Bhagya

Students can download Maths Chapter 4 Geometry Ex 4.4 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 4 Geometry Ex 4.4

Miscellaneous Practise Problems

Question 1.
Find the type of lines marked in thick lines (Parallel, intersecting or perpendicular)

Solution:
(i) Parallel lines
(ii) Parallel lines
(iii) Parallel and Perpendicular lines
(iv) Intersecting lines

Question 2.
Find the parallel and intersecting line segments in the picture given below.

Solution:
(a) Parallel line segments

• \(\overline{\mathrm{YZ}} \text { and } \overline{\mathrm{DE}}\)
• \(\overline{\mathrm{EA}} \text { and } \overline{\mathrm{ZV}}\)
• \(\overline{\mathrm{VW}} \text { and } \overline{\mathrm{AB}}\)
• \(\overline{\mathrm{WX}} \text { and } \overline{\mathrm{BC}}\)
• \(\overline{\mathrm{YX}} \text { and } \overline{\mathrm{DC}}\)
• \(\overline{\mathrm{YD}} \text { and } \overline{\mathrm{XC}}\)
• \(\overline{\mathrm{XC}} \text { and } \overline{\mathrm{WB}}\)
• \(\overline{\mathrm{WB}} \text { and } \overline{\mathrm{VA}}\)
• \(\overline{\mathrm{VA}} \text { and } \overline{\mathrm{ZE}}\)
• \(\overline{\mathrm{ZE}} \text { and } \overline{\mathrm{YD}}\)

(b) Intersecting line segments

• DE and ZV
• WX and DC

Question 3.
Name the following angles as shown in the figure.

(i) ∠1 =
(ii) ∠2 =
(iii) ∠3 =
(iv) ∠1 + ∠2 =
(v) ∠2 + ∠3 =
(vi) ∠1 + ∠2 + ∠3 =
Solution:
(i) ∠1 = ∠DBC or ∠CBD
(ii) ∠2 = ∠DBE or ∠EBD
(iii) ∠3 = ∠ABE or ∠EBA
(iv) ∠1 + ∠2 = ∠EBC or ∠CBE
(v) ∠2 + ∠3 = ∠ABD or ∠DHA
(vi) ∠1 + ∠2 + ∠3 = ∠ABC or ∠B or ∠CBA

Question 4.
Measure the angles of the given figures using a protractor and identify the type of angle as acute, obtuse, right or straight.

Solution:
(i) right angle
(ii) acute angle
(iii) straight angle
(iv) obtuse angle

Question 5.
Draw the following angles using the protractor.
(i) 45°
(ii) 120°
(iii) 65°
(iv) 135°
(v) 0°
(vi) 180°
(vii) 38°
(viii) 90°
Solution:

Question 6.
From the figures given below, classify the following pairs of angles into complementary and non-complementary.

Solution:
We know that the two angles are complementary if they add up to 90°.
Therefore (a) (i) is complementary.
In (v) ∠ABC and ∠CBD are complementary
(b) (ii), (iii), (iv) and (v) are non-complementary

Question 7.
From the figures given below, classify the following pairs of angles into supplementary and non supplementary.

Solution:
Ans: (ii) and (iv) are supplementary angles.
(i), and (iii) non-supplementary angles.

Question 8.
From the figure

(i) name a pair of complementary angles.
(ii) name a pair of supplementary angles.
Solution:
(i) ∠FAE; ∠EAD
(ii) ∠FAD; ∠DAC
∠BAC; ∠CAE
∠FAB; ∠BAC
∠FAB; ∠FAE

Question 9.
Find the complementary angle of
(i) 30°
(ii) 26°
(iii) 85°
(iv) 0°
(v) 90°
Solution:

Question 10.
Find the supplementary angle of
(i) 70°
(ii) 35°
(iii) 165°
(iv) 90°
(v) 0°
(vi) 180°
(vii) 95°
Solution:

Challenging Problems

Question 11.
Think and write an object having
(i) Parallel lines (1) ……….. (2) ………. (3) ………..
(ii) Perpendicular lines (1) ……… (2) ……… (3) ………..
(iii) Intersecting lines (1) ……….. (2) ………. (3) ……….
Solution:
(i) Legs of the table, railway tracks, edges of the scale
(ii) Adjacent sides of a Board, Crossbars of windows, Adjacent sides of the textbook
(iii) Crossbars of windows, Ladder, blades of a scissor.

Question 12.
Which angle is equal to twice its complement.
Solution:
Let the angle be x
According to the problem, x = 2 × (90 – x)
x = 180 – 2x
x + 2x = 180
3x = 180
x = \(\frac{180}{3}\)
x = 60
∴ The angle is 60°

Question 13.
Which angle is equal to two-thirds of its supplement.
Solution:
Let the angle be x
According to the problem,
x = \(\frac{2}{3}\) × (180° – x)
3x = 2(180 – x)
3x = 360 – 2x
3x + 2x = 360°
5x = 360°
x = \(\frac{360°}{5}\)
x = 72°
∴ The angle is 72°

Question 14.
Given two angles are supplementary and one angle is 20° more than the other. Find the two angles.
Solution:
Let the angles be x and x + 20°
According to the problem,
x + x + 20 = 180°
2x + 20° = 180°
2x = 180° – 20°
2x = 160°
x = \(\frac{160°}{2}\)
x = 80°
x + 20 = 80° + 20°
= 100°
∴ The two angles are 80° and 100°

Question 15.
Two complementary angles are in ratio 7 : 2. Find the angles.
Solution:
Let the angles be 7x, 2x
According to the problem,
7x + 2x = 90
9x = 90
x = \(\frac{90}{9}\)
x = 10
7x = 7 × 10
= 70
2x = 2 × 10
= 20
∴ Two angles are 70° and 20°

Question 16.
Two supplementary angles are in ratio 5 : 4. Find the angles.
Solution:
Total of two supplementary angles = 180°
Given they are in the ratio 5 : 4
Dividing total angles to 5 + 4 = 9 equal parts.
One angle \(=\frac{5}{9} \times 180=100^{\circ}\)
Another angle \(=\frac{4}{9} \times 180=80^{\circ}\)
Two angles are 100° and 80°.

Students can download Maths Chapter 1 Numbers Ex 1.5 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 1 Numbers Ex 1.5

Question 1.
Fill in the blanks.
(i) The difference between the smallest natural number and the smallest whole number is ……….
(ii) 17 × …….. = 34 × 17
(iii) When ……… is added to a number, it remains the same.
(iv) Division by ………. is not defined.
(v) Multiplication by ……… leaves a number unchanged.
Solution:
(i) 1
(ii) 34
(iii) Zero
(iv) 0
(v) 1

Question 2.
Say True or False.

1. 0 is the identity for multiplication of whole numbers.
2. The Sum of two whole numbers is always less than their product.
3. Both addition and multiplication are associative for whole numbers.
4. Both addition and multiplication are commutative for whole numbers.
5. Multiplication is distributive over addition for whole numbers.

Solution:

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

Question 3.
Name the property being illustrated in each of the cases given below.

1. 75 + 34 = 34 + 75
2. (12 × 4) × 8 = 12 × (4 × 8)
3. 50 + 0 = 50
4. 50 × 1 = 50
5. 50 × 42 = 50 × 40 + 50 × 2

Solution:

1. Commutativity for addition
2. Associativity for multiplication
3. Zero is the additive identity
4. One is the multiplicative identity
5. Distributivity of multiplication over addition

Question 4.
Use the properties of whole numbers and simplify.
(i) 50 × 102
(ii) 500 × 689 – 500 × 89
(iii) 4 × 132 × 25
(iv) 196 + 34 + 104
Solution:
(i) 50 × 102
= 50 × (100 + 2)
= (50 × 100) + (50 × 2)
= 5000 + 100 = 5100

(ii) 500 × 689 – 500 × 89
= 500 × (689 – 89)
= 344500 – 44500
= 300000
= 500 × (689 – 89)
= 500 × 600
= 3,00000

(iii) (4 × 132) × 25
= 4 × (132 × 25)
= (4 × 132) × 25
= 528 × 25
= 13200
= 4 × (132 × 25)
= 4 × 3300
= 13200

(iv) (196 + 34) + 104 = 196 + (34 + 104)
(196 + 34) + 104 = 230 + 104 = 334
196 + (34 + 104) = 196 + 138 = 334

Objective Type Questions

Question 5.
(53 + 49) × 0 is
(a) 102
(b) 0
(c) 1
(d) 53 + 49 × 0
Solution:
(b) 0

Question 6.
\(\frac{59}{1}\) is
(a) 1
(b) 0
(c) \(\frac{1}{59}\)
(d) 59
Solution:
(d) 59

Question 7.
The product of a non-zero whole number and its successor is always
(a) an even number
(b) an odd number
(c) zero
(d) none of these
Solution:
(a) an even number

Question 8.
The whole number that does not have a predecessor is
(a) 10
(b) 0
(c) 1
(d) none of these
Solution:
(b) 0

Question 9.
Which of the following expressions is not zero?
(a) 0 × 0
(b) 0 + 0
(c) \(\frac{2}{0}\)
(d) \(\frac{0}{2}\)
Solution:
(c) \(\frac{2}{0}\)
Dividing by 0 is not defined.

Question 10.
Which of the following is not true?
(a) (4237 + 5498) + 3439 = 4237 + (5498 + 3439)
(b) (4237 × 5498) × 3439 = 4237 × (5498 × 3439)
(c) 4237 + 5498 × 3439 = (4237 + 5498) × 3439
(d) 4237 × (5498 + 3439) = (4237 × 5498) + (4237 × 3439)
Solution:
(c) 4237 + 5498 × 3439 = (4237 + 5498) × 3439

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.2 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.2

Question 1.
What is the slope of a line whose inclination with positive direction of x -axis is
(i) 90°
(ii) 0°
Solution:
Here θ = 90°
Slope (m) = tan θ
Slope = tan 90°
= undefined.

(ii) Here θ = 0°
Slope (m) = tan θ
Slope = tan 0°
= 0

Question 2.
What is the inclination of a line whose slope is
(i) 0
(ii) 1
Solution:
(i) m = 0
tan θ = 0 ⇒ θ = 0°
(ii) m = 1 ⇒ tan θ = tan 45° ⇒ 0 = 45°

Question 3.
Find the slope of a line joining the points
(i) (5,\(\sqrt { 5 }\)) with the origin
(ii) (sin θ, -cos θ) and (-sin θ, cos θ)
Solution:
(i) The given points is (5,\(\sqrt { 5 }\)) and (0, 0)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) = \(\frac{0-\sqrt{5}}{0-5}\)
= \(\frac{\sqrt{5}}{5}=\frac{1}{\sqrt{5}}\)

(ii) The given points is (sin θ, -cos θ) and (-sin θ, cos θ)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{\cos \theta+\cos \theta}{-\sin \theta-\sin \theta}\)
= \(\frac{2 \cos \theta}{-2 \sin \theta}\) = – cot θ

Question 4.
What is the slope of a line perpendicular to the line joining A(5,1) and P where P is the mid-point of the segment joining (4,2) and (-6,4).
Solution:
Mid point of XY = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\) = (\(\frac { 4-6 }{ 2 } \),\(\frac { 2+4 }{ 2 } \))
= (\(\frac { -2 }{ 2 } \),\(\frac { 6 }{ 2 } \)) = (-1, 3)

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) = (\(\frac { 3-1 }{ -1-5 } \))
= \(\frac { 2 }{ -6 } \) = – \(\frac { 1 }{ 3 } \)

Question 5.
Show that the given points are collinear: (-3, -4), (7,2) and (12, 5)
Solution:
The vertices are A(-3, -4), B(7, 2) and C(12, 5)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { 2+4 }{ 7+3 } \) = \(\frac { 6 }{ 10 } \) = \(\frac { 3 }{ 5 } \)
Slope of BC = \(\frac { 5-2 }{ 12-7 } \) = \(\frac { 3 }{ 5 } \)
Slope of AB = Slope of BC = \(\frac { 3 }{ 5 } \)
∴ The three points A,B,C are collinear.

Question 6.
If the three points (3, -1), (a, 3) and (1, -3) are collinear, find the value of a.
Solution:
The vertices are A(3, -1), B(a, 3) and C(1, -3)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { 3+1 }{ a-3 } \) = \(\frac { 4 }{ a-3 } \)
Slope of BC = \(\frac { 3+3 }{ a-1 } \) = \(\frac { 6 }{ a-1 } \)
Since the three points are collinear.
Slope of AB = Slope BC
\(\frac { 4 }{ a-3 } \) = \(\frac { 6 }{ a-1 } \)
6 (a – 3) = 4 (a – 1)
6a – 18 = 4a – 4
6a – 4a = -4 + 18
2a = 14 ⇒ a = \(\frac { 14 }{ 2 } \) = 7
The value of a = 7

Question 7.
The line through the points (-2, a) and (9,3) has slope –\(\frac { 1 }{ 2 } \) Find the value of a.
Solution:
The given points are (-2, a) and (9, 3)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
– \(\frac { 1 }{ 2 } \) = \(\frac { 3-a }{ 9+2 } \) ⇒ – \(\frac { 1 }{ 2 } \) = \(\frac { 3-a }{ 11 } \)
2(3 – a) = -11 ⇒ 6 – 2a = -11
-2a = -11 – 6 ⇒ -2a = -17 ⇒ a = – \(\frac { 17 }{ 2 } \)
∴ The value of a = \(\frac { 17 }{ 2 } \)

Question 8.
The line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (8,12) and (x, 24). Find the value of x.
Solution:
Find the slope of the line joining the point (-2, 6) and (4, 8)
Slope of line (m1) = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 8-6 }{ 4+2 } \) = \(\frac { 2 }{ 6 } \) = \(\frac { 1 }{ 3 } \)
Find the slope of the line joining the points (8, 12) and (x, 24)
Slope of a line (m2) = \(\frac { 24-12 }{ x-8 } \) = \(\frac { 12 }{ x-8 } \)
Since the two lines are perpendicular.
m₁ × m₂ = -1
\(\frac { 1 }{ 3 } \) × \(\frac { 12 }{ x-8 } \) = -1 ⇒ \(\frac{12}{3(x-8)}=-1\)
-1 × 3 (x – 8) = 12
-3x + 24 = 12 ⇒ – 3x = 12 -24
-3x = -12 ⇒ x = \(\frac { 12 }{ 3 } \) = 4
∴ The value of x = 4

Question 9.
Show that the given points form a right angled triangle and check whether they satisfies Pythagoras theorem.
(i) A(1, -4) , B(2, -3) and C(4, -7)
(ii) L(0, 5), M(9,12) and N(3,14)
Solution:
(i) The vertices are A(1, -4), B(2, -3) and C(4, -7)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -3+4 }{ 2-1 } \) = \(\frac { 1 }{ 1 } \) = 1
Slope of BC = \(\frac { -7+3 }{ 4-2 } \) = \(\frac { -4 }{ 2 } \) = -2
Slope of AC = \(\frac { -7+4 }{ 4-1 } \) = – \(\frac { 3 }{ 3 } \) = -1
Slope of AB × Slope of AC = 1 × -1 = -1

∴ AB is ⊥^r to AC
∠A = 90°
∴ ABC is a right angle triangle
Verification:

20 = 2 + 18
20 = 20 ⇒ Pythagoras theorem verified

(ii) The vertices are L(0, 5), M(9, 12) and N(3, 14)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of LM = \(\frac { 12-5 }{ 9-0 } \) = \(\frac { 7 }{ 9 } \)

Slope of MN = \(\frac { 14-12 }{ 3-9 } \) = \(\frac { 2 }{ -6 } \) = – \(\frac { 1 }{ 3 } \)
Slope of LN = \(\frac { 14-5 }{ 3-0 } \) = \(\frac { 9 }{ 3 } \) = 3
Slope of MN × Slope of LN = – \(\frac { 1 }{ 3 } \) × 3 = -1
∴ MN ⊥ LN
∠N = 90°
∴ LMN is a right angle triangle
Verification:


130 = 90 + 40
130 = 130 ⇒ Pythagoras theorem is verified

Question 10.
Show that the given points form a parallelogram:
A (2.5,3.5), B(10, -4), C(2.5, -2.5) and D(-5, 5).
Solution:
Let A(2.5, 3.5), B(10, -4), C(2.5, -2.5) and D(-5, 5) are the vertices of a parallelogram.


Slope of AB = Slope of CD = -1
∴ AB is Parallel to CD ……(1)

Slope of BC = Slope of AD
∴ BC is parallel to AD
From (1) and (2) we get ABCD is a parallelogram.

Question 11.
If the points A(2, 2), B(-2, -3), C(1, -3) and D(x, y) form a parallelogram then find the value of x and y.
Solution:
Let A(2, 2), B(-2, -3), C(1, -3) and D(x, y) are the vertices of a parallelogram.

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -3-2 }{ -2-2 } \) = \(\frac { -5 }{ -4 } \) = \(\frac { 5 }{ 4 } \)
Slope of BC = \(\frac { -3+3 }{ -2-1 } \) = \(\frac { 0 }{ -3 } \) = 0
Slope of CD = \(\frac { y+3 }{ x-1 } \)
Slope of AD = \(\frac { y-2 }{ x-2 } \)
Since ABCD is a parallelogram
Slope of AB = Slope of CD
\(\frac { 5 }{ 4 } \) = \(\frac { y+3 }{ x-1 } \)
5(x – 1) = 4 (y + 3)
5x – 5 = 4y + 12
5x – 4y = 12 + 5
5x – 4y = 17 ……(1)
Slope of BC = Slope of AD
0 = \(\frac { y-2 }{ x-2 } \)
y – 2 = 0
y = 2
Substitute the value of y = 2 in (1)
5x – 4(2) = 17
5x -8 = 17 ⇒ 5x = 17 + 18
5x = 25 ⇒ x = \(\frac { 25 }{ 5 } \) = 5
The value of x = 5 and y = 2.

Question 12.
Let A(3, -4), B(9, -4) , C(5, -7) and D(7, -7). Show that ABCD is a trapezium.
Solution:
Let A(3, -4), B(9, -4), C(5, -7) and D(7, -7) are the vertices of a quadrilateral.

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -4+4 }{ 9-3 } \) = \(\frac { 0 }{ 6 } \) = 0
Slope of BC = \(\frac { -7+4 }{ 5-9 } \) = \(\frac { -3 }{ -4 } \) = \(\frac { 3 }{ 4 } \)
Slope of CD = \(\frac { -7+7 }{ 7-5 } \) = \(\frac { 0 }{ 2 } \) = 0
Slope of AD = \(\frac { -7+4 }{ 7-3 } \) = \(\frac { -3 }{ 4 } \) = – \(\frac { 3 }{ 4 } \)
The slope of AB and CD are equal.
∴ AB is parallel to CD. Similarly the slope of AD and BC are not equal.
∴ AD and BC are not parallel.
∴ The Quadrilateral ABCD is a trapezium.

Question 13.
A quadrilateral has vertices at A(-4, -2), B(5, -1) , C(6, 5) and D(-7, 6). Show that the mid-points of its sides form a parallelogram.
Solution:
Let A(-4, -2), B(5, -1), C(6, 5) and D(-7, 6) are the vertices of a quadrilateral.





Slope of EF = Slope of GH = \(\frac { 7 }{ 10 } \)
∴ EF || GH …….(1)
Slope of FG= Slope of EH = – \(\frac { 7 }{ 12 } \)
∴ FG || EH ……(2)
From (1) and (2) we get EFGH is a parallelogram.
The mid point of the sides of the Quadrilateral ABCD is a Parallelogram.

Question 14.
PQRS is a rhombus. Its diagonals PR and QS intersect at the point M and satisfy QS = 2PR. If the coordinates of S and M are (1, 1) and (2, -1) respectively, find the coordinates of P.
Solution:
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of SM = \(\frac { 1+1 }{ 1-2 } \) = \(\frac { 2 }{ -1 } \) = -2
Slope of PM = \(\frac { 1 }{ 2 } \) (Since SM and PM are ⊥^r)
Let the point p be (a,b)
Slope of PM = \(\frac { 1 }{ 2 } \)
\(\frac { b+1 }{ a-2 } \) = \(\frac { 1 }{ 2 } \) ⇒ a – 2 = 2b + 2

a – 2b = 4
a = 4 + 2b ……(1)
Given QS = 2PR
\(\frac { QS }{ 2 } \) = PR
∴ SM = PR
SM = 2PM (PR = 2PM)

Squaring on both sides

∴ (b + 1)² = \(\frac { 1 }{ 4 } \) ⇒ b + 1 = ± \(\frac { 1 }{ 2 } \)
b = \(\frac { 1 }{ 2 } \) – 1 (or) b = – \(\frac { 1 }{ 2 } \) – 1
= – \(\frac { 1 }{ 2 } \) – 1 (or) b = –\(\frac { 1 }{ 2 } \) – 1
= – \(\frac { 1 }{ 2 } \) (or) – \(\frac { 3 }{ 2 } \)
a = 4 + 2b
a = 4 + 2 (\(\frac { -1 }{ 2 } \))
a = 3
a = 4 + 2 (\(\frac { -3 }{ 2 } \))
a = 4 – 3
a = 1
The point of p is (3,\(\frac { -1 }{ 2 } \)) (or) (1,\(\frac { -3 }{ 2 } \))

Students can download Maths Chapter 1 Numbers Ex 1.4 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 1 Numbers Ex 1.4

Question 1.
Fill in the blanks.
(i) The nearest 100 of 843 is _____
(ii) The nearest 1000 of 756 is ______
(iii) The nearest 10,000 of 85654 is ______
Solution:
(i) 800.
The digit in tens place is 4 < 5
(ii) 1000.
The digit in hundred places is 7 ≥ 5
(iii) 90,000.
The digit in a thousand places is 5 ≥ 5

Question 2.
Say True or False
(i) 8567 is rounded off as 8600 to the nearest 10.
(ii) 139 is rounded off as 100 to the nearest 100.
(iii) 1,70,51,972 is rounded off as 1,70,00,000 to the nearest lakh.
Solution:
(i) False
Hint: In one’s place the digit is 7 ≥ 5. So 8580
(ii) True
Hint: In tens place, we have 3 < 5. So 100
(iii) False
Hint: In ten thousand places the digit is 5 ≥ 5. So 1,71,000,000

Question 3.
Round off the following to the given nearest place.
(i) 4,065; hundred
(ii) 44,555; thousand
(iii) 86,943; ten thousand
(iv) 50,81,739; lakh
(v) 33,75,98,482; ten crore
Solution:
(i) 4100
(ii) 45,000
(iii) 90,000
(iv) 51,00,000
(v) 30,00,00,000

Question 4.
Estimate the sum of 157826 and 32469 rounded off to the nearest ten thousand.
Solution:
= 157826 + 32469
= 190295
When rounded off to nearest ten thousand = 1,90,000

Question 5.
Estimate by rounding off each number to the nearest hundred.
(i) 8074 + 4178
(ii) 1768977 + 130589
Solution:
(i) 8074 + 4178 = 12,252
When rounded off to nearest hundred 12,300

(ii) 1768977 + 130589 = 18,99,566
When rounded off to nearest hundred = 18,99,600

Question 6.
The population of a city was 43,43,645 in the year 2001 and 46,81,087 in the year 2011. Estimate the increase in population by rounding off to the nearest thousands.
Solution:
Population in 2001 = 43,43,645
Population in 2011 = 46,81,087
Increase in population = 46,81,087 – 43,43,645 = 3,37,442
When rounded off to the nearest thousand = 3,37,000

Objective Type Questions

Question 7.
The number which on rounding off to the nearest thousand gives 11000 is
(a) 10345
(b) 10855
(c) 11799
(d) 10056
Solution:
(b) 10855

Question 8.
The estimation to the nearest hundredth of 76812 is
(a) 77000
(b) 76000
(c) 76800
(d) 76900
Solution:
(c) 76800
In tens place the digit is 1 < 5, So 76800

Question 9.
The number 9785764 is rounded off to the nearest lakh as
(a) 9800000
(b) 9786000
(c) 9795600
(d) 9795000
Solution:
(a) 9800000

Question 10.
The estimated difference of 167826 and 2765 rounded off to the nearest thousand is
(a) 180000
(b) 165000
(c) 140000
(d) 155000
Solution:
(b) 165000

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 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 1 Chapter 1 Numbers Ex 1.3

Question 1.
Fill in the blanks.

1. If Arulmozhi saves ₹ 12 per day then she saves ₹ ____ in 30 days.
2. If a person ‘A’ earns ₹ 1800 in 12 days, then he earns ₹ ____ in a day.
3. 45 ÷ (7 + 8) – 2 = _____

Solution:

1. 12 × 30 = ₹ 360
2. \(\frac{1800}{12}=150\)
3. 45 ÷ 15 – 2 = 3 – 2 = 1

Question 2.
Say True or False

1. 3 + 9 × 8 = 96
2. 7 × 20 – 4 = 136
3. 40 + (56 – 6) ÷ 2 = 45

Solution:

1. False
   Hint: 3 + 9 × 8 = 3 + 72 = 75
2. True
   Hint: 7 × 20 – 4 = 140 – 4 = 136
3. False
   Hint: 40 + 50 ÷ 2 = 40 + 25 = 65

Question 3.
The number of people who visited the Public Library for the past 5 months was 1200, 2000, 2450, 3060 and 3200 respectively. How many people visited the library in the last 5 months.
Solution:
People visited the library for past 5 months = 1200 + 2000 + 2450 + 3060 + 3200 = 11910
Total people visited = 11910

Question 4.
Cheran had a bank savings of ₹ 7,50,250. He withdrew ₹ 5,34,500 for educational purposes. Find the balance amount in his account.
Solution:
Bank Savings of Cheran = ₹ 7,50,250
Withdrew Amount = ₹ 5,34,500
Balance Amount = ₹ 2,15,750

Question 5.
In a cycle factory, 1560 bicycles were manufactured every day. Find the number of bicycles manufactured in 25 days.
Solution:
Bicycles manufactured in one day = 1560
Bicycles manufactured in 25 days = 1560 × 25
= 39,000 bicycles

Question 6.
₹ 62,500 was equally distributed as a New Year bonus for 25 employees of a company. How much did each receive?
Solution:
Total amount = Rs 62500
Total number of employees = 25
Bonus amount received by each employee = Rs 62500 ÷ 25
= RS \(\frac{62500}{25}\)
= Rs 2500

Question 7.
Simplify the following numerical expressions:
(i) (10 + 17) ÷ 3
(ii) 12 – [3 – {6 – (5 – 1)}]
(iii) 100 + 8 ÷ 2 + {(3 × 2) -6 ÷ 2}
Solution:
(i) (10 + 17) ÷ 3
= 27 ÷ 3
= 9

(ii) 12 – [3 – {6 – (5 – 1)}]
= 12 – [3 – (6 – 4)]
= 12 – [3 – 2]
= 12 – 1
= 11

(iii) 100 + 8 ÷ 2 + {(3 × 2) – 6 ÷ 2}
= 100 + 8 ÷ 2 + {6 – 3}
= 100 + 8 ÷ 2 + 3
= 100 + 4 + 3
= 107

Objective Type Questions

Question 8.
The value of 3 + 5 – 7 × 1 is _____
(a) 5
(b) 7
(c) 8
(d) 1
Answer:
(d) 1
3 + 5 – 7 × 1 = 3 + 5 – 7 = 8 – 7 = 1

Question 9.
The value of 24 ÷ {8 – (3 × 2)} is ______
(a) 0
(b) 12
(c) 3
(d) 4
Answer:
(b) 12
24 ÷ {8 – 3 × 2} = 24 ÷ {8 – 6} = 24 ÷ 2 = 12

Question 10.
Use BIDMAS and put the correct operator in the box.
2 ….. 6 – 12 ÷ (4 + 2) = 10
(a) +
(b) –
(c) ×
(d) ÷
Answer:
(c) ×
2 ….. 6 – 12 ÷ 6 = 10
⇒ 2 ….. 6 – 2 = 10
⇒ 2 × 6 – 2 = 10

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 1 Numbers Ex 1.2 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 1 Numbers Ex 1.2

Question 1.
Fill in the blanks with > or < or =.

1. 48792 ………. 48972
2. 1248654 ……… 1246854
3. 658794 ……… 658794

Solution:

1. <
2. >
3. =

Question 2.
Say True or False.

1. The difference between the smallest number of seven digits and the largest number of six digits is 10.
2. The largest 4 digit number formed by the digits 8, 6, 0, 9 using each digit only once is 9086
3. The total number of 4 digit number is 9000

Solution:

1. False
   Hint: 1000000 – 999999 = 1
2. False
   Hint: 9999 – 999 = 9000
3. True

Question 3.
Of the numbers 1386787215, 137698890, 86720560, which one is the largest? Which one is the smallest?
Solution:
We know that the number with more digits is greater.
The greatest number is 1386787215
The smallest number is 86720560

Question 4.
Arrange the following numbers in the descending order:
128435, 10835, 21354, 6348, 25840
Solution:
128435 > 25840 > 21354 > 10835 > 6348

Question 5.
Write any eight-digit number with 6 in ten lakhs place and 9 in ten-thousandth place.
Solution:
76594231
86493725

Question 6.
Rajan writes a 3-digit number, using the digits 4, 7, and 9. What are the possible numbers he can write?
Solution:
974, 947, 479, 497, 749, 794

Question 7.
The password to access my ATM card includes the digits 9, 4, 6, and 8. It is the smallest 4 digit even number. Find the password of my ATM card.
Solution:
Given digits are 9, 4, 6, and 8.
The smallest number with these digits is 4689 Given that it is an even number.
It may be 4698.
So password of the ATM card is 4698.

Question 8.
Postal Index Number consists of six digits. The first three digits are 6, 3, and 1. Make the largest and the smallest Postal Index Number by using the digits 0, 3, and 6 each only once.
Solution:
Largest Postal Index Number = 631603
Smallest Postal Index Number = 631036

Question 9.
The heights (in meters) of the mountains in Tamil Nadu are as follows:

(i) Which is the highest mountain listed above?
(ii) Order the mountains from the highest to the lowest.
(iii) What is the difference between the heights of the mountains Anaimudi and Mahendragiri?
Solution:
(i) Anaimudi (2695 m)
(ii) Anaimudi, Doddabetta, Velliangiri, Mahendragiri
(iii) 2695 m – 1647 m = 1048 m

Objective Type Questions

Question 10.
Which list of numbers is in order from the smallest to the largest?
(a) 1468, 1486, 1484
(b) 2345, 2435, 2235
(c) 134205, 134208, 154203
(d) 383553, 383548, 383642
Solution:
(c) 134205, 134208, 154203

Question 11.
The Arabian Sea has an area of 1491000 square miles. This area lies between which two numbers?
(a) 1489000 and 1492540
(b) 1489000 and 1490540
(c) 1490000 and 1490100
(d) 1480000 and 1490000
Solution:
(a) 1489000 and 1492540

Question 12.
The chart below shows the number of newspapers sold as per the Indian Readership Survey in 2018. Which could be the missing number in the table?

(a) 8
(b) 52
(c) 77
(d) 26
Solution:
(d) 26

Students can download Maths Chapter 5 Information Processing Ex 5.1 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 5 Information Processing Ex 5.1

Question 1.
Convert the following numerical expressions into Tree diagrams
(i) 8 + (6 × 2)
(ii) 9 – (2 × 3)
(iii) (3 × 5) – (4 – 2)
(iv) [(2 × 4) + 2] × (8 – 2)]
(v) [(6 + 4) × 7] – [2 × (10 – 5)]
(vi) [(4 × 3) – 2] + [8 × (5 – 3)]
Solution:

Question 2.
Convert the following tree diagrams into numerical expressions.

Solution:
(i) The numerical Expression is 9 × 8
(ii) The numerical expression is (7 + 6) – 5
(iii) The numerical expression is (8 + 2) – (6 + 1)
(iv) The numerical expression is (5 × 6) – (10 ÷ 2)

Question 3.
Convert the following algebraic expressions into tree diagrams.
(i) 10 v
(ii) 3a – b
(iii) 5x + y
(iv) 20t × p
(v) 2(a + b)
(vi) (x × y) – (y × z)
(vii) 4x + 5y
(viii) (Im – n) ÷ (pq + r)
Solution:

Question 4.
Convert Tree diagrams into Algebraic expressions.

Solution:
(i) Algebraic Expression is p + q
(ii) Algebraic Expression is l – m
(iii) Algebraic Expression is (a × b) – c (or) (ab) – c
(iv) Algebraic Expression is (a + b) – (c + d)
(v) Algebraic Expression is (8 ÷ a) + [ (6 ÷ 4) + 3]

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) \(\overleftrightarrow {AB} \)