Category: Class 6

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.3

Miscellaneous Practice Problems

Question 1.
What are the angles of an isosceles right-angled triangle?
Solution:
Since it is a right-angled triangle
One of the angles is 90°
Other two angles are equal because it is an isosceles triangle.
Other two angles must be 45° and 45°
Angles are 90°, 45°, 45°.

Question 2.
Which of the following correctly describes the given triangle.

(a) It is a right isosceles triangle.
(b) It is an acute isosceles triangle.
(c) It is an obtuse isosceles triangle.
(d) it is an obtuse scalene triangle.
Solution:
(c) It is an obtuse isosceles triangle.

Question 3.
Which of the following is not possible?
(a) An obtuse isosceles triangle
(b) An acute isosceles triangle
(c) An obtuse equilateral triangle
(d) An acute equilateral triangle
Solution:
(c) an obtuse equilateral triangle

Question 4.
If one angle of an isosceles triangle is 124°, then find the other angles.
Solution:
In an isosceles triangle, any two sides are equal. Also, two angles are equal.
Sum of three angles of a triangle = 180°
Given one angle = 124°
Sum of other two angles = 180° – 124° = 56°
Other angles are = \(\frac{56}{2}\) = 28°
28° and 28°.

Question 5.
The diagram shows a square ABCD. If the line segment joints A and C, then mention the type of triangles so formed.

Solution:
For a square all sides are equal and each angle is 90°.
∆ABC and ∆ADC are isosceles right-angled triangles.

Question 6.
Draw a line segment AB of length 6 cm. At each end of this line segment AB, draw a line perpendicular to the line AB. Are these lines parallel?
Solution:
Here CA and DB are perpendicular to AB.
Yes CA and DB are parallel.

Construction:
(i) Drawn a line segment AB of length 6 cm.
(ii) Place the set square on the line in such a way that the vertex of its right angle coincides with B first and A next and one arm of the right angle coincides with the line AB.
(iii) Drawn lines DB and CA through B and A, the other arm of the right angle of the set square.
(iv) The line CA and DB are perpendicular to AB at A and B.

Challenge Problems

Question 7.
Is a triangle possible with the angles 90°, 90° and 0°? Why?
Solution:
No, a triangle cannot have more than one right angle

Question 8.
Which of the following statements is true. Why?
(a) Every equilateral triangle is an isosceles triangle.
(b) Every isosceles triangle is an equilateral triangle.
Solution:
(a) It is true
In an equilateral triangle, all three sides are equal.
It can be an isosceles triangle also, which has two sides equal.
(b) But every isosceles triangle need not be an equilateral triangle.

Question 9.
If one angle of an isosceles triangle is 70°, then find the possibilities for the other two angles.
Solution:
70°, 40° (or) 55°, 55°

Question 10.
Which of the following can be the sides of an isosceles triangle?
(a) 6 cm, 3 cm, 3 cm
(b) 5 cm, 2 cm, 2 cm
(c) 6 cm, 6 cm, 7 cm
(d) 4 cm, 4 cm, 8 cm
Solution:
In a triangle sum of any two sides greater than the third side
(a), (b) and (d) cannot form a triangle.
(c) can be the sides of an isosceles triangle.

Question 11.
Study the given figure and identify the following triangles.

(a) equilateral triangle
(b) isosceles triangles
(c) scalene triangles
(d) acute triangles
(e) obtuse triangles
(f) right triangles
Solution:
(a) BC = 1 + 1 + 1 + 1 = 4 cm
AB = AC = 4 cm
∆ABC is an equilateral triangle.
(b) ∆ABC and ∆AEF are isosceles triangles.
Since AB = AC = 4 cm Also AE = AF.
(c) In a scalene triangle, no two sides are equal.
∆AEB, ∆AED, ∆ADF, ∆AFC, ∆ABD, ∆ADC, ∆ABF and ∆AEC are scalene triangles.
(d) In an acute-angled triangle all the three angles are less than 90°.
∆ABC, ∆AEF, ∆ABF and ∆AEC are acute-angled triangles.
(e) In an obtuse-angled triangle any one of the angles is greater than 90°.
∆AEB and ∆AFC are obtuse angled triangles.
(f) In a right triangle, one of the angles is 90°.
∆ADB, ∆ADC, ∆ADE and ∆ADF are right-angled triangles.

Question 12.
Two sides of the triangle are given in the table. Find the third side of the triangle?

Solution:

Question 13..
Complete the following table.

Solution:

(i) Always acute angles
(ii) Acute angle
(iii) Obtuse angle

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 3 Ratio and Proportion Ex 3.2

Question 1.
Fill in the blanks of the given equivalent ratios.
(i) 3 : 5 = 9 : ___
(ii) 4 : 5 = ___ : 10
(iii) 6 : ____ = 1 : 2
Solution:
(i) 15
Hint: \(\frac{3}{5}=\frac{3 \times 3}{5 \times 3}=\frac{9}{15}\)
(ii) 8
Hint: \(\frac{4}{5}=\frac{4 \times 2}{5 \times 2}=\frac{8}{10}\)
(iii) 12
Hint: \(\frac{1}{2}=\frac{1 \times 6}{2 \times 6}=\frac{6}{12}\)

Question 2.
Complete the Table:

Solution:

Question 3.
Say True or False.
(i) 5 : 7 is equivalent to 21 : 15.
(ii) If 40 is divided in the ratio 3 : 2, then the larger part is 24.
Solution:
(i) False
(ii) True

Question 4.
Give two equivalent ratios for each of the following.
(i) 3 : 2
(ii) 1 : 6
(iii) 5 : 4
Solution:

Question 5.
Which of the two ratios is larger?
(i) 4 : 5 or 8 : 15
(ii) 3 : 4 or 7 : 8
(iii) 1 : 2 or 2 : 1
Solution:

Question 6.
Divide the numbers given below in the required ratio.
(i) 20 in the ratio 3 : 2
(ii) 27 in the ratio 4 : 5
(iii) 40 in the ratio 6 : 14
Solution:
(i) Ratio = 3 : 2
Sum of the ratio = 3 + 2 = 5
5 parts = 20
1 part = \(\frac{20}{5}\)
= 4
3 parts = 3 × 4 = 12
2 parts = 2 × 4 = 8
20 can be divided in the form as 12, 8.

(ii) Ratio = 4 : 5
Sum of the ratio = 4 + 5 = 9
9 parts = 27
1 part = \(\frac{27}{9}\) = 3
4 parts = 4 × 3 = 12
5 parts = 5 × 3 =15
27 can be divided in the form as 12, 15.

(iii) 40 in the ratio 6 : 14
Ratio = 6 : 14
Sum of the ratio = 6 + 14 = 20
20 parts = 40
1 part = \(\frac{40}{20}\) = 2
6 parts = 2 × 6 = 12
14 parts = 2 × 14 = 28
40 can be divided in the form as 12, 28.

Question 7.
In a family, the amount spent in a month for buying Provisions and Vegetables are in the ratio 3 : 2. If the allotted amount is ₹ 4000, then what will be the amount spent for
(i) Provisions and
(ii) Vegetables?
Solution:
Dividing the total amount ₹ 4000 into 3 + 2 = 5 equal parts then
(i) For Provisions:
3 out of 5 parts are spent for provisions and 2 out of 5 parts for vegetables.
\(4000 \times \frac{3}{5}=2400\) for provisions
(ii) For vegetables:
\(4000 \times \frac{2}{5}=1600\) for Vegetables.
₹ 2400 spend on provisions and ₹ 1600 spend on Vegetables.

Question 8.
A line segment 63 cm long is to be divided into two parts in the ratio 3 : 4. Find the length of each part.
Solution:
Total length = 63 cm Ratio = 3 : 4
Sum of the ratio = 3 + 4 = 7
7 parts = 63 cm
1 part = \(\frac{63}{7}\) = 9 cm
3 parts = 3 × 9 cm = 27 cm
4 parts = 4 × 9 cm = 36 cm
∴ 63 cm can be divided into the parts as 27 cm and 36 cm.

Objective Type Questions

Question 9.
If 2 : 3 and 4 : ___ or equivalent ratios, then the missing term is ____
(a) 6
(b) 2
(c) 4
(d) 3
Solution:
(a) 6
Hint: \(\frac{2}{3}=\frac{2 \times 2}{3 \times 2}=\frac{4}{6}\)

Question 10.
An equivalent ratio of 4 : 7 is
(a) 1 : 3
(b) 8 : 15
(c) 14 : 8
(d) 12 : 21
Solution:
(d) 12 : 21

Question 11.
Which is not an equivalent ratio of \(\frac{16}{24}\) ?
(a) \(\frac{6}{9}\)
(b) \(\frac{12}{18}\)
(c) \(\frac{10}{15}\)
(d) \(\frac{20}{28}\)
Solution:
(d) \(\frac{20}{28}\)
Hint: \(\frac{16}{24}=\frac{8 \times 2}{8 \times 3}=\frac{2}{3}\)

Question 12.
If Rs 1600 is divided
(a) Rs 480
(b) Rs 800
(c) Rs 1000
(d) Rs 200
Solution:
(c) Rs 1000

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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

Question 1.
Observe the diagram and fill 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) end points/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:

L, M, N are collinear points

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

X, Y, Z and A are non-collinear points.

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

l₁, l₂ and l₃ are the concurrent lines.
‘O’ is the point of concurrency.

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

l₁, l₂ and l₃ are non concurrent lines.
A, B and C are the 3 points of intersection.
There are two points of intersection X and Y

Objective Type Questions

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

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
Hint: Non-collinear points are points which are not on a line.

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

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 1 Numbers Ex 1.1

Question 1.
Fill in the blanks.

1. The smallest 7 digit number is _______
2. The largest 8 digit number is _______
3. The place value of 5 in 7005380 is ________
4. The expanded form of the number 76,70,905 is _______

Solution:

1. 10,00,000
2. 9,99,99,999
3. 5 × 1000 = 5000
4. 7 × 10,00,000 + 6 × 1,00,000 + 7 × 10,000 + 0 + 9 × 100 + 0 + 5 × 1
   (or)
   70,00,000 + 6,00,000 + 70,000 + 900 + 5

Question 2.
Say True or False.

1. In the Indian System of Numeration, the number 67999037 is written as 67999037
2. The successor of a one-digit number is always a one-digit number.
3. The predecessor of a 3-digit number is always a 3 or 4-digit number.
4. 88888 = 8 × 10000 + 8 × 100 + 8 × 10 + 8 × 1

Solution:

1. True
2. False
3. False
4. False

Question 3.
Complete the given order.
Ten crore, crore, ten lakh, ____, ____, _____ , _____, _____.
Solution:
Ten crores, Crore, Ten lakh, Lakh, Ten Thousand, Thousand, Hundred, Ten, One

Question 4.
How many ten thousands are there in the smallest 6 digit number?
Solution:
Smallest six-digit number is 1,00,000
1 lakh = Ten Thousand
Another Method:
Lakh is only one place to the left of Ten thousand
1 lakh is 10 times ten thousand

1 lakh = Ten-Ten Thousand

Question 5.
Using the digits 5, 2, 0, 7, 3 forms the largest 5 digit number and the smallest 5 digit number.
Solution:
Given digits = 5, 2, 0, 7, 3
Largest 5 digit number – 75320
Smallest 5 digit number – 20357

Question 6.
Observe the commas and write down the place value of 7.

1. 56,74,56,345
2. 567,456,345

Solution:

1. 56,74,56,345
   Place value of 7 is 7 × 10,00,000 = 70,00,000 = Seventy Lakhs.
2. 567,456,345
   Place value of 7 is 7 × 1,000,000 = 7,000,000 = Seven Million.

Question 7.
Write the following numbers in the International system by using commas.
(i) 347056
(ii) 7345671
(iii) 634567105
(iv) 1234567890
Solution:

Question 8.
Write the largest six-digit number and put commas in the Indian and the International Systems.
Solution:
The largest six-digit number is 999999

Question 9.
Write the number names of the following numerals in the Indian System.
(i) 75,32,105
(ii) 9,75,63,453
Solution:
(i) 75,32,105

Seventy-Five Lakhs Thirty-Two Thousand One Hundred and Five
(ii) 9,75,63,453

Nine crores Seventy Five Lakhs Sixty Three Thousand Four Hundred and Fifty-Three.

Question 10.
Write the number names in words using the International System
(i) 345,678
(ii) 8,343,710
(iii) 103,456,789
Solution:
(i) 345,678

Three Hundred and Forty-Five Thousand Six Hundred and Seventy-Eight
(ii) 8,343,710

Eight Million Three Hundred and Forty-Three Thousand Seven Hundred and Ten.
(iii) 103,456,789

One Hundred Three Million Four Hundred Fifty-Six Thousand Seven Hundred and Eighty’ Nine.

Question 11.
Write the number name in numerals.

1. Two crores thirty lakhs fifty-one thousand nine hundred eighty.
2. Sixty-six million three hundred forty-five thousand twenty-seven.
3. Seven hundred eighty-nine million, two hundred thirteen thousand four hundred fifty six.

Solution:

1. 2,30,51,980
2. 66,345,027
3. 789,213,456

Question 12.
Tamil Nadu has about twenty-six thousand three hundred forty-five square kilometre of Forest land. Write the number mentioned in the statement in the Indian System.
Solution:
26,345 sq km.

Question 13.
The number of employees in the Indian Railways is about 10 lakhs. Write this in the International System of numeration.
Solution:
1.000,000 (one million)

Objective Type Questions

Question 14.
1 billion is equal to
(a) 100 crore
(b) 100 million
(c) 100 lakh
(d) 10000 lakh
Solution:
(a) 100 crore

Question 15.
The successor of 10 million is
(a) 1000001
(b) 10000001
(c) 9999999
(d) 100001
Solution:
(b) 10000001

Question 16.
The difference between successor and predecessor of 99999 is
(a) 90000
(b) 1
(c) 2
(d) 99001
Solution:
(c) 2

Question 17.
The expanded form of the number 6,70,905 is
(a) 6 × 10000 + 7 × 1000 + 9 × 100 + 5 × 1
(b) 6 × 10000 + 7 × 1000 + 0 × 100 + 9 × 100 + 0 × 10 + 5 × 1
(c) 6 × 1000000 + 7 × 10000 + 0 × 1000 + 9 × 100 + 0 × 10 + 5 × 1
(d) 6 × 100000 + 7 × 10000 + 0 × 1000 + 9 × 100 + 0 × 10 + 5 × 1
Solution:
(d) 6 × 100000 + 7 × 10000 + 0 × 1000 + 9 × 100 + 0 × 10 + 5 × 1

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.1

Question 1.
Fill in the blanks.
(i) 250 ml + \(\frac{1}{2}\) ml = _____ l.
(ii) 150 kg 200 g + 55 kg 750 g = ____ kg ____ g.
(iii) 20 l – 1 l 500 ml = ____ l ___ ml
(iv) 450 ml × 5 = ____ l ____ ml.
(v) 50 Kg ÷ 100 g = ______
Solution:
(i) \(\frac{3}{4}\) l
(ii) 205 kg 950 g
(iii) 18 l 500 ml
(iv) 2l 250 ml
(v) 500

Question 2.
True or False
(i) Pugazhenthi ate 100 g of nuts which is equal to 0.1 kg.
(ii) Meena bow it 250 ml of butter milk which is equal to 2.5 l.
(iii) Karkuzhali’s bag 1 kg 250 g and poong- kodi’s bag 2 kg 750 g. The total weight of their bags 4 kg.
(iv) Vanmathi bought 4 books each weighing 500 g. Total weight of 4 books is 2 kg.
(v) Gayathiri bought 1 kg of birthday cake. She shared 450 g with her friends.
The weight of cake remaining is 650 g
Solution:
(i) True
(ii) False
(iii) True
(iv) True
(v) False

Question 3.
Convert into indicated units:
(i) 10 l and 5 ml into ml
(ii) 4 km and 300 m into m
(iii) 300 mg into g
Solution:
(i) 10 l and 5 ml
= 10 × 1000 ml + 5 ml
= 10,000 ml + 5 ml [∴ 1l = 1000 ml]
= 10,005 ml.
∴ 10 l and 5 ml = 10,005 ml.
(ii) 4 km and 300 m into m.
4 km and 300 m
= 4 × 1000 m + 300 m
= 4000 m + 300 m [∴ 1 km = 1000 m]
= 4300 m
∴ 4 km and 300 m = 4300 m
(iii) 300 mg into g.

Question 4.
Convert into higher units:
(i) 13000 mm (km, m, cm)
(ii) 8257 ml (kl, l)
Solution:
(i) 13000 mm (km, m, cm)
(ii) 8257 ml (kl, l)

Question 5.
Convert into lower units:
(i) 15 km (m, cm, mm)
Solution:
15 km = 15 × 1000 m = 15000 m
15 km = 15 × 100000 cm
= 1500000 cm
15 km = 15 × 1000000 mm
= 15000000 mm

(ii) 12 kg (g, mg)
Solution:
12 kg = 12 × 1000 g
= 12000 g
12 kg = 12 × 1000000 mg
= 12000000 mg

Question 6.
Compare and put > or < or = in the following:
(i) 800 g + 150 g ____ 1 kg
(ii) 600 ml + 400 ml ____ 1 l
(iii) 6 m 25 cm ____ 600 cm + 25 cm
(iv) 88 cm ____ 8 m 8 cm
(v) 55 g ____ 550 mg
Solution:
(i) 800 g + 150g < 3kg
(ii) 600 ml + 400 ml = 1 l
(iii) 6 m 25 cm = 600 cm + 25 cm
(iv) 88 cm < 8 m 8 cm
(v) 55 g > 550 mg

Question 7.
Geetha brought 2 l and 250 ml of water in a bottle. Her friend drank 300 ml from it. How much of water is remaining in the bottle?
Solution:
Total Capacity of water = 2 l 250 ml
= (2 x 1000 + 250) l
= 2000 + 250 ml
= 2250 ml
Water Consumed = 300 ml
Water remaining in the bottle = (2250 – 300) ml
= 1950 ml
= 1 litre 950 ml

Question 8.
Thenmozhi’s height is 1.25 m now. She grows for 5 cm every year. What would be her height after 6 years?
Solution:
Thenmozhi’s present height = 1.25 m
Rate of growth per year = 5 cm
Her growth in 6 years = 5 cm × 6 = 30 cm.
After 6 years her height = 1.25 m + 30 cm
= 1.25 × 100 + 30 cm
= 125 + 30 cm
= 155 cm.
∴ After 6 years Thenmozhi’s height will be 155 cm.

Question 9.
Priya bought 22 \(\frac{1}{2}\) kg of onion/ Krishna bought 18 \(\frac{3}{4}\) kg of onion and sethu bought 9 kg 250 g of onion, what is the total weight of onion did they buy?
Solution:
Total weight of onion bought
= 22 \(\frac{1}{2}\) + 18 \(\frac{3}{4}\) + 9 \(\frac{1}{4}\) kg
= 22 kg 500 g + 18 kg 750 g + 9 kg 250 g
= 49 kg 1500 g
= 50 kg 500 g

Question 10.
Maran walks 1.5 km every day to reach the school while Mahizhan walks 1400 m. Who walks more distance and by how much?
Solution:
Distance which Maran walks = 1.5 km = 1.5 × 1000 m = 1500 m
The distance which Mahizhan walks = 1400 m.
Here 1500 > 1400
∴ Difference = 1500 – 1400 = 100 m.
∴ Maran walks more distance = 100 m.

Question 11.
In a JRC one day camp, 150 gm of rice and 15 ml oil are needed for a student. If there are 40 students to attend the camp how much of rice and oil are needed?
Solution:
Rice needed for a student = 150 gm
Rice needed for 40 students = 40 × 150 gm = 6000 gm = 6 kg
Oil needed for a student = 15 ml
Oil needed for 40 students = 40 × 15 ml = 600 ml

Question 12.
In a school, 200 litres of lemon juice is prepared. If 250 ml lemon juice is given to each student, how many students get the juice?
Solution:
Total lemon juice prepared = 200 l = 200 × 1000 ml = 2,00,000 ml.
∴ Quantity of Lemon juice given to one student = 250 ml.
∴ Number of students can get = \(\frac{2,00,000}{250}\) = 800
∴ 800 students can get the lemon juice.

Question 13.
How many glasses of the given capacity will fill a 2 litre jug?
(i) 100 ml ……….
(ii) 50 ml ……….
(iii) 500 ml ………
(iv) 1 l ………
(v) 250 ml ………
Solution:
(i) 20
(ii) 40
(iii) 4
(iv) 2
(v) 8

Objective Type Questions

Question 14.
9 m 4 cm is equal to _____
(a) 94 cm
(b) 904 cm
(c) 9.4 cm
(d) 0.94 cm
Solution:
(b) 904 cm

Question 15.
1006 g is equal to ………
(i) 1 kg 6 g
(ii) 10 kg 6 g
(iii) 100 kg 6 g
(iv) 1 kg 600 g
Solution:
(i) 1 kg 6 g

Question 16.
Every day 150 l of water is sprayed in the garden. Water sprayed in a week is ____
(a) 700 l
(b) 1000 l
(c) 950 l
(d) 1050 l
Solution:
(d) 1050 l

Question 17.
Which is the greatest? 0.007 g, 70 mg, 0.07 cg ……….
(i) 0.07 cg
(ii) 0.007 g
(iii) 70 mg
(iv) all are equal
Solution:
(iii) 70 mg

Question 18.
7 km – 4200 m is equal to
(a) 3 km 800 m
(b) 2 km 800 m
(c) 3 km 200 rn
(d) 2 km 200 m
Solution:
(b) 2 km 800 m

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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.

Solution:

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

Solution:

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

Solution:
(i), (iii) and (v) are Right Angles.

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

Solution:
(i), (iii) and (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.

1. 20° and 70° are complementary.
2. 88° and 12° are complementary.
3. 80° and 180° are supplementary.
4. 0° and 180° are supplementary.

Solution:

1. True
2. False
3. False
4. True

Question 8.
Draw and label each of the angles
(i) ∠NAS = 90°
(ii) ∠BLG = 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 ‘l’ complementary angles in each case.

Solution:
(i) If two angles add up to 180°, they are supplementary.
We know that the straight angle ∠CBC = 180°
Given ∠CBD = 25°
∴ ∠DBA = 180° – 25° = 155°
∴ Supplementary ∠DBA = 155°

(ii) If two angles add upto 90°, they are complementary
Given ∠ABC = 25°
∠CBD = 30°
∴ ∠DBA = 90° – 30° = 60°
∴ Complementary angle ∠DBA = 60°

(iii) Given ∠CBA = 90°
∠CBD = 46°
∴ ∠DBA = 90° – 46° = 44°
We know that the sum of two angles is 90°, then they are complementary
∴ Complementary angle ∠DBA = 44°

(iv) Given ∠DBE = 180°
∠DBC = 67°
∴ ∠CBE = 180° – 67° = 113°
Here ∠DBC + ∠DBE = 180°
∴ Supplementary angle ∠CBE = 113°

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 the 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°
Hint: ∠XYZ = ∠BYZ = ∠AYZ

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 3 Ratio and Proportion Ex 3.1

Question 1.
Fill in the blanks.
(i) Ratio of ₹ 3 to ₹ 5 = ____
(ii) Ratio of 3 m to 200 cm = ______
(iii) Ratio of 5 km 400 m to 6 km = ____
(iv) Ratio of 75 paise to ₹ 2 = ____
Solution:
(i) 3 : 5
(ii) 3 : 2
Hint: 3m = 300 cm
(iii) 9 : 10
Hint: 5km 400 m = 5400m and 6 km = 6000 m
(iv) 3 : 8
Hint: ₹ 2 = 200 paise

Question 2.
Say whether the following statements are True or False.
(i) The ratio of 130 cm to 1 m is 13 : 10.
(ii) One of the terms in a ratio cannot be 1.
Solution:
(i) True
(ii) False

Question 3.
Find the simplified form of the following ratios.
(i) 15 : 20
(ii) 32 : 24
(iii) 7 : 15
(iv) 12 : 27
(v) 75 : 100
Solution:

Question 4.
Akilan walks 10 km in an hour while Selvi walks 6 km in an hour. Find the simplest ratio of the distance covered by Akilan to that of Selvi.
Solution:
Ratio of the distance covered by Akilan to that of Selvi = 10 km : 6 km
= \(\frac{10}{6}\)
= \(\frac{5}{3}\)
= 5 : 3

Question 5.
The cost of parking a bicycle is ₹ 5 and the cost of parking a scooter is ₹ 15. Find the simplest ratio of the parking cost of a bicycle to that of a scooter.
Solution:
Parking cost of bicycle = ₹ 5
Parking cost of Scooter = ₹ 15
\(\frac{\text { Parking cost of bicyle }}{\text { Parking cost of scooter }}=\frac{5}{15}=\frac{1}{3}\)
Parking cost of bicycle : scooter = 1 : 3

Question 6.
Out of 50 students in class 30 are boys. Find the ratio of
(i) number of boys to the number of girls
(ii) number of girls to the total number of students
(iii) number of boys to the total number of students
Solution:
The total number of students = 50.
The number of boys = 30.
Then number of girls = 50 – 30 = 20.

Objective Type Questions

Question 7.
The ratio of Rs 1 to 20 paise is
(a) 1 : 5
(b) 1 : 2
(c) 2 : 1
(d) 5 : 1
Solution:
(d) 5 : 1

Question 8.
The ratio of 1 m to 50 cm is _____
(a) 1 : 50
(b) 50 : 1
(c) 2 : 1
(d) 1 : 2
Solution:
(c) 2 : 1
Hint: 1 m = 100 cm

Question 9.
The length and breadth of a window are in 1 m and 70 cm respectively. The ratio of the length to the breadth is
a) 1 : 7
(b) 7 : 1
(c) 7 : 10
(d) 10 : 7
Solution:
(d) 10 : 7

Question 10.
The ratio of the number of sides of a triangle to the number of sides of a rectangle is ____
(a) 4 : 3
(b) 3 : 4
(c) 3 : 5
(d) 3 : 2
Solution:
(b) 3 : 4
Triangle has three sides and a rectangle has four sides.

Question 11.
If Azhagan is 50 years old and his son is 10 years old then the simplest ratio between the age of Azhagan to his son is
(a) 10 : 50
(b) 50 : 10
(c) 5 : 1
(d) 1 : 5
Solution:
(c) 5 : 1

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 5 Information Processing Ex 5.2

Miscellaneous Practice problems

Question 1.
Find HCF of 188 and 230 by Euclid’s game.
Solution:
By Euclid’s game HCF (a, b) = HCF (a, a – b) if a > b.
Here HCF (188, 230) = HCF (230, – 188) because 230 > 188
= HCF (188, 42) = HCF (146, 42)
= HCF (104, 42) = HCF (62, 42)
= HCF (42, 20) = HCF (22, 20)
= HCF (20,2) = HCF (18, 2) = 2
∴ HCF (230, 188) = 2

Question 2.
Write the numbers from 1 to 50. From that find the following.
(i) The numbers which are neither divisible by 2 nor 7.
(ii) The prime numbers between 25 and 40.
(iii) All square number upto 50.
Solution :
(i) 9, 11, 13, 15, 17, 19, 23, 25, 27, 29, 31, 33, 37, 39, 41, 43, 45, 47
(ii) 29, 31, 37
(iii) 1, 4, 9, 16, 25, 36, 49

Question 3.
Complete the following pattern




Solution:
(i)

(ii)

 

Question 4.
Complete the table by using the following instructions.
A : It is the 6th term in the Fibonacci sequence.
B : The predecessor of 2.
C : LCM of 2 and 3.
D : HCF of 6 and 20.
E : The reciprocal of 1/5.
F : The opposite number of -7.
G : The first composite number.
H : Area of a square of side 3 cm.
I : The number of lines of symmetry of an equilateral triangle. After completing the table, what do you observe? Discuss.

Solution:
A : 6th term in Fibonacci sequence is 8.
B : Predecessor of 2 is 1.
C : LCM of 2 and 3 is 6.
D : HCF of 6 and 20 is 2.
E : Reciprocal of \(\frac{1}{5}\) is 5.
F : Opposite number of – 7 is 7.
G : The first composite number is 4.
H : Area of square of side 3 cm is 3 × 3 = 9 cm².
I : The number of lines of symmetry of an equilateral triangle is 3.
∴ The table becomes

From the table we observe that the numbers are from 1 to 9

Question 5.

Solution:

Question 6.
Replace the letter by symbols as + for A, – for B, × for C and ÷ for D. Find the answer for the pattern 4B3C5A30D2 by doing the given operations.
Solution:
4

Question 7.
Observe the pattern and find the word by hiding the numbers
1H2O3W 4A5R6E 7Y809U?
Solution:
When hiding the numbers we get

HOW ARE YOU?

Question 8.
Arranging from eldest to the youngest. What do you get ?

Solution:
Arranging from eldest to the youngest we get
F – refers to grandparents
A – refers to parents
M – refers to uncle
I – refers to elder sister
L – refers to me
Y – refers to younger brother
So we get FAMILY

Challenge Problems

Question 9.
Prepare a daily time schedule for evening study at home.
Solution:
5.30 pm – arrival
5.30 pm – 6.30 pm – Tea, Tv programme
6.30 pm – 7.30 pm – Maths
7.30 pm – 8.30 pm – Supper, Tv news
8.30 pm – 9.00 pm – English
9.00 pm – 9.30 pm – Science
9.30 pm – 10.00 pm – Social science
10.00 pm – Going to bed.

Question 10.
Observe the geometrical pattern and answer the following questions

i) Write down the number of sticks used in each of the iterative pattern.
ii) Draw the next figure in the pattern also find the total number sticks used in it.
Solution:
Number of sticks used in first pattern = 3
Number of sticks in second pattern = 9
Number of sticks in third pattern =18
ii) Next pattern

Question 11.
Find HCF of 28y, 35, 42 by Euclid’s game.
Solution:

Question 12.
Follow the given instructions to fill your name in the OMR sheet.
* The name should be written in capital letters from left to right.
* One alphabet is to be entered in each box.
* If any empty boxes are there at the end they should be left blank.
* Ball point pen is to be used for shading the bubbles for the corresponding alphabets.

Solution:

Question 13.
Consider the Postal Index Number (PIN) written on the letters as follows: 604506; 604516; 604560; 604506; 604516; 604516; 604560; 604516; 604505; 604470; 604515; 604520; 604303; 604509; 604470. How the letters can be sorted as per Postal Index Numbers?
Solution:
604 is common for all postal index numbers. Compare the remaining 3 digits, 303, 470, 505, 506 (two) 509, 510. 515, 516 (Four), 520, 560 (two).

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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:
(i) There are 10 line segments.
(ii) They are \(\overline{\mathrm{PA}}, \overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CQ}}, \overline{\mathrm{PB}}, \overline{\mathrm{AC}}, \overline{\mathrm{BQ}}, \overline{\mathrm{PC}}, \overline{\mathrm{AQ}} \text { and } \overline{\mathrm{PQ}}\)

Question 3.
Measure the following line segments.

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

Question 4.
Construct a line segment using ruler and compass.
(i) \(\overline{\mathrm{AB}}=7.5 \mathrm{cm}\)
(ii) \(\overline{\mathrm{CD}}=3.6 \mathrm{cm}\)
(iii) \(\overline{\mathrm{QR}}=10 \mathrm{cm}\)
Solution:
(i) \(\overline{\mathrm{AB}}=7.5 \mathrm{cm}\)

Construction:
Step 1: Drawn a line l and marked point A.
Step 2: Measured 7.5 cm using compass placing the pointer at ‘0’ and pencil pointer at 7.5 cm.
Step 3: Placing a compass pointer at A, drawn an arc on l with the pencil pointer. It cut l at B.
Step 4: AB is the required segment of length 7.5 cm.

(ii) \(\overline{\mathrm{CD}}=3.6 \mathrm{cm}\)

Construction:
Step 1: Drawn a line l and marked the point ‘C’ on it.
Step 2: Measured 3.6 cm using a compass by placing the pointer at ‘O’ and pencil at 3.6 cm
Step 3: Placing pointer at C drawn the arc on ‘l’ with pencil pointer
Step 4: CD is the required line segment of length 3.6 cm

(iii) \(\overline{\mathrm{QR}}=10 \mathrm{cm}\)

Construction:
Step 1: Drawn a line ‘l’ and marked the point Q on it
Step 2: Measured 10 cm using a compass by placing the pointer at ‘0’ and pencil pointer at 10 cm.
Step 3: Placing pointer at Q drawn the arc on ‘l’ with pencil pointer and named the point R
Step 4: QR is the required line segments of 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.

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

Solution:
(i) Parallel lines:
\(\overrightarrow{\mathrm{CD}} \text { and } \overrightarrow{\mathrm{EF}}\); \(\overrightarrow{\mathrm{CD}} \text { and } \overrightarrow{\mathrm{IJ}}\); \(\overrightarrow{\mathrm{EF}} \text { and } \overrightarrow{\mathrm{IJ}}\) are parallel lines.
(ii) Intersecting lines:
(a) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{CD}}\)
(b) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{EF}}\)
(c) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{GH}}\)
(d) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{IJ}}\)
(e) \(\overrightarrow{\mathrm{GH}} \text { and } \overrightarrow{\mathrm{IJ}}\)
Points of Intersection:
P, Q and R are the points of intersection.

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  is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3
Hint: \(\overline{\mathrm{AB}}, \overline{\mathrm{AC}}, \overline{\mathrm{BC}}\)

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

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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

Question 1.
Suppose, you have two shorts, one is black and the other one is blue; three shirts which are in 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:
We have given two shorts which are black and blue in colour. Take it as T black and T blue.
Also, we have 3 shirts, coloured white, blue and red denoted by S white, S blue and S red.
Now fix T black and then T blue the different combinations are

Thus we get a total of 6 combinations as
Black short and White shirt
Black short and Blue shirt
Black short and Redshirt
Blue short and White shirt
Blue short and Blue shirt
Blue short and Redshirt.
But it is given short and the shirt is of different colours.
We give up Blue short and Blue shirt combination. So we have 5 different combinations.

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