Category: Class 6

Students can download Maths Chapter 3 Perimeter and Area Ex 3.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.1

Question 1.
The table given below contains some measures of the rectangle. Find the unknown values.

Solution:

(i) Area of the rectangle = (length × breadth) sq unit.
Perimeter of a rectangle = 2(1 + b) units.
l = 5 cm
b = 8 cm
∴ p = 2 (l + b) cm = 2 (5 + 8) cm = 2 × 13 cm
p = 26 cm
Area = (l × b) cm² = (5 × 8) cm²
A = 40 cm²

(ii) l = 13 cm
p = 54 cm
Perimeter = 2 (l + b) units
54 = 2 (13 + b) cm
\(\frac{54}{2}\) = 13 + b
27 = 13 + b
b = 27 – 13
b = 14 cm
Area = l × b sq. unit = 13 × 14 cm²
A = 182 cm²

(iii) b = 15 cm
p = 60 cm
p = 2 (l + b) units
60 = 2 (l + 15) cm
\(\frac{60}{2}\) = l + 15
30 = l + 15
l = 30 – 15 .
l = 15 cm
Area = l × b unit² = 15 × 15 cm² = 225 cm²
A = 225 cm²

(iv) l = 10 m
Area = 120 sq metre
Area = l × b sq.m
120 = 10 × 6
b = \(\frac{120}{10}\)
b = 12 m
Perimeter =2 (l + b) units = 2(10 + 12) units = 2 × 22 m
A = 44 m

(v) b = 4 feet.
Area = 20 sq. feet
Area = l × b sq .feet
20 = l × 4
l = \(\frac{20}{4}\) feet
l = 5 feet
Perimeter = 2 (l + b) units.
p = 2 (5 + 4) feet = 2 × 9
p = 18 feet

Question 2.
The table given below contains some measures of the square. Find the unknown values.

Solution:
(i) 24 cm, 36 cm²
(ii) 25 m, 625 m²
(iii) 7 feet, 28 feet

Question 3.
The table given below contains some measures of the right angled triangle. Find the unknown values.

Solution:

Area of the right triangle = \(\frac{1}{2}\) × (base × height) unit²
(i) b = 20 cm
h = 40 cm
Area = \(\frac{1}{2}\) (b × h) cm² = \(\frac{1}{2}\) × 20 × 40 = 400 cm²
A = 400 cm²

(ii) b = 5 feet
Area = \(\frac{1}{2}\) × b × h unit²
= 20 = \(\frac{1}{2}\) × 5 × h sq. feet
\(\frac{20 \times 2}{5}\) = h
h = 8 feet

(iii) Area = \(\frac{1}{2}\) × (base × height) unit²
24 = \(\frac{1}{2}\) × b × 12 m²
base = \(\frac{24 \times 2}{12}\) m = 4 m
Base = 4m

Question 4.
The table given below contains some measures of the triangle. Find the unknown values.

Solution:
(i) 13 cm
(ii) 6 m
(iii) 8 feet

Question 5.
Fill in the blanks.
(i) 5 cm² = ______ mm²
(ii) 26 m² = ______ cm²
(iii) 8 km² = ______ m²
Solution:
(i) 500
(ii) 260000
(iii) 8000000

Question 6.
Find the perimeter and area of the following shapes.

Solution:
(i) Perimeter = (4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4)
= 48 cm
a = 4 cm
Area of 1 square = 4 × 4 cm²
= 16 cm²
Area of 5 squares = 5 × 16 cm²
= 80 cm²

(ii) Perimeter = (4 + 5 + 4 + 5 + 4 + 5 + 4 + 5)
= 36 cm
Area of 1 triangle = \(\frac{1}{2}\) × b × h sq units
= \(\frac{1}{2}\) × 4 × 5 cm²
= 10 cm²
Area of 4 triangles= 4 × 10 cm²
= 40 cm²
Area of the square = 3 × 3 cm²
= 9 cm²
Total area = (40 + 9) cm²
= 49 cm²

(iii) Perimeter = (15 + 50 + 12 + 13 + 10 + 10 + 40)
= 150 cm
Area of the square = 10 × 10 cm²
= 100 cm²
= 250 cm²
Area of the triangle = \(\frac{1}{2}\) × 12 × 5 cm²
= \(\frac{1}{2}\) × 12⁶ x 5 cm²
= 30 cm²
Total area = (100 + 250 + 30) cm²
= 380 cm²

Question 7.
Find the perimeter and the area of the rectangle whose length is 6 m and breadth is 4m?
Solution:
l = 6 m, b = 4 m Perimeter of the rectangle
= 2 (l + b) units
= 2 (6 + 4) m
= 2 (10) m
= 20 m
Area of the rectangle = l × b sq units
= 4 × 6 m²
= 24 m²

Question 8.
Find the perimeter and area of a square whose side is 8 cm.
Solution:
a = 8 cm
The perimeter of a square
= 4a units
= 4 × 8 cm
= 32 cm
Area of the square = a × a sq units
= 8 × 8 cm²
= 64 cm²

Question 9.
Find the perimeter and the area of right angled triangle whose sides are 6 feet, 8 feet and 10 feet.
Solution:
Perimeter of the triangle
= (a + b + c) units
= (6 + 8 + 10) feet
= 24 feet
Area of the triangle = \(\frac{1}{2}\) × b × h sq units
\(\frac{1}{2}\) × 6³× 8 feet square = 24 sq. feet

Question 10.
Find the perimeter of
(i) A scalene triangle with sides 7 m, 8 m, 10 m.
(ii) An isosceles triangle with equal sides 10 cm each and third side is 7 cm.
(iii) An equilateral triangle with a side of 6 cm.
Solution:
(i) Perimeter of the triangle
= (a + b + c) units
= (7 + 8 + 10) m
= 25

(ii) Perimeter of the triangle
= (10 + 10 + 7) cm
= 27 cm

(iii) Perimeter of the triangle
= (6 + 6 + 6) cm
= 18 cm

Question 11.
The area of a rectangular shaped photo is 820 sq. cm. and its width is 20 cm. What is its length? Also find its perimeter.
Solution:
Given Area = 820 cm²
Width = 20 cm
Area of the rectangle
= l × b sq. units
820 = l × 20
\(\frac{820}{20}\) = l
41 = l
length l = 41 cm
Perimeter = 2(l + b) units
= 2(41 + 20) cm
= 2(61) cm
= 122 cm

Question 12.
A square park has 40 m as its perimeter. What is the length of its side? Also find its area.
Solution:
perimeter = 40 m
4a = 40 m
a = \(\frac{40}{4}\)
Side a = 10 m
Area = a × a sq units
= 10 × 10 m²
= 100 m²

Question 13.
The scalene triangle has 40 cm as its perimeter and whose two sides are 13 cm and 15 cm, find the third side.
Solution:
Let the third side be C
perimeter = (a + b + c) units
40 = 13 + 15 + C
40 = 28 + C
C = 40 – 28
C = 12 units
C = 12 cm

Question 14.
A field is in the shape of a right angled triangle whose base is 25 m and height 20 m. Find the cost of levelling the field at the rate of Rs 45/- per sq.m²
Solution:
b = 25 m, h = 20 m
Area of the triangle = \(\frac{1}{2}\) × bh sq.units
= \(\frac{1}{2}\) × 25 × 20 m²
= 250 m²
Cost of levelling 1 m² = Rs 45
Cost of levelling 250 m² = Rs 45 × 250
= Rs. 11250

Question 15.
A square of side 2 cm is joined with a rectangle of length 15 cm and breadth 10 cm. Find the perimeter of the combined shape.
Solution:

Perimeter of the combined shape = Lengths of the outer boundaries
= (15 + 10 + 2 + 2 + 2 + 13 + 10) cm = 54 cm
Perimeter = 54 cm

Objective Type Questions

Question 16.
The following figures are of equal area. Which figure has the least perimeter?

Solution:
(b)

Question 17.
If two identical rectangles of perimeter 30 cm are joined together, then the perimeter of the new shape will be
(a) equal to 60 cm
(b) less than 60 cm
(c) greater than 60 cm
(d) equal to 45 cm
Solution:
(b) less than 60 cm

Question 18.
If every side of a rectangle is doubled, then its area becomes _____ times
(a) 2
(b) 3
(c) 4
(d) 6
Solution:
(c) 4
2l × 2b = 4l × b

Question 19.
The side of the square is 10 cm. If its side is tripled, then by how many times will its perimeter increase?
(a) 2 times
(b) 4 times
(c) 6 times
(d) 3 times
Solution:
(d) 3 times

Question 20.
The length and breadth of a rectangular sheet of paper are 15 cm and 12 cm respectively. A rectangular piece is cut from one of its corners. Which of the following statement is correct for the remaining sheet?
(a) Perimeter remains the same but the area changes
(b) Area remains the same but the perimeter changes
(c) There will be a change in both area and perimeter
(d) Both the area and perimeter remains the same.
Solution:
(a) Perimeter remains the same but the area changes

Students can download Maths Chapter 5 Statistics Ex 5.4 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Ex 5.4

Miscellaneous Practice Problems

Question 1.
The heights (in centimeters) of 40 children are.

Prepare a tally marks table.
Solution:

Question 2.
There are 1000 students in a school. Data regarding the mode of transport of the students is given below. Draw a pictograph to represent the data.

Solution:
Mode of transport of the students
Scale: 1 Unit=100 Students

Question 3.
The following pictograph shows the total savings of a group of friends in a year. Each .picture represents a saving of Rs.100. Answer the following questions.

(i) What is the Ratio of Ruby’s saving to that of Thasnim’s?
(ii) What is the ratio of Kuzhali’s savings to that of others?
(iii) How much is Iniya’s savings?
(iv) Find the total amount of savings of all the friends?
(v) Ruby and Kuzhali save the same amount. Say True or False.
Solution:
(i) Ratio of Ruby’s saving to that of Thasnim’s
\(=\frac{\text { Ruby’s saving }}{\text { Thasnim’s saving }}=\frac{5 \times 100}{4 \times 100}=\frac{5}{4}=5: 4\)
Ratio of Ruby’s saving to that of Thasnims = 5 : 4
(ii) Ratio of Kuzhali’s savings to that of others
\(=\frac{\text { kuzhali’s saving }}{\text { others saving }}=\frac{5 \times 100}{19 \times 100}=\frac{5}{19}=5: 19\)
Ratio of Kuzhali’s saving to that of others = 5 : 19
(iii) Iniya’s saving = 3 × 100 = ₹ 300
(iv) Saving of all the friends = (5 + 7 + 4 + 5 + 3) × 100 = 24 × 100 = ₹ 2400.
Total savings = ₹ 2400
(v) True.

Challenging Problems

Question 4.
The table shows the numbers of moons that orbit each of the planets in oar solar system.

Make a Bar graph for the above data.
Solution:
Number of moons that orbit each of the planets in our solar system
Scale: 1 Unit = 2 moons

Question 5.
The prediction of the weather in the month of September is given below.

(i) Make a frequency table of the types of weather by reading the calendar.
(ii) How many days are either cloudy or partly cloudy?
(iii) How many days do not have rain? Give two ways to find the answer?
(iv) Find the ratio of the number of Sunny days to Rainy days.
Solution:
(i)
(ii) 14 days
(iii) 24 days (30 – 6 = 24 days)
(iv) 10 : 6

Question 6.
26 students were interviewed to find out what they want they to become in future. Their responses are given in the following table.

Represent this data using pictograph.
Solution:
Students responses in an interview about their future profession
Scale: 1 Unit = 1 student

Question 7.
Yasmin of class VI was given a task to count the number of books which are biographies, in her school library. The information collected by her is represented as follows.

Observe the pictograph and answer the following questions.
(i) Which title has the maximum number of biographies?
(ii) Which title has the minimum number of biographies?
(iii) Which title has exactly half the number of biographies as Novelists?
(iv) How many biographies are there on the title of sportspersons?
(v) What is the total number of biographies in the library?
Solution:
(i) ‘The title Novelists’ have the maximum number of biographies
(ii) ‘The title Scientists’ have a minimum number of biographies.
(iii) ‘Sportspersons’ title has exactly half the number of biographies as Novelist.
(iv) \((1 \times 20)+\frac{20}{4}=20+5=25\) biographies are there in the title sportsperson.
(v) 8 × 20 = 160 biographies are there in the library.

Question 8.
The bar graph illustrates the results of a survey conducted on vehicles crossing over a Toll Plaza in one hour.

Observe the bar graph carefully and fill up the following table.

Solution:
Vans = 50; Buses = 40; Cars = 65; Others = 15
Total Vehicles = 245

Question 9.
The lengths (in the nearest centimeter) of 30 drumsticks are given as follows.

Draw the bar graph showing the same information.
Solution:
The lengths (in nearest cm) of drumsticks
Scale : 1 Unit = 1 drumstick

Students can download Maths Chapter 6 Information Processing Ex 6.3 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Ex 6.3

Question 1.
How many Triangles are there in each of the following figures?

Solution:
(i) 12 triangles
(ii) 16 triangles
(iii) 32 triangles
(iv) 35 triangles

Question 2.
Find the number of dots in the tenth figure of the following patterns.

Solution:
(i) 55
(ii) 100

Question 3.

(i) Draw the next pattern.
(ii) Prepare a table for the number of dots used for each pattern.
(iii) Explain the pattern.
(iv) Find the number of dots in the 25th pattern.
Solution:

(iv) 350

Question 4.

Solution:
(i) 20 squares
(ii) 10 squares

Question 5.
How many circles are there in the following figure?

Solution:
7 circles

Question 6.
Find the minimum number of straight lines used in forming the following figures.

Solution:
(i) 10
(ii) 12

Students can download Maths Chapter 6 Information Processing Ex 6.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Ex 6.2

Question 1.
In the following magic triangle, arrange the numbers from 1 to 6, so that you get the same sum on all its sides.

Solution:
One of the answers is.

Question 2.
Using the numbers from 1 to 9
(i) Can you form a magic triangle?
(ii) How many magic triangles can be formed?
(iii) What are the sums of the sides of the magic triangle?

Solution:
(i) Yes
(ii) 5
(iii) 17, 19, 20, 21, 23

Question 3.
Arrange the odd numbers from 1 to 17 without repetition to get a sum of 30 on each side of the magic triangle.

Solution:

Question 4.
Put the numbers 1, 2, 3, 4, 5, 6 & 7 in the circles so that each straight line of three numbers add up to the same total.

Solution:

Question 5.
Place the number 1 to 12 in the 12 circles so that the sum of the numbers in each of the six lines of the star is 26. Use each number from 1 to 12 exactly once. Find more possible ways?

Solution:

Students can download Maths Chapter 6 Information Processing Ex 6.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Ex 6.1

Question 1.
Suppose, you have two pairs of shorts, one is black and the other one is blue; three shirts which are white, blue and red. You again wish to make different combinations, but you always want to make sure that the shorts and shirt that you wear are of different colours. List and check how many combinations are possible now.
Solution:
6 combinations are possible
Black-White
Black-Blue
Black-Red
Blue-White
Blue-Blue
Blue-Red

Question 2.
You have two red and two blue blocks. How many different towers can you build that are four blocks high using these blocks? List all the possibilities.
Solution:
6 Possibilities,
R – B – R – B
R – R – B – B
B – R – R – B
B – R – B – R
B – B – R – R
R – B – B – R

Students can download Maths Chapter 4 Geometry Ex 4.3 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.3

Question 1.
Observe the diagram and fill in the blanks.

(i) ‘A’, ‘O’ and ‘B’ are ……… points.
(ii) ‘A’, ‘O’ and ‘C’ are ……….. points.
(iii) ‘A’,‘B’ and ‘C’are ……… points.
(iv) ……… is the point of concurrency.
Solution:
(i) collinear points
Hint: Points on a line.
(ii) non-collinear points
Hint: Points not on a line
(iii) endpoints/non-collinear points
(iv) O is the point of concurrency.
Hint: A points where lines meet

Question 2.
Draw any line and mark any 3 points that are collinear.
Solution:

Question 3.
Draw any line and mark any 4 points that are not collinear.
Solution:

Question 4.
Draw any 3 lines to have a point of concurrency.
Solution:

Question 5.
Draw any 3 lines that are not concurrent. Find the number of points of intersection.
Solution:

Number of points of intersection = 3

Objective Type Questions

Observe the Diagram and give answers

Question 6.
A set of collinear points in the figure are
(a) A, B, C
(b) A, F, C
(c) B, C, D
(d) A, C, D
Solution:
(b) A, F, C
Hint: Collinear points are points on a line.

Question 7.
A set of non-collinear points in the figure are ……….
(a) A, F, C
(b) B, F, D
(c) E, F, G
(d)A,D,C
Solution:
(d) A, D, C

Question 8.
A point of concurrency in the figure is ______
(a) E
(b) F
(c) G
(d) H
Solution:
(b) F

Students can download Maths Chapter 4 Geometry Ex 4.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.2

Question 1.
Use any number of the given dots to make different angles.
(i) An Accute angle

(ii) An Obtuse Angle

(iii) A Right Angle

(iv) A Straight Angle

Solution:
(i) An Accute angle

(ii) An Obtuse Angle

(iii) A Right Angle

(iv) A Straight Angle

Question 2.
Name the vertex and sides that form each angle.

Solution:
(i) D, DE and DF
(ii) D, DE and DC
(iii) P, PQ and PR
(iv) S, SV and ST

Question 3.
Pick out the Right angles from the given figures.

Solution:
(i), (iii), (v)

Question 4.
Pick out the Accute angles from the given figures.

Solution:
(i), (iii), (iv) are the Acute Angles.

Question 5.
Pick out the Obtuse angles from the given figures.

Solution:
(i) and (ii) are the Obtuse Angles.

Question 6.
Name the angle in each figure given below in all the possible ways.

Solution:
(i) ∠M or ∠LMN or ∠NML
(ii) ∠Q or ∠PQR or ∠RQP
(iii) ∠N or ∠MNO or ∠ONM
(iv) ∠A or ∠TAS or ∠SAT
(v) ∠Y or ∠XYZ or ∠ZYX
(vi) There are 3 angles in (vi)

• ∠ADC or ∠CDA
• ∠ CDB or ∠BDC
• ∠D or ∠ADB or ∠BDA

Question 7.
Say True or False.
(i) 20° and 70° are complementary.
(ii) 88° and 12° are complementary.
(iii) 80° and 180° are supplementary.
(iv) 0° and 180° are supplementary.
Solution:
(i) True
Hint: 20°+ 70° = 90°
(ii) False
Hint: 88° + 180° = 260° ≠ 1
(iii) False
Hint: 80° + 180° = 260° ≠ 1
(iv) True
Hint: 0° + 180° = 180°

Question 8.
Draw and label each of the angles.
(i) ∠NAS = 90°n
(ii) ∠BIG = 35°
(iii) ∠SMC = 145°
Solution:

Question 9.
Identify the types of angles shown by the hands of the given clock.

Solution:
(i) Obtuse angle
(ii) Zero angle
(iii) Straight angle
(iv) Acute angle
(v) Right angle

Question 10.
Find the supplementary/complementary angles in each case.

Solution:

Objective Type Questions

Question 11.
In this figure, which is not the correct way of naming an angle?

(a) ∠Y
(b) ∠ZXY
(c) ∠ZYX
(d) ∠XYZ
Solution:
(b) ∠ZXY

Question 12.
In this figure, ∠AYZ = 45. If point ‘A’ is shifted to point ‘B’ along the ray, then the measure of ∠BYZ is

(a) more than 45°
(b) 45°
(c) less than 45°
(d) 90°
Solution:
(b) 45°

Students can download Maths Chapter 4 Geometry Ex 4.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.1

Question 1.
Fill in the blanks.
(i) A line through two endpoints ‘A’ and ‘B’ is denoted by ______
(ii) Aline segment from point ‘B’ to point ‘A’ is denoted by ______
(iii) A ray has ______ endpoint(s).
Solution:
(i) \(\overleftrightarrow { AB }\)
(ii) \(\bar { BA } \)
(iii) one

Question 2.
How many line segments are there in the given line? Name them.

Solution:
10, \(\overline { PQ } \), \(\overline { PA } \), \(\overline { PB } \), \(\overline { PC } \), \(\overline { AB } \), \(\overline { BC } \), \(\overline {CQ} \), \(\overline { AQ } \), \(\overline { BQ } \), \(\overline { AC } \).

Question 3.
Measure the following line segments.

Solution:
\(\overline { XY } \) = 2.4 cm, \(\overline { AB } \) = 3.4 cm, \(\overline { EF } \) = 4 cm, \(\overline { PQ } \) = 3 cm.

Question 4.
Construct a line segment using a ruler and compass.
(1) \(\overline { AB } \) = 7.5 cm
(2) \(\overline { CD } \) = 3.6 cm
(3) \(\overline { QR } \) = 10 cm
Solution:
(1)
(i) Draw line 1 and mark a point A on it.
(ii) Measure 7.5 cm using a compass, placing the pointer at ‘O’ and the pencil pointer at 7.5 cm.
(iii) Place the pointer of the compass at A then draw a small arc on the line 1 with the pencil pointer. It cuts line 1 at a point and name that point as B.
(iv) Now \(\overline { AB } \) is the required line segment of length 7.5 cm.

(2)
(i) Draw a line 1 and mark a point C on it.
(ii) Measure 3.6 cm using a compass, placing the pointer at O and the pencil pointer at 3.6 cm.
(iii) Place the pointer of the compass at C then draw a small arc on the line 1 with the pencil pointer. It cuts line 1 at a point and names the point as D.
(iv) Now \(\overline { CD } \) is the required line segment of length 3.6 cm.

(3)
(i) Draw a line 1 and mark a point Q on it.
(ii) Measure 10 cm using compass placing the pointer at O and the pencil pointer at 10 cm.
(iii) Place the pointer of the compass at Q then draw a small arc on the line 1 with the pencil pointer. It cuts the line 1 at a point and name that point as R.
(iv) Now \(\overline { QR }\) is the required line segment of length 10 cm.

Question 5.
From the given figure
(i) identify the parallel lines
(ii) identify the intersecting lines
(iii) name the points of intersection.

Solution:
(i) Parallel lines:
(a) \(\overrightarrow{\mathrm{CD}} \text { and } \overrightarrow{\mathrm{AB}}\)
(b) \(\overrightarrow{\mathrm{EF}} \text { and } \overrightarrow{\mathrm{GH}}\)
(ii) Intersecting lines:
(a) \(\overrightarrow{\mathrm{CD}} \text { and } \overrightarrow{\mathrm{EF}}\)
(b) \(\overrightarrow{\mathrm{CD}} \text { and } \overrightarrow{\mathrm{GH}}\)
(c) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{EF}}\)
(d) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{GH}}\)
(iii) Point of Intersection:
P, Q, R and S are the points of Intersection.
(iii) P, Q, R and S

Question 6.
From the given figure, name the
(i) parallel lines
(ii) intersecting lines
(iii) points of intersection.

Solution:
(i) \(\overleftrightarrow { CD } \) and \(\overleftrightarrow { EF } \), \(\overleftrightarrow { CD } \) and \(\overleftrightarrow { IJ } \), \(\overleftrightarrow { EF } \) and \(\overleftrightarrow { IJ } \)
(ii) \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { CD } \), \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { EF } \), \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { IJ } \), \(\overleftrightarrow { GH } \) and \(\overleftrightarrow { IJ } \), \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { GH } \)
(iii) P, Q and R

Question 7.
From the given figure, name
(i) all pairs of parallel lines.
(ii) all pairs of intersecting lines.
(iii) pair of lines whose point of intersection is ‘V’.
(iv) point of intersection of the lines ‘l₂‘ and ‘l₃‘.
(v) point of intersection of the lines ‘l₁‘, and ‘l₅

Solution:
(i) Pairs of parallel lines:

• l₃ and l₄
• l₄ and l₅
• l₃ and l₅

(ii) Pairs of intersecting lines:

• l₁ and l₂
• l₁ and l₃
• l₁ and l₄
• l₁ and l₅
• l₂ and l₃
• l₂ and l₄
• l₂ and l₅

(iii) l₁ and l₂ intersect at ‘V’
(iv) point of intersection of the lines ‘l₂‘ and; l₅‘ is ‘Q’
(v) point of intersection of the lines ‘l₁‘ and ‘l₅‘ is ‘U’

Objective Type Questions

Question 8.
The number of line segments in
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3

Question 9.
A line is denoted as __________
(a) AB
(b) \(\overrightarrow{AB}\)
(c) \(\overleftrightarrow {AB} \)
(d) \(\overline { AB }\)
Solution:
(c) \(\overleftrightarrow {AB} \)

Students can download Maths Chapter 1 Numbers Ex 1.3 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Ex 1.3

Miscellaneous Practice Problems

Question 1.
Every even number greater than 2 can be expressed as the sum of two prime numbers. Verify this statement for every even number upto 16.
Solution:
Even numbers greater then 2 upto 16 are 4, 6, 8, 10, 12, 14 and 16
4 = 2 + 2
6 = 3 + 3
8 = 3 + 5
10 = 3 + 7 (or) 5 + 5
12 = 5 + 7
14 = 7 + 7 (or) 3 + 11
16 = 5 + 11 (or) 3 + 13

Question 2.
Is 173 a prime? Why?
Solution:
Yes, 173 is a prime. Because it has only 1 and itself as factors.

Question 3.
For which of the numbers, from n = 2 to 8.
Is 2n – 1 a prime?
Solution:
n = 2 ⇒ 2n – 1 = 2 × 2 – 1
= 4 – 1
= 3 (prime)
n = 3 ⇒ 2n – 1 = 2 × 3 – 1
= 6 – 1
= 5 (prime)
n = 4 ⇒ 2n – 1 = 2 × 4 – 1
= 8 – 1
= 7 (prime)
n = 5 ⇒ 2n – 1 = 2 × 5 – 1
= 10 – 1
= 9 (Not prime)
n = 6 ⇒ 2n – 1 = 2 × 6 – 1
= 12 – 1
= 11 (prime)
n = 7 ⇒ 2n – 1 = 2 × 7 – 1
= 14 – 1
= 13
n = 8 ⇒ 2n – 1 = 2 × 8 – 1
= 16 – 1
= 15 (Not prime)

Question 4.
Explain your answer with the reason for the following statements.
(a) A number is divisible by 9 if it is divisible by 3.
(b) A number is divisible by 6 if it is divisible by 12.
Solution:
(i) False, 42 is divisible by 3 but it is not divisible by 9
(ii) True, 36 is divisible by 12. Also divisible by 6.

Question 5.
Find A as required
(i) The greatest 2 digit number 9 A is divisible by 2.
(ii) The least number 567A is divisible by 3.
(iii) The greatest 3 digit number 9A6 is divisible by 6.
(iv) The number A08 is divisible by 4 and 9.
(v) The number 225A85 is divisible by 11.
Solution:
(i) 98 A = 8
(ii) 5670 A = 0
(iii) 996 A = 9
(iv) 108 A = 1
(v) 225885 A = 8

Question 6.
Numbers divisible by 4 and 6 are divisible by 24. Verify this statement and support your answer with an example.
Solution:
False 12 is divisible by both 4 and 6. But not divisible by 24

Question 7.
The sum of any two successive odd numbers is always divisible by 4. Justify this statement with an example.
Solution:
True 3 + 5 = 8 is divisible by 4.

Question 8.
Find the length of the longest rope that can be used to measure exactly the ropes of length 1 m 20cm, 3m 60 cm and 4 m.
Solution:
1 m 20 cm = 120 cm
3 m 60 cm = 360 cm
4 m = 400 cm
This is a HCF related problem. So, we need to find the HCF of 120,360 and 400.

120 = 2 × 2 × 2 × 3 × 5
360 = 2 × 2 × 2 × 3 × 3 × 5
400 = 2 × 2 × 2 × 2 × 5 × 5
HCF = 2 × 2 × 2 × 5 = 40
The length of the longest rope = 40 cm

Challenge Problems

Question 9.
The sum of three prime numbers is 80. The difference between the two of them is 4. Find the numbers.
Solution:
Given the sum of three prime numbers is 80
The numbers will be one or two-digit prime numbers. Also one of them is 2
Sum of the remaining 2 numbers = 78 [∵ one number must be even]
Also their difference = 4 (given) [Otherwise sum of three odd numbers is odd]
The numbers will be 37 and 41
The required numbers are 2, 37, 41

Question 10.
Find the sum of all the prime numbers between 10 and 20 and check whether that sum is divisible by all the single-digit numbers.
Solution:
Prime numbers between 10 and 20 are 11, 13, 17 and 19
Sum = 11 + 13 + 17 + 19 = 60
60 is divisible by 1, 2, 3, 4, 5 and 6.

Question 11.
Find the smallest number which is exactly divisible by all the numbers from 1 to 9.
Solution:
2520

Question 12.
The product of any three consecutive numbers is always divisible by 6. Justify this statement with an example.
Solution:
Yes. Because one of every two consecutive integers is even and so the product of three consecutive integers is even and divisible by 2.
Also one of every 3 consecutive integers is divisible by 3.
Product of any three consecutive integers is divisible by 6.
Example: 5 × 6 × 7

Question 13.
Malarvizhi, Karthiga, and Anjali are friends and natives of the same village. They work in different places. Malarvizhi comes to her home once in 5 days. Similarly, Karthiga and Anjali come to their homes once in 6 days and 10 days respectively, Assuming that they met each other on the 1st of October, when will all the three meet again?
Solution:
This is an LCM related problem. So, we need to find the LCM of 5, 6, and 10.

LCM = 5 × 2 × 1 × 3 × 1
= 30
All the three will meet again once in 30 days.

Question 14.
In an apartment consisting of 108 floors, two lifts A & B starting from the ground floor, stop at every 3rd and 5th floors respectively. On which floors, will both of them stop together?
Solution:
LCM of 3 and 5 = 3 × 5 = 15
The lifts stop together at floors 15, 30, 45, 60, 75, 90, and 105.

Question 15.
The product of 2 two-digit numbers is 300 and their HCF is 5. What are the numbers?
Solution:
15 × 20 = 300
HCF of 15 and 20 is 5
The numbers are 15 and 20

Question 16.
Find whether the number 564872 is divisible by 88. (use of the test of divisibility rule for 8 and 11 will help)
Solution:
564872 Divisibility by 8
564872 It is divisible by 8
Divisibility by 11
5 + 4 + 7 = 16
6 + 8 + 2 = 16
16 – 16 = 0
It is divisible by both 8 and 11 and hence divisible by 88.

Question 17.
Wilson, Mathan, and Guna can complete one round of a circular track in 10, 15, and 20 minutes respectively. If they start together from at the starting point of 7 am, at what time will they meet together again at the same starting point?
Solution:
This is an LCM related problem. So, we need to find the LCM of 10, 15, and 20.

LCM = 5 × 2 × 1 × 3 × 2 = 60 min
They will meet together again after 60 minutes.
ie. 7.am + 60 minutes = 8 am.

Students can download Maths Chapter 4 Geometry Ex 4.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.1

Question 1.
Fill in the blanks.
(a) Every triangle has at least _____ acute angles.
(b) A triangle in which none of the sides equal is called a _____.
(c) In an isosceles triangle ______ angles are equal.
(d) The sum of three angles of a triangle is ______.
(e) A right-angled triangle with two equal sides is called ______.
Solution:
(a) Two
(b) Scalene Triangle
(c) Two
(d) 180°
(e) Isosceles right-angled triangle

Question 2.
Match the following

Solution:
(i) Scalene triangle
(ii) Right-angled triangle
(iii) Obtuse angled triangle
(iv) Isosceles triangle
(v) Equilateral triangle

Question 3.
In ΔABC, name the

a) Three sides: ……….., ………., ……….
b) Three Angles: ………., ………, ……….
c) Three Vertices: ………., ………., ……….
Solution:
(a) \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CA}}\)
(b) ∠ABC, ∠BCA, ∠CAB or ∠A, ∠B, ∠C
(c) A, B, C

Question 4.
Classify the given triangles based on its sides as scalene, isosceles or equilateral.

Solution:
(i) Equilateral triangle
(ii) Scalene triangle
(iii) Isosceles triangle
(iv) Scalene triangle

Question 5.
Classify the given triangles based on their angles as acute-angled, right-angled, or obtuse-angled.

Solution:
(i) Acute angled triangle
(ii) Right-angled triangle
(iii) Obtuse angled triangle
(iv) Acute angled triangle

Question 6.
Classify the following triangles based on its sides and angles?

Solution:
(i) Isosceles Acute angled triangle
(ii) Scalene Right-angled triangle
(iii) Isosceles Obtuse angled triangle
(iv) Isosceles Right-angled triangle
(v) Equilateral Acute angled triangle
(vi) Scalene Obtuse angled triangle

Question 7.
Can a triangle be formed with the following sides? If yes, name the type of triangle.
(i) 8 cm, 6 cm, 4 cm
(ii) 10 cm, 8 cm, 5 cm
(iii) 6.2 cm, 1.3 cm, 3.5 cm
(iv) 6 cm, 6 cm, 4 cm
(v) 3.5 cm, 3.5 cm, 3.5 cm
(vi) 9 cm, 4 cm, 5 cm
Solution:
(i) Sum of two smaller sides of the triangle
= 6 + 4 = 10 cm > 8 cm
It is greater than the third side. So, a triangle can be formed a scalene triangle.

(ii) Sum of two smaller sides of the triangle
= 8 + 5 = 13 cm > 10 cm
It is greater than the third side. So, a triangle can be formed a scalene triangle.

(iii) Sum of two smaller sides of the triangle
= 1.3 + 3.5 = 4.8 cm < 6.2 cm
It is not greater than the third side. So, a triangle cannot be formed.

(iv) Two sides are equal.
So, a triangle can be formed. Isosceles triangle.

(v) Three sides are equal.
So, a triangle can be formed equilateral triangle.

(vi) Sum of two smaller sides of the triangle
= 4 + 5 = 9 cm = 9 cm
It is equal to the third side. No, a triangle cannot be formed.

Question 8.
Can a triangle be formed with the following angles? If yes, name the type of triangle.
(i) 60°, 60°, 60°
(ii) 60°, 40°, 42°
(iii) 90°, 55°, 35°
(iv) 60°, 90°, 90°
(v) 70°, 60°, 50°
(vi) 100°, 50°, 30°
Solution:
(i) 60°, 60°, 60°
Sum of the angles = 60°+ 60°+ 60° = 180°
A triangle can be formed as an acute-angled triangle.

(ii) 60°, 40°, 42°
Sum of the angles = 60° + 40° + 42° = 142°
A triangle cannot be formed.

(iii) 90°, 50°, 35°
Sum of the angles = 90° + 55° + 35° = 180°
A triangle can be formed Right-angled triangle.

(iv) 60°, 90°, 90°
No, a triangle can not be formed.
A triangle cannot have more than one right angle.

(v) 70°, 60°, 50°
Sum of the angles = 70° + 60° + 50° = 180°
A triangle can be formed as an acute-angled triangle.

(vi) 100°, 50°, 30°
Sum of the angles = 100° + 50° + 30° = 180°
A triangle can be formed as an obtuse-angled triangle.

Question 9.
Two angles of the triangles are given. Find the third angle
(i) 80°, 60°
(ii) 75°, 35°
(iii) 52°, 68°
(iv) 50°, 90°
(v) 120°, 30°
(vi) 55°, 85°
Solution:
(i) 80°, 60°
Let the third angle be x.
Sum of the angles = 180°
80° + 60° + x = 180°
140 + x = 180°
x = 180°- 140°
x = 40°
Third angle = 40°

(ii) 52°, 68°
Let the third angle be x.
Sum of the angles = 180°
52° + 68° + x = 180°
120 + x = 180°
180° – 120°
x = 60°
Third angle = 60°

(iii) 75°, 35°
Let the third angle be x.
Sum of the angles 180°
75° + 35° + x = 180°
110 + x = 180°
x = 180° – 110°
x = 70°
Third angle = 70°

(iv) 50°, 90°
Let the third angle be x. Sum of the angles = 180°
50° + 90° + x = 180°
140 + x = 180°
x = 180° – 140°
x = 40°
Third angle = 40°

(v) 120°, 30°
Let the third angle be x.
Sum of the angles = 180°
120° + 30° + x = 180°
150 + x = 180°
x = 180° – 150°
x = 30°
Third angle = 30°

(vi) 55°, 85°
Let the third angle be x.
Sum of the angles = 180°
55° + 85° + x = 180°
140 + x = 180°
x = 180° – 140°
x = 40°
Third angle = 40°

Question 10.
I am a closed figure with each of my three angles is 60°. Who am I?
Solution:
Equilateral triangle

Question 11.
Using the given information, write the type of triangle in the table given below.

Solution:
(ii) Acute angled triangle, Isosceles triangle
(iii) Right-angled triangle, Isosceles triangle
(iv) Acute angled triangle, Scalene triangle
(v) Acute angled triangle, Scalene triangle
(vi) Right-angled triangle. Scalene triangle
(vii) Obtuse angled triangle, Scalene triangle
(viii) Obtuse angled triangle, Isosceles triangle

Objective Type Questions

Question 12.
The given triangle is ……….

(a) a right-angled triangle
b) an equilateral triangle
(c) a scalene triangle
(d) an obtuse-angled triangle
Solution:
(b) an equilateral triangle

Question 13.
If all angles of a triangle are less than a right angle, then it is called _____.
(a) an obtuse-angled triangle
(b) a right-angled triangle
(c) an isosceles right-angled triangle
(d) an acute-angled triangle
Solution:
(d) an acute-angled triangle

Question 14.
If two sides of a triangle are 5 cm and 9 cm, then the third side is
(a) 5 cm
(b) 3 cm
(c) 4 cm
(d) 14 cm
Solution:
(a) 5 cm

Question 15.
The angles of a right-angled triangle are
(a) acute, acute, obtuse
(c) acute, right, right
(c) right, obtuse, acute
(d) acute, acute, right
Solution:
(d) acute, acute, right

Question 16.
An equilateral triangle is
(a) an obtuse-angled triangle
(b) a right-angled triangle
(c) an acute-angled triangle
(d) a scalene triangle
Solution:
(c) an acute-angled triangle