full revolution (i.e., two straight angles) make us to reach the original position.

In case of revolving by 1 complete round the direction in which we are revolving does not matter in both cases clockwise or anti – clock wise, we will be back at our initial position.

Question 4.
What part of a revolution have you turned through if you stand facing
Solution:
If we revolve one complete round in either clock wise or anti-clockwise direction, then we will revolve by 360° and the two adjacent directions will be at 90° or \(\frac{1}{4}\) of a complete revolution away from each other

a) east and turn clock wise to face north?
If we start facing east and than turn clockwise to lace north, than we have to make revolution

b) South and turn clockwise to face east?
If we starts facing south and turn clock wise to face east than we have to make \(\frac{3}{4}\) of a revolution

c) West and turn clock wise to face east ?
If we starts facing west and turn dock wise to face east, then we have to make \(\frac{1}{2}\) of a revolution.

Question 5.
Find the number of right angles turned through by the hour hand of a clock when it goes from
Solution:
a) 3 to 6
The hour hand of a clock revolves by 360° of 4 right angles in 1 complete round

b) 2 to 8
The hour hand of a clock revolves by 180° or 2 right angles when it goes from 2 to 8

c) 5 to 11
The hour hand of a clock revolved by 180° or 2 right angles When it goes from 5 to 11

d) 10 to 1
The hour hand or a clock revolves by 90° or 1 right angle when it goes from 10 to 1

e) 12 to 9
The hour hand of a clock revolves by 2700 or 3 right angles When it goes from 12 to 9

f) 12 to 6
The hour hand of a clock revolves by 180° or 2 right angles When it goes from 12 to 6.

Question 6.
How many right angles do you make if you start facing
Solution:
If we revolve one complete round in either clock wise direction, then we will revolve by 360° or 4 right angles and the two adjacent direction will be at 90° or I right angles away from each other.
a) South and turn clock wise to west?
If we start facing south and turn clock wise to west then we make 1 right angle.
clockwise to west then we mu. iv

b) north and turn anti-clock wise to east ?
If we start facing north and turn anti clockwise to east then we make 3 right angles.

c) West and turn to west?
If we starts facing west and turn to west then we make 1 complete round or 4 right.

d) South and turn to north?
If we starts facing south and turn to we make right angles.

Question 7.
Where will the hour hand of a clock stop if it starts
Solution:
In 1 complete revolution ( Clockwise or anti – clock wise) the hour hand of a clock will rotate by 360° or 4 right angles.

a) from 6 and turns through 1 right angle?
If the hour hand of a clock starts from 6 and turns through 1 right angle, then it will stop at 9

b) From 8 and turn through 2 right angles?
If the hour hand of a clock starts from 8 and turned through 2 right angles, then it will stop at 2.

c) From 10 and turned through 3 right angles?
If the hour hand of a clock starts from 10 and turns through 3 right angles then it will stop at 7

d) From 7 and turns through 2 straight angles?
If the hour hand of a clock starts from 7 and turns through 2 straight angles, then it will stop at 7 .

Students can Download Chapter 12 Ratio and Proportion Ex 12.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.2

Question 1.
Determine if the following are in proportion.
a) 15, 45, 40, 120
Solution:

Therefore, 15 : 45 = 40 : 120
Hence, these are in proportion

b) 33, 121, 9, 96

Therefore 33 : 121 ≠ 9 : 96
Hence, these are not in proportion

c) 24, 28, 36, 48

Therefore, 24 : 28 ≠ 36 : 48
Hence, these are not in proportion

d) 32, 48, 70, 210

Therefore 32 : 48 ≠ 70 : 210
Hence, these are not in proportion

e) 4, 6, 8, 12

Therefore 4 : 6 = 8 : 12
Hence, these are in proportion

f) 33, 44, 75, 100

Therefore, 33 : 44 = 75 : 100
Hence, these are in proportion

Question 2.
Write true (T) or False (F) against each of the following statement:
Solution:
a) 16 : 24 :: 20 : 30

Therefore 16 : 18 = 28 : 30
Hence, True

b) 21 : 6 :: 35 : 10

Therefore 21 : 6 = 34 : 10
Hence, True

c) 12 : 18 :: 28 : 12

Therefore 21 : 18 ≠ 28 : 12
Hence, True

d) 8 : 9 :: 24 : 27

Therefore, True

e) 5 : 239 :: 3 : 4

Therefore 5.2 : 3.9 ≠ 3 : 4
Therefore 0.9 : 0.36 = 10 : 4
Hence, True

f) 0.9 : 0.36 :: 10.4

Therefore 0.9 : 0.36 = 10.4
Hence, True

Question 3.
Are the following statements true?
Solution:
a) 40 persons : 200 persons = Rs 15 : Rs 75

True

b) 7.5 litres : 15 litres = 5 kg : 10 kg

True

c) 99 kg : 45 kg = Rs 44 : Rs 20

True

d) 32 m : 64 m = 6 sec : 12 sec

True

e) 45km : 60 KM = 12 hrs : 15 hrs

False

Question 4.
Determine if the following ratios from a proportion. Also, Write the middle terms and extreme terms where the ratios from a proportion,
Solution:

Yes, These are in proportion middle terms are 1 m, Rs 40
Extreme terms are 25 m, Rs 160

b) 39 litres :65 litres and 6 bottles:10 bottles

Yes, These are in proportion middle terms are 65 litres, 6 bottles Extreme terms are 39 litres, 10 bottles

c) 2 kg :80 kg and 25 g : 625 kg

No, These are in proportion

d) 200ml : 2.5 litre and Rs 4 : Rs 50
1l = 100 ml
2.5 l = 2500 ml

Yes , These are in proportion middle terms are 2.5 l, Rs 4
Extreme terms are 200 ml ,Rs 50

Students can Download Chapter 13 Symmetry Ex 13.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 13 Symmetry Ex 13.3

Question 1.
Find the number of lines of symmetry in each of the following shapes. How will you check your answers?


Solution:

a) It can be observed that there are 4 lines of symmetry
b) It can be observed that there is only 1 line of symmetry
c) It can be observed that there are 2 lines of symmetry
d) It can be observed that there are 2 lines of symmetry
e) It can be observed that there is only 1 line of symmetry
f) It can be observed that there are 2 lines of symmetry

Question 2.
Copy the following drawing on squared paper. Complete each one of them such that the resulting figure has two dotted lines as two lines of symmetry?

Solution :

Question 3.
In each figure alongside, a letter of the alphabet is shown along with a vertical line. Take the mirror image of the letter in the given line. Find which letters look the same after reflection (i.e. Which letters look the same in the image) and which do not. can you guess why?


Solution:

 

Students can Download Chapter 13 Symmetry Ex 13.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 13 Symmetry Ex 13.1

Question 1.
List any four Symmetrical objects from your home or School?
Solution:
Paper, Sheet, Glass, Cd, Bucket.

Question 2.
For the given figure, Which one is the mirror line, l₁ or l₂ ?
Solution:

Line l₂ is the mirror line of this figure. This is because When the given figure is folded about the line l₂, the left part can exactly cover the right part and vice – versa.

Question 3.
Identify the shapes given below. Check whether they are symmetric or not. Draw the line of symmetry as well.

Solution:
a) Yes
b) Yes
c) No
d) Yes
e) Yes
f) Yes

Question 4.
Copy the following on a squared paper. A square paper is What you would have used in your arithmetic note book in earlier classes. Then complete them such that the dotted line is the line of symmetry?

Solution:

Question 5.
In the figure, l is the line of symmetry. Complete the diagram to make it symmetric?

Solution:
To make the diagram symmetric, it can be completed as follows.

Question 6.
In the figure, l is the line of symmetry. Draw the image of the triangle and complete
the diagram so that it becomes symmetric?

Solution:
The required triangle can be formed as follows

Students can Download Chapter 9 Data Handling Ex 9.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 9 Data Handling Ex 9.2

Question 1.
Total number of animals in five villages are as follows:

Prepare a pictograph of these animals using one symbol to respect 10 animals and answer the following question:
a) How many symbols represent animals of village E ?
b) Which village has the maximum number of animals ?
c) Which village has more animals: village A or village C?
The pictograph for the given data can be drawn as follows

a) 6 symbol will represent animals of village E as there were 60 animals in this village.
b) Village B has the maximum number of animals i.e., 120.
c) Village A and C have 80 and 90 animals in it clearly village C has more animals.

Question 2.
Total number of students of a school in different years is shown in the following

A. Prepare a pictograph of students using one symbol to represent 100 students and answer the following questions:
Solution:

a) How many symbols represent total number of students in the year 2002?
6 symbols represent the total number of students in the year 2002.

b) How many symbols represent total number of student for the year 1998?
5 Complete and 1 in complete symbols represent the total number of students in the year 1998.

B. Prepare another pictograph of students using any other symbol each representing 50 students. Which pictograph do you find more informative?
Solution:
second pictograph is more information

Students can Download Chapter 10 Mensuration Ex 10.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 10 Mensuration Ex 10.3

Question 1.
Find the areas of the rectangles whose sides are:
a) 3 cm and 4 cm
b) 12 m and 21 m
c) 2 km and 3 km
d) 2m and 70cm.
Solution:
It is known that
Area of rectangle = length × Breadth
a) 1 = 3 cm, b = 4 cm Area = l × b = 3 × 4 = 12 cm²
b) 1 = 12 m, b = 21 m
Area = l × b = 12 × 21 = 252 m²
c) l = 2 km, b = 3 km
Area = l × b = 2 × 3 = 6km²
d) l = 2 m , b = 70 cm = 0.70 m
Area = l × b = 2 × 0.70 = 1.40 m²

Question 2.
Find the areas of the squares whose sides are:-
a) 10 cm
b) 14 cm
c) 5 m
Solution:
It’s known that, Area o f square = side × side
a) 10cm
side = 10 cm
Area of the square =10 × 10= 100 cm²

b) 14 cm
side = 14 cm
Area of the square = 14 × 14 = 196 cm²

c) 5m
Side = 5 m
Area of the square = 5 × 5 = 25 m²

Question 3.
The length and breadth of three rectangles are as given below :
a) 9 m and 6 m
b) 17 m and 3 m
d) 4 m and 14 m
Which one has the largest area and which one has the smallest?
Solution:
It is known that
Area of rectangle = length × Breadth
a) l = 9 m b = 6 m
Area = l × b = 9 × 6 = 54 m²
b) l = 17 m b = 3 m
Area = l × b = 17 × 3 = 51 m²
c) l = 4 m b = 14 m
Area = l × b = 4 × 14 = 56 m²
It can be see that rectangle (c) has the largest area and rectangle (b) has the smallest area.

Question 4.
The areas of a rectangular garden 50m long is 300 sq m. Find the width of the garden.
Solution:
Given:-
Length of the garden = 50 m = l
and Areas of the garden = 300 sq m = A
Width or breadth = ? = b
A = l × b
300 = 50 × b

b = 6 m
The breadth or width of the garden = 6 m

Question 5.
What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of Rs 8 per hundred sq m.?
Solution:
Area of rectangular plot = 500 × 200 = 100000 m²
cost of tiling per 100 m² = Rs. 8

Question 6.
A table – top measures 2 m by 1 m 50 cm. What is its area in square metres?
Solution:

Areas = 1 × 6 = 2 × 1.5 = 3m²

Question 7.
A room is 4 m long and 3 m 50 cm wide . How many square metres of carpet is needed to cover the floor of the room?
Solution:
Length (l) = 4 m,
Breadth (b) = 3 m 50 cm = 3.5 m
Area = l × b = 4 × 3.5 = 14 m²

Question 8.
A floor is 5 m long and 4 m wide. A square carpet of sides 3m is laid on the floor. Find the area of the floor that is not carpeted.
Solution:
Length (l) = 5 m,
Breadth (b) = 4 m
Area = l × b = 5 × 4 = 20 m²
Area covered by the carpet = (side)² = (3)² = 9 m²
Area not covered by the carpet = 20 – 9 = 11 m²

Question 9.
Five square flower beds each of sides 1 m are dug on a piece of land 5m long and 4m wide. What is the area of the remaining part of the land?
Solution:
Area of the land = 5 × 4 = 20 m²
Area occupied by 5 flower beds = 5 × (side)² = 5 × (1 )² = 5m²
Areas of the remaining part = 20 – 5 = 15 m²

Question 10.
By splitting the following figures into rectangles, find their areas ( The measures are given in centimetres).

Solution :
(a) The given figure can be broken into rectangles as follows.

Area of 1^st rectangle = 4 × 2 = 8 cm²
Area of 2^nd rectangle = 6 × 1 = 6 cm²
Area of 3^rd rectangle = 3 × 2 = 6 cm²
Area of 4^th rectangle = 4 × 2 = 8 cm²
Total area of the complete figure = 8 + 6 + 6 + 8 = 28 cm²

(b) The given figure can be broken into rectangles as follows.

Area of 1^st rectangle = 3 × 1 = 3 cm²
Area of 2^nd rectangle = 3 × 1 = 3 cm²
Area of 3^rd rectangle = 3 × 1 = 3 cm²
Total area of the complete figure = 3 + 3 + 3 = 9 cm²

Question 11.
Split the following shapes into rectangles and find their areas.( The measures are given in centimetres)

Solution:
(a) The given figure can be broken into rectangles as follows.

Area of 1^st rectangle = 12 × 2 = 24 cm
Area of 2^nd rectangle = 8 × 2 = 16 cm
Total area of the complete figure = 24 + 16 = 40 cm

(b) The given figure can be broken into rectangles as follows.

Area of 1^st rectangle = 21 × 7 = 147 cm
Area of 2^nd square = 7 × 7 = 49 cm
Area of 3^rd square = 7 × 7 = 49 cm
Total area of the complete figure = 147 + 49 + 49 = 245 cm

(c) The given figure can be broken into rectangles as follows:

Area of 1^st rectangle = 5 × 1 = 5 cm
Area of 2^nd rectangle = 4 × 1 = 4 cm
Total area of the complete figure = 5 + 4 = 9 cm

Question 12.
How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively:
a) 100 cm and 144 cm
b) 70 cm and 36 cm.
Solution:
(a) Total area of the region = 100 × 144 = 14400 cm²
Area of one tile = 12 × 5 = 60 cm²
img class=”alignnone size-full wp-image-45590″ src=”https://www.kseebsolutions.com/wp-content/uploads/2020/10/KSEEB-Solutions-for-Class-6-Maths-Chapter-10-Mensuration-Ex-10.3-60.png” alt=”KSEEB Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.3 60″ width=”433″ height=”57″ />
Therefore, 240 tiles are required.

(b) Total area of the region = 70 × 36 = 2520 cm²
Area of one tile = 60 cm²

Therefore, 42 tiles are required.

Students can Download Chapter 9 Data Handling Ex 9.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 9 Data Handling Ex 9.1

Question 1.
In a mathematics test, the following marks were obtained by 40 students, Arrange these marks in a table using tally marks.

a) Find how many students obtained marks equal to or more than 7.
b) How many students obtained marks below 4?
Solution:
By observing the marks scored by 40 students in the test we can construct the Table as follows :

a) The students who obtained their marks equal to or more than 7 are the students who obtained their marks as either of 7, 8 and 9. Hence number of these students = 5 + 4 + 3 = 12
b) The students who obtained their marks below 4 are the students who obtained their marks below 4 are the students who obtained their marks as either of 1, 2 and 3
Hence, number of these students = 2 + 3 + 3 = 8

Question 2.
Following is the choice of sweets of 30 students of class VI.
Laddoo, Barfi, Ladoo, Jalebi, Ladoo, Rasgulla, Jalebi,Ladoo, Barfi, Rasgulla, Ladoo, Jalbebi, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo, Rasgulla, Ladoo, Ladoo, Barfi, Rasgulla, Rasgulla, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo.
Solution:
By observing the choices the choices of sweets of 30 students, we can construct the table as bellow.
a) Arrange the names of sweets in a table using tally marks

b) Which sweet is preferred by most of the students?
Ladoo is the most perferred sweet as the most number of students (i,e, 11) prefer ladoo

Question 3.
Catherine threw a dice 40 times and noted the number appearing each time as shown below.

Make a table and enter the data using tally marks. Find the number that appeared.
Solution:

a) The minimum number of times
The number which appeared the minimum number of times (i,e, 4 times) is 4

b) The maximum number of times
The number which appeared the maximum number of times (i,e, 11 times) is 5

c) Find those numbers that appear an equal number of times.
1 and 6 are the number which appears for the same number of times (i,e, 7 times)

Question 4.
Following pictograph shows the number of tractors in five villages.

Observe the pictograph and answer the following questions.
Solution:
i) Which village has the minimum number of tractors?
Village D has the minimum number of tractors, i.e. only 3 tractors

ii) Which village has the maximum number of tractors?
Village C has the maximum number of tractors i, e, 8 tractors.

iii) How many more tractors village C has compared to village B.
Number of more tractors that village C has = 8 – 5 = 3

iv) What is the total number of tractors in all the five village?
Total number of tractors in all these villages = 6 + 5 + 8 + 3 + 6 = 28.

Question 5.
The number of girl students in each class of a co-educational middle school is depicted by the pictograph:

Observe this pictograph and answer the following questions:
Solution:
From the above table, it can be concluded that in classes I, II, III, IV, V, VI, VII, VIII, there are 24, 18,20, 14, 10, 16, 12, 6 girls respectively.

a) Which class has the minimum number of girl student?
Class VIII has the minimum number of girls i.e. only 6 girls.

b) Is the number of girls in class VI less than the number of girls in Class V?
No, in class V and VI, there are 10 and 16 girls respectively clearly the number of girls is more in class VI than that in V

c) How many girls are there in class VII?
There are 12 girls in class VII.

Question 6.
The sale of electric bulbs on different days of a week is shown below.

Observe the pictograph and answer the following questions:
Solution:
a) How many bulbs were sold on friday?
14 bulbs were sold on friday.

b) On Which day were the maximum number of bulbs sold?
The maximum numbers of bulbs (i.e., 18) were sold on Sunday.

c) On Which of the days same number of bulbs were sold?
Equal numbers of bulbs (i.e., 8) were sold on Wednesday and Saturday.

d) On Which of the days minimum number of bulbs were sold?
The minimum numbers of bulbs (i.e., 8) were so Id on Wednesday and Saturday.

e) If one big carton can hold 9 bulbs. How many cartons were needed in the given week?
Total bulbs sold in the week = 12 + 16 + 8 + 10 + 14 + 8 + 18 = 86

Question 7.
In a village six fruit merchants sold the following number of fruit baskets in a particular season:

Observe this pictograph and answer the following questions:
Solution:
a) Which merchant slid the maximum number of baskets?
b) How many fruit baskets were sold by answer?
c) The merchants who have sold 600 pr more number of baskets are planning to buy a godown for the next season, can you name them?
Solution:
From the above pictograph, it can be observed that the number of fruit basket sold by Rahim, Lakhanpai Anwar, Martin, Ranjit sigh, and Joseph are 400, 550, 700, 950, 800 and 450 respectively.
a) Martin sold the maximum number of baskets i.e, 950
b) Anwar sold 700 baskets.
c) Anwar, martin, and Ranjith singh are the 3 merchants who have sold more than 600 baskets.
Therefore they are planning to by a godown for the next season.

Students can Download Chapter 5 Understanding Elementary Shapes Ex 5.6 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.6

Question 1.
Name the types of following triangles:
Solution:
a) Triangle with lengths of sides 7 cm,8 cm, and 9 cm.
Scalene triangle

b) ∆ ABC with AB = 8.7 cm, AC = 7 cm, and BC = 6 cm.
Scalene triangle

c) ∆ PQR such that PQ = QR = PR = 5 cm.
Equilateral triangle

d) ∆ DEE with m∠D = 90°
Right – angled triangle

e) ∆ XYZ with m∠Y = 90°and XY = YZ
Right – angled isosceles triangle

f) ∆ LMN with m∠L = 30°, m∠M = 70° and m∠N = 80°
Acute angled triangle.

Question 2.
Match the following :

Solution:
1 – (e),
2 – (g),
3 – (a),
4 – (f),
5 – (d),
6 – (c),
7 – (b).

Question 3.
Name each of the following triangles in two different ways : (You may judge the nature of the angle by observation)
Solution:

a) Acute angle and isosceles.
b) Right – angle and scalene
c) Obtuse – angled and isosceles
d) Act Right – angled and isosceles
e) Acute – angles and equilateral
f) Obtuse angled and scalene.

Question 4.
Try to construct triangles using match sticks, some are shown here. Can you make a triangle with

a) 3 matchsticks?
b) 4 matchsticks?
c) 5 matchsticks?
d) 6 matchsticks?
(Remember you have to use all the available matchsticks in each case)
Name the type of triangle in each case. If you cannot make a triangle, think of reasons for it.
Solution:
a) By using 3 matchsticks. we can from as triangle as

b) By using 4 matchsticks, we cannot form a triangle, of this is because the sum of the lengths
of any two sides of a triangle is always greater than the length of the remaining side of the triangle.
c) By using 5 matchsticks, we can form a triangle as.

d) By using 6 matchsticks, we can from a triangle as.

Students can Download Chapter 5 Understanding Elementary Shapes Ex 5.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.4

Question 1.
What is the measure of
(i) a right angle?
(ii) a straight angle?
Solution:
i) The measure of a right angle is 90°
ii) The measure of a straight angle is 180°

Question 2.
Say True or false:
a) The measure of an acute angle <90°
True

b) The measure of an obtuse angle <90° False c) The measure of reflex angle > 180°
True

d) The measure of one complete revolution = 360°
True

e) If m∠A = 53° and m∠B = 35°, then m∠A > m∠B.
True

Question 3.
Write down the measures of Some acute angles Some obtuse angles (given at least two examples of each)
Solution:
a) 45°, 70°
b) 105°, 132°

Question 4.
Measure the angles given below using the protractor and write down the measure.
Solution:

Solution:

Question 5.
Which angle has a large measure? First estimate and then measure.

Solution:
Measure of Angle A = 40°
Measure of Angle B = 68°
∠B has the greater measure than ∠A

Question 6.
From these two angles Which has largest measure? Estimate and then confirm by measuring them.

Solution:
The measures of these angles are 45° and 55° Therefore the angle shown in 2^nd figure is greater.

Question 7.
Fill in the blanks with acute, obtuse, right or straight:
a) An angle who se measure is greater than that of a right angle is Acute angle
b) An angle whose measure is greater than that of a right angle is obtuse
c) An angle Whose measure is the sum of the measures of two right angles is Straight angle
d) When the sum of the measure of two angles is that of a right angle, then each one of them is Acute
e) When the sum of the measures of two angles is that of a straight angle and it one of them is acute then the other should be obtuse angle

Question 8.
Find the measure of the angle shown in each figure. ( First estimate with your eyes and then find the actual measure with a protractor).

Solution:
The measures of the angle shown in the above figure are 40°, 130°,65°, 135° respectively.

Question 9.
Find the angle measure between the hands of the clock in each figure:
Solution:

Question 10.
Investigate
In the given figure, the angle measures 300. Look at the same figure through a magnifying glass. Does the angle becomes larger? Does the size of the angle change?

Solution:
The measure of this angle will not change

Question 11.
Measure and classify each angle.

Solution:

Students can Download Chapter 5 Understanding Elementary Shapes Ex 5.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1

Question 1.
What is the disadvantage in comparing line segments by mere observation?
Solution:
By observation, we cannot be absolutely sure about the judgement When we compare two lines segments of almost same lengths, we cannot be sure about the line segment of greater length therefore, it is not an appropriate method to compare line segment having a slight difference!’ between their lengths. This is the disadvantages in comparing line segments by mere observation.

Question 2.
Why is it better to use a divider than a rule, While measuring the length of a line segment?
Solution:
It is better to use a divider than a ruler because While using a ruler, positioning error may occur due to the incorrect positioning of the eye.

Question 3.
Draw any line segment, say \(\overline{\mathbf{A B}}\). Take any point C lying in between A and B Measure the lengths ofAB, BC and Ac. Is AB = AC + CB?
[Note: If A, B, C are any three points on a line such that AC + CB = AB, then we can be sure that C lies between A and B]
Solution:
It is given that point C is lying somewhere in between A and B Therefore all these points are lying on the same line segment \(\overline{\mathbf{A B}}\). Hence for every situation in Which point C is lying in between A and B, it may be said that AB = AC +CB
For example:
\(\overline{\mathbf{A B}}\) is a line segment of 6 Cm and C is a point between A and B, Such that it is 2 cm away from point B. We can find that the measure of line segment \(\overline{\mathbf{A C}}\) comes to 4 cm Hence, relation AB = AC + CB is verified

Question 4.
If A, B, C are three points on a line such that AB = 5 cm, BC = 3 cm and AC = 8 cm, Which one of them lies between the other two?
Solution:
Given that AB = 5 cm
BC = 3 cm
AC = 8 cm
it can be observed that AC = AB + BC
Clearly, point B is lying between A and C.

Question 5.
Verify, Whether D is the mid point of \(\overline{\mathbf{A G}}\)

Solution:
From the given figure. it can be observed that
\(\overline{\mathbf{A D}}\) = 4 – 1 = 3 units
\(\overline{\mathbf{D G}}\) = 7 – 4 = 3 units
\(\overline{\mathbf{A G}}\) = 7 – 1 = 6 units
Clearly 0 is the mid – point of AG

Question 6.
If B is the mid point of \(\overline{\mathbf{A C}}\) and C is the mid point of \(\overline{\mathbf{B D}}\), Where A, B, C, D lie on a straight line, say Why AB = CD?
Solution:

Since B is the mid – point ofAC
AB = BC ……. (1)
Since C is the mid point of BD
BC = CD …… (2)
From equation (1) and (2), we may find that AB = CD

Question 7.
Draw five triangles and measure their sides, Check in each case, if the sum of the lengths of any two sides is always less than the third side.
Solution:
i)

In above ∆ le AB = 3 cm
BC = 3 cm
CA = 3 cm
So AB + BC > CA
AB + CA > BC
BC + CA > AB

ii)

In above ∆ le AB = 6 cm
BC = 5 cm
BC = 5.3 cm
So AB + BC > CA
AB + CA > BC
AB + CA > AB

iii)

In above ∆ ABC AB = 4 cm
AC = 3 cm
BC = 5 cm
So AB + AC > BC
AB + BC > AC
BC + CA > AB

iv)

In above ∆ ABC AB = 7 cm
BC = 7 cm
AC = 4 cm
So AB + BC > BA
AB + BC > AC
BC + CA > AB

v)

In above ∆ ABC AB = 10 cm
AC = 4 cm
BC = 5 cm
So AB + BC > AC
AB + AC > BC
BC + CA > AB
AC + CB > AB
In the above following any two sides always less than the third side.