Category: Class 8

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.7 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.7

Question 1.
Fill in the blanks:
(i) The solution of the equation ax + b = 0 is _______ .
Answer:
\(-\frac{b}{a}\)
Hint:
ax + b = 0
ax = – b
∴ x = \(-\frac{b}{a}\)

(ii) If a and b are positive integers then the solution of the equation ax = b has to be always _______ .
Answer:
Positive
Hint:
Since a & b are positive integers, b
The solution to the equation ax = b is x = \(\frac{b}{a}\) is also positive.

(iii) One-sixth of a number when subtracted from the number itself gives 25. The number is _______ .
Answer:
30
Hint:
Let the number be x.
As per question, when one sixth of number is subtracted from itself it gives 25

(iv) If the angles of a triangle are in the ratio 2:3:4 then the difference between the greatest and the smallest angle is _______ .
Answer:
40°
Hint:
Given angles are in the ratio 2:3:4
Let the angles be 2x, 3x & 4x
Since sum of the angles of a triangle is 180°,
We get 2x + 3x + 4x = 180
∴ 9x = 180
∴ x = \(\frac{180}{9}\) = 20°
∴ The angles are 2x = 2 × 20 = 40°
3x = 3 × 20 = 60°
4x = 4 × 20 = 80°
∴ Difference between greatest & smallest angle is
80° – 40° = 40°

(v) In an equation a + b = 23. The value of a is 14 then the value of b is _______ .
Answer:
b = 9
Hint:
Given equation is a + b = 23, a = 14
14 + b = 23
∴ b = 23 – 14 = 9
b = 9

Question 2.
Say True or False
(i) “Sum of a number and two times that number is 48” can be written as y + 2y = 48
Answer:
True
Hint:
Let the number be ‘y’
∴ Sum of number & two times that number is 48
Can be written as y + 2y = 48 – True

(ii) 5(3x + 2) = 3(5x – 7) is a linear equation in one variable.
Answer:
True
Hint:
5 (3x + 2) = 3 (5x – 7) is a linear equation in one variable – ‘x’ – True

(iii) x = 25 is the solution of one third of a number is less than 10 the original number.
Answer:
False
Hint:
One third of number is 10 less than original number.
Let number be ‘x’. Therefore let us frame the equation
\(\frac{x}{3}\) = x – 10
∴ x = 3x – 30
3x – x = 30
2x = 30
x = 15 is the solution

Question 3.
One number is seven times another. If their difference is 18, find the numbers.
Answer:
Let the numbers be x & y
Given that one number is 7 times the other & that the difference is 18.
Let x = 7y
also, x – y = 18 (given)
Substituting for x in the above
We get 7y – y = 18
∴ 6y = 18
y = \(\frac{18}{6}\) = 3
∴ x = 7y = 7 × 3 = 21
The number are 3 & 21

Question 4.
The sum of three consecutive odd numbers is 75. Which is the largest among them?
Answer:
Given sum of three consecutive odd numbers is 75
Odd numbers are 1,3,5,7,9, 11, 13
∴ The difference between 2 consecutive odd numbers is always 2. or in other words, if one odd number is x, the next odd number would be x + 2 and the next number would be x + 2 + 2x + 4
i.e x + 4
Since sum of 3 consecutive odd nos is 75
∴ x + x + 2 + x + 4 = 75
∴ 3x + 6 = 75 ⇒ 3x = 75 – 6
∴ 3x = 69
x = \(\frac{69}{3}\) = 23
∴ The odd numbers are 23, 23 + 2, 23 + 4
i.e 23, 25, 27
∴ Largest number is 27.

Question 5.
The length of a rectangle is \(\frac{1}{3}\) rd of its breadth. If its perimeter is 64 m, then find the length and breadth of the rectangle.
Answer:
Let length & breadth of rectangle be ‘l’ and ‘b’ respectively
Given that length is \(\frac{1}{3}\) of breadth,
∴ l = \(\frac{1}{3}\) × b ⇒ l = \(\frac{b}{3}\) ⇒ b = 3l ……. (1)

Also given that perimeter is 64 m
Perimeter = 2 × (l + b)
2 × 1 + 2 × b = 64
Substituting for value of b from (1), we get
2l + 2(3l) = 64
∴ 2l + 6l = 64
8l = 64
∴ l = \(\frac{64}{8}\) = 8m
b = 3l = 3 × 8 = 24m
Ienglh l = 8 m in & breadth b = 24 m

Question 6.
A total of 90 currency notes, consisting only of ₹ 5 and ₹ 10 denominations, amount to ₹ 500. Find the number of notes in each denomination.
Answer:
Let the number of ₹ 5 notes be ‘x’
And number of ₹ 10 notes be ‘y’
Total numbers of notes is x + y = 90 (given)
The total value of the notes is 500 rupees.
Value of one ₹ 5 rupee note is 5
Value of x ₹ 5 rupee notes is 5 × x = 5x
∴ Value of y ₹ 10 rupee flotes is 10 × y = lOy
∴ The total value is 5x + 10y which is 500
∴ we have 2 equations:
x + y = 90 ….(1)
5x + 10y = 500 ….(2)
Multiplying both sides of(1) by 5, we get
5 × x + 5 × y = 90 × 5
5x + 5y = 450
Subtracting (3) from (2), we get

∴ y = \(\frac{50}{5}\) = 10
Substitute y = 10 in equation (1)
x + y = 90 ⇒ x + 10 = 90 ⇒ x = 90 – 10 ⇒ x = 80
There are ₹5 denominations are 80 numbers and ₹10 denominations are 10 numbers

Question 7.
At present, Thenmozhi’s age is 5 years more than that of Murali’s age. Five years ago, the ratio of Then mozhi’s age to Murali’s age was 3 : 2. Find their present ages.
Answer:
Let present ages of Thenmozhi & Murali be ‘t’ & ‘m’
Given that at present
Then mozhi’s age is 5 years more than Murali
∴ t = m + 5 …… (1)
5 years ago, Thenmozhi’s age would be t – 5
& Murali’s age would be m – 5
Ratio of their ages is given as 3 : 2

2(t – 5) = 3(m – 5)
2 × t – 2 × 5 = 3 × m – 3 × 5 ⇒ 2t – 10 = 3m – 15
Substituting for t from (1)
2(m + 5) – 10 = 3m – 15

2m = 3m – 15
3m – 2m = 15
m = 15
t = m + 5 = 15 + 5 = 20
∴ Present ages of Thenmozhi & Murali are 20 & 15

Question 8.
A number consists of two digits whose sum is 9. If 27 is subtracted from the original number, its djgit.s are interchanged. Find the original number.
Answer:
Let the units/digit of a number be ‘u’ & tens digit of the number be ‘t’
Given that sum of it’s digits is 9
∴ t + u = 9 ……. (1)
If 27 is subtracted from original number, the digits are interchanged
The number is written as 10t + u
[Understand: Suppose a 2 digit number is 21
it can be written as 2 × 10 + 1
∴ 32 = 3 × 10 + 2
45 = 4 × 10 + 5
tu = t × 10 + u = 1ot + u]
Given that when 27 is subtracted, digits interchange
10t + u – 27 = 10u + t (number with interchanged digits)
∴ By transposition & bringing like variables together
10t – t + u – 10u = 27
∴ 9t – 9u = 27
Dividing by ‘9’ throughout , we get
\(\frac{9 t}{9}-\frac{9 u}{9}=\frac{27}{9}\) ⇒ t – u = 3 ……. (2)
Solving (1) & (2):

t = 6 substitute in (1)
t + u = 9 ⇒ 6 + u = 9 ⇒ u = 9 – 6 = 3
Hence the number is 63.

Question 9.
The denominator of a fraction exceeds Its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, we get \(\frac{3}{2}\). Find the original fraction.
Answer:
Let the numerator & denominator be ‘n’ & ‘d’
Given that denominator exceeds numerator by 8
∴ d = n + 8 ……. (1)
If numerator increased by 17 & denominator decreased by 1,
it becomes (n + 17) & (d – 1), fraction is \(\frac{3}{2}\).

2(n + 17) = 3(d – 1)
2n + 2 × 17 = 3d – 3

∴ 34 + 3 = 3d – 2n
∴ 3d – 2n = 37 …….. (2)
Substituting eqn. (1) in (2), we get,
3 × (n + 8) – 2n = 37
3n + 3 × 8 – 2n = 37

∴ n = 37 – 24 = 13
d = n + 8 = 13 + 8 = 21
The fraction is \(\frac{n}{d}=\frac{13}{21}\)

Question 10.
If a train runs at 60 km/hr it reaches its destination late by 15 minutes. But, if it runs at 85 kmph it is late by only 4 minutes. Find the distance to be covered by the train.
Answer:
Let the distance to be covered by train be ‘d’

Case 1:
If speed = 60km/h
The time taken is 15 minutes more than usual (t + \(\frac{15}{60}\))
Let usual time taken be ‘t’ hrs.
Caution: Since speed is given in km/hr, we should take care to maintain all units such as time should be in hour and distance should be in kin.
Given that in case 1, it takes 15 min. more
15m = \(\frac{15}{60}\) hr = \(\frac{1}{4}\)hr.
∴ Substituting in formula,

Since usually it takes ‘t’ hr, but when running at 60 k, it kes 15 min (\(\frac{1}{4}\)hr) extra.
Multiplying by 60 on both sides
d = 60 × t + 60 × \(\frac{1}{4}\) = 6ot + 15 …… (1)

Case 2:
Speed is given as 85 km/h
Time taken is only 4 min (\(\frac{4}{60}\)hr) more than usual time
∴ time taken = (t + \(\frac{1}{15}\)) hr. \(\left(\frac{4}{60}=\frac{1}{15}\right)\)
Using the formula,

Multiplying by 85 on both sides
\(\frac{d}{85}\) × 85 = 85 × t + 85 × \(\frac{1}{15}\)
∴ d = 85t + \(\frac{17}{3}\) ….. (2)
From (1) & (2), we will solve for ‘r’
Equating & eliminating ‘d’ we get

∴ By transposing, we get

Substituting this value of ‘t’ in eqn. (1), we get
d = 60t + 15
= 60 × \(\frac{28}{75}\) + 15 = \(\frac{1680}{75}\) + 15 = 22.4 + 15
= 37.4 km

Question 11.
Sum of a number and its half is 30 then the number is ______
(a) 15
(b) 20
(c) 25
(d) 40
Answer:
(b) 20
Hint:
Let number be ‘x’
∴ half of number is \(\frac{x}{2}\)
Sum of number and it’s half is given by
x + \(\frac{x}{2}\) = 30 [Multiplying by 2 on both sides]
2x + x = 30 × 2
3x = 60
x = \(\frac{60}{3}\) = 20

Question 12.
The exterior angle of a triangle is 1200 and one of its interior opposite angle 58°, then the other opposite interior angle is _________
(a) 62°
(b) 72°
(c) 78°
(d) 68°
Answer:
(a) 62°
As per property of A. exterior angle is equals to sum of interior opposite angles
Let the other interior angle to be found be ‘x’
∴ x + 58 = 120°
∴ x = 120 – 58 = 62°

Question 13.
What sum of money will earn 500 as simple interest in 1 year at 5% per annum?
(a) 50000
(b) 30000
(c) 10000
(d) 5000
Answer:
(c) 10000
Hint:
Let sum of money be P’
Time period (n) is given as 1 yr.
Rate of simple interest (r) is given as 5% p.a
∴ As per formula for simple interest.
S.I = \(\frac{\mathrm{P} \times r \times n}{100}=\frac{\mathrm{P} \times 5 \times 1}{100}\) = 500
∴ P × 5 × n = 500 × 100
∴ p = \(\frac{500 \times 100}{5}\) = 100 × 100 = 10,000

Question 14.
The product of LCM and HCF of two numbers is 24. If one of the number is 6, then the other number is ________
(a) 6
(b) 2
(c) 4
(d) 8
Answer:
(C) 4
Hint:
Product of LCM & HCF of 2 numbers is always product of the numbers. [this is property]
Product of LCM & HCF is given as 24
∴ Product of the 2 nos. is 24
Given one number is 6. Let other number be ‘x’
∴ 6 × x = 24
x = \(\frac{24}{6}\) = 4

Question 15.
The largest number of the three consecutie numbers is x+ 1, then the smallest number is ________ .
(A) x
(B) x + 1
(C) x + 2
(D) x – 1
Answer:
(D) x – 1
Hint:
The 3 consecutive numbers are: x – 1, x, x + 1

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.10 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.10

Question 1.
The sum of three numbers is 58. The second number is three times of two-fifth of the first number and the third number is 6 less than the first number. Find the three numbers.
Answer:
Here what we know
a + b + c = 58 (sum of three numbers is 58)
Let the first number be b ‘x’
b = a + 3 (the second number is three times of of the first \(\frac{2}{5}\) number)
b = 3 × \(\frac{2}{5}\)x \(\frac{6}{5}\)x
Third number = x – 6
Sum of the numbers is given as 58.
∴ x + \(\frac{6}{5}\)x + (x – 6) = 58
Multiplying by 5 throughout, we get
5 × x + 6x + 5 × (x – 6) = 58 × 5
5x + 6x + 5x – 30 = 290
∴ 16x = 290 + 30
∴ 16x = 320
∴ x = \(\frac{320}{16}\)
x = 20
1^st number = 20

3^rd number = 24 – 16 = 14

Question 2.
In triangle ABC, the measure of ∠B is two-third of the measure of ∠A. The measure of ∠C is 200 more than the measure of ∠A. Find the measures of the three angles.
Answer:
Let angle ∠A be a°
Given that ∠B = \(\frac{2}{3}\) × ∠A = \(\frac{2}{3}\)a
& given ∠C = ∠A + 20 = a + 20
Since A, B & C are angles of a triangle, they add up to 180° (∆ property)
∴∠A + ∠B + ∠C = 180°
⇒a + \(\frac{2}{3}\)a + a + 20 = 180°
\(\frac{3 a+2 a+3 a}{3}\) + 20 = 180°
\(\frac{8 a}{3}\) = 180 – 20 = 160
∴ a = \(\frac{160 \times 3}{8}\) = 60°

∠C = 80°

Question 3.
Two equal sides of an isosceles triangle are 5y – 2 and 4y + 9 units. The third side is 2y + 5 units. Find ‘y’ and the perimeter of the triangle.
Answer:
Given that 5y – 2 & 4y + 9 are the equal sides of an isosceles triangle.
∴ The 2 sides are equal

∴5y – 4y = 9 + 2 (by transposing)
∴ y = 11
∴ 1^st side = 5y – 2 = 5 × 11 – 2 = 55 – 2 = 53
2^ndside = 53 .
3^rdside = 2y + 5 = 2 × 11 + 5 = 22 + 5 = 27
Perimeter is the sum of all 3 sides
∴ P = 53 + 53 + 27 = 133 units

Question 4.
In the given figure, angle XOZ and angle ZOY form a linear pair. Find the value of x.

Answer:
Since ∠XOZ & ∠ZOY form a linear pair, by property, we have their sum to be 180°
∴ ∠XOZ + ∠ZOY 180°
∴ 3x – 2 + 5x + 6 = 180°
8x + 4 = 180 = 8x = 180 – 4
∴ 8x = 76 ⇒ x = \(\frac{176}{8}\) ⇒ x = 22°
XOZ = 3x – 2 = 3 × 22 – 2 = 66 – 2 = 64°
YOZ = 5x + 6 = 5 × 22 + 6
= 110 + 6 = 116

Question 5.
Draw a graph for the following data:

Answer:
Graph between side of square & area

When we plot the graph, we observe that it is not a linear relation.

Challenging problems

Question 6.
Three consecutive integers, when taken in increasing order and multiplied by 2, 3 and 4 respectIvely, total up to 74. Find the three numbers.
Answer:
Let the 3 consecutive integers be ‘x’, ‘x + 1’ & ‘x + 2’
Given that when multiplied by 2, 3 & 4 respectively & added up, we get 74

Simplifying the equation, we get
2x + 3x + 3 + 4x + 8 = 74
9x + 11 = 63
9x = 63 ⇒ x = \(\frac{63}{9}\) = 7
First number = 7
Second numbers = x + 1 ⇒ 7 + 1 = 8
Third numbers = x + 2 ⇒ 7 + 2 = 9
∴ The numbers are 7, 8 & 9

Question 7.
331 students went on a field trip. Six buses were filled to capacity and 7 students had to travel in a van. How many students were there in each bus?
Answer:
Let the number of students in each bus be ‘x’
∴ number of students in 6 buses = 6 × x = 6x
A part from 6 buses, 7 students went in van
A total number of students is 331
∴ 6x + 7 = 331
∴ 6x = 331 – 7 = 324
∴ x = \(\frac{324}{6}\) = 54
∴ There are 54 students in each bus.

Question 8.
A mobile vendor has 22 items, some which are pencils and others are ball pens. On a particular day, he is able to sell the pencils and ball pens. Pencils are sold for ₹ 15 each and ball pens are sold at ₹ 20 each. If the total sale amount with the vendor is ₹ 380,
how many pencils did he sell?
Answer:
Let vendor have ‘p’ number of pencils & ‘b’ number of ball pens
Given that total number of items is 22
∴ p + b = 22
Pencils are sold for ₹ 15 each & ball pens for ₹ 20 each
total sale amount = 15 × p + 20 × b
= 15p + 20b which is given to be 380.
∴ 15p + 20b = 380
Dividing by 5 throughout,
\(\frac{15 p}{5}+\frac{20 b}{5}\) = \(\frac{380}{5}\) ⇒ – 3p + 4b = 76
Multiplying equation (1) by 3 we get
3 × p + 3 × b = 22 × 3
⇒ 3p + 3b = 66
Equation (2) – (3) gives

∴ b = 10
∴ p = 12
He sold 12 pencils

Question 9.
Draw the graph of the lines y = x, y = 2x, y = 3x and y = 5x on the same graph sheet. Is there anything special that you find in these graphs?
Answer:
(i) y = x
(ii) y = 2x,
(iii) y = 3x
(iv) y = 5x

(i) y = x
When x = 1, y = 1
x = 2, y = 2
x = 3, y = 2

(ii) y = 2x
When x = 1, y = 2
x = 2, y = 4
x = 3, y = 6

(iii) y = 3x
When x = 1, y = 3
x = 2, y = 6
x = 3, y = 9

(i) y = 5x
When x = 1, y = 5
x = 2, y = 10
x = 3, y = 15
When we plot the above points & join the points to form line, we notice that the lines become progressively steeper. In other words, the slope keeps increasing.

Question 10.
Consider the number of angles of a convex polygon and the number of sides of that polygon. Tabulate as follows:

Use this to draw a graph illustrating the relationship between the number of angles and the number of sides of a polygon.
Answer:
Angles

Name of Polygon	No of angles	No. of Sides
Triangle	3	3
Rectangle	4	4
Pentagon	5	5
Hexagon	6	6
Deptagon	7	7
Octagon	8	8

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.5

Question 1.
Subtract: -2(xy)² (y³ + 7x²y + 5) from 5y² (x²y³ – 2x⁴y + 10x²)
Answer:

Question 2.
Multiply (4x² + 9) and (3x – 2).
Answer:
(4x² + 9)(3x – 2) = 4x²(3x – 2) + 9(3x – 2)
= (4x²)(3x) – (4x²) (2) + 9(3x) – 9(2)
= (4 × 3 × x × x²) – (4 × 2 × x²) + (9 × 3 × x) – 18
= 12x³ – 8x² + 27x – 18(4x³ + 9)(3x – 2)
= 12x³ – 8x² + 27x – 18

Question 3.
Find the simple interest on Rs. 5a²b² for 4ab years at 7b% per annum.
Answer:

Question 4.
The cost of a note book is Rs. 10ab. If Babu has Rs. (5a²b + 20ab² + 40ab). Then how many note books can he buy?
Answer:
For ₹ 10 ab the number of note books can buy = 1.

Number of note book he can buy = \(\frac { 1 }{ 2 }\)a + 2b + 4

Question 5.
Factorise: (7y² – 19y – 6)
Answer:
7y² – 19y – 6 is of the form ax² + bx + c where a = 7; b = – 19; c = – 6

Product = – 42	Sum = -19
1 × – 42 = -42	1 + (-42) = – 41
2 × – 21 = – 42	2 + (-21) = – 19

The product a × c = 7 × – 6 = – 42
sum b = – 19

The middle term – 19 y can be written as – 21y + 2y
7y² – 19y – 6 = 7y² – 21y + 2y – 6
= 7y(y – 3) + 2(y – 3)
= (y – 3)(7y + 2)
7y² – 19y – 6 = (y – 3)(7y + 2)

Question 6.
A contractor uses the expression 4x² + 11x + 6 to determine the amount of wire to order when wiring a house. If the expression comes from multiplying the number of rooms times the number of outlets and he knows the number of rooms to be (x + 2), find the number of outlets in terms of ’x’. [Hint : factorise 4x² + 11x + 6]
Answer:
Given Number of rooms = x + 2
Number of rooms × Number of outlets = amount of wire.
(x + 2) × Number of outlets = 4x² + 11x + 6
Number of outlets = \(\frac{4 x^{2}+11 x+6}{x+2}\) … (1)
Now factorising 4x² + 11x + 6 which is of the form ax² + bx + c with a = 4 b = 11 c = 6.
The product a × c = 4 × 6 = 24
sum b = 11

Product = 24	Sum = 11
1 × 24 = 24	1 + 24 = 25
2 × 12 = 24	2 + 12 = 14
3 × 8 = 24	3 + 18 = 11

The middle term 11x can be written as 8x + 3x
∴ 4x² + 11 x + 6 = 4x² + 8x + 3x + 6
= 4x(x + 2) + 3 (x + 2)
4x² + 11x + 6 = (x + 2)(4x + 3)
Now from (1) the number of outlets

∴ Number of outlets = 4x + 3

Question 7.
A mason uses the expression x² + 6x + 8 to represent the area of the floor of a room. If the decides that the length of the room will be represented by (x + 4), what will the width of the room be in terms of x?
Answer:
Given length of the room = x + 4 .
Area of the room = x² + 6x + 8
Length × breadth = x² + 6x + 8
breadth = \(\frac{x^{2}+6 x+8}{x+4}\) ….. (1)
Factorizing x² + 6x + 8, it is in the form of ax² + bx + c
Where a =1 b = 6 c = 8.
The product a × c = 1 × 8 = 8
sum = b = 6

Product = 8	Sum = 6
1 × 8 = 8	1 + 8 = 9
2 × 4 = 8	2 + 4 = 6

The middle term 6x can be written as 2x + 4x
∴ x² + 6x + 8 = x² + 2x + 4x + 8
= x(x + 2) + 4(x + 2)
x² + 6x + 8 = (x + 2)(x + 4)
Now from (1)

∴ Width of the room = x + 2

Question 8.
Find the missing term: y² + (-)x + 56 = (y + 7)(y + -)
Answer:
We have (x + a)(x + b) = x² + (a + b)x + ab
56 = 7 × 8
∴ y² + (7 + 8)x + 56 = (y + 7) (y + 8)

Question 9.
Factorise : 16p⁴ – 1
Answer:
16p⁴ – 1 = 2⁴p⁴ – 1 =(2²)² (p²)² – 1²
= (2²p²)² – 1²
Comparing with a² – b² (a + b)(a – b) where a = 2²p² and b= 1
∴ (2²p²)² – 1² = (2²p² + 1)(2²p² – 1)
= (4p² + 1)(4p² – 1)
∴ 16p⁴ – 1 = (4p² + 1)(4p² – 1) = (4p² + 1)(2²p² – 1²)
= (4p² + 1) [(2p)² – 1²] = (4p² + 1) (2p + 1)(2p – 1) [∵ using a² – b² = (a + b)(a – b)]
∴ 16p⁴ – 1 = (4p² + 1)(2p + 1)(2p – 1)

Question 10.
Factorise : 3x³ – 45x²y + 225xy² – 375y³
Answer:
= 3x³ – 45x²y + 225xy² – 375y³
= 3(x³ – 15x²y + 75xy² – 125y³)
= 3(x³ – 3x²(5y) + 3x(5y)² – (5y)³)
= 3(x – 5y)³

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.9 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.9

Question 1.
Fill in the blanks:
(i) y = p x where p ∈ Z always passes through the _________ .
Answer:
Origin (0,0)
Hint:
[When we substitute x = 0 in equation, y also becomes zero. (0,0) is a solution]

(ii) The intersecting point of the line x = 4 and y = -4 is _________ .
Answer:
4, -4
Hint:
x = 4 is a line parallel to the y – axis and
y = -4 is a line parallel to the x – axis. The point of intersection is a point that lies on both lines & which should satisfy both the equations. Therefore, that point is (4, -4)

(iii) Scale for the given graph,
On the x-axis 1 cm = _________ units
y-axis 1 cm = _________ units

Answer:
3 units, 25 units
Hint:
With reference to given graph,
On the x – axis. 1 cm = 3 units
y axis, 1 cm = 25 units

Question 2.
Say True or False.
(i) The points (1,1) (2,2) (3,3) lie on a same straight line.
Answer:
True
Hint:
The points (1, 1), (2, 2), (3, 3) all satisfy the equation y = x which is straight line. Hence, it is true

(ii) y = -9x not passes through the origin.
Answer:
False
Hint:
y = -9x substituting for x as zero, we get y = -9 × 0 = 0
∴ for x = 0, y = 0. Which means line passes through (0, 0), hence statement is false.

Question 3.
Will a line pass through (2, 2) if it intersects the axes at (2, 0) and (0, 2).
Answer:
Given a line intersects the axis at (2, 0) & (0, 2)
Let line intercept form be expressed as
ax + by = 1 Where a & b are the x & y intercept respectively.
Since the intercept points are (2. 0) & (0, 2)
a = 2, b = 2
∴ 2x + 2y = 1
When the point (2. 2) is considered & substituted in the equation
2x + 2y = 1, we get
2 × 2 + 2 × 2 = 4 ≠ 1
∴ the point (2. 2) does not satisfy the equation. Therefore the line does not pass through (2, 2)
Alternatively graphical method

as we can see the line doesn’t pass through (2, 2)

Question 4.
A line passing through (4, – 2) and intersects the Y-axis at (0, 2). Find a point on the line in the second quadrant.
Answer:
Line passes through (4, – 2)
y – axis intercept point – (0, 2) using 2 point formula.



Any point in II quadrant will have x as negative & y as positive.
So let us take x value as – 2
∴ -2 + y = 2
∴ y = 2 + 2 = 4
∴ Point in II Quadrant is (-2, 4)

Question 5.
If the points P(5, 3) Q(-3, 3) R (-3, -4) and S form a rectangle then find the coordinate of S.
Answer:
Plotting the points on a graph (approximately)
Steps:

• Plot P, Q, R approximately on a graph.
• As it is a rectangle, RS should be parallel to PQ & QR should be paraHel to PS
• S should lie on the straight line from R parallel to x-axis & straight line from P parallel to y-axis
• Therefore, we get S to be (5, -4)

[Note: We don’t need graph sheet for approximate plotting. This is just for graphical understanding]

Question 6.
A line passes through (6, 0) and (0, 6) and an another line passes through (-3, 0) and (0, -3). What are the points to be joined to get a trapezium?
Answer:
In a trapezium. there are 2 opposite sides that are parallel. The other opposite sides are non-parallel.
Now, let us approximately plot the points for our understanding
[no need of graph sheet]

• Plot the points (0, 6), (6, 0), (-3, 0) & (0, -3)
• Join (0, 6) & (6, 0)
• Join (-3,0) & (0, – 3)
• We find that the lines formed by joining the points are parallel lines.
• So, for forming a trapezium, we should join (0, 6), (-3, 0) & (0, -3), (6, 0)

Question 7.
Find the point of intersection of the line joining points (- 3, 7) (2, – 4) and (4, 6) (- 5, – 7).Also find the point of intersection of these lines and also their intersection with the axis.
Answer:

Equation of line joining 2 points by 2 point formula is given by

Cross multiplying, we get

Transposing the variables, we get
11 x + 5 y = 35 – 33 = 2
11 x + 5y = 2 – Line 1
Similarly, we should find out equation of second line


∴ 9y – 54 = 13x – 52
∴ 9y – 13x = 2 – Line 2
For finding point of intersection, we need to solve the 2 line equation to find a point that will satisfy both the line equations.
∴ Solving for x & y from line 1 & line 2 as below
11x + 5y = 2 ⇒ multiply both sides by 13,
11 × 13x + 5 × 13y = 26 …….. (3)
Line 2: 9y – 13x = 2 ⇒ multiply both sides by 11
9 × 11y – 13 × 11x = 22 ……… (4)

∴ 164 y = 48
∴ y = \(\frac{48}{164}=\frac{12}{41}\)
Substituting this value ofy in line I we get
11 x + 5 y = 2
11 x + 5 × \(\frac{12}{41}\) = 2
11 x = 2 – \(\frac{60}{41}=\frac{82-60}{41}=\frac{22}{41}\)
∴ x = \(\frac{2}{41}\)
[∴ Point of intersection is \(\left(\frac{2}{41}, \frac{12}{41}\right)\)]
To find point of intersection of the lines with the axis, we should substitute values & check
Line 1: 11 x + 5 y = 2
Point of intersection of line with x – axis, i.e y coordinate is ‘0’
∴ put y = 0 in above equation
∴ 11 x – 5 × 0 = 2
∴ 11x + 0 = 2
∴ x = \(\frac{2}{11}\)
∴ [Point is \(\left(\frac{2}{11}, 0\right)\)]
Similarly, Point of intersection of line with y – axis is when x-coordinate becomes ‘0’
∴ put x = 0 in above equation
∴ 11 × 0 + 5y = 2
∴ 0 + 5y = 2
y = \(\frac{2}{5}\)
∴ [Point is \(\left(0, \frac{2}{5}\right)\)]
Similarly for line 2,
9y – 13x = 2
For finding x intercept, i.e point where line meets x axis, we know that y coordinate becomes ‘0’
∴ Substituting y = 0 in above eqn. we get
9 × 0 – 13x = 2
∴ 0 – 13x = 2
∴ x = \(\frac{-2}{13}\)
∴ [Point: \(\left(\frac{-2}{13}, 0\right)\)]
Similarly for y – intercept, x – coordinate becomes ‘0’,
∴ Substituting for x = 0 in above equation, we get
9 y – 13 × 0 = 2
9y – 0 = 2
9y = 2
y = \(\frac{2}{9}\)
[Point \(\left(0, \frac{2}{9}\right)\)]

Question 8.
Draw the graph of the following equations: (i) x = – 7 (ii) y = 6
Answer:

Question 9.
Draw the graph of
(i) y = – 3x
(ii ) y = x – 4
(ii) y = 2x + 5
Answer:
To draw graph, we need to find out some points.
(i) y = – 3x
for y = -3x, let us first substituting values & check
put x = 0
y = 3 × 0 = 0
∴ (0,0) is a point
put x = 1
y = -3 × 1 = – 3
∴ (1, – 3) is a point
If join these 2 points, we will get the line

(ii) y = x – 4
for y = x – 4
put x = 0
y = 0 – 4 = – 4
∴ (0, – 4) is a point
x = 4
y = 4 – 4 = 0
∴ (4, 0) is a point

(iii) y = 2x + 5
for y = 2x + 5
put x = – 1
y = 2(-1) + 5 = – 2 + 5 = 3
∴ (-1, 3) is a point
put x = – 2
y = 2(-2) + 5 = – 4 + 5 = 1
∴ (-2, 1) is a point
Now let us plot the points & join them on graph

Question 10.
Find the values
(a) y = x + 3

Answer:
Let y = x+3
(i) if x = 0, y = 0 + 3 = 3,
∴ y = 3
(ii) y = 0, 0 = x + 3,
∴ x = – 3
(iii) x = -2, y = -2 + 3,
∴ y = 1
(iv) y = – 3, – 3 = x + 3,
∴ x = – 6

y = x + 3

(b) 2x + y – 6 = 0

Answer:
Let 2x + y – 6 = 0
(i) x = 0 2 × 0 + y – 6 = 0 ∴ y = 6
(ii) y = 0, 2x + 0 – 6 = 0, ∴ 2x = 6
x = 3 ,
(iii) x = -1, 2 × (-1) + y – 6 = 0, 8 + y = 0
y = 8
(iv) y = -2, 2x – 2 – 6 = 0, 2x = 8
x = 4

2x + y – 6 = 0

(c) y = 3x + 1

Answer:
(i) x = -1, y = 3(-1) + 1 = 0
∴ y = -2
(ii) x = 0, y = 3(0) + 1 = 0
∴ y = 1
(iii) x = 1, y = 3(1) + 1 = 0
∴ y = 4
(iv) x = 2, y = 3(2) + 1 = 0
∴ y = 7

y = 3x + 1

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.8 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.8

Question 1.
Fill in the blanks:
(i) X- axis and Y-axis intersect at _________ .
Answer:
Origin (0,0)

(ii) The coordinates of the point in third quadrant are always _________ .
Answer:
negative

(iii) (0, -5) point lies on _________ axis.
Answer:
Y-axis

(iv) The x- coordinate is always ______ on the y-axis.
Answer:
Zero

(v) ___________ coordinates are the same for a line parallel to Y-axis.
Answer:
X

Question 2.
Say True or False:
(i) (-10,20) lies in the second quadrant.
Answer:
True
Hint:
(-10, 20)
x = -10, y = 20
∴ (-10, 20) lies in second quadrant – True

(ii) (-9, 0) lies on the x-axis.
Answer:
True
Hint:
(-9, 0) on x – axis. Y- coordinate is always zero.
∴ (-9, 0) lies on x axis – True

(iii) The coordinates of the origin are (1,1).
Answer:
False
Hint:
Coordinate of origin is (0, 0), not (1, 1). Hence – False

Question 3.
Find the quadrants without plotting the points on a graph sheet.
(3, -4), (5, 7), (2, 0), (-3, -5), (4, -3), (-7, 2), (-8, 0), (0, 10), (-9, 50).
Answer:

If X & y coordinate are positive – I quad
If x is positive,y is negative – IV quad
If x is negative, y is positive – II quad
If both are negative, then – III quad

Question 4.
Plot the following points in a graph sheet.
A(5, 2), B(-7, -3), C(-2, 4), D(-1, -1), E(0, -5), F(2, 0), G(7, -4), H(-4, 0), I(2, 3), J(8, -4), K(0, 7).
Answer:

Question 5.
Use the graph to determine the coordinates where each figure is located.

a) Star	____________
b) Bird	____________
c) Red circle	____________
d) Diamond	____________
e) Triangle	____________
f) Ant	____________
g) Mango	____________
h) Housefly	____________
i) Medal	____________
j) Spider	____________

Answer:

a) Star	(3, 2)
b) Bird	(-2, 0)
c) Red circle	(-2, -2)
d) Diamond	(-2, 2)
e) Triangle	(-1, 1)
f) Ant	(3, -1)
g) Mango	(0, 2)
h) Housefly	(2, 0)
i) Medal	(-3, 3)
j) Spider	(0, -2)

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.2

Question 1.
Fill in the blanks
(i)

Answer:
\(\frac{18 m^{4}\left(n^{8}\right)}{2 m^{(3)} n^{3}}\) = 9 mn⁵

(ii)

Answer:
\(\frac{l^{4} m^{5} n^{(7)}}{2 l m^{(3)} n^{6}}=\frac{l^{3} m^{2} n}{2}\)

(iii)

Answer:
\(\frac{42 a^{4} b^{5}\left(c^{2}\right)}{6(a)^{4}(b)^{2}}\) = (7)b³c²

Question 2.
Say True or False
(i) 8x³y ÷ 4x² = 2xy
Answer:
True

(ii) 7ab³ ÷ 14 ab = 2b²
Answer:
False

Question 3.
Divide
(i) 27 y³ by 3y
(ii) x³ y² by x²y
(iii) 45x³ y² z⁴ by (-15 xyz)
(iv) (3xy)² by 9xy
Answer:
(i) 27 y³ by 3y
\(\frac{27 y^{3}}{3 y}\) = \(\frac{27}{3} y^{3-1}\) = 9y²

(ii) x³ y² by x²y
\(\frac{x^{3} y^{2}}{x^{2} y}\) = x^3-2 y^2-1 = x¹ y¹ = xy

(iii) 45x³ y² z⁴ by (-15 xyz)
\(\frac{45 x^{3} y^{2} z^{4}}{-15 x y z}\) = \(\frac{45}{-15}\) x^3-2 y^2-1 y^4-1 z^4-1 = -3x² yz³

(iv) (3xy)² by 9xy
\(\frac{(3 x y)^{2}}{3 \times(3 x y)}\) = \(\frac{(3 x y)^{2}}{3 \times(3 x y)}\) = \(\frac{1}{3}\) (3xy)^2-1 = \(\frac{1}{3}\) 3xy = xy

Question 4.
Simplify
(i) \(\frac{3 m^{2}}{m}+\frac{2 m^{4}}{m^{3}}\)
(ii) \(\frac{14 p^{5} q^{3}}{2 p^{2} q} \frac{12 p^{3} q^{4}}{3 q^{2}}\)
Answer:
(i) \(\frac{3 m^{2}}{m}+\frac{2 m^{4}}{m^{3}}\)
\(\frac{3 m^{2}}{m}+\frac{2 m^{4}}{m^{3}}\) = 3m^2-1 + 2m^4-3
= 3m + 2m
= (3 + 2) m
= 5m

(ii) \(\frac{14 p^{5} q^{3}}{2 p^{2} q} \frac{12 p^{3} q^{4}}{3 q^{2}}\)
\(\frac{14 p^{5} q^{3}}{2 p^{2} q} \frac{12 p^{3} q^{4}}{3 q^{2}}\) = \(\frac{14}{2}\)p^5-2q^3-1 – \(\frac{12}{3}\) p³q^4-3
= 7p³q² – 4p³q

Question 5.
Divide:
(i) 32y² – 8yz by 2y
(ii) (4m²n³ + 16m⁴ n² – mn) by 2mn
(iii) 5xy² – 18x²y³ + 6xy by 6xy
(iv) 81(p⁴ q² r³ + 2p³q³ r² – 5p²q²r²) by (3pqr)²
Answer:
(i) 32y² – 8yz by 2y

(ii) (4m²n³ + 16m⁴ n² – mn) by 2mn

(iii) 5xy² – 18x²y³ + 6xy by 6xy

(iv) 81(p⁴ q² r³ + 2p³q³ r² – 5p²q²r²) by (3pqr)²

= \(\frac{81}{9}\)(p²q²r²)^1-1 (p²r + 2pq – 5)
= 9(p²r + 2pq – 5) = 9 p²r + 18pq – 45

Question 6.
Identify the errors and correct them.
(i) 7y² – y² + 3y² = 10y²
Answer:
7y² – y² + 3y² = 10y² = (7 – 1 + 3)y²
= (6 + 3)y²
= 9y²

(ii) 6xy + 3xy = 9x²y²
Answer:
6xy + 3xy = (6 + 3) xy
= 9 xy

(iii) m(4m – 3) = 4m² – 3
Answer:
m(4m – 3) = m(4m) + m(-3)
= 4m² – 3m

(iv) (4n)² – 2n + 3 = 4n² – 2n + 3
Answer:
(4n)² – 2n + 3 = 16n² – 2n + 3

(v) (x – 2)(x + 3) = x² – 6
Answer:
(x – 2)(x + 3) = x(x + 3) – 2 (x + 3)
= x(x) + (x) × 3 + (-2) (x) + (-2) (3)
= x² + 3x – 2x – 6
= x² + x – 6

(vi) -3p² + 4p – 7 = -(3p² + 4p – 7)
Answer:
-3p² + 4p – 7 = -(3p² – 4p + 7)

Question 7.
Statement A: If 24p²q is divided by 3pq, then the quotient is 8p.
Statement B: Simplification of \(\frac{(5 x+5)}{5}\) is 5x.
(i) Both A and B are true
(ii) A is true but B is false
(iii) A is false but B is true
(iv) Both A and B are false
Answer:
(ii) A is true but B is false
Hint:

Question 8.
Statement A: 4x² + 3x – 2 = 2(2x² + \(\frac{3 x}{2}\) – 1)
Statement B: (2m – 5) – (5 – 2m) = (2m – 5) + (2m – 5)
(i) Both A and B are true
(ii) A is true but B is false
(iii) A is false but B is true
(iv) Both A and B are false
Answer:
(i) Both A and B are true
Hint:
(2m – 5) – (5 – 2m) = 2m – 5 – 5 + 2m = 4m – 10
(2m – 5) + (2m – 5) = 4m – 10

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.1

Question 1.
Complete the table.

Answer:

Question 2.
Find the product of the terms.
(i) -2mn, (2m)², -3mn
(ii) 3x²y , -3xy³, x²y²
Answer:
(i) (-2mn) × (2m)² × (-3mn) = (-2mn) × 2²m² × (-3mn) = (- 2mn) × 4m² × (- 3mn)
= (-) (+)(-) (2 × 4 × 3) (m × m² × m) (n × n)
= +24 m⁴ 4n²

(ii) (3x²y) × (-3xy³) × (x²y²) = (+) × (-) × (+) × (3 × 3 × 1)(x² × x × x²) × (y × y³ × y²)
= -9x⁵ y⁶

Question 3.
If l = 4pq², b = -3p²q h = 2p³q³ then, find the value of l × b × h.
Answer:
Given l = 4pq²
b = -3p²q
h = 2p³q³
l × b × h = (4pq²) × -3p²q × 2p³q³
= (+) (-) (+) (4 × 3 × 2) (p × p² × p³) (q × q² × q³)
= – 24p⁶q⁶

Question 4.
Expand
(i) 5x(2y – 3)
(ii) -2p(5p² – 3p + 7)
(iii) 3mn(m³n³ – 5m²n + 7mn²)
(iv) x²(x + y + z)+ y²(x+ y + z) + z²(x – y – z)
Answer:
(i) 5x(2y – 3)
5x(2y -.3) = (5x)(2y) – (5x)(3)
= (5 × 2) (x × y) – (5 × 3)x
= 10xy – 15x

(ii) -2p(5p² – 3p + 7)
-2p (5p² -3p + 7) = (-2p) (5p²) + (-2p) (-3p) + (-2p) (7)
= [(-) (+) (2 × 5) (p × p²)] + [(-) (-) (2 × 3) (p × p)] + (-)(+)(2 × 7)p
= -10p³ + 6p² – 14p

(iii) 3mn(m³n³ – 5m²n + 7mn²)
3mn(m³n³ – 5m²n + 7 mn²) = (3mn) (m³n³) + (3mn) (-5m³n) + (3mn)(7mn³)
= (3)(m ×m³) (n × n³) + (+)(-)(3 × 5)(m × m²)(n × n)+ (3 × 7) (m × m)(n × n²)
= 3m⁴n⁴ – 15m³n² + 21m²n³

(iv) x²(x + y + z)+ y²(x+ y + z) + z²(x – y – z)
x²(x + y + z) + y²(x + y + z) + z²(x – y – z) = (x² × x) + (x² × y) + (x² × z) + (y² × x) + (y² × y) + (y² × z) + (z² × x) + z² (-y) + z² (-z)
= x³ + x²y + x²z + xy² + y³ + y²z + xz² – yz² – z³
= x³ + y³ – z³ + x²y + x²z + xy² + zy² + xz² – yz²

Question 5.
Find the product of
(i) (2x + 3)(2x – 4)
(ii) (y² – 4)(2y² + 3y)
(iii) (m² – n)(5m²n² – n²)
(iv) 3(x – 5) × 2(x – 1)
Answer:
(i) (2x + 3)(2x – 4)
(2x + 3)(2x – 4) = (2x)(2x – 4) + 3(2x – 4)
= (2x × 2x) – 4(2x) – 3(2x) – 3(4)
= 4x² – 8x + 6x – 12
= 4x² + (- 8 + 6)x – 12
= 4x² – 2x – 12

(ii) (y² – 4)(2y² + 3y)
(y² – 4)(2y² + 3y) = y²(2y² + 3y) – 4(2y²) – 4(3y)
= y²(2y²) + y² (3y) – 4(2y²) – 4 (3y)
= 2y⁴ + 3y³ – 8y² – 12y

(iii) (m² – n)(5m²n² – n²)
(m² – n)(5m² n² – n²) = m² (5m²n² – n²) – n (5m²n² – n²)
= m² (5m²n²) + m² (-n²) – n (5m²n² ) + (-) (-) n (n²)
= 5m⁴n² – m²n² – 5m²n³ + n³

(iv) 3(x – 5) × 2(x – 1)
3(x – 5) × 2(x – 1) = (3 × 2) (x – 5) (x – 1)
= 6 × [x (x – 1) – 5 (x – 1)]
6 [x.x – x.1 – 5x + (-1)(-)5 1]
= 6[x² – x – 5x + 5] = 6[x² + (-1 – 5)x + 5]
= 6[x² – 6x + 5] = 6x² – 36 x + 30

Question 6.
Find the missing term.
(i) 6xy × ______ = -12x³y
Answer:
6xy × (-2x²) = -12x³y

(ii) ______ × (-15m²n³p) = 45m³n³p²
Answer:
-3mp × (-15m²n³p) = 45m³n³p²

(iii) 2y(5x²y – ____ + 3____ ) = 10x²y² – 2xy + 6y³
Answer:
2y(5x²y – x + 3y²) = 10x²y² – 2xy + 6y³

Question 7.
Match the following

(A) iv, v, ii, i, iii
(B) v, iv, iii, ii, i
(C) iv, v, ii, iii, i
(D) iv, v, iii, ii, i
Answer:
(C) iv, v, ii, iii, i
a) – iv
b) – v
c) – ii
d) – iii
e) – i

Question 8.
A car moves at a uniform speed of (x + 30) km/hr. Find the distance covered by the car in (y + 2)hours. (Hint: distance = speed × time).
Answer:
Speed of the car = (x + 30) km/hr.
Time = (y + 2) hours
Distance = Speed × time
= (x + 30) (y + 2) = x(y + 2) + 30(y + 2)
= (x) (y) + (x) (2) + (30) (y) + (30) (2)
= xy + 2x + 30y + 60
Distance covered = (xy + 2x + 30y + 60) km

Objective Type Questions

Question 9.
The product of 7p³ and (2p²)² is
(A) 14p¹²
(B) 28p⁷
(C) 9p⁷
(D) 11p¹²
Answer:
(B) 28p⁷

Question 10.
The missing terms in the product -3m³n × 9(_) = _______ m⁴n³ are
(A) mn², 27
(B) m²n, 27
(C) m²n², – 27
(D) mn², – 27
Answer:
(A) mn², 27

Question 11.
If the area of a square is 36x⁴y² then, its side is _________ .
(A) 6x⁴y²
(B) 8x²y²
(C) 6x²y
(D) -6x²y
Answer:
(C) 6x²y

Question 12.
If the area of a rectangle is 48m²n³ and whose length is 8mn² then, its breadth is _____ .
(A) 6 mn
(B) 8m²n
(C) 7m²n²
(D) 6m²n²
Answer:
(A) 6 mn

Question 13.
If the area of a rectangular land is (a² – b²) sq.units whose breadth is(a – b) then, its length is_________
(A) a – b
(B) a + b
(C) a² – b
(D) (a + b)²
Answer:
(B) a + b

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 2 Measurements InText Questions Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 2 Measurements InText Questions

Think (Text Book Page No. 51)

Question 1.
\(\frac{22}{7}\) and 3.14 are rational numbers. Is ‘π’ a rational number? Why?
Answer:
\(\frac{22}{7}\) and 3.14 are rational numbers π has non-terminating and non-repoating decimal expansion. So it is not a rational number. it is an irrational number.

Question 2.
When is the’π’day celebrated? Why?
Answer:
March 14^th i.e. 3/14 every year. Approximately value of’ ‘π’ is 3.14.

Think (Text Book Page No. 53)

The given circular figure is di4dd into six equal parts. Can we call the parts as sectors? Why?

Answer:
No, the equal parts are not sectors. Because a sector is a plane surface that is enclosed between two radii and the circular arc of the circle. Here the boundaries are not radii.

Try These (Text Book Page No. 53)

Fill the central angle of the shaded sector (each circle is divided into equal sectors)

Answer:

Think (Text Book Page No. 54)

Question 1.
Instead of multiplying by \(\frac{1}{2}\), \(\frac{1}{3}\) and \(\frac{1}{4}\), we shall multiply by \(\frac{180^{\circ}}{360^{\circ}}\), \(\frac{120^{\circ}}{360^{\circ}}\) and \(\frac{90^{\circ}}{360^{\circ}}\) respectively. why?
Answer:
So, \(\frac{180^{\circ}}{360^{\circ}}\) = \(\frac{1}{2}\)
\(\frac{120^{\circ}}{360^{\circ}}\) = \(\frac{1}{3}\)
\(\frac{90^{\circ}}{360^{\circ}}\) = \(\frac{1}{4}\)

Think (Text Book Page No. 57)

If the radius of a circle is doubled, what will happen to the area of the new circle so formed?
Answer:
If r = 2r₁ ⇒ Area of the circle = πr² = π(2r₁)² = π4r₁²
Area = 4 × old area

Think (Text Book Page No. 61)

All the sides of a rhombus are equal. It is a regular polygon?
Answer:
For a regular polygon all sides and all the angles must be equal. But in a rhombus all the sides are equal. But all the angles are not equal
∴ It is not a regular polygon.

Try This (Text Book Page No. 64)

In the above example split the given mat into two trapeziums and verify your answer.
Answer:
Area of the mat = Area of I trapezium + Area of II trapezium
= [\(\frac { 1 }{ 2 }\) × h₁ × (a₁ + b₁)] + [\(\frac { 1 }{ 2 }\) × h₂ × (a₂ + b₂)] sq. units
= [\(\frac { 1 }{ 2 }\) × 2 × (7 + 5)] + \(\frac { 1 }{ 2 }\) × 2 × (9 + 7) sq.feet
= 12 + 16 = 28 sq.feet
∴ Cost per sq.feet = ₹ 20
Cost for 28sq. feet = ₹ 20 × 28 = ₹ 560
∴ Total cost for the entire mat = ₹ 560
Both the answers are the same.

Try This (Text Book Page No. 68)

Tabulate the number of faces (F), vertices (V) and edges (E) for the following polyhedrons. Also find F + V – E

What do you observe from the above table? We observe that, F + V – E = 2 in all the cases. This is true for any polyhedron and this relation F + V – E = 2 is known as Euler’s formula.
Answer:

From the table F + V – E = 2 for all the solid shapes.

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 2 Measurements Ex 2.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 2 Measurements Ex 2.3

Question 1.
Fill ini the blanks:

(i) The three dimensions of a cuboid are ______ , ______ and ______ .
Answer:
length, breadth, height

(ii) The meeting point of more than two edges in a polyhedron is called as ______ .
Answer:
Vertex

(iii) A cube has _________ faces.
Answer:
six

(iv) The cross section of a solid cylinder is ______ .
Answer:
circle

(v) If a net of a 3-D shape has six plane squares, then it is called ______ .
Answer:
cube

Question 2.
Match the following

Answer:
(i) – b
(ii) – a
(iii) – d
(iv) – c

Question 3.
Which 3 – D shapes do the following nets represents? Draw them.
(i)

Answer:
The net represents cube, because it has 6 squares.

(ii)

Answer:
The net represents cuboid

(iii)

Answer:
The net represents Triangular prism

(iv)

The net represents square pyramid

(v)

Answer:
Ine net represents cylinder

Question 4.
For each solid, three views are given. Identify for each solid, the corresponding Top, Front and Side (T, F and S) views.

Answer:

Question 5.
Verify Euler’s formula for the table given below.

Answer:
Euler’s formula is given by F + V – E = 2
(i) F = 4; V = 4; E = 6
F + V – E = 4 + 4 – 6 = 8 – 6
F + V – E = 2
∴ Euler’s formula is satisfied.

(ii) F = 10; V = 6; E = 12
F + V – E = 10 + 6 – 12 = 16 – 12 = 4 ≠ 2
∴ Euler’s formula is not satisfied.

(iii) F = 12; V = 20; E = 30
F + V – E = 12 + 20 – 30 = 32 – 30 = 2
∴Euler’s formula is satisfied.

(iv) F = 20; V = 13; E = 30
F + V – E = 20 + 13 – 30 = 33 – 30 = 3 ≠ 2
∴Euler’s formula is not satisfied.

(v) F = 32; V = 60; E = 90
F + V – E = 32 + 60 – 90 = 92 – 90 = 2
∴ Euler’s formula is satisfied.

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 2 Measurements Ex 2.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 2 Measurements Ex 2.2

Question 1.
Find the perimeter and area of the figures given below. (π = \(\frac{22}{7}\))
(i)

Answer:

Length of the arc of the semicircle = \(\frac{1}{2}\) × 2πr units
= \(\frac{22}{7} \times \frac{7}{2}\) m =11 m
∴ Perimeter = Sum of all lengths of sides that form the closed boundary
P = 11 + 10 + 7 + 10 m
Perimeter = 38 m
Area = Area of the rectangle – Area of semicircle
= (l × b) – \(\frac{1}{2}\) πr² sq. units

= 50.75 m² (approx)
Area of the figure = 50.75 m² approx.

(ii)

Answer:

Perimeter = sum of outside lengths
Length of the arc of quadrant circle = \(\frac{1}{4}\) × 2πr units
= \(\frac{1}{2} \times \frac{22}{7}\) × 3.5cm
= 11 × 0.5 cm = 5.5cm
∴ Length of arc of 2 sectors 2 × 5.5 cm
= 11 cm
∴ Perimeter P = 11 + 6 + 3.5 + 6 + 35 cm
P = 30 cm
Area Area of 2 quadrant circle + Area of a rectangle.
= 2 × \(\frac{1}{4}\)πr² + lb sq. units
= (\(\frac{1}{2} \times \frac{22}{7}\) × 3.5 × 3.5) + (6 × 3.5) cm²
= (11 × 3.5 × 0.5) + 21 cm²
= (19.25 + 21) cm² = 40.25 cm²
∴ Area = 40.25 cm² approx

Question 2.
Find the area of the shaded part in the following figures. (π = 3.14)
(i)

Answer:
Area of the shaded part = Area of 4 quadrant circles of radius \(\frac { 10 }{ 2 }\) cm
= 4 × \(\frac { 1 }{ 4 }\) × πr² = 3.14 × \(\frac { 10 }{ 2 }\) × \(\frac { 10 }{ 2 }\) cm²
= \(\frac { 314 }{ 4 }\) cm² = 78.5 cm²
Area of the shaded part = 78.5 cm²
Area of the unshaded part = Area of the square – Area of shaded part
= a² – 78.5 cm² = (10 × 10) – 78.5 cm²
= 100 – 78.5cm² = 21.5cm²
Area of the unshaded part = 21.5 cm² (approximately)

(ii)

Answer:
Area of the shaded part = Area of semicircle – Area of the triangle

= \(\frac { 1 }{ 2 }\) × 3.14 × 7 × 7 – \(\frac { 1 }{ 2 }\) × 14 × 7cm²
= \(\frac{153.86}{2}\) – 49 cm² = 76.93 – 49 cm²
= 27.3 cm²
∴ Area of the shaded part = 27.93 cm² (approximately)

Question 3.
Find the area of the combined figure given which is got by joining of two parallelograms

Answer:
Area of the figure = Area of 2 parallelograms with base 8 cm and height 3 cm
= 2 × (bh) sq. units
= 2 × 8 × 3cm² = 48 cm²
∴ Area of the given figure = 48 cm²

Question 4.
Find the area of the combined figure given, formed by joining a semicircle of diameter 6 cm with a triangle of base 6 cm and height 9 cm. (π = 3.14)

Answer:
Area of the figure = Area of the semicircle of radius 3 cm + 2 (Area of triangle with b = 9 cm and h = 3 cm)
= \(\left(\frac{1}{2} \pi r^{2}\right)+\left(2 \times \frac{1}{2} b h\right)\) sq. units
= \(\frac { 1 }{ 2 }\) × 3.14 × 3 × 3 + (2 × \(\frac { 1 }{ 2 }\) × 9 × 3)cm²
= \(\frac{28.26}{2}\) + 27 cm² = 14.13 + 27 cm² = 41.13 cm²
∴ Area of the figure = 41.13 cm² (approximately)

Question 5.
The door mat which is in a hexagonal shape has the following measures as given in the figure. Find its area.

Answer:
Area of the doormat = Area of 2 trapezium
Height of the trapezium h =\(\frac { 70 }{ 2 }\) cm: a = 90 cm; b = 70 cm
∴ Area of the trapezium = \(\frac { 1 }{ 2 }\) h (a + b) sq. units
Area of the door mat = 2 × \(\frac { 1 }{ 2 }\) × 35 (90 + 70)cm²
= 35 × 160 cm² = 5600 cm²
∴ Area of the door mat = 5600 cm²

Question 6.
A rocket drawing has the measures as given in the figure. Find its area.

Answer:
Area = Area of a rectangle + Area of a triangle + Area of a trapezium
For rectangle length l = 120 – 20 – 20 cm = 80 cm
Breadth b = 30 cm
For the triangle base = 30 cm
Height = 20 cm
For the trapezium height h = 20 cm
Parallel sided a = 50 cm
b = 30cm
∴ Area of the figure (l × b) + (\(\frac { 1 }{ 2 }\) × base × height) + \(\frac { 1 }{ 2 }\) × h × (a + b)sq. units
= (80 × 30) + (\(\frac { 1 }{ 2 }\) × 30 × 20) + \(\frac { 1 }{ 2 }\) × 20 × (50 + 30) cm²
= 2400 + 300 + 800 cm² = 3500 cm²
Area of the figure = 3500 cm²

Question 7.
Find the area of the irregular polygon shaped fields given below.

Answer:
Area of the field = Area of trapezium FBCH + Area of ∆DHC + Area of ∆EGD + Area of ∆EGA + Area of ∆BFA
Area of the triangle = \(\frac { 1 }{ 2 }\) bhsq.units
Area of the trapezium = \(\frac { 1 }{ 2 }\) × h × (a + b) sq.units
Area of the trapezium FBCH = \(\frac { 1 }{ 2 }\) × (10 + 8) × (8 + 3)m² = 9 × 11 = 99 m² ….. (1)
Area of the ∆DHC = \(\frac { 1 }{ 2 }\) × 8 × 5 m² = 20 m² ….. (2)
Area of ∆EGD = \(\frac { 1 }{ 2 }\) × 8 × 15m² = 60 m² …….. (3)
Area of ∆EGA = \(\frac { 1 }{ 2 }\) × 8 × (8 + 6)m² = 4 × 14 m²
= 56m²
Area of ∆BFA = \(\frac { 1 }{ 2 }\) × 3 × 6m² = 9 m²
∴ Area of the field = 99 + 20 + 60 + 56 + 9 m²
= 244 m²
Area of the field = 244 m²