Class 8 Maths Chapter 6 New Numbers Questions and Answers Kerala Syllabus

Students often refer to Kerala State Syllabus SCERT Class 8 Maths Solutions and Class 8 Maths Chapter 6 New Numbers Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 8 Maths Chapter 6 Solutions New Numbers

Class 8 Kerala Syllabus Maths Solutions Chapter 6 New Numbers Questions and Answers

New Numbers Class 8 Questions and Answers Kerala Syllabus

Coloured Numbers & Below Zero (Page No. 98)

Question 1.
Calculate the following.
(i) 4 – 9
(ii) 14 – 29
(iii) 5 – 10
(iv) 25 – 65
(v) \(\frac{1}{2}-\frac{3}{4}\)
(vi) \(\frac{1}{3}-\frac{1}{2}\)
Answer:
(i) 4 – 9 = -(9 – 4) = -5

(ii) 14 – 29 = -(29 – 14) = -15

(iii) 5 – 10 = -(10 – 5) = -5

(iv) 25 – 65 = -(65 – 25) = -40

(v) \(\frac{1}{2}-\frac{3}{4}=-\left(\frac{3}{4}-\frac{1}{2}\right)=-\left(\frac{3}{4}-\frac{2}{4}\right)=-\left(\frac{3-2}{4}\right)=-\frac{1}{4}\)

(vi) \(\frac{1}{3}-\frac{1}{2}=-\left(\frac{1}{2}-\frac{1}{3}\right)=-\left(\frac{3}{6}-\frac{2}{6}\right)=-\frac{1}{6}\)

Addition and Subtraction (Page No. 99)

Question 1.
Calculate the following.
(i) -4 + 9
(ii) -9 + 4
(iii) -15 + 8
(iv) -8 + 15
(v) \(-\frac{1}{2}+\frac{3}{4}\)
(vi) \(-\frac{3}{4}+\frac{1}{2}\)
Answer:
(i) -4 + 9 = 9 – 4 = 5

(ii) -9 + 4 = 4 – 9 = -(9 – 4) = -5

(iii) -15 + 8 = 8 – 15 = -(15 – 8) = -7

(iv) -8 + 15 = 15 – 8 = 7

(v) \(-\frac{1}{2}+\frac{3}{4}=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\)

(vi) \(-\frac{3}{4}+\frac{1}{2}=\frac{1}{2}-\frac{3}{4}=-\left(\frac{3}{4}-\frac{1}{2}\right)=-\frac{1}{4}\)

Class 8 Maths Chapter 6 New Numbers Questions and Answers Kerala Syllabus

Subtracting Again (Page No. 100)

Question 1.
Do the computations below:
(i) -3 – 1
(ii) -9 + 4
(iii) -10 – 4
(iv) -7 – 8
(v) -1 – 1
(vi) -10 + 20
(vii) 8 – 12
(viii) \(1 \frac{1}{2}-7 \frac{1}{2}\)
(ix) -25 – 3\(\frac {1}{2}\)
(x) \(-\frac{1}{2}-\frac{1}{4}\)
(xi) \(-2 \frac{1}{2}-1 \frac{1}{2}\)
(xii) \(-3 \frac{1}{2}+3 \frac{1}{2}\)
Answer:
(i) -3 – 1 = -(3 + 1) = -4

(ii) -9 + 4 = 4 – 9
= -(9 – 4)
= -5

(iii) -10 – 4 = (10 + 4) = -14

(iv) -7 – 8 = -(7 + 8) = -15

(v) -1 – 1 = -(1 + 1) = -2

(vi) -10 + 20 = 20 – 10 = 10

(vii) 8 – 12 = -(12 – 8) = -4

(viii) \(1 \frac{1}{2}-7 \frac{1}{2}\) = (1 – 7) + \(\left(\frac{1}{2}-\frac{1}{2}\right)\)
= -(7 – 1)
= -6

(ix) -25 – 3\(\frac {1}{2}\)= -(25 + 3\(\frac {1}{2}\)) = -28\(\frac {1}{2}\)

(x) \(-\frac{1}{2}-\frac{1}{4}=-\left(\frac{1}{2}+\frac{1}{4}\right)=-\frac{3}{4}\)

(xi) \(-2 \frac{1}{2}-1 \frac{1}{2}\) = -(2 + 1) – \(\left(\frac{1}{2}+\frac{1}{2}\right)\) = -4

(xii) \(-3 \frac{1}{2}+3 \frac{1}{2}=3 \frac{1}{2}-3 \frac{1}{2}\) = 0

Class 8 Maths Chapter 6 Kerala Syllabus New Numbers Questions and Answers

Class 8 Maths New Numbers Questions and Answers

Question 1.
The maximum temperature for seven consecutive days in a city is 26°C, and the minimum temperature is 21°C. The temperature difference is,
(a) 6°
(b) 5°
(c) 10°
(d) 9°
Answer:
(b) 5°
26 – 21 = 5

Question 2.
If x, y are any two positive numbers, and x < y, then x – y is
(a) Zero
(b) Negative numbers
(c) Positive numbers
(d) Impossible to express exactly
Answer:
(b) Negative number
If x < y then x – y = -(y – x)

Question 3.
If x = -1, y = 3, then x + y is
(a) -1 – 3
(b) 3 – 1
(c) 3 + 1
(d) -3 – 1
Answer:
(b) 3 – 1
If x and y be positive numbers, -x + y = y – x

Question 4.
Considering the temperature at which the water becomes ice is 0°C, then the temperature which represents 7 degrees less than this is,
(a) 0 – 1
(b) 0 – 7
(c) 7 + 7
(d) 7 – 0
Answer:
(b) 0 – 7

Class 8 Maths Chapter 6 New Numbers Questions and Answers Kerala Syllabus

Question 5.
\(3 \frac{1}{3}-1 \frac{1}{3}\) is equal to,
(a) 1
(b) 2
(c) 3
(d) -2
Answer:
(b) 2

Question 6.
Calculate the following.
(i) 3 – 5
(ii) 11 – 12
(iii) 18 – 27
(iv) 37 – 73
(v) \(\frac{1}{3}-\frac{1}{4}\)
(vi) \(\frac{1}{5}-\frac{1}{10}\)
Answer:
(i) 3 – 5 = -(5 – 3) = -2
(ii) 11 – 12 = -(12 – 11) = -1
(iii) 18 – 27 = -(27 – 18) = -9
(iv) 37 – 73 = -(73 – 37) = -36
(v) \(\frac{1}{3}-\frac{1}{4}=\frac{4}{12}-\frac{3}{12}=\frac{4-3}{12}=\frac{1}{12}\)
(vi) \(\frac{1}{5}-\frac{1}{10}=\frac{2}{10}-\frac{1}{10}=\frac{2-1}{10}=\frac{1}{10}\)

Question 7.
Calculate the following.
(i) -7 + 11
(ii) -11 + 17
(iii) -20 + 28
(iv) -11 + 15
(v) \(-\frac{1}{3}+\frac{3}{5}\)
(vi) \(-\frac{3}{5}+\frac{1}{4}\)
Answer:
(i) -7 + 11 = 11 – 7 = 4
(ii) -11 + 17 = 17 – 11 = 6
(iii) -20 + 28 = 28 – 20 = 8
(iv) -11 + 15 = 15 – 11 = 4
(v) \(-\frac{1}{3}+\frac{3}{5}=-\frac{5}{15}+\frac{9}{15}=\frac{9}{15}-\frac{5}{15}=\frac{4}{15}\)
(vi) \(-\frac{3}{5}+\frac{1}{4}=-\frac{12}{20}+\frac{5}{20}=\frac{5}{20}-\frac{12}{20}=-\left(\frac{12}{20}-\frac{5}{20}\right)=-\frac{7}{20}\)

Question 8.
Calculate the following.
(i) -10 – 11
(ii) -8 – 4
(iii) -20 – 8
(iv) -6 – 9
(v) -3 – 3
(vi) -17 + 23
(vii) 19 – 27
Answer:
(i) -10 – 11 = -(10 + 11) = -21
(ii) -8 – 4 = -(8 + 4) = -12
(iii) -20 – 8 = -(20 + 8) = -28
(iv) -6 – 9 = -(6 + 9) = -15
(v) -3 – 3 = -(3 + 3) = -6
(vi) -17 + 23 = 23 – 17 = 6
(vii) 19 – 27 = -(27 – 19) = -8

Question 9.
(a) What is 1 – 2?
(b) Find 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10.
Answer:
(a) -(2 – 1) = -1
(b) -1 – 1 – 1 – 1 – 1 = -5

Question 10.
Using this equation x2 – y2 = (x + y)(x – y)
(a) What is 22 – 12?
(b) What is 92 – 102?
(c) Find 12 – 22 + 32 – 42 + …… + 92 – 102?
Answer:
(a) (2 + 1)(2 – 1) = 3 × 1 = 3
(b) -19
(c) -3 – 7 – 9 – 11 – 15 – 19 = -(3 + 7 + 9 + 11 + 15 + 19) = -64

Class 8 Maths Chapter 6 New Numbers Questions and Answers Kerala Syllabus

Question 11.
Write the following in the simplified form.
\(\frac{12-11+10-9+8-7}{6-5+4-3+2-1}\)
Answer:
\(\frac{12-11+10-9+8-7}{6-5+4-3+2-1}=\frac{1+1+1}{1+1+1}=\frac{3}{3}\) = 1

Question 12.
Look at the pattern given below:
1 + 2 + 3 = 2 × 3 = 6
1 + 2 + 3 + 4 + 5 = 3 × 5 = 15
1 + 2 + 3 + 4 + 5 + 6 + 7 = 4 × 7 = 28
(a) Write the sum of the first 9 natural numbers?
(b) Write the sum of the first 25 natural numbers?
(c) What is -1 – 2 – 3 …. -49?
Answer:
(a) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 5 × 9 = 45
(b) 13 × 25 = 325
(c) -(1 + 2 + 3 + …… + 49) = -(12 × 49) = -637

Question 13.
Look at the pattern given below:
1 + 3 = 4 = 22
1 + 3 + 5 = 9 = 32
1 + 3 + 5 + 7 = 16 = 42
(a) Complete the next line.
(b) Find the sum of the first 10 odd numbers.
(c) What is the 20th odd number?
(d) What is -1 – 3 – 5 – …….. – 39?
Answer:
(a) 1 + 3 + 5 + 7 + 9 = 25 = 52
(b) 102 = 100
(c) 39
(d) -(1 + 3 + 5 + ……… + 39) = -400

Question 14.
Given below is an unfinished magic square made up of positive numbers and negative numbers. The sum of the numbers in rows, columns, and diagonals is equal.
Class 8 Maths Chapter 6 New Numbers Questions and Answers Kerala Syllabus Extra Questions Q14
Find x + y?
Answer:
Consider the number that can be written in the top right side of the column, be A.
-7 + 6 + A = x + y + A
⇒ -7 + 6 = x + y
⇒ x + y = 6 – 7
⇒ x + y = -1

Question 15.
Calculate \(\frac{1}{7}-\frac{3}{7}+\frac{5}{7}-\frac{7}{7}+\cdots+\frac{21}{7}\)
Answer:
\(\frac{1-3+5-7+9-11+13-15+17-19+21}{7}=\frac{-10+21}{7}\) = \(\frac {11}{7}\)

Class 8 Maths Chapter 6 Notes Kerala Syllabus New Numbers

→ On a number line, the numbers written on the right side of 0 are called positive numbers.

→ Subtracting a positive number from a smaller positive number gives the negative of the number obtained by subtracting the smaller from the larger.
For example: we can write, 2 – 3 = -(3 – 2) = -1

→ x – y = -(y – x) for all positive numbers x and y and x < y

→ Adding the negative of a positive number to a positive number means subtracting the first number from the second.

→ -x + y = y – x for all positive numbers x and y.

→ If from the negative of a positive number, we subtract a positive number, we get the negative of the sum of these positive numbers

→ -x – y = -(x + y) for any two positive numbers x and y.

Class 8 Maths Chapter 6 New Numbers Questions and Answers Kerala Syllabus

Positive numbers, along with the negative signs, are used to describe some specific situations. This type of representation is needed in certain situations. Such as, to express the temperature when the water turns into ice, is less than zero degrees Celsius. To represent these types of numbers, 1°C, -5 °C, we use negative signs with positive numbers. The theoretical minimum temperature is -273.15°C. On the Kelvin scale, this is called absolute zero temperature. In this unit, a new type of number is introduced for the first time. These numbers are the negatives of the positive numbers. Such numbers written with a minus sign are called negative numbers. This makes the word new numbers meaningful.

Coloured Numbers
In a calculating machine, the calculations are done in the order given below:
Class 8 Maths Chapter 6 New Numbers Questions and Answers Kerala Syllabus Notes 1
The values given as input and the values obtained as output are natural numbers.
What is the smallest input value that can be given to get the very smallest output?
Answer:
The smallest output is 1
1 + 10 = 11
11 × 3 = 33
33 + 10 = 43
43 × 3 = 129

Arun and Vinu are in a game. The one who wins the game gets a score of 3, and the one who fails in the game gets a score of 1. Arun wins 6 games. Vinu receives 18 points. How many games are there?
Answer:
Arun wins 6 games. That means Vinu fails in 6 games.
That means, in the score of 18 scored by Vinu, 6 points are obtained from failure.
12 is the score obtained by Vinu in winning the game. That means the score of 4 games.
The total number of games is 6 + 4 = 10.

Below Zero
On a number line, the numbers written on the right side of 0 are called positive numbers. There are different types of numbers, such as natural numbers, integers.
When a – symbol is written along with 2, we can read it as -2. It is a negative number.

Subtracting a positive number from a smaller positive number gives the negative of the number obtained by subtracting the smaller from the larger.
In algebraic form: x – y = -(y – x) for all 0 < x < y

From the two positive numbers, subtract the larger from the smaller.
10 and 13 are two positive numbers.
From the smaller number 10, the larger number is subtracted.
When 13 is subtracted from 10, first 10 is subtracted and then 3 is also subtracted.
When 10 is subtracted from 10, we get 0. And 3 is subtracted from 0, we get -3.
If 13 is subtracted from 10, we get -3. Another way we can say it is that, to subtract 13 from 10, subtract 10 from 13, and put the negative sign with the value.

For example, the normal temperature in a place is 10°C. But in the case of extreme cold, it decreases to 20°C. In the case of extreme cold, what is the temperature?
Answer:
10 – 20 = -(20 – 10) = -10
To subtract 20 from 10, we can subtract 10 from 20 and put the negative sign along with the value.

Worksheet – 1

Question 1.
Find \(7 \frac{1}{2}-10 \frac{1}{2}\)?
Answer:
\(7 \frac{1}{2}-10 \frac{1}{2}=-\left(10 \frac{1}{2}-7 \frac{1}{2}\right)\) = -3

Question 2.
Complete the following calculations.
(a) \(\frac{1}{3}-\frac{1}{2}\)
(b) \(3 \frac{1}{3}-5 \frac{1}{2}\)
Answer:
Class 8 Maths Chapter 6 New Numbers Questions and Answers Kerala Syllabus Notes 2

Worksheet – 2

Question 1.
What should be subtracted from 11\(\frac {1}{3}\) to set 14\(\frac {1}{3}\)?
Answer:
Number should be subtracted = \(11 \frac{1}{3}-14 \frac{1}{3}\)
= \(-\left(14 \frac{1}{3}-11 \frac{1}{3}\right)\)
= \(-\left(3+11 \frac{1}{3}-11 \frac{1}{3}\right)\)
= -3

Class 8 Maths Chapter 6 New Numbers Questions and Answers Kerala Syllabus

Question 2.
What should be subtracted from 7\(\frac {1}{3}\) to get 11\(\frac {2}{3}\)?
Answer:
Number should be subtracted = \(7 \frac{1}{3}-11 \frac{2}{3}\)
= \(-\left(11 \frac{2}{3}-7 \frac{1}{3}\right)\)
= \(-\left((11-7)+\left(\frac{2}{3}-\frac{1}{3}\right)\right)\)
= \(-4 \frac{1}{3}\)

Adding and Subtracting
Adding the negative of a positive number to a positive number means subtracting the first number from the second.
In algebraic form: -x + y = y – x for all positive numbers x and y.

Worksheet – 3

Question 1.
Find \(-1 \frac{1}{2}+3 \frac{1}{2}\)?
Answer:
\(-1 \frac{1}{2}+3 \frac{1}{2}=3 \frac{1}{2}-1 \frac{1}{2}\)
= (3 – 1) + \(\left(\frac{1}{2}-\frac{1}{2}\right)\)
= 2 + 0
= 2

Question 2.
Find \(-3 \frac{1}{5}+7 \frac{2}{5}\)?
Answer:
\(-3 \frac{1}{5}+7 \frac{2}{5}=7 \frac{2}{5}-3 \frac{1}{5}=(7-3)+\left(\frac{2}{5}-\frac{1}{5}\right)=4 \frac{1}{5}\)

Worksheet – 4

Question 1.
Calculating the following:
(a) -6 + 5
(b) \(-3 \frac{1}{2}+1 \frac{1}{2}\)
(c) -11 + 7 – 12 + 4
(d) -10 + 11 – 12 + 13 – …… + 19 – 20 + 21
Answer:
(a) -6 + 5 = 5 – 6
= -(6 – 5)
= -1
(b) \(-3 \frac{1}{2}+1 \frac{1}{2}=1 \frac{1}{2}-3 \cdot \frac{1}{2}=-\left(3 \frac{1}{2}-1 \frac{1}{2}\right)\) = -2
(c) -11 + 7 – 12 + 4 = (7 – 11) + (4 – 12)
= -(11 – 7) – (12 – 4)
= -4 – 8
= -12
(d) -10 + 11 – 12 + 13 …. + 19 – 20 + 21 = (11 – 10) + (13 – 12) + ….. + (21 – 20)
= 1 + 1 + 1 + 1 + 1 + 1
= 6

Subtracting Again
If from the negative of a positive number, we subtract a positive number, we get the negative of the sum of these positive numbers.
In algebraic form: -x – y = -(x + y) for any two positive numbers x and y.

Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus

Students often refer to Kerala State Syllabus SCERT Class 8 Maths Solutions and Class 8 Maths Chapter 7 Bisectors Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 8 Maths Chapter 7 Solutions Bisectors

Class 8 Kerala Syllabus Maths Solutions Chapter 7 Bisectors Questions and Answers

Bisectors Class 8 Questions and Answers Kerala Syllabus

Bisectors of Lines (Page No. 107)

Draw the figures below using only a ruler and a compass.

Question 1.
Square of sides 4\(\frac {1}{4}\) centimetres.
Answer:
Draw a line 8.5 cm long.
Draw its perpendicular bisector.
Each portion is of length 4\(\frac {1}{4}\) cm.
Marking half the length of the line on the perpendicular bisector.
Complete the square.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 107 Q1

Question 2.
Rectangle of sides 5\(\frac {1}{4}\) centimeters and 3\(\frac {1}{4}\) centimeters.
Answer:
Draw a line AB of length, \(5 \frac{1}{4}+3 \frac{1}{4}=8 \frac{1}{2}\) cm
Draw the perpendicular bisector of the line.
Half of the length of the line is 4.25 cm.
Mark the midpoint as M.
Mark a point P on AB which is 1cm away from the point M.
AP = 5\(\frac {1}{4}\) cm and BP = 3\(\frac {1}{4}\) cm.
Draw a perpendicular from P to AB and mark a point C on it such that PB = PC.
Draw a square APCD.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 107 Q2

Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus

Question 3.
Equilateral triangle of side 2.75 centimetres.
Answer:
Draw a line of length 2.75 × 4 = 11 cm.
Draw the perpendicular bisector of the line.
Divides it into two equal parts.
Agai,n draw the perpendicular bisector for the one part, \(\frac {1}{4}\) is a part of 11 cm.
The length of one part is 2.75 cm.
Draw an equilateral triangle with this length as its sides.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 107 Q3

Question 4.
Triangle of area 9 square centimetres and one side 4.5 centimetres.
Answer:
The area of the square is 2 × 9 = 18 sq.cm.
One side is 4.5 cm, and the other side is 4 cm.
One of its diagonals makes two equal triangles.
The area of a triangle is 9 sq cm, and one side is 4.5 cm.
A line AP of length 8.5 cm.
Mark the point B such that AB = 4.5 cm
Draw a perpendicular to the line through B and mark a point C on it such that BP = BC.
Draw the square ABCD.
Draw the diagonal of AC and mark the triangle as ABC.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 107 Q4

Rhombuses (Page No. 111, 112)

Question 1.
Draw a rhombus with each pair of lengths given below for the diagonals:
(i) 6 centimetres, 4 centimetres
(ii) 6.5 centimetres, 4 centimetres
(iii) 6 centimetres, 4.5 centimetres
Answer:
(i) Draw a line 6 cm long.
Draw its perpendicular bisector.
The diagonals of a rhombus are perpendicular bisectors of each other.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 111 Q1

(ii) Draw a line 6.5 cm long.
Draw its perpendicular bisector.
Draw the rhombus.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 111 Q1.1

(iii) Draw a line 6 cm long.
Draw its perpendicular bisector.
Draw the rhombus.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 111 Q1.2

Question 2.
The picture shows the quadrilateral formed by joining the midpoints of a rectangle:
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 111 Q2
(i) Are the diagonals of this quadrilateral parallel to the sides of the rectangle? Why?
(ii) Is this quadrilateral a rhombus? Why?
Answer:
(i) Square ABCD. AD and BC are equal and parallel.
These are perpendicular to AB.
So the line PR is parallel to AB and CD.
Similarly, the line SQ is parallel to AD and BC.
The diagonals of PQRS are parallel to the sides of ABCD.

(ii) Since P, Q, R, and S are the midpoints of the sides, the diagonals of PQRS are perpendicular bisectors of each other.
PQRS is a rhombus.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 111 Q2.1

Question 3.
Draw a rhombus with diagonals 6.5 centimetres and 4.5 centimetres.
Answer:
Draw a square with side lengths of 6.5 cm and 4.5 cm.
Draw the perpendicular bisectors of length and breadth.
These lines pass through the midpoint of the sides.
Join the midpoints of the sides.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 111 Q3

Question 4.
Prove that each diagonal of a rhombus bisects the angles at the vertices it joins.
Answer:
Consider the rhombus ABCD.
AC and BD are the diagonals.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 111 Q4
Diagonals are the perpendicular bisectors of each other.
The four sides are equal.
In triangle ACD, AD = CD.
So the angles opposite to these sides are equal.
That is x = y.
Diagonals bisect the angle at the vertices they join.

Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus

Question 5.
Draw a square with diagonals 7 centimetres.
Answer:
Draw a line 7 cm long.
Draw the perpendicular bisector of the line.
Mark the midpoint as O.
Draw a circle considering the line as the diameter.
Complete the square.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 111 Q5

Question 6.
Draw this picture.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 111 Q6
Answer:
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 111 Q6.1

Bisector of an Angle (Page No. 115)

Question 1.
Draw the angles below:
(i) \(37 \frac{1}{2}^{\circ}\)
(ii) \(62 \frac{1}{2}^{\circ}\)
Answer:
Draw an angle of 75°.
Draw the arc in such a way that it cuts the sides at a fixed radius with the corner at the centre.
An isosceles triangle is obtained by joining the points that intersect the sides by the arc.
The perpendicular from the vertex is the angle bisector.
For this, consider the centre as the points that intersect the sides and draw arcs of the same radius that intersect each other.
Complete the bisector by drawing a line that joins this point with the vertex.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 115 Q1

Question 2.
Use only a ruler and a compass to draw the following:
(i) The angles below.
(a) 45°
(b) 135°
(c) 75°
(d) 15°
(ii) The triangle with one side of length 6 centimetres and angle \(67 \frac{1}{2}^{\circ}\) and \(22 \frac{1}{2}^{\circ}\) at its ends.
Answer:
(i) (a) 45°
Draw a line and mark a point on it.
Draw the perpendicular to the line through the point.
Draw the angle bisector for the angle formed at 90°.
This forms a 45° angle.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 115 Q2

(b) 135°
Draw a line and mark a point on it.
Draw the perpendicular to the line through the point.
Draw the angle bisector for the angle formed at 90°.
Draw an angle 45°
90 + 45 = 135.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 115 Q2.1

(c) 75°
First, draw a 90° angle.
Inside this angle, draw a 60° angle.
Draw the angle bisector for the 30° angle formed along with these.
15 + 60 = 75
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 115 Q2.2

(d) 15°
Draw a 60° angle and mark the angle bisector of this.
Draw a 30° angle and mark the angle bisector of this.
We get a 15° angle.

(ii) Draw a 6 cm long line.
Draw an angle of 135° on one end and 45° on the other end.
Use only the compass and scale for the construction (above construction).
Draw a triangle with the angle bisectors of this.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Page 115 Q2.3

Class 8 Maths Chapter 7 Kerala Syllabus Bisectors Questions and Answers

Class 8 Maths Bisectors Questions and Answers

Question 1.
Which of the following is a quadrilateral in which all the angles are equal?
(a) Rhombus
(b) Rectangle
(c) Parallelogram
(d) Square
Answer:
(d) Square
All angles are 90 °.

Question 2.
Two statements are given below:
(i) The diagonals of all parallelograms bisect each other perpendicularly.
(ii) The diagonal of a rhombus bisects each other perpendicularly.
(a) (i) and (ii) are true
(b) Only (i) is true
(c) Only (ii) is true
(d) (i) and (ii) are not true
Answer:
(c) Only (ii) is true.

Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus

Question 3.
If a 145° angle is divided equally, then what is the measure of one part?
(a) 75°
(b) 70°
(c) \(40 \frac{1^{\circ}}{2}\)
(d) \(72 \frac{1^{\circ}}{2}\)
Answer:
(d) \(72 \frac{1^{\circ}}{2}\)

Question 4.
What are the measures needed to draw a fixed rhombus?
(a) Only the length of its sides
(b) One diagonal and one side
(c) Two diagonals
(d) Perimeter
Answer:
(c) Two diagonals

Question 5.
The line l1 is perpendicular to l2. The line l2 is perpendicular to l3.
(i) l1 and l3 are perpendicular lines.
(ii) l1 and l3 are parallel lines.
(a) (i) and (ii) are true
(b) (i) and (ii) are false
(c) (i) is true (ii) is false
(d) (i) is false (ii) is true
Answer:
(d) (i) is false (ii) is true

Question 6.
Draw a 7.5-centimeter-long line and construct a right triangle whose perpendicular sides are 3.75 centimeters long.
Answer:
Draw a 7.5 cm long line.
Draw the perpendicular bisector of this line.
Complete the right triangle by drawing the half length of the line on the perpendicular line.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Extra Questions Q6

Question 7.
Draw an isosceles triangle whose length of one side is 9.5 centimetres, and the height of this side is 4.75 centimetres.
Answer:
Draw a line of length 9.5 cm.
Draw the perpendicular bisector of this line.
Mark the half-length of the line on the top of its perpendicular bisector.
From this point, join both ends of the line and form a triangle.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Extra Questions Q7

Question 8.
Draw a square with a perimeter is 23 centimeters.
Answer:
Draw a line of length 11.5 centimeters.
Draw the perpendicular bisector of the line.
Mark the half-length of the line on the perpendicular bisector.
Complete the square.
One side = \(\frac {11.5}{2}\) cm
Perimeter = 4 × \(\frac {11.5}{2}\) = 23 cm
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Extra Questions Q8

Question 9.
Draw a line 7.5 cm long and divide it into two equal parts. Draw an equilateral triangle with a perimeter of 11.25 cm.
Answer:
Draw a line of length 7.5 cm.
Draw a perpendicular bisector of the line.
Length of one side = \(\frac {7.5}{2}\) cm
Draw an equilateral triangle by considering the half length as the side.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Extra Questions Q9

Question 10.
Draw a square with the length of its diagonals as 6.5 centimetres.
Answer:
Draw a line of length 6.5 cm.
Draw the perpendicular bisector of this line.
Take the midpoint of the line as the center and draw a circle with half the length of the line as its radius.
Draw a square by joining the ends of the line with the points on the perpendicular bisector where the circle touches it.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Extra Questions Q10

Question 11.
Draw a rhombus with the length of its diagonals as 6 centimetres, 8.5 centimetres.
Answer:
Draw the perpendicular bisector for one of its diagonals.
Draw the rhombus by marking the length of the other diagonal on the perpendicular bisector.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Extra Questions Q11

Question 12.
Draw \(22 \frac{1}{2}^{\circ}\) cone using scale and compass.
Answer:
Draw a line and mark a point on it.
Draw a perpendicular line through the point.
A 90° angle is formed. Draw the angle bisector.
A 45° angle is formed. Draw the angle bisector.
\(22 \frac{1}{2}^{\circ}\) angle is formed.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Extra Questions Q12

Question 13.
Draw a right triangle with two angles of 45° and a length of its hypotenuse of 4.75 centimetres.
Answer:
Draw a line of length 9.5 cm.
Draw its perpendicular bisector.
The part of the line on one side of the perpendicular bisector will be 4.75 cm.
The angle between the line and the perpendicular bisector is 90°.
Draw the angle bisector.
Draw a perpendicular from one end of the line to the angle bisector.
It forms a right triangle.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Extra Questions Q13

Question 14.
Draw a right triangle with one angle 30° and the length of its hypotenuse as 4.25 centimetres.
Answer:
Draw a line of length 8.5 cm.
Draw its perpendicular bisector.
The half length of the line is 4.25 cm.
Draw an angle of 60° to the vertex that intersects the bisector.
Draw a perpendicular from one end of the line to the side of its angle.
It forms a 30° – 60° – 90° right triangle.
The hypotension will be 4.25 cm.

Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus

Question 15.
The length of the side of an isosceles triangle is 4.25 centimetres. Draw the triangle with the height of its side as 7 centimetres.
Answer:
Draw a line of length 9.5 cm.
Draw the perpendicular bisector of the line.
One side of the bisector is 4.25 cm.
Draw the perpendicular bisector for the 4.25 cm line.
Mark the height on the perpendicular bisector and complete the triangle.

Class 8 Maths Chapter 7 Notes Kerala Syllabus Bisectors

→ A perpendicular line that passes through the midpoint of a line is called a perpendicular bisector. Any point on the perpendicular bisector of a line is at the same distance to its endpoints.

→ The perpendicular bisector of the line is the line joining the points that are at equal distance from the endpoints of a line.

→ Rhombuses are the quadrilaterals with opposite sides parallel. A parallelogram with four sides of the same length is called rhombus.

→ For a rhombus, the diagonals are perpendicular bisectors of each other. The converse is also true. The quadrilateral with its diagonals are perpendicular bisectors of each other is a rhombus.

→ If the diagonals of a quadrilateral are of the same length and are perpendicular bisectors of each other, then it forms a square.

→ The diagonals of a rectangle are equal and bisect each other. But they will not be a perpendicular bisector.

→ Any point on the bisector of an angle is at the same perpendicular distance from the sides of the angle.

→ Similarly, any point at the same perpendicular distance from two sides of an angle lies on the bisector of that angle.

Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus

As the word implies, bisectors divide equally. All lines that divide a line into equal parts are called bisectors of the line. We can draw many bisectors to a line, but a line which is perpendicular to a line is called the perpendicular bisector. In this chapter, we are discussing the geometrical concepts related to the construction of a perpendicular bisector. This concept is used in many situations. The basic idea behind the construction of the geometrical shape rhombus is that the diagonals are perpendicular bisectors of each other. In this chapter, we also discuss the bisector of the angle. The geometrical concepts of bisectors are essential for the further study of triangles and circles.

Bisectors of Lines
The point on a line at the same distance from both ends of a line is the midpoint of the line.
For example, a line of length 6 cm is equally spaced from both ends of the line, and the point on the line is 3 cm from both ends.
But what about the point that is 4 centimetres away from both sides?
To find that point, take 4 cm on a compass and mark it from both ends.
The point where they meet is 4 cm from the endpoints.
In this way, a point can be drawn at a distance of 5 cm from both ends of the line.
In this way, many points at the same distance can be drawn from both sides.
The line they are attached to will go through the midpoint of the line drawn first.

What can we understand from this?
To draw the perpendicular bisector of the line, mark the points at the same distance from both endpoints.
Let’s see how we can express this concept geometrically.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Notes 1
P is a point on the perpendicular bisector of the line AB.
M is the midpoint of AB.
AMP, BMP are right-angled triangles.
Using Pythagoras’ theorem,
AM2 + MP2 = AP2, BM2 + MP2 = BP2
AM = BM, and MP is a common line, so AP2 = BP2, AP = BP.

Worksheet – 1

Question 1.
Draw a 7.5-centimetre-long line and construct the perpendicular bisector.
Answer:
Draw a 7.5 cm line AB.
Considering the endpoints A and B as the centres, mark two arcs that cut each other.
Take a little distance more and draw another two arcs that intersect each other.
Draw a line through the point where the arcs intersect each other.
This is the perpendicular bisector of the first line.

Rhombuses
The shapes with four sides are called quadrilaterals.
The quadrilaterals with opposite sides that are parallel will be one of the following.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Notes 2
In this, we learn about rhombuses. And understand the characteristics of others.
The opposite sides of all these forms are parallel. But a different shape is formed when certain other characteristics are combined.

A diagonal is a line that joins the opposite corners.
The diagonals of a square and a rhombus are equal.
The diagonals of a quadrilateral and a rhombus are not equal, but they bisect each other.
Diagonals of a rhombus bisect each other perpendicularly.

Let us look at the characteristics of a square and a rhombus.
The sides are equal. Diagonals are perpendicular bisectors.
A rhombus can be drawn using the idea that the diagonals are perpendicular bisectors.

Worksheet – 2

Draw a rhombus with one angle of 40° and one side 5 centimetres.
Answer:
Draw an angle of 40°.
Draw a circle with a radius of 5 centimetres and consider the vertex as the centre.
Mark the points that cut the sides of the angle by the circle.
Draw a line parallel to the side of the angle through these points.
The sides of the angle and the lines together form a parallelogram.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Notes 3

Worksheet – 3

Draw a rhombus with the length of its diagonal as 6 centimetres.
Answer:
Draw a 6 cm long line AC.
Draw the perpendicular bisector of this line.
Draw a circle by considering the midpoint of the line as its centre.
Mark the points where the circle divides the perpendicular bisectors as C and D.
Draw a rectangle ABCD.
ABCD is a rhombus.
When did it become a square?
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Notes 4

Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus

Bisector of an Angle
It’s a view from the perpendicular bisector to the angle bisector.
We know about the isosceles triangles. Two sides are equal.
If we draw a perpendicular bisector to the third side, it will pass through the opposite corner.
This line will divide the opposite angle into two equal parts. So this line is also called the angle bisector.

Worksheet – 4

Draw an angle of 40° and construct its angle bisector.
Answer:
Draw a 40° angle.
Draw a circle by considering the vertex A as the centre.
The circle cuts the sides of the angle at the points B and C.
Draw the triangle ABC. The triangle ABC is an isosceles triangle.
Draw the perpendicular bisector of BC. This line is the bisector of angle A.
Class 8 Maths Chapter 7 Bisectors Questions and Answers Kerala Syllabus Notes 5

Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus

Students often refer to Kerala State Syllabus SCERT Class 8 Maths Solutions and Class 8 Maths Chapter 4 Polygons Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 8 Maths Chapter 4 Solutions Polygons

Class 8 Kerala Syllabus Maths Solutions Chapter 4 Polygons Questions and Answers

Polygons Class 8 Questions and Answers Kerala Syllabus

Names & Angles (Page No. 65)

Question 1.
The sum of the angles of a polygon is 1980°. What is the sum of the angles of a polygon with one side more? And for a polygon with one side less?
Answer:
When the number of sides increases by 1 angle sum increases by 180°.
Angle sum is 1980 + 180 = 2160°
When the number of sides decreases by 1 angle sum decreases by 180°.
Angle sum 1980 – 180 = 1800°

Question 2.
What is the sum of the angles of a 27-sided polygon?
Answer:
Angle sum of a polygon having n sides is 180° × (n – 2)
Angle sum 180 × 25 = 4500°

Question 3.
The sum of the angles of a polygon is 8100°. How many sides does it have?
Answer:
Sum of the angles = (n – 2) × 180°
Where n is the number of sides.
∴ (n – 2) × 180 = 8100
⇒ n – 2 = \(\frac {8100}{180}\) = 45
⇒ n = 45 + 2
⇒ n = 47

Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus

Question 4.
Is the sum of the angles of any polygon equal to 1000°? Explain.
Answer:
Inner angle sum is always a multiple of 180.
1000 is not a multiple of 180.
Angle sum cannot be 1000°.

Question 5.
A 20-sided polygon has equal angles. How much is each angle?
Answer:
Angle sum of a 20 sided polygon = 180 × (20 – 2)
= 180 × 18
= 3240°
One inner angle = \(\frac {3240}{20}\) = 162

Outer Angles (Page No. 71)

Question 1.
We have seen that in a triangle, the outer angle at any vertex is equal to the sum of the inner angles at the other two vertices; and in a quadrilateral, the sum of the outer angles at any two vertices is equal to the sum of the inner angles at the other two vertices. So, here are some possible investigations:
(i) Is there any such relation between the inner and outer angles of a pentagon?
(ii) And in a hexagon?
(iii) Is there a general relation that is true for all polygons?
Answer:
(i) Angles of a pentagon are a, b, c, d, e.
Angle sum = a + b + c + d + e = 180 × 3 = 540°.
Outer angles are 180 – a, 180 – b, 180 – c.
Sum = 540 – (a + b + c)
Sum of two inner angles = 540 – (540 – c + d)) = c + d
The sum of two inner angles is equal to the sum of the other three outer angles.

(ii) Angles are x1, x2, x3, x4, x5, x6
Two inner angles are x1, x2
Sum of four outer angles = 180 – x3 + 180 – x4 + 180 – x5 + 180 – x6 = 720 – (x3 + x4 + x5 + x6)
Sum of four outer angles = 720 – (720 – (x1 + x2) = x1 + x2
The sum of two inner angles is equal to the sum of the other outer angles.

(iii) The angles of a polygon having n sides: x1, x2,…, xn
Excluding these two inner angles, there will be n – 3 outer angles.
Sum of the outer angles = 180 – x3 + 180 – x4 + ……. + 180 – xn
Sum of n – 3 outer angles = 180 × (n – 2) – (x3 + x4 + …… + xn)
= 180(n – 2) – (180(n – 2) – (x1 + x2))
= x1 + x2

Textbook Page No. 72

Question 1.
All inner angles of an 18-sided polygon are equal.
(i) How much is each outer angle?
(ii) How much is each inner angle?
Answer:
(i) Outer angles are equal.
One outer angle = \(\frac {360}{18}\) = 20°
(ii) One inner angle = 180 – 20 = 160°

Question 2.
(i) In which polygon is the sum of outer angles of a polygon, one from each vertex, equal to the sum of its inner angles?
(ii) In which polygon is the sum of the outer angles twice the sum of the inner angles?
(iii) In which polygon is the sum of the outer angles half the sum of the inner angles?
(iv) In which polygon is the sum of the outer angles one-third the sum of the inner angles?
Answer:
(i) Sum of the outer angles 360°.
Since the outer angle sum is 360°.
∴ 180 × (n – 2) = 360
⇒ n – 2 = 2
⇒ n = 4
Number of sides 4.
The polygon is a Quadrilateral.

(ii) Sum of the outer angles 360°.
The sum of the inner angles is 180°.
The polygon is a Triangle.

(iii) Sum of the outer angles = 2 × 360 = 720
∴ 180 × (n – 2) = 720
⇒ n – 2 = 4
⇒ n = 6
The polygon is a Hexagon.

(iv) The sum of the inner angles is three times the sum of the outer angles.
180 × (n – 2) = 3 × 360
⇒ n – 2 = \(\frac{3 \times 360}{180}\) = 6
⇒ n = 8
Number of sides = 8
The polygon is an Octagon.

Regular Polygons (Page No. 77, 78)

Question 1.
(i) Draw a hexagon of equal sides with angles different.
(ii) Draw a hexagon of equal angles with different sides.
Answer:
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Page 77 Q1

Question 2.
The picture shows a regular hexagon with vertices on a circle. Prove that the length of its sides is equal to the radius of the circle.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Page 77 Q2
Answer:
Inner angle sum = 180 × (6 – 2)
= 180 × 4
= 720°
All angles are equal.
One angle = \(\frac {720}{6}\) = 120°
Bisectors of angles pass through the centre.
That is angle around the centre will be divided into six equal parts.
Length of side is equal to radius.

Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus

Question 3.
Draw a regular octagon (eight sides) of sides 3 centimetres.
Answer:
Angle sum of octagon = 180 × (8 – 2) = 1080
One angle of regular octagon = \(\frac {1080}{8}\) = 135°
Draw a line AB of length 3 cm.
With B as the centre, draw a 135° angle and mark C at a distance of 3 cm away.
Continue this process.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Page 77 Q3

Question 4.
Draw a circle of radius 3 centimetres and draw a regular octagon with all vertices on this circle.
Answer:
Draw the circle, its diameter, and its perpendicular bisector.
Now the angle around the centre is divided into four equal parts.
Bisect the angles again and complete the octagon.

Question 5.
The picture below shows a regular hexagon and a triangle joining its alternate vertices:
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Page 77 Q5
Is this an equilateral triangle? Why?
Answer:
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Page 77 Q5.1
Triangles around the middle triangles are equal.
Sum of the small angles = 180 – 120 = 60°
One small angle = 30°
Angles of middle triangle are 30 + 30 = 60°
This is an equilateral triangle.

Question 6.
The picture below shows a regular hexagon and a quadrilateral joining four of its vertices:
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Page 77 Q6
Is this quadrilateral a rectangle? Why?
Answer:
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Page 77 Q6.1
One angle of a regular hexagon = 180 × (6 – 2) ÷ 6 = 120°
Look at the triangles on the sides of the quadrilateral.
These are isosceles triangles.
Angles are 30°, 30°, 120°
One angle of the quadrilateral = 120 – 30 = 90°
This is a rectangle.

Question 7.
Calculate the inner angle and the outer angle of a regular polygon of 15 sides.
Answer:
Measure of one outer angle = \(\frac {360}{15}\) = 24°
Measure of one inner angle = 180 – 24 = 156°

Question 8.
An outer angle of a regular polygon is 20°. How many sides does it have?
Answer:
Outer angle sum = 360°
All outer angles are equal.
One angle is \(\frac {360}{20}\) = 18°

Question 9.
An inner angle of a regular polygon is 168°. How many sides does it have?
Answer:
One outer angle = 180 – 168 = 12°
Number of sides = \(\frac {360}{12}\) = 30

Class 8 Maths Chapter 4 Kerala Syllabus Polygons Questions and Answers

Class 8 Maths Polygons Questions and Answers

Question 1.
A regular polygon has an inner and outer angle sum equal. Which of the following is that polygon?
(a) Equilateral triangle
(b) Square
(c) Regular pentagon
(d) Regular hexagon
Answer:
(b) Square

Question 2.
Two statements are given below.
p1: Whatever the number of sides of a polygon, the outer angle sum is 360°.
p2: inner angle sum increases by 180° if the number of sides increases by 1.
(a) p1 and p2 are true
(b) p1 is true p2 is false
(c) p1 is false p2 is true
(d) p1 and p2 are false
Answer:
(a) p1 and p2 are true

Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus

Question 3.
A polygon has 7 sides
(a) What is the sum of its inner angles?
(b) What is the sum of the outer angles?
Answer:
(a) Angle sum of a polygon of n sides is 180° × (n – 2)
Angle sum is 180° × 5 = 900°
(b) 360°

Question 4.
A polygon has an inner angle sum that is two times the outer angle sum.
(a) What is its inner angle sum?
(b) Find the number of sides of the polygon.
Answer:
(a) 2 × 360 = 720°
(b) 180 × (n – 2) = 720
⇒ n – 2 = 4
⇒ n = 6

Question 5.
Find the sum of the angles marked in the star.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Extra Questions Q5
Answer:
Diagram
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Extra Questions Q5.1
In triangle BQD, the sum of the marked angles is the outer angle at Q.
In triangle PCE, the marked angle sum is the outer angle at P.
All marked angles make the sum 180°.

Question 6.
A polygon has all inner angles equal. There are 18 sides.
(a) What is the inner angle sum?
(b) Find the measure of an inner angle?
(c) What is the measure of one outer angle?
Answer:
(a) 180 × (18 – 2) = 2880°
(b) \(\frac {2880}{18}\) = 160°
(c) 180 – 160 = 20°

Question 7.
This is the section of a regular polygon.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Extra Questions Q7
(a) What are the angles of a triangle?
(b) What is an outer angle?
(c) How many sides does this polygon have?
Answer:
(a) 18°, 18°, 144°
(b) 180 – 144 = 36°
(c) \(\frac {360}{6}\) = 10

Question 8.
Draw a circle of radius 3 cm and construct an equilateral triangle with corners on the circle.
Answer:
Divide the angle around the centre into three equal parts by drawing radii.

Question 9.
A regular polygon has 15 sides.
(a) What is the measure of an outer angle?
(b) What is the inner angle of this polygon?
(c) Find the inner angle sum.
Answer:
(a) \(\frac {360}{15}\) = 24
(b) 180 – 24 = 156°
(c) 15 × 156 = 2340°

Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus

Question 10.
The diagram shows an equilateral triangle, a square, and one diagonal of the square. What is the value of?
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Extra Questions Q10
Answer:
An equilateral triangle has one inner angle = 60°
Square has inner angle = 90°
A diagonal divides it into 45° each.
x = 60 + 45 = 105°

Question 11.
Squares are drawn on the sides of a regular hexagon. Find x.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Extra Questions Q11
Answer:
One inner angle of a regular hexagon is 120°.
Two squares in a corner make the angles 90°.
Angle around the corner is 360°.
x = 360° – (120° + 90° + 90°) = 60°

Question 12.
The measure of one outer angle of a regular polygon is 40°.
(a) What is the measure of one inner angle?
(b) Find the number of sides of this polygon?
Answer:
(a) 140°
(b) \(\frac {360}{40}\) = 9

Question 13.
The measure of an outer angle of a regular polygon is 2x, and the measure of an inner angle is 4x.
(a) Use the relationship between inner and outer angles to find x.
(b) Find the measure of one inner and outer angle.
(c) Find the number of sides in the polygon and the type of polygon.
Answer:
(a) The sum of the inner angles at a corner is 180°.
⇒ 6x = 180
⇒ x = 30
(b) Angles are 60°, 120°
(c) Regular hexagon

Question 14.
The measure of one interior angle of a regular polygon is 144°. How many sides does it have?
Answer:
One outer angle is 180 – 144 = 36°
Number of sides = \(\frac {360}{36}\) = 10

Question 15.
Five angles of a hexagon have measures 100°, 110°, 120°, 130°, 140°. What is the measure of the sixth angle?
Answer:
The inner angle sum of the hexagon is 720°.
Sixth angle = 720 – (100 + 110 + 120 + 130 + 140)
= 720 – 600
= 120°

Class 8 Maths Chapter 4 Notes Kerala Syllabus Polygons

→ Sum of the angles of a triangle is 180°. Polygons can be divided into triangles by drawing diagonals.

→ Inner angle sum of a polygon is the product of 180° and 2 less than the number of sides.

→ Using algebra for the number of sides n, then the inner angle sum is 180° × (n – 2).

→ When the number of sides increases by 1, the sum of the inner angles increases by 180°.

→ If the number of sides decreases by 1, then the angle sum decreases by 180°.

→ The sum of the inner angle and outer angle at the corner of a polygon is 180°.

Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus

→ Outer angle of a triangle when one side is extended is the sum of the inner angles of the other two corners.

→ Outer angle sum of a polygon is 360°.

→ One angle of a regular polygon of n sides is 180° × \(\frac{n-2}{n}\).

→ Outer angles are 360° × \(\frac {1}{n}\).

Polygons are closed figures with lines. The polygon having 3 sides is called a triangle. A polygon having 4 sides is a quadrilateral, a polygon with five sides pentagon.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Notes 1
Polygons are named based on the number of sides. The relation between sides and inner angle sum, sum of the inner angle and outer angle at a corner, sum of the outer angles, and angle around a point are discussed in this unit.

Names
A polygon is defined as a two-dimensional shape formed with straight lines. A polygon cannot have curves, and its lines should be fully connected. For example, triangles, rectangles, pentagons etc.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Notes 2
The polygons depend upon their sides. Triangles have three sides, quadrilaterals have four sides, and pentagons have five sides. There can be any number of sides for a polygon.

Regular polygon
If the polygon has sides of the same length and angles of the same size is called a regular polygon.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Notes 3

Angles
The sum of the three angles of a triangle is 180°.
The sum of all four angles of a rectangle is 360°.

In the case of a quadrilateral,
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Notes 4
If we draw a diagonal, then two of its angles are split into two. The quadrilateral is also now split into two triangles.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Notes 5
If we add up all the angles of both the triangles, it amounts to adding all the angles of the quadrilateral.
Therefore, the sum of the angles of a quadrilateral = 2 × the Sum of the angles of a triangle
= 2 × 180°
= 360°
That means the sum of the angles of a quadrilateral is 360°.
The sum of the angles of a pentagon is 360° + 180° = 540°
The sum of the angles of a hexagon is 540° + 180° = 720°

The sum of the angles of a polygon is 180° multiplied by two less than the number of sides.

The sum of the angles of a polygon of ‘n’ sides is (n – 2) × 180°.

For example: Find the sum of the angles of a polygon of 20 sides.
Answer:
(20 – 2) × 180° = 18 × 180°
= (182 × 10)°
= 3240°

Worksheet – 1

Question 1.
What is the sum of the angles of a polygon with 102 sides?
Answer:
Sum of angles = (n – 2) × 180°
= (102 – 2) × 180°
= 100 × 180°
= 18000°

Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus

Question 2.
The sum of angles of a polygon is 10440°. Then
(a) The sum of angles of a polygon with one more side = 10440 + ________ = ________
(b) The sum of angles of a polygon with one side less = 10440 – ________ = ________
Answer:
(a) 10440 + 180 = 10620°
(b) 10440 – 180 = 10260°

Question 3.
The sum of angles of a polygon is 6300°. How many sides does it have?
Answer:
Sum of angles = (n – 2) × 180°
∴ (n – 2) × 180° = 6300
⇒ n – 2 = \(\frac {6300}{180}\) = 35
⇒ n = 35 + 2
⇒ n = 37

Question 4.
The number of triangles formed by drawing diagonals from one vertex of a polygon is 18. Then
(a) What is the sum of the angles of the polygon?
(b) What is the number of sides of the polygon?
Answer:
(a) 18 × 180° = 3240°
(b) Number of sides = 18 + 2 = 20

Outer Angles
An angle formed outside the triangle at a vertex by extending a side is called an outer angle of the triangle. The angle at this vertex of the triangle is called an interior angle.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Notes 6
The sum of the inner and outer angles of a triangle is 180°.
In any triangle, the sum of the inner and outer angles at each vertex is 180°.

In any triangle, since the sum of all three inner angles is 180°, the sum of two inner angles is equal to the third inner angle subtracted from 180°. And this equals the outer angle at the third vertex.

The same reason holds, even if we change the inner angles. That is, in any triangle, the outer angle at any vertex is equal to the sum of the inner angles at the other two vertices.

If we take one outer angle at each vertex,
The sum of the outer angles of a triangle is 360°.
The sum of the outer angles of a quadrilateral is 360°.
The sum of the outer angles of a polygon is 360°.

In general, for a polygon with n vertices,
The sum of all angles is n × 180°
Sum of inner angles is (n – 2) × 180°
The sum of the outer angles is 360°.

The sum of the outer angles at any two vertices of a quadrilateral is equal to the sum of the inner angles at the other two vertices.

Worksheet – 2

Question 1.
Compute all outer angles and inner angles of the triangle and the sum of outer angles.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Notes 7
Answer:
Remaining inner angle = 180° – (77° + 58°)
= 180° – 135°
= 45°
Outer angles are = 180° – 58°, 180° – 77°, 180° – 45° = 122°, 103°, 135°
Sum of outer angle is = 122° + 103° + 135° = 360°

Question 2.
Compute all outer angles and inner angles of the quadrilateral and the sum of outer angles.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Notes 8
Answer:
Remaining inner angle of quadrilateral is = 360° – (87° + 97° + 90°)
= 360° – 274°
= 86°
or
180° – 94° = 86°
Outer angles are 180° – 86°, 180° – 87°, 180° – 97°, 180° – 90° = 94°, 93°, 83°, 90°
Sum of outer angles = 93° + 83° + 90° + 94° = 360°

Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus

Question 3.
Find x from the figure.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Notes 9
Answer:
From the figure.
∠ACB = 180° – 130° = 50°
∠BAC = 180° – (41 + 50) = 89°
∴ x = 180° – 89° = 91°
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Notes 10

Regular Polygons
Polygons with the lengths of sides equal and the sizes of angles equal are called regular polygons.
In a regular polygon of n sides
Each inner angle is \(\frac{n-2}{n}\) × 180°
Each outer angle is \(\frac {1}{n}\) × 360°

Worksheet – 3

Question 1.
In the figure, ABCD is a square. ‘O’ is the centre of the circle. Find the measure of ∠ACD, ∠AOD.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Notes 11
Answer:
∠ACD = ∠CAD
= \(\frac{\left(180^{\circ}-\angle \mathrm{ADC}\right)}{2}\)
= \(\frac{\left(180^{\circ}-90^{\circ}\right)}{2}\)
= \(\frac {90}{2}\)
= 45°
OA = OD
∴ ∠OAD = ∠ODA = 45°
∴ ∠AOD = 180° – (45° + 45°) = 90°

Question 2.
In the figure ABCDE is a regular pentagon ‘O’ is the centre of the circle.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Notes 12
Find
(i) ∠ABD
(ii) ∠ACD
(iii) ∠AOD
(iv) Is there any relation between these angles?
(v) Find ∠AED.
Answer:
(i) ABCDE is a regular pentagon.
So each angle will be \(\frac {540}{5}\) = 108°
also BC = DC means in ∠CBD = ∠CDB
= \(\frac{180^0-108^0}{2}\)
= \(\frac {72}{2}\)
= 36°
∴ ∠ABD = 108 – 36 = 72°

(ii) Considering ΔBAC, we will get
∠BAC = ∠BCA
= \(\frac{180^0-108^0}{2}\)
= \(\frac {72}{2}\)
= 36°
∴ ∠ACD = 108° – 36° = 72°

(iii) In a regular pentagon, if a circle passes through all the vertices, as from each vertex, if we join to the centre of the circle, all angles formed in the centre will be equal.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Notes 13
It will be \(\frac {360}{5}\) = 72°
So ∠AOD = 2 × 72° = 144°

(iv) Here ∠ABD = ∠ACD
also ∠AOD = 2 × ∠ABD = 2 × ∠ACD

(v) ∠AED = \(\frac {540}{5}\) = 108°

Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus

Question 3.
In the figure, ABCDEF is a regular hexagon. ‘O’ is the centre of the circle. Find
(i) ∠ABF
(ii) ∠ACF
(iii) ∠ADF
(iv) ∠AEF
(v) ∠AOF
(vi) Is there is any relation between these angles? Find ∠BDF and ∠BAF also.
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Notes 14
Answer:
(i) ∠BAF = \(\frac {720}{6}\) = 120° because ABCDEF is regular hexagon.
AB = AF, so ∠ABF = ∠AFB
= \(\frac{180^{\circ}-120^{\circ}}{2}\)
= 30°
∴ ∠ABF = 30°

(ii) From the figure
∠BCF = \(\frac {120}{2}\) = 60°
∠BCA = \(\frac{\left(180^{\circ}-120^{\circ}\right)}{2}\) = 30°
∴ ∠ACF = 60° – 30° = 30°

(iii) As above ∠ADF = 30°

(iv) ∠AEF = \(\frac{\left(180^{\circ}-120^{\circ}\right)}{2}\) = 30

(v) As we know, in the case of a regular pentagon, here
∠AOF = \(\frac {360}{6}\) = 60°
Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus Notes 15

(vi) ∠ABF = ∠ACF = ∠ADF
also ∠AOF = 2 × ∠ABF = 2 × ∠ACF = 2 × ∠ADF
∠BDF = 2 × 30° = 60°
∠BAF = \(\frac {720}{6}\) = 120°.

Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters

The comprehensive approach in SCERT Class 8 Basic Science Textbook Solutions Chapter 9 Cells and Cell Clusters Important Questions ensure conceptual clarity.

Cells and Cell Clusters Extra Questions and Answers Class 8 Basic Science Chapter 9 Kerala Syllabus

Cells and Cell Clusters Class 8 Important Questions

Question 1.
Choose the correct answer.
Statement 1: Theodor Schwann discovered that all plants are made of cells.
Statement 2: Matthias Jakob Schleiden discovered that all animals are made of cells.
(A) Statement 1 and 2 are correct
(B) Statement 1 and 2 are incorrect
(C) Statement 1 is correct Statement 2 is incorrect
(D) Statement 1 is incorrect Statement 2 is correct
Answer:
(B) Statement 1 and 2 are incorrect

Question 2.
Which statement is not true regarding the cell theory?
(A) Only animals are made of cells
(B) All living things are made of one or more cells.
(C) The cell is the basic unit of life.
(D) New cells are formed from pre-existing cells.
Answer:
(A) Only animals are made of cells

Question 3.
Complete the statement.
In sufficient light, the human eye can distinguish between two points 0.2 millimetres apart. This distance is known as the eye’s ………………………..
Answer:
Resolution

Question 4.
Which of the following statements is correct regarding electron microscopes?
(A) It helps in seeing up to a thousand times larger.
(B) It helps in detailed observation of living cells, viruses and molecular structures.
(C) It uses electron beam.
(D) It uses light.
Answer:
(B) and (C).

Question 5.
Choose the correct answer and write it down.
Which cannot be observed under a normal microscope.
(a) Mitochondria
(b) Virus
(c) Bacterium
(d) Paramecium
Answer:
(b) Virus

Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters

Question 6.
Identify and fill in the word pairs. Also, write the relationship between the word pairs.
The energy production centre of the cell: Mitochondria
Protein production centre: ………………..
Answer:
Ribosome, cell organelles and their functions.

Question 7.
Which one is isolated? Write the common characteristic of the others.
Mitochondria, lysosomes, ribosomes, and Golgi apparatus
Answer:
Lysosome, other organelles found in both plant and animal cells

Question 8.
What is the function of the plant tissue shown in the picture?
Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters 1
Answer:
Providing strength and support to plant parts (Sclerenchyma)

Question 9.
Who invented the electron microscope?
Answer:
German scientists Ernest Raska and Max Knoll invented the electron microscope

Question 10.
Choose and write the correct pairs.
(A) Matthias Jacob Schleiden discovered that all plants are made of cells
(B) Theodor Schwann – introduced the idea that new cells arise from existing cells
(C) Rudolf Virchow discovered that all animals are made of cells
(D) Ernst Ruska and Max Knoll invented the electron microscope.
Answer:
(A) Matthias Jacob Schleiden discovered that all plants are made of cells and (D) Ernst Ruska and Max Knoll invented the electron microscope.

Question 11.
Complete the illustration.
Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters 2
Answer:
(a) All organisms are made up of one or more cells.
(b) New cells originate from pre-existing cells.

Question 12.
List the statements given below.
(A) Multiple lenses are used.
(B) Objects can be seen up to ten times larger.
(C) Only one lens is used.
(D) Objects can be seen up to a thousand times larger.

Simple microscope Compound microscope

Answer:
Simple microscope: (B) and (C).
Compound microscope: (A) and (D)

Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters

Question 13.
Completed the table.

Feature Compound microscope Electron microscope
The need for light ……(a)…… An electron beam is used instead of light.
Magnification Power Up to 1000 times. ……(b)……
Lens ……(c)…… ……(d)……

Answer:
(a) Light is needed,
(b) Magnifies objects more than a million times,
(c) Using multiple lenses,
(d) Electromagnets are used.

Question 14.
What is the role of the following in preparing the object of observation?
(a) Stain
(b) Glycerin
Answer:
(a) Stain – to give colour to the object of observation
(b) Glycerin – to prevent the object of observation from drying out

Question 15.
If there is a mistake in the statements given below, correct the mistake by changing only the underlined word.
a) The cell wall is the hard outer layer of the cell membrane.
b) The cell membrane provides protection and shape to the cell.
c) The cell wall in plants is mainly made of cellulose.
d) Substances enter and leave the cell through the Golgi apparatus.
Answer:
b) The cell wall provides protection and shape to the cell.
d) Substances enter and leave the cell through the cell membrane.

Question 16.
Write the differences between cytoplasm and protoplasm.
Answer:
Cytoplasm is a jelly-like fluid that fills the cell. It maintains all the organelles in their place and serves as the medium for chemical reactions.
Protoplasm consists of all the components inside the cell membrane, including the nucleus and cytoplasm.

Question 17.
Observe the picture and answer the questions below.
Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters 3
(a) Which organelle is represented by the picture?
(b) What is the function of this organelle?
Answer:
(a) Endoplasmic reticulum

(b) These organelles appear as a network of tubules within the cell, serve as pathways for conducting materials. It helps in the synthesis and removal of materials required by the cell.

Question 18.
Observe the picture and answer the questions below.
Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters 4
(a) Which is the complex tissue represented by the picture?
(b) What is the function of this complex tissue?
Answer:
(a) Xylem
(b) The conduction of water and salt to the leaves takes place through the xylem.

Question 19.
Choose the answer to the statements given below from the box and write it down.

Muscle tissue, Connective tissue, Epithelial tissue, Neural tissue

a) Connects different parts of the body.
b) Controls and coordinates bodily functions.
Answer:
a) Connective tissue
b) Neural tissue

Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters

Question 20.
Observe the picture and answer the questions below.
Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters 5
(a) Which are the organelles indicated by A and B in the picture?
(b) What is the function of these organelles?
Answer:
(a) A – Mitochondria, B – Centriole
(b) A – Mitochondria – Energy production centre of the cell. It stores the energy obtained from the oxidation of glucose and distributes it as needed.
B – Centriole – Helps in cell division

Question 21.
Copy the picture and identify and mark the parts according to the statements given below.
Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters 6
(a) The control center of the cell.
(b) The energy production center of the cell.
Answer:
Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters 7

Question 22.
Arrange the items in the box in order of their size.

Red blood cell, protein, animal cell, mitochondria, paramecium, virus

Answer:
Protein, virus, Mitochondria, red blood cell, animal cell, paramecium.

Question 23.
Choose the answer to the statements given below from the box and write it down.

Theodor Schawnn, Ernst Ruska, Anton van Leeuwenhoek, Max Knoll, Robert Hooke, Rudolf Virchow, Matthias Jakob Schleiden

(a) Invented the electron microscope.
(b) Small parts like boxes that are stacked are called ‘cells’, meaning chambers.
(c) Found that all plants are made of cells.
(d) Introduced the idea that new cells arise from existing cells.
(e) Observed water taken from a pond with the help of a better microscope and discovered microorganisms in it.
(f) Found that all animals are made of cells
Answer:
(a) Ernst Ruska, Max Knoll
(b) Robert Hooke
(c) Matthias Jakob Schleiden
(d) Rudolf Virchow
(e) Anton van Leeuwenhoek
(f) Theodor Schwann.

Question 24.
From the statements given below, select and write the one related to the compound microscope.
(a) Requires light.
(b) Magnifies objects more than a million times.
(c) Uses electron beams
(d) Magnifies up to a thousand times.
(e) Uses multiple lenses
(f) Uses electromagnets.
Answer:
(a) Requires light, (d) Magnifies up to a thousand times,(e) Uses multiple lenses

Question 25.
Observe the picture and answer the questions below.
Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters 8
(a) What does the picture indicate?
(b) What are the parts indicated by A and B?
(c) What are the characteristics of the part indicated by A?
Answer:
(a) Nucleus

(b) A – Chromatin reticulum B – Nucleolus

(c) Characteristics of chromatin reticulum: Chromatin network is the structure that appears like a network of threads within the nucleoplasm. During cell division, these chromatin networks condense into chromosomes.

Question 26.
Complete the illustration.
Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters 9
Answer:
(a) Chloroplast,
(b) They contain pigments that impart red, orange and yellow colours. They attract the animals for pollination and seed dispersal,
(c) Leucoplasts.

Question 27.
List the statements given below.
(a) Thin cell wall. Intercellular spaces are present.
(b) Cell wall made of substances like cellulose and pectin.
(c) Helps maintain the shape of plant parts.
(d) Cells in some parts have the ability to divide.
(e) Performs functions like photosynthesis and food storage.
(f) The cell wall is thicker only in some parts.

Collenchyma Parenchyma

Answer:

Collenchyma Parenchyma
(b), (c), (f), (a), (d), (e).

Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters

Question 28.
What are the different types of meristematic tissues?
Answer:
Meristems are actively dividing cells that are of three major types: apical meristem, lateral meristem and intercalary meristem. The apical meristem is seen at the tip of the root and stem. Those seen at the lateral sides are called the lateral meristem. In monocot plants, the meristem seen between two nodes is called the intercalary meristem.

Question 29.
Observe the picture and answer the questions below.
Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters 10
(a) Which animal tissue is represented in the picture?
(b) What are the functions of this animal tissue?
Answer:
(a) Neural tissue
(b) Makes impulse transmission possible. It controls and coordinates bodily activities.

Question 30.
Stem cells have special importance in modern medicine.
(a) What are stem cells?
(b) What are the characteristics of stem cells?
Answer:
(a) Stem cells are specialised cells that can develop into various types of cells such as muscle cells, nerve cells, blood cells, etc.

(b) Stem cells have the ability to create new types of cells. Hence, they are referred to as the body’s master cells. Stem cells help to eliminate damaged cells, promote the growth of new cells and maintain healthy tissues. In modem medicine, stem cells hold significant importance.

Question 31.
What was the contribution of Robert Hooke and Anton van Leeuwenhoek to cell biology?
Answer:
Robert Hooke, a scientist, was observing a thin piece of cork through his microscope. He observed small parts like boxes stacked together. He called them ‘cells’, meaning chambers. Anton van Leeuwenhoek observed water taken from a pond with the help of a much better microscope and discovered microorganisms in it.

Question 32.
What is cell biology?
Answer:
Cell biology is the study of cells.

Question 33.
Write down the scientists who made important discoveries in the field of cell biology and their discoveries.
Answer:
HISTORY OF CELL BIOLOGY

  • Cell biology is the study of cells.
  • Matthias Jakob Schleiden discovered that all plants are composed of cells.
  • Theodor Schwann discovered that all animals are composed of cells.
  • Rudolf Virchow introduced the concept that new cells are formed from pre-existing cells.

Cell Theory

    • All organisms are made up of one or more cells.
    • The cell is the basic unit of life.
    • New cells originate from pre-existing cells.
  • Different types of microscopes are used to observe very small or tiny cells. In simple microscopes, lenses are used to magnify objects.
  • The human eye can distinguish two spots having a distance of 0.2 mm in adequate light. This distance is called the resolution of the eye.
  • A lens is required to distinguish spots having a distance of less than 0.2 mm.
  • A simple microscope in which a single lens is used can magnify an object upto 10 times than its original size.
  • A compound microscope, in which more than one lens is used, can magnify an object up to 1000 times.
  • The magnification power of a lens is its ability to magnify objects. If eye piece lens magnifies the object to 10 x (10 times) and objective lens magnifies it into 40 x (40 times), the magnification power of the lens is 400 x.

Electron microscope

  • An instrument that magnifies objects more than a million times.
  • It helps to observe cells, viruses and molecular structure in detail.
  • In an electron microscope electron beam is used instead of light.
  • In 1934, German scientists Ernest Raska and Max Knoll invented the electron microscope.
  • In an electron microscope, electromagnetic lenses are used to focus a beam of electrons onto the specimen, allowing for much higher resolution imaging compared to light microscopes.

Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters

Question 34.
What is cell theory?
Answer:
In the 19th century, the cell theory was formed by synthesizing all the observations by various scientists. It is the basic theory of cell biology.

Question 35.
What are the ideas in cell theory?
Answer:
Cell Theory

    • All organisms are made up of one or more cells.
    • The cell is the basic unit of life.
    • New cells originate from pre-existing cells.
  • Different types of microscopes are used to observe very small or tiny cells. In simple microscopes, lenses are used to magnify objects.
  • The human eye can distinguish two spots having a distance of 0.2 mm in adequate light. This distance is called the resolution of the eye.
  • A lens is required to distinguish spots having a distance of less than 0.2 mm.
  • A simple microscope in which a single lens is used can magnify an object upto 10 times than its original size.
  • A compound microscope, in which more than one lens is used, can magnify an object up to 1000 times.
  • The magnification power of a lens is its ability to magnify objects. If eye piece lens magnifies the object to 10 x (10 times) and objective lens magnifies it into 40 x (40 times), the magnification power of the lens is 400 x.

Question 36.
What are the uses of lenses in microscopes?
Answer:
Various types of microscopes are used to observe microscopic cells. In ordinary microscopes, lenses help to magnify objects.

Question 37.
What is resolution?
Answer:
In sufficient light, the human eye can distinguish between two points 0.2 millimetres apart. This distance is the resolution of the eye. A lens is required to distinguish points less than 0.2 millimetres apart.

Question 38.
How is a simple microscope different from a compound microscope?
Answer:
A simple microscope, which uses only a single lens, can magnify objects up to ten times. However, a compound microscope, which uses multiple lenses, can magnify objects up to a thousand times.

Question 39.
What is the magnification power?
Answer:
The Magnification power of a lens is its ability to magnify objects. If eye piece lens magnifies the object to 10 x (10 times) and objective lens magnifies it into 40 x (40 times), the magnification power of the lens is 400 x.

Question 40.
What are the features of an electron microscope?
Answer:
An electron microscope is an instrument that magnifies objects more than a million times. It helps to observe cells, viruses and molecular structure in detail. In an electron microscope electron beam is used instead of light. In 1934, German scientists Ernest Raska and Max Knoll invented the electron microscope. In an electron-microscope, electromagnetic lenses are used to focus a beam of electrons onto the specimen, allowing for much higher resolution imaging compared to light microscopes.

Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters

Question 41.
What are the different types of electron microscopes in use today?
Answer:
a) Transmission electron microscope (TEM): Sends electrons through a very thin sample. Provides 2D images.

b) Scanning electron microscope (SEM): Sends electrons across the surface of an object. Provides 3D images.

c) Scanning Transmission Electron Microscope (STEM): Combines TEM and SEM techniques. Provides high-quality images.

Question 42.
What is a cell wall? Write its characteristics.
Answer:
The cell wall is the tough outer layer of the cell membrane. It provides protection and shape to the cell. The cell wall in plants is made primarily of a substance called cellulose.

Question 43.
Which is known as the cell membrane or plasma membrane?
Answer:
The cell membrane is a thin, flexible layer that surrounds the cell.

Question 44.
What is the function of the cell membrane?
Answer:
Substances enter and exit the cell through the cell membrane.

Question 45.
Why plasma membrane is known as a semi-permeable membrane?
Answer:
The plasma membrane does not allow all substances to pass through. Therefore, the plasma membrane is known as the semi-permeable membrane.

Question 46.
What are cell organelles?
Answer:
Organelles are the parts found in the cytoplasm that perform functions necessary for the survival of the cell. The main organelles are the cytoplasm, Golgi apparatus, endoplasmic reticulum, and mitochondria.

Question 47.
Prepare a note about the nucleus of a cell.
Answer:
The Nucleus is the centre that controls the cell. Chromatin network is the structure that appears like a network of threads within the nucleoplasm. During cell division, these chromatin networks condense into chromosomes. There is also a part called nucleolus within the nucleus.

Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters

Question 48.
What are the characteristics and functions of parenchyma?
Answer:
They contain living cells, a thin cell wall, and intercellular spaces are present. Perform functions such as photosynthesis and food storage. Cells in some parts have the ability to divide.

Question 49.
List the characteristics and functions of collenchyma.
Answer:
Collenchyma tissues are made up of living cells. The cell wall is made of substances such as cellulose and pectin. The cell wall is thicker only in some parts. Its function is to help maintain the shape of plant parts.

Question 50.
Describe the characteristic features and functions of sclerenchyma tissues.
Answer:
Sclerenchyma is a tissue composed of dead cells, characterised by a thick cell wall. The cell wall is of uniform thickness throughout, providing strength and support to plant parts.

Question 51.
Make a note of the differences between simple tissues and complex tissues.
Answer:
Tissues that are composed of similar types of cells are known as simple tissues. Parenchyma, collenchyma, and sclerenchyma are examples of simple tissues.

Xylem and phloem are made up of cells of different shapes and sizes. Therefore, they are called complex tissues. The transport of water and salts to the leaves takes place through the xylem. The phloem tissues transport the food produced by the leaves to various parts.

Question 52.
What are permanent tissues? Give some examples.
Answer:
Cells that have lost the ability to divide are called permanent tissues. Parenchyma, sclerenchyma, collenchyma, xylem, and phloem are examples of permanent tissues.

Question 53.
Plants like bamboo do not gain weight. Why?
Answer:
Lateral meristem helps in gaining weight. Lateral meristem is not found in plants like bamboo. Therefore, plants like bamboo do not gain weight.

Question 54.
What are the different types of animal tissues?
Answer:
Epithelial tissue, connective tissue, muscle tissue, and neural tissue.

Class 8 Basic Science Chapter 9 Important Questions Kerala Syllabus Cells and Cell Clusters

Question 55.
What are stem cells? What is their speciality?
Answer:
Stem cells are specialized cells in the body that can develop into different types of cells, such as muscle cells, nerve cells, and blood cells. They are called the master cells of the body because they can create new types of cells. Stem cells help new cells to grow by replacing worn-out cells and heal wounds.

Question 56.
Write the characteristic features of different types of animal tissues.
Answer:

Animal tissues Characteristic features
Epithelial tissue • Covers and protects the surfaces of the body and internal organs.
• Helps in the absorption of various substances.
• It produces secretions like mucus.
Connective tissue • Provides support to various parts of the body.
• Connects various parts together.
• Bone, blood, fibrous tissue, etc. are connective tissues.
Muscle tissue • Helps in body movement and locomotion.
• Composed of cells capable of contraction and relaxation.
Neural tissue • Makes impulse transmission possible.
• Controls and coordinates bodily activities.

Class 8 Basic Science Chapter 8 Important Questions Kerala Syllabus Origin of Life, Origin of Living Things

The comprehensive approach in SCERT Class 8 Basic Science Textbook Solutions Chapter 8 Origin of Life, Origin of Living Things Important Questions ensure conceptual clarity.

Origin of Life, Origin of Living Things Extra Questions and Answers Class 8 Basic Science Chapter 8 Kerala Syllabus

Origin of Life, Origin of Living Things Class 8 Important Questions

Question 1.
Choose the correct answer.
Statement 1: Fossil studies also show that Pakicetus, the ancestor of the whales, had the physical characteristics of a wolf, as well as a body structure suitable for swimming.
Statement 2: The ancestor of the Whales was an ancient mammal called Pakicetus, which lived 60 million years ago.
(a) Statements 1 and 2 are correct.
(b) Statements 1 and 2 are wrong
(c) Statement 1 is true and Statement 2 is false
(d) Statement 1 is false and Statement 2 is true
Answer:
(c) Statement 1 is true and Statement 2 is false

Question 2.
Scientists who advanced the theory of chemical evolution
(a) Harold Urey, Stanley Miller
(b) Oparin, Haldane
(c) Sidney Fox, Joan Oro
(d) Watson, Crick
Answer:
(b) Oparin, Haldane

Question 3.
Complete the statement.
The ancestor of all whale species, including the blue whale, was an ancient mammal called …………………….
Answer:
Pakicetus

Question 4.
From the given below, select and write the one related to prokaryotes.
(a) Fewer organelles
(b) Organelles with membrane
(c) membrane bound nucleus
(d) membrane bound nucleus absent
(e) More organelles
Answer:
(a) and (d)

Question 5.
Identify and complete the word pairs. Also, write the relationship between the word pairs.
4 – 4.6 billion years : Origin of the Earth:: 2.5 – 541 million years : ……………………….
Answer:
Multicellular organisms, such as eukaryotes, timeline and animals

Class 8 Basic Science Chapter 8 Important Questions Kerala Syllabus Origin of Life, Origin of Living Things

Question 6.
How did life originate on Earth?
Answer:
Scientists believe that life originated on Earth about 3.5 billion years ago.

Question 7.
What are the theories related to the origin of life?
Answer:
Theories related to the origin of life: Panspermia theory, Chemical evolution theory

Question 8.
What are the characteristics of prokaryotes?
Answer:
Few organelles, no membrane-bound organelles, and no membrane-bound nucleus.

Question 9.
Over time, complex organisms evolved into the plants and animals we see today.
(a) When is life thought to have originated on Earth?
(b) What conditions led to the formation of complex organisms?
Answer:
(a) Scientists believe that life originated on Earth about 3.5 billion years ago.
(b) Some of the simpler organisms released oxygen through photosynthesis, which led to the formation of more complex organisms.

Question 10.
Complete the list.

Theory Explanation
Panspermia hypothesis …………..(a)…………………
…………….(b)…………… Life emerged as a result of the chemical reactions that took place in the ocean under unique conditions on the early Earth.

Answer:
(a) Life originated on another planet and arrived on Earth by chance as microorganisms or as particles. These tiny particles of life are called panspermia.
(b) Chemical evolution theory

Question 11.
The names of the scientists who made contributions that validate how the primitive cell was formed through the process of chemical evolution are given below. Write their contributions.
(a) Sidney Fox
(b) Joan Oro
Answer:
Sydney Fox: Demonstrated that molecules similar to proteins can be artificially produced
Joan Oro: The nitrogenous base adenine, one of the main building blocks of nucleic acids, was artificially produced.

Question 12.
List the statements given below.
(a) More organelles
(b) Organelles are not membrane-bound
(c) No membrane-bound nucleus
(d) There is a membrane-bound nucleus

Prokaryotic cell Eukaryotic cell

Answer:
Prokaryotic cell: (b) and (c), Eukaryotic cell: (a) and (d).

Class 8 Basic Science Chapter 8 Important Questions Kerala Syllabus Origin of Life, Origin of Living Things

Question 13.
Explain the theory of chemical evolution?
Answer:
The theory of chemical evolution explains that life arose as a result of the combination of chemicals in the ocean under unique conditions on early Earth. Although many theories have emerged over time to explain the origin of life, the theory of chemical evolution is the one that-is most supported by evidence and has received the most acceptance in the scientific world.

Question 14.
Observe the illustration and answer the questions below.
Class 8 Basic Science Chapter 8 Important Questions Kerala Syllabus Origin of Life, Origin of Living Things 1
(a) Which process does the illustration indicate?
(b) Explain the process that the illustration indicates.
Answer:
(a) Formation of organelles found in cells.

(b) Eukaryotic cell engulfs small aerobic bacteria. Instead of digesting the small cell, it protects it. Gradually, these become mitochondria.
Eukaryotic cell engulfs small photosynthetic bacteria. Instead of digesting the photosynthetic bacteria, it – protects it. Gradually, these become chloroplasts.

Question 15.
The stages in which various living things evolved from prokaryotes are given below. Write them in chronological order.
(a) Dinosaurs
(b) Origin of the Earth
(c) Emergence of mammals, origin of humans
(d) Multicellular organisms called eukaryotes
(e) Unicellular prokaryotes with simple structures
(f) Plants and animals on land
Answer:
(b), (e), (d), (f), (a), (c)

Question 16.
What is the Cosmic Calendar?
Answer:
The cosmic calendar is a depiction of the chronology from the creation of the universe to the emergence of man in an easy-to-understand way. If a calendar were prepared assuming the age of the universe to be just one year, one second in that calendar would be equivalent to approximately 438 years.

Question 17.
Choose and write the correct pairs.
(a) 4 – 4.6 billion years ago: Origin of the Earth
(b) 3.5 – 2.5 billion years ago: Multicellular organisms called eukaryotes
(c) 541 – 252 million years ago: Plants and animals on land
(d) 252 – 66 million years ago: Single-celled prokaryotes with simple structures
(e) 66 million years ago to the present: Emergence of mammals, origin of humans
Answer:
(a), (c), (e).

Question 18.
Observe the illustration and answer the questions below.
Class 8 Basic Science Chapter 8 Important Questions Kerala Syllabus Origin of Life, Origin of Living Things 2
(a) Who performed the experiment shown in the illustration?
(b) What is the source of energy in this experiment?
(c) How does this experiment validate the theory of chemical evolution?
Answer:
(a) Harold Urey and Stanley Miller

(b) Electric spark

(c) This experiment was conducted by artificially recreating the conditions of the primitive Earth in the laboratoiy. This experiment was able to produce amino acids, which are organic molecules. This experiment proved that organic molecules can be formed from inorganic components under suitable conditions.

Question 19.
Observe the illustration and answer the questions below.
Class 8 Basic Science Chapter 8 Important Questions Kerala Syllabus Origin of Life, Origin of Living Things 3
(a) Which theory related to the origin of life is indicated in the illustration?
(b) Explain the theory.
(c) Complete the illustration.
Answer:
(a) Chemical evolution theory
(b) The chemical evolution theory explains that life emerged as ,a result of the combination of chemicals in the ocean under special conditions on the early Earth.
(c) A – Sunlight, lightning, ultraviolet rays, volcanic eruptions B – The complex molecule called nucleic acid and the lipid envelope are formed.

Class 8 Basic Science Chapter 8 Important Questions Kerala Syllabus Origin of Life, Origin of Living Things

Question 20.
Complete the list.

Scientist Contributions
…………..(a)………….. They together proved that the fundamental units responsible for the origin of life can form from simple gases.
Sidney Fox …………..(b)…………..
Joan Oro …………..(c)…………..

Answer:
(a) Harold Urey, Stanley Miller
(b) Proved that molecules similar to proteins can be synthesised artificially.
(c) Adenine, one of the key building blocks of nucleic acids, was artificially synthesised.

Question 21.
Observe the picture and answer the questions below.
Class 8 Basic Science Chapter 8 Important Questions Kerala Syllabus Origin of Life, Origin of Living Things 4
(a) Which cell is indicated in the figure?
(b) What are the characteristics of this cell?
Answer:
(a) Eukaryotic cell
(b) More organelles, membrane-bound organelles, and a membrane-bound nucleus.

Question 22.
Arrange based on the cosmic calendar

Land plants, eukaryotes, mammals, vertebrates, humans, dinosaurs

Answer:
Eukaryotes, vertebrates, land plants, dinosaurs, mammals, humans

Question 23.
Explain the evolutionary history of the whale.
Answer:
The ancestor of whales, including the blue whale, was an ancient mammal called Pakicetus, which lived 50 million years ago. Fossil studies show that this creature had the physical characteristics of a wolf, as well as a body structure suitable for swimming. As its descendants spent more time in the water, their legs evolved to resemble paddles. The evolutionary history of whales is evidence that organisms change as part of their adaptation to changing conditions. All living things, including humans, have a fascinating history of this kind of evolution.

Question 24.
Who are the scientists who made contributions that validate the formation of the primitive cell through the process of chemical evolution?
Answer:
Harold Urey, Stanley Miller: They together proved that the fundamental units responsible for the origin of life can form from simple gases.
Sydney Fox: Proved that molecules similar to proteins can be synthesised artificially.
Joan Oro: Adenine, one of the key building blocks of nucleic acids, was artificially synthesised.

Class 8 Basic Science Chapter 8 Important Questions Kerala Syllabus Origin of Life, Origin of Living Things

Question 25.
Explain the Harold Urey and Stanley Miller experiments.
Answer:
Among the many experimental evidence supporting the theory of chemical evolution, the most notable is the Urey-Miller experiment. This experiment was conducted in a laboratory by artificially recreating the conditions of the primitive Earth. This experiment was able to produce amino acids, which are organic molecules. This experiment proved that organic molecules can form from inorganic components under suitable conditions.
Class 8 Basic Science Chapter 8 Important Questions Kerala Syllabus Origin of Life, Origin of Living Things 5

Question 26.
What changes occurred in the primitive cell?
Answer:
The primitive cell is composed of nucleic acids capable of self-replication and a lipid membrane that surrounds it. Cells such as bacteria evolved from this primitive cell. These first, simple organisms are called prokaryotes.

Question 27.
How did complex organisms evolve from primitive cells?
Answer:
About 3.8 billion years ago, primitive cells evolved from molecules in Earth’s oceans. Prokaryotic cells evolved 3.5 billion years ago. About 2.5 billion years ago, photosynthesis began, bringing oxygen to the atmosphere. Later, eukaryotic cells with organelles such as mitochondria and chloroplasts evolved. Then came simple multicellular organisms, and about 800 million years ago, more complex organisms had formed.

Class 8 Basic Science Chapter 7 Important Questions Kerala Syllabus Let’s Cultivate and Reap Goodness

The comprehensive approach in SCERT Class 8 Basic Science Textbook Solutions Chapter 7 Let’s Cultivate and Reap Goodness Important Questions ensure conceptual clarity.

Let’s Cultivate and Reap Goodness Extra Questions and Answers Class 8 Basic Science Chapter 7 Kerala Syllabus

Let’s Cultivate and Reap Goodness Class 8 Important Questions

Question 1.
Choose the correct answer.
Statement 1: Statement 1: Vertical farming is an innovative agricultural method in which crops are grown in vertically stacked layers.
Statement 2: Vertical farming can be implemented even in urban areas with limited space.
(a) Statements 1 and 2 are correct.
(b) Statements 1 and 2 are wrong
(c) Statement 1 is true and Statement 2 is false
(d) Statement 1 is false and Statement 2 is true
Answer:
(a) Statements 1 and 2 are correct

Question 2.
Which one is isolated? Write the common characteristic of the others.
Potassium, zinc, copper, manganese
Answer:
Potassium, others are micronutrients

Question 3.
Complete the statement.
The taste, colour, smell, and nutritional value of the products will vary depending on the characteristics of the land where they are cultivated. Based on this, agricultural products produced in certain regions receive …………………..
Answer:
Geographical indication status

Question 4.
Identify and match the word pairs. Also, write the relationship between the word pairs.
Azospirillum: Biofertiliser:: Urea: ……………………..
Answer:
Artificial fertiliser, fertilisers and their characteristics.

Question 5.
Complete the table.

Irrigation method Peculiarity
Drip irrigation ………….(a)……………
………….(b)………….. A method of directly supplying water to the roots of plants through a cotton cloth filter from a water source is called wick irrigation. It requires less water than drip irrigation.

Answer:
(a) This is an irrigation method using pipes and valves to water the roots in drops. By doing this, the loss of water is reduced to maximum. Another significance of this method is that the availability of water to all plants is ensured.

(b) Wick Irrigation

Class 8 Basic Science Chapter 7 Important Questions Kerala Syllabus Let’s Cultivate and Reap Goodness

Question 6.
Which are the correct pairs?
a) Pod borer – Legume
b) Leaf-rolling Caterpillar – Brinjal
c) Stem Borer – Okra
d) Pod borer – Brinjal
e) Leaf-rolling Caterpillar – Okra
Answer:
(a), (e).

Question 7.
Which pair is correct?
a) Online system – Disease and pest information collection, solution – Up-to-date knowledge about climate change – Market opportunities
b) Family members – financial assistance to farmers, procurement, marketing
c) Local self Government Institutions – Financial Assistance, Employment Assistance, Psychological Support.
d) Cooperative societies – Use of barren lands – Approval – Government assistance
e) Krishi Bhavan – Distribution of best planting materials – Scientific guidance and training programs for agriculture
Answer:
(a), (e).

Question 8.
Select and write down the possibilities of overcoming space constraints from those given below.

Sack fanning, smart farming, aquaponics, greenhouse, vertical farming, terrace farming, tissue culture, mulching

Answer:
Sack farming, aquaponics, vertical farming, and terrace farming.

Question 9.
Observe the picture and answer the questions below.
Class 8 Basic Science Chapter 7 Important Questions Kerala Syllabus Let’s Cultivate and Reap Goodness 1
(a) Which farming method is shown in the picture?
(b) What is the advantage of this farming method?
Answer:
(a) Pet bottle farming
(b) To maximise the use of space

Question 10.
Select the ones related to artificial fertilisers from the ones given below.
a) Does not harm soil structure or decomposers.
b) A much smaller amount is sufficient compared to other fertilisers.
c) Plants can easily absorb it.
d) Urea is an example
e) Plants get the nutrients released from these bio residues by the decomposers.
f) It is artificially produced and used to increase the nutrient value of the soil.
g) Excessive fertilizer application can harm soil structure and decomposers.
h) Microorganisms are added to the soil to help fix nitrogen and phosphorus.
Answer:
(c), (d), (f), (g)

Question 11.
Observe the newspaper article and answer the questions below.

GM crops as hope for farmers…
Kochi – GM crops are reaching farmers to increase their income…..

(a) What are GM crops?
(b) Write examples of GM crops?
Answer:
(a) GM crops are genetically modified crops. Genetic engineering allows for the incorporation of new traits into crops by altering their genetic structure.

(b) Examples of such crops include cotton, which is resistant to pests, soybeans, which are resistant to herbicides, and rice, which is rich in vitamin A.

Question 12.
What is tissue culture?
Answer:
This is a technology that helps produce large numbers of plants with the same characteristics as the parent plant. Seedlings are produced by separating tissues from suitable parts of the plant and growing them in a special nutrient medium.

Class 8 Basic Science Chapter 7 Important Questions Kerala Syllabus Let’s Cultivate and Reap Goodness

Question 13.
Differentiate between drip irrigation and wick irrigation.
Answer:
Drip irrigation – This is an irrigation method that uses pipes and valves to deliver water in droplets to the root zone. This minimises water loss and ensures that all plants receive the same amount of water.
Wick irrigation – This method delivers water directly from the water source to the root zone of plants using a cotton wick. It requires less water than drip irrigation.

Question 14.
(a) What are the characteristics of the following agricultural products?

Palakkadan Matta rice, Malabar pepper, Kuttiyattur mango

(b) On what basis are such agricultural products special?
Answer:
(a) They have been granted geographical indication status.

(b) Depending on the characteristics of the land where it is cultivated, the taste, colour, smell, and nutritional value of the products will vary. Based on this, agricultural products produced in certain regions are given Geographical Indication status.

Question 15.
What is the role of the following in making agriculture a success story?
(a) Online system
(b) Family members
(c) Local self-government bodies
Answer:
(a) Online system – disease and pest information collection, solutions, real-time information on climate change, market opportunities
(b) Family members – Financial assistance – Employment assistance – Psychological support
(c) Local self-government bodies – Use of barren lands, approval, government assistance.

Question 16.
List the following. Give a title to the list.
Magnesium, chlorine, nickel, sulfur

…………………………………. ……………………………………

Answer:

Macronutrients Micronutrients
Magnesium, Sulfur Chlorine, Nickel

Question 17.
Write the similarities and differences.
• Nano fertilisers
• Artificial fertilisers
Answer:
Similarity – Both are artificial fertilisers. They can be easily absorbed by plants.

Difference – Nano fertilisers: Artificial fertilisers in the form of nanoparticles. Much smaller amounts are required compared to other fertilisers. They do not change the chemical composition of the soil like artificial fertilisers.

Artificial fertilisers: Excessive application of fertilisers can harm the soil structure and disrupt the activities of decomposers.
Class 8 Basic Science Chapter 7 Important Questions Kerala Syllabus Let’s Cultivate and Reap Goodness 2

Question 18.
If there is a mistake in the underlined part of the statements given below, correct it and write it down.
(a) Nanofertilisers added microorganisms to the soil that help in the fixation of nitrogen and phosphorus.
(b) When using organic fertilisers, plants receive nutrients that decomposers release from them.
(c) Organic fertilisers are sufficient in very small quantities compared to other fertilisers.
(d) Ammonium sulphate is an artificial fertiliser.
(e) Excessive application of biofertilisers can harm soil structure and decomposers.
Answer:
(a) Biofertilizers are fertilizers that add microorganisms to the soil that help fix nitrogen and phosphorus.
(c) Nano-fertilisers are only needed in very small quantities compared to other fertilisers.
(e) Excessive application of artificial fertilisers can harm soil structure and decomposers.

Question 19.
Native varieties are beneficial for the environment. Justify the statement.
Answer:
Native varieties are adapted to the local environment and can be grown in harmony with it. They are naturally resistant to pests and diseases in the area, which reduces the use of pesticides. They also help maintain the ecological balance by supporting local pollinators. They also require less fertiliser, which can reduce costs.

Class 8 Basic Science Chapter 7 Important Questions Kerala Syllabus Let’s Cultivate and Reap Goodness

Question 20.
A farmer raises cattle, chicken, duck, fish, and rice.
(a) What is this fanning method known as?
(b) What are the advantages of this fanning method?
Answer:
(a) Integrated farming

(b) Multiple sources of income – Farmers can earn money from multiple activities, not just from one crop. Even if one crop fails, they can earn income from other crops.
Efficient use of land and resources – Waste from one activity can be used for another by doing different activities in the same area.

Question 21.
Complete the illustration.
Answer:
(a) By sharing images and information related to crops, you can find solutions to problems with the help of artificial intelligence.

(b) Plants are grown in nutrient solutions, and with the help of sensors, the amount of nutrients is detected and provided as needed.

(c) Aeroponics

Question 22.
What is the government aiming to achieve by giving agricultural awards to children?
Answer:
The government gives such awards to encourage children to develop an interest in agriculture and understand the importance of agriculture in our lives. Such agricultural awards /make children proud of learning and participating in agriculture. Recognise children who farm at school or at home and motivate them to do more farming. It helps students understand why agriculture is important for food and the environment. Young minds may have new and creative ideas for agriculture. Awards help encourage such ideas. It teaches teamwork, responsibility, patience, and care for nature. Agriculture is a part of our culture and heritage. These awards help children appreciate it.

Question 23.
What are some agricultural awards in Kerala?
Answer:

  • Karshakottama Award for the best farmer
  • Karshaka Tilakam Award for the best woman farmer
  • Kerakesari Award for the best coconut farmer
  • Samashakti Award for the best agricultural worker
  • Yuva Karshaka Award for young farmers
  • Karshaka Pratibha Award for the best student farmer

Question 24.
What are the government-led programs to help new entrepreneurs?
Answer:
Government of India:

  • Startup India
    Objective – Supporting new business ideas
  • Pradhan Mantri Mudra Yojana
    Objective – Providing financial assistance to start small enterprises.
  • Atal Innovation Mission
    Objective – Encouraging creative and innovative ideas among children in schools and colleges.
  • Made in India
    Objective – Developing India as a manufacturing hub.

Government of Kerala

  • Kerala Start-up Mission
    Aim – To make Kerala a start-up friendly state
  • Young Innovators Program
    Aim – To identify and encourage innovative ideas among school and college students
  • Kudumbashree Mission
    Aim – Women empowerment

Question 25.
What are the Various farming methods to maximise the use of space for farming?
Answer:
Sack farming, aquaponics, Pet bottle farming, vertical farming, pot cultivation, and terrace farming.

Question 26.
What are the limitations and possibilities of various farming methods to maximise the use of land for agriculture?
Answer:

  • Sack farming: Potential: Helps grow vegetables in small spaces using old sacks.
    Limitations: Requires regular watering, cannot grow crops with deep roots.
  • Aquaponics: Potential: Grows plants and fish together, saves water and space.
    Limitations: Requires technical knowledge and regular maintenance.
  • Pet bottle farming: Potential: Reuses waste bottles to grow small plants in limited spaces.
    Limitations: Only suitable for small plants with shallow roots.
  • Vertical Farming: Potential: Grows more crops in less space using vertically stacked layers.
    Limitations: Requires proper sunlight or artificial lighting. Costs more
  • Pot farming: Potential: An Easy way to grow vegetables and flowers at home.
    Limitations: Limited soil and space can reduce plant growth.
  • Terrace Farming: Potential: Converts rooftops into mini-farms for farming.
    Limitations: There is a possibility of water stagnation on the terrace. The ground should be provided with a drainage system.

Class 8 Basic Science Chapter 7 Important Questions Kerala Syllabus Let’s Cultivate and Reap Goodness

Question 27.
Prepare a note on vertical farming.
Answer:
Vertical farming uses shelves or stacked layers to grow plants upwards, allowing more crops to be grown in a smaller space. The plants are arranged in such a way that all layers receive sufficient sunlight. Since water drips from the top layer to the ones below, many plants reuse the same water, which saves water.

Question 28.
A vertical farming model that can be built at school and home at low cost.
Answer:
Arrange 4 – 5 used plastic bottles or grow bags vertically on a wooden frame or hanging ropes. Make holes in each bottle for water to drain. Fill the bottle with soil and compost. Grow small plants like mint, spinach, coriander, and chili.

Question 29.
What are the different fertilisation methods?
Answer:
Some fertilisers are spread on the field before or after planting. Others are applied near the roots of plants. Liquid fertilisers are sprayed on the leaves, while others are mixed with water and sprayed on the crops. Using different methods, depending on the type of crop and soil, ensures better yields.

Question 30.
What are the different types of fertilisers used in agriculture?
Answer:
Organic fertilisers, artificial fertilisers, nano fertilisers, and biofertilisers.

Question 31.
What are the benefits of biofertiliser?
Answer:
Microorganisms that help in the fixation of nitrogen and phosphorus are added to the soil. They do not harm the soil structure and decomposers. Examples include Rhizobium, Azospirillum, and Mycorrhiza.

Question 32.
What are the characteristics of organic fertilisers?
Answer:
They are Bio-residues. It will not harm either the texture of the soil or the decomposers. Plants get the nutrients released from these bioresidues by the decomposers. Nutrients may be less, and the process of decomposition takes a long time.

Question 33.
What are the characteristics of nano-fertilisers?
Answer:
They are artificial fertilisers in the form of nanoparticles. They can be easily absorbed by plants. They are required in very small quantities compared to other fertilisers. They do not change the chemical composition of the soil like artificial fertilisers. Examples of nano-fertilisers include nano-urea, nano-phosphate, and nano-magnesium sulfate.

Question 34.
Prepare a note on artificial fertilisers.
Answer:
Artificial fertilisers are artificially produced and used to increase the nutrient value of the soil. These can be easily absorbed by plants. Excessive application of fertilisers can also harm the soil structure and decomposers. Ammonium phosphate, urea, and ammonium sulfate are examples of artificial fertilisers.

Class 8 Basic Science Chapter 7 Important Questions Kerala Syllabus Let’s Cultivate and Reap Goodness

Question 35.
What is the importance of indigenous varieties?
Answer:
Indigenous varieties are those that are adapted to the local environment. They are naturally resistant to pests and diseases in the area, so the use of pesticides can be reduced. They also help maintain ecological balance by supporting local pollinators. They also require less fertiliser and reduce costs.

Question 36.
What are the various methods of producing improved varieties of planting material?
Answer:
Grafting, budding, layering and tissue culture are used to produce seedlings with the same characteristics as the parent plant.

Question 37.
What would be the advantage of a rice plant that produces rice containing many nutrients?
Answer:
This will be very helpful as this rice plant can provide a complete diet. This means that, especially for people who cannot afford to buy a variety of foods, all the nutrients like proteins, vitamins, minerals, and energy can be obtained only from rice. This will reduce malnutrition, help keep people healthy, and make agriculture more useful. Scientists are already working on crops that contain vitamin A, such as golden rice.

Question 38.
Gather additional information about tissue culture.
Answer:
Uses
To grow large numbers of plants quickly. To produce exact copies (clones) of a healthy plant. To preserve rare or endangered plants. To grow disease-free plants.

Varieties
In agriculture – crops such as bananas, sugarcane, and orchids
In horticulture – for flowers and ornamental plants.
In forests – to grow more trees quickly.
In medicine and research – to study plant diseases and develop drugs.

Advantages
Rapid production of thousands of plants. All plants have the same quality, making it possible to grow plants even in places with limited space. Can be done all year round regardless of weather conditions.

Limitations
Requires a separate lab and trained people. The cost of setting up a lab is high.

Question 39.
How do GM crops threaten indigenous crops?
Answer:
GM crops can spread rapidly and take over farmland, reducing the number of traditional crops. This can lead to a loss of biodiversity. Native plants and their natural resistance to local pests or conditions may be lost over time.

Question 40.
How does a greenhouse help in agriculture?
Answer:
This system helps in cultivating crops in both the rainy season and summer. The greenhouse is made of sheets like plastic, nylon, and polyethene. It also helps in reducing the incidence of pests and diseases as it is covered on all sides.

Question 41.
What is the advantage of drip irrigation?
Answer:
This is an irrigation method that uses pipes and valves to help water drip down to the root zone. Doing this minimises water loss and ensures equal water availability to all plants.

Question 42.
What is wick irrigation?
Answer:
This is a method of delivering water directly from the water source to the root zone of plants using a cotton wick. It requires less water than drip irrigation.

Class 8 Basic Science Chapter 7 Important Questions Kerala Syllabus Let’s Cultivate and Reap Goodness

Question 43.
How does mulching help farms?
Answer:
This is the traditional method of covering the soil in the fiejds using dry leaves, hay, etc, to reduce water loss due to evaporation. Through this method, it is possible to reduce the growth of weeds and to enrich the soil.

Question 44.
Prepare a note on Integrated Pest Management.
Answer:
PEST CONTROL

  • Although there are different methods for pest control, the population density of pests and the nature off the crops should be taken into account when choosing the pest control methods.
  • Farmers do not want to kill all the pests, but rather control their growth in a way that does not damage the crops.
  • Integrated pest control is a method that minimises the use of pesticides through automated pest control methods using various types of nets and traps, friendly insects, and the cultivation of seeds that are resistant to pests.

Question 45.
What is integrated farming?
Answer:
Agriculture is the cultivation of birds, fish, and animals along with plant care. Thus, integrated farming is the care of diverse organisms together.

Question 46.
Prepare a note on smart farming.
Answer:
Smart farming is the effective use of modern technologies in agriculture.

  • Artificial intelligence: By sharing images and information related to crops, you can find solutions to problems with the help of artificial intelligence.
  • Internet of Things (IoT): With the help of sensors, moisture, fertility, and pest infestations are identified, and a problem-solving system is prepared using machines.
  • Hydroponics: Plants are grown in nutrient solutions, and with the help of sensors, the amount of nutrients is detected and proyided as needed.
  • Aeroponics: Water and nutrients are delivered to the roots that grow into the air in a timely manner . with the help of sensors.

Question 47.
How to make agricultural product marketing smart?
Answer:
In this era of online marketing, agricultural marketing also needs to become smart. By providing a system that allows consumers to see everything from sowing to harvesting, the reliability and market value of the products can be increased. This can be easily achieved through sensors, drones, and AI cameras in the fields. This can also help deliver products directly to consumers worldwide. This can also increase the global marketing potential of products that have been granted geographical indication status.

Class 8 Basic Science Chapter 7 Important Questions Kerala Syllabus Let’s Cultivate and Reap Goodness

Question 48.
Prepare a note on geographical indication.
Answer:
The Ministry of Commerce and Industry has to approve the geographical indication status. Depending on the characteristics of the land in which it is cultivated, there will be differences in the taste, colour, smell and nutritional value of the products. On the basis of this, agricultural products produced in certain areas are given geographical indication status. Such products get added value in the market.

Question 49.
What are some agricultural products that have received Geographical Indication status?
Answer:
Palakkadan Matta rice, Malabar pepper, Kuttiattoor Mango, Kodungallur pottuvellari, Tirur Vettila,
Benganapally mango, Kashmiri saffron, Nagpur orange, Jeeraksala rice, Arunachal orange.

Question 50.
What are the advantages of mobile apps for farmers?
Answer:
There are apps that provide weather alerts, pest and disease warnings, expert advice on agriculture, market price levels, and information about benefits for farmers. If such facilities are implemented properly, the agriculture sector will become stronger.

Class 8 Basic Science Chapter 6 Important Questions Kerala Syllabus Metals and Non Metals

The comprehensive approach in SCERT Class 8 Basic Science Textbook Solutions Chapter 6 Metals and Non Metals Important Questions ensure conceptual clarity.

Metals and Non Metals Extra Questions and Answers Class 8 Basic Science Chapter 6 Kerala Syllabus

Metals and Non Metals Class 8 Important Questions

Question 1.
Which metal has the highest malleability?
a) Aluminium
b) Iron
c) Copper
d) Gold
Answer:
d) Gold

Question 2.
Which of the following metals is the most electrically conductive?
a) Aluminium
b) Silver
c) Copper
d) Gold
Answer:
b) Silver

Question 3.
Find the odd one out
a) Aluminium
b) Sodium
c) Potassium
d) Lithium
Answer:
a) Aluminium (Others are soft metals)

Question 4.
Find and fill in the relationship
The shining of metal surfaces: Metallic lustre
The ability to produce sound when struck on the surface: ____________
Answer:
Sonority

Question 5.
What is the basic substance formed when magnesium oxide dissolves in water?
Answer:
Magnesium Hydroxide

Class 8 Basic Science Chapter 6 Important Questions Kerala Syllabus Metals and Non Metals

Question 6.
Why do iron appliances in houses near the sea rust quickly?
Answer:
Iron rusts quickly due to the presence of salt (sodium chloride) in the air.

Question 7.
What metal is present in haemoglobin in the blood?
Answer:
Iron

Question 8.
Which of the following metals is subject to corrosion the fastest?
a) Gold
b) Aluminium
c) Iron
d) Sodium
Answer:
d) Sodium

Question 9.
Generally metals can be hammered and flattened into thin sheets. This characteristic property of metals is known as _____________ .
Answer:
Malleability

Question 10.
Write aluminium, copper, gold, and silver in the order of decreasing electrical conductivity.
Answer:
Silver > Copper > Gold > Aluminium

Question 11.
Metals can be drawn and stretched into wires. This property is called ductility. Write any two cases where ductility is used.
Answer:

  • Filament of bulb
  • Electric wires
  • Gold ornaments

Question 12.
Which properties of metals are utilised in the following cases?
a) Iron is used to make weapons.
b) Aluminium is used to make food packaging.
Answer:
a) Hardness
b) Malleability

Question 13.
Give examples of non-metals that are added to metals to make alloys.
Answer:

  1. Carbon
  2. Phosphorous

Class 8 Basic Science Chapter 6 Important Questions Kerala Syllabus Metals and Non Metals

Question 14.
Alloys generally have high melting points. Give examples of metals with low melting points.
Answer:

  1. Gallium
  2. Caesium

Question 15.
Which properties of metals are used in the following situations?
a) Copper is used to make bells.
b) Tungsten is used to make the filament of a light bulb.
Answer:
a) Sonority – the ability to produce a specific sound when struck with a hard object.
b) Ductility – the ability to be formed into very thin wires.

Question 16.
What factors increase the corrosion of iron?
Answer:

  • Presence of air
  • Presence of moisture
  • Presence of salt

Question 17.
Copper nails are used instead of iron nails in coastal areas. Why?
Answer:
In coastal areas, there is a presence of salt in the air. Iron rusts quickly in the presence of air and moisture. Therefore, copper nails are used instead of iron nails.

Question 18.
Classify the following as metals or non-metals.
• Carbon
• Phosphorous
• Sulphur
• Gallium
• Helium
• Sodium
• Caesium
Answer:

Metals Non-Metals
Sodium
Gallium
Caesium
Calcium
Carbon
Sulphur
Phosphorous
Helium

Question 19.
Magnesium reacts with oxygen in the air to form magnesium oxide.
a) Write the chemical equation for this reaction.
b) What is the product of magnesium oxide dissolved in water?
Answer:
a) 2Mg + O2 → 2MgO
b) Magnesium Hydroxide

Question 20.
Write any two metals found in the human body and their importance.
Answer:
Iron – is an important component of haemoglobin in the blood.
Calcium – strengthens bones and teeth. Regulates blood pressure.

Question 21.
Compare and list any three properties of metals and non-metals.
Answer:

Metals Non metals
Generally, they are hard. Generally, they are less hard.
They exhibit sonority. They don’t exhibit sonority.
Generally, they are electrically conductive. Not thermal conductors.

Question 22.
What are the properties of gold that make it suitable for making jewellery?
Answer:
a) Gold is used to make jewellery by alloying it with copper or silver.
b) Malleability, metallic lustre, corrosion resistance, ductility.

Class 8 Basic Science Chapter 6 Important Questions Kerala Syllabus Metals and Non Metals

Question 23.
Give reasons for the following:
a) The handles of electrical tools and cooking utensils are made of non-metals.
b) Metals like sodium, potassium etc. are stored in kerosene.
c) Pickles and curds are not stored in metal containers.
Answer:
a) Electric tools and cookware are made of metals. Metals are good conductors of electricity and heat. Therefore, their handles are made of non-metals such as wood, plastic, rubber, etc. to ensure safe handling without injury.

b) Sodium and potassium are very reactive metals. They react rapidly with air and moisture and are destroyed. They are stored in kerosene to avoid contact with the environment.

c) Acids are present in sour foods like pickles and curd. Acids can react with metals to form other substances and can corrode metal containers.

Question 24.
a) What is an alloy?
b) What are the benefits of alloying metals?
c) Give examples of alloys.
Answer:
a) Metals are combined with other metals or non-metals to make alloys.

b) It is done to obtain materials with desirable properties and to resist corrosion. A very small amount of non-metals like carbon, phosphorus, etc., are also added to such alloys.

c) Stainless steel, brass, bronze

Question 25.
a) What is corrosion of metals?
b) Suggest ways to prevent corrosion of metals
Answer:
a) Metals react with various elements in the atmosphere to form compounds. This process is palled corrosion.

b)

  • Paint metal surfaces.
  • Apply oil and grease.
  • Apply non-metallic coatings such as rubber and plastic.
  • Convert into alloys.

Question 26.
Take three test tubes. Put a rust-free and shiny iron nail in each of them. Place a wet cloth in the first test tube, and add a cloth soaked in salt water to the second test tube.. Iron nails should be fully immersed in these three test tubes. Take anhydrous calcium chloride or quicklime in the third test tube. All three test tubes should be closed immediately with a cork after adding the materials.
What will be the observation after a few days? Write the reason for the observation in each test tube.
Answer:
Test tube 1: The nail in the test tube is rusty
This is due to the presence of air and moisture in the test tube.
Test tube 2: The nail in the test tube is rustier. In addition to the presence of air and moisture, the presence of salt increases the rate of rusting.
Test tube 3: The nail in the test tube is not rusty. The nail absorbs the moisture inside the test tube. In the absence of moisture, the rusting process does not take place.

Question 27.
Observe the picture.
Class 8 Basic Science Chapter 6 Important Questions Kerala Syllabus Metals and Non Metals 1
Calcium chloride is taken in a test tube. A smooth iron nail is moistened and passed through a cork and the test tube is closed.
a) What will be the observation after a few days?
b) Write the reason for the observation.
c) What is the conclusion drawn from this?
Answer:
a) The part of the nail inside the test tube is not rusted. The part of the nail outside the test tube is rusted. The calcium chloride absorbs the moisture inside the test tube. Therefore, the iron does not rust.
b) The presence of water and moisture outside the test tube causes the iron to rust rapidly.
c) The presence of water and air affects the rusting of iron.

Question 28.
a) How is soda water made?
b) Write the chemical equation for this reaction.
c) What is observed when blue litmus paper is dipped in soda water? What is the reason?
Answer:
a) Soda water is made by dissolving carbon dioxide in water under high pressure.
b) CO2 + H2O → H2CO3
c) Soda water is carbonic acid. The presence of acid turns blue litmus red.

Class 8 Basic Science Chapter 6 Important Questions Kerala Syllabus Metals and Non Metals

Question 29.
Observe the diagram of the electric circuit.
Class 8 Basic Science Chapter 6 Important Questions Kerala Syllabus Metals and Non Metals 2
a) Which of the following objects will make the bulb light up if you connect it to the part marked PQ?
• Rolled aluminum foil
• Rubber band
• Glass rod
• Plastic wire
• Iron nail
• Silver necklace
b) What is the reason for this difference?
Answer:

Bulb Glows The bulb does not glow
Rolled Aluminium foil
Iron nail
Silver necklace
Glass rod
Rubber band
Plastic wire

b) Aluminium, iron, and silver are electrical conductors. Glass, rubber, and plastic are not electrical conductors.

Question 30.
Select and arrange the items from columns B and C that match those in section A.

A B C
Aluminium Sonority Filaments
Tungsten Thermal Conductivity Church Bells
Copper Ductility Cooking utensils

Answer:

A B C
Aluminium Thermal Conductivity Cooking utensils
Tungsten Ductility Filaments
Copper Sonority Church bells

Class 8 Basic Science Chapter 5 Important Questions Kerala Syllabus Elements and Compounds

The comprehensive approach in SCERT Class 8 Basic Science Textbook Solutions Chapter 5 Elements and Compounds Important Questions ensure conceptual clarity.

Elements and Compounds Extra Questions and Answers Class 8 Basic Science Chapter 5 Kerala Syllabus

Elements and Compounds Class 8 Important Questions

Question 1.
All matter is made up of minute particles called _____________ .
a) Atom
b) Molecule
c) Element
d) Compound
Answer:
a) Atom

Question 2.
____________ discovered that water could be split into hydrogen and oxygen by passing electricity through it.
a) Mendeleev
b) Henry Moseley
c) Cavendish
d) Humphry Davy
Answer:
d) Humphry Davy

Question 3.
Cr is the symbol for which element?
a) Calcium
b) Chromium
c) Chlorine
d) Carbon
Answer:
b) Chromium

Question 4.
Who invented the modern periodic table?
Answer:
Henry Moseley

Question 5.
An ammonia molecule contains one nitrogen atom and three hydrogen atoms. How would you represent five ammonia molecules?
Answer:
5NH3

Class 8 Basic Science Chapter 5 Important Questions Kerala Syllabus Elements and Compounds

Question 6.
Find the relation and fill up the blanks.
Sodium – (Na): Natrium – Latin name
Potassium – (K): ______________ .
Answer:
Kalium – Latin name

Question 7.
Find the monoatomic element from the following.
a) Hydrogen
b) Helium
c) Oxygen
d) Chlorine
Answer:
b) Helium

Question 8.
Find the Odd one out.
a) Rutherfordium
b) Nobelium
c) Rubidium
d) Einsteinium
Answer:
Rubidium (Others are synthetic elements)

Question 9.
Which group is not included in Dobereiner’s triads?
a)

Li, Na, k

b)

He, Ne, Ar

c)

Ca, sr, Ba

d)

CI, Br, I

Answer:
b)

He, Ne, Ar

Question 10.
Give an example of a polyatomic element.
Answer:
Phosphorous (P4)
Sulphur (S8)

Question 11.
a) What is the white powder formed when magnesium burns in air?
b) Write its chemical formula.
Answer:
a) Magnesium Oxide
b) MgO

Class 8 Basic Science Chapter 5 Important Questions Kerala Syllabus Elements and Compounds

Question 12.
a) Which substances are obtained when sugar is heated strongly in a dry boiling tube?
b) Which of these is an element? Which is a compound?
Answer:
a) Carbon – water
b) Element – Carbon
Compound – Water

Question 13.
What is the basis for naming the elements given below?
a) Mendelevium
b) Rubidium
Answer:
a) From the name of the scientist Mendeleev
b) From the basis of red colour

Question 14.
Write the names of two elements that are found in liquid state at normal temperature.
Answer:
Mercury, Bromine

Question 15.
How do elements and compounds differ?
Answer:
Elements are pure substances made up of the same type of atoms. They cannot be further divided into components through chemical reactions.

Compounds are pure substances formed by the combination of atoms of two or more elements in a fixed ratio. These can be dissociated into constituent elements through chemical reactions.

Question 16.
The symbol for the element hydrogen is H. What is the difference between 2H and H2?
Answer:
2H refers the two atoms of Hydrogen.
H2 refers to a nitrogen molecule formed by two atoms.

Question 17.
Find the number of molecules and the total number of atoms in the following.
a) O2
b) 5F2
Answer:
a) No. of molecules – 1
Total no. of atoms – 2

b) No. of molecules – 5
Total no. of atoms – 10

Question 18.
It is said that oxygen is an element and carbon dioxide is a compound. Why?
Answer:
In an oxygen molecule, there are only the same type of atoms.
In carbon dioxide, there is the combination of atoms of the elements oxygen and carbon in a fixed ratio.

Question 19.
a) What are the components of water?
b) How can we separate the components of water?
Answer:
a) Hydrogen, Oxygen
b) Water can be split into its components by passing electricity through it.

Question 20.
a) What are periods and groups in the periodic table?
b) How many of these are there in the modern periodic table?
Answer:
a) The horizontal rows in the table are called periods and the vertical columns are called groups. Elements in a group show similar chemical properties.

b) Periods – 7
Groups – 18

Class 8 Basic Science Chapter 5 Important Questions Kerala Syllabus Elements and Compounds

Question 21.
When the element carbon burns in air, it combines with oxygen to form carbon dioxide gas.
a) Write what are the reactants and products in this reaction.
b) Write the chemical equation for this reaction using symbols and chemical formulas.
Answer:
a) Reactants: – Carbon (C)
Oxygen (O2)
Products – Carbon dioxide (CO2)
b) C + O2 → CO2

Question 22.
Classify the following into elements and compounds and list them.
• Carbon
• Carbon dioxide
• Ammonia
• Sodium chloride
• Oxygen
• Sodium
Answer:

Elements Compounds
• Carbon
• Oxygen
• Sodium
• Carbon dioxide
• Ammonia
• Sodium Chloride

Question 23.
Write any three ideas in John Dalton’s atomic theory.
Answer:

  • All matter is made up of minute particles called atoms.
  • An atom is the smallest particle that can participate in a chemical reaction.
  • An atom cannot be divided during a chemical reaction.
  • An atom cannot be created or destroyed.
  • All atoms of an element are of the same size, property and mass.
  • Atoms of different elements differ in their size, property and mass, (any three)

Question 24.
The molecular formula of a compound is H2SO4.
a) What elements are contained in this compound?
b) Find the total number of molecules contained in 5H2SO4 and also the total number of atoms.
Answer:
a) Hydrogen (H)
Sulphur (S)
Oxygen (O)

b) Number of molecules = 5
Total number of atoms = 35
(H = 10, S = 5, O = 20)

Question 25.
What are the advantages of the periodic table prepared by Mendeleev?
Answer:

  • Mendeleev arranged the 63 elements known till that day in ascending order of their atomic masses and prepared a table.
  • Blank spaces were left for elements to be discovered and their properties were predicted.
  • Elements with similar properties were grouped together to simplify the study of elements and their compounds.

Question 26.
What are the limitations of Mendeleev’s periodic table?
Answer:

  • The order of atomic mass was not strictly followed in the case of all elements. Hydrogen could not be given an exact position.
  • Elements with very different properties had to be placed in the same group.

Question 27.
The names of some scientists and their contributions are given in the table. Write them in the correct order.

A B
Splitting of water by passing electricity through it Henry Moseley
Representation of elements by letters Sir Humphry Davy
Modern Periodic table Berzelius

Answer:

A B
Splitting of water by passing electricity through it Sir Humphry Davy
Representation of elements by letters Berzelius
Modern Periodic table Henry Moseley

Question 28.
Sugar is heated strongly in a dry boiling tube.
a) What will be the observation?
b) What products are formed?
c) What are the elements in sugar?
Answer:
a) Observation – A black substance remains in the boiling tube.
b) Carbon (C), Water (H2O)
c) Carbon (C), Hydrogen (H), Oxygen(O)

Class 8 Basic Science Chapter 5 Important Questions Kerala Syllabus Elements and Compounds

Question 29.
Below are some statements related to early attempts at element classification. Match the statements with the ones in brackets that match them.
(Mendeleev, Dobereiner, Berzelius, Lavoisier, Newlands, Henry Moseley)
a) Classified the then-known elements into metals and non-metals.
b) Classified the elements of similar properties into groups of three.
c) Arranged the elements in ascending order of their atomic masses.
Answer:
a) Lavoisier
b) Dobereiner
c) Mendeleev

Question 30.
The names, nomenclature, and symbols of some elements are given in A, B, C columns. Match these in the correct order and write them.

A
Element
B
Nomenclature
C
Symbol
Chromium From the Latin name Cr
Copernicium From the Greek word, which means colour Cn
Copper From the name of a scientist Cu

Answer:

A
Element
B
Nomenclature
C
Symbol
Chromium From the Greek word, which means colour Cr
Copernicium From the name of a scientist Cn
Copper From the Latin name Cu

 

Class 8 Basic Science Chapter 4 Important Questions Kerala Syllabus Chemistry of Changes

The comprehensive approach in SCERT Class 8 Basic Science Textbook Solutions Chapter 4 Chemistry of Changes Important Questions ensure conceptual clarity.

Chemistry of Changes Extra Questions and Answers Class 8 Basic Science Chapter 4 Kerala Syllabus

Chemistry of Changes Class 8 Important Questions

Question 1.
Burning of wax is a ……………………… change.
(Chemical change / Physical change)
Answer:
Chemical change

Question 2.
What is the form of energy exchanged during the state transition of matter?
Answer:
Heat energy

Question 3.
The amount of space required for matter to exist is …………………………
Answer:
Volume

Question 4.
What gas is produced when magnesium reacts with hydrochloric acid?
a) Chlorine
b) Hydrogen
c) Oxygen
d) Nitrogen
Answer:
b) Hydrogen

Question 5.
An example of a substance that undergoes sublimation is:
a) Water
(b) Talc
(c) Wax
(d) Camphor
Answer:
(d) Camphor

Class 8 Basic Science Chapter 4 Important Questions Kerala Syllabus Chemistry of Changes

Question 6.
Find the relationship and complete.
Gas that aids combustion: Oxygen
Gas that burns: _____________
Answer:
Hydrogen

Question 7.
What is the energy change that occurs in a Dry cell?
Answer:
Chemical energy is converted into electrical energy.

Question 8.
What is the chemical product formed when electricity is passed through water containing a small amount of acid?
Answer:
Hydrogen and Oxygen

Question 9.
Which solution is used for electroplating copper onto an iron nail?
Answer:
Copper sulphate solution

Question 10.
To gold-plate a copper bangle, what should be connected to the positive terminal of the battery?
Answer:
Gold

Question 11.
Select the correct statements related to gases from the ones given below:
a) Less distance between particles.
b) More energy of particles.
c) Higher speed of particle movement.
d) More attractive force between particles.
Answer:
b) More energy of particles. and c) Higher speed of particle movement.

Question 12.
When a candle burns, both chemical and physical changes occur. What are these?
Answer:
Chemical Change: Burning of a candle
Physical Change: Melting of a candle

Question 13.
Magnesium is taken in a test tube, and a liquid is added to it. A vigorous chemical reaction occurs, and a gas is released. If a burning splinter is shown to the gas, it burns with a “pop” sound.
a) Identify the probable liquid added to the test tube.
b) Which gas was produced?
Answer:
a) Dilute Hydrochloric acid
b) Hydrogen

Class 8 Basic Science Chapter 4 Important Questions Kerala Syllabus Chemistry of Changes

Question 14.
What is the energy change occurring in the following?
a) Combustion of fuel
b) Photosynthesis
Answer:
a) Chemical energy is converted into heat energy
b) Light energy is converted into chemical energy.

Question 15.
Classify the following into chemical changes and physical changes:
a) Wax melts
b) Wax burns
c) Milk turns into curd
d) Milk thickens (or solidifies) on cooling
Answer:

Chemical changes Physical changes
Wax burns Wax melts
Milk turns into curd. Milk thickens (or solidifies) on cooling.

Question 16.
Two processes are given below:
Quicklime is produced by strong heating limestone (or shells).
Slaked lime is produced by adding water to quicklime.
a) Which of these is an endothermic reaction?
b) Which is an exothermic reaction?
Answer:
a) Quicklime is produced by strong heating limestone (or shells)
b) Slaked lime is produced by adding water to quicklime.

Question 17.
What are the suitable solutions for electroplating the metals given below?
a) Silver
b) Gold
Answer:
(a) Sodium cyanide + Silver cyanide
(b) Sodium cyanide + Gold cyanide

Question 18.
What are the forms of energy associated with chemical reactions?
Answer:
Heat, Light, Electricity

Question 19.
What are the main properties of matter
Answer:
It occupies space
It has mass

Question 20.
Quicklime is taken in a steel cup. A little water is added to it.
(a) What is the substance formed when water is added to quicklime?
(b) Why do people handling quicklime wear gloves?
Answer:
a) When water is added to quicklime (calcium oxide), slaked lime (calcium hydroxide) is formed.

b) The reaction of quicklime with water is highly exothermic. If quicklime comes into contact with moisture on the skin, it will react and release a lot of heat. This can cause severe burns to the skin.

Class 8 Basic Science Chapter 4 Important Questions Kerala Syllabus Chemistry of Changes

Question 21.
Suggest an experiment to prove that matter requires space to exist.
Answer:
Aim: To demonstrate that matter occupies space.
Materials: Beaker, water, and a small stone.
Procedure & Observation:

  1. Fill a beaker partially with water and note the initial water level.
  2. Carefully lower the stone into the water.
  3. Observe that the water level rises.

Conclusion: The rise in water level proves that the stone, being matter, takes up space (has volume)

Question 22.
Write what type of chemical reaction each of the following is:
a) Fireflies emit light.
b) When zinc and copper nails are inserted into a lemon and connected to an LED through a copper wire, the LED glows.
c) Ammonium chloride decomposes when heated strongly.
Answer:
a) Bioluminescence
b) Electrochemical Reaction
c) Thermochemical reactions

Question 23.
Complete the following chemical reaction
Class 8 Basic Science Chapter 4 Important Questions Kerala Syllabus Chemistry of Changes 1
Answer:
(a) Silver (Ag)
(b) Chlorine gas (Cl2)
(c) Water (H2O)
(d) Oxygen (O2)
(e) Water (H2O)
(f) Slaked lime (Ca(OH)2)

Question 24.
A substance is strongly heated in a test tube. When a burning incense stick was held near the mouth of the test tube, the incense stick flared up.
a) What caused the incense stick to flare up?
b) What might be the substance that was heated?
c) What type of chemical reaction is this?
Answer:
a) When a substance is heated, oxygen gas is released. Oxygen is a gas that aids in burning/combustion.
b) Potassium permanganate
c) Endothermic reaction

Question 25.
What is the reason for the following?
a) Water kept in the freezer turns into ice.
b) Silver salts are stored in colored bottles.
c) Some specific types of medicines are stored in refrigerators.
Answer:
a) Water turns into ice in a freezer due to the cooling/freezing process (Phase Transition). When water is cooled below its freezing point (0°C), its molecules lose kinetic energy and arrange themselves into a more ordered, crystalline structure. This change of state from liquid to solid is a physical process that occurs when enough heat energy is removed from the water.

b) Silver salts (like silver bromide, silver chloride) are stored in colored or opaque bottles because they are sensitive to light and undergo photochemical reactions when exposed to it. Light energy can decompose silver salts into elemental silver, causing them to blacken or lose their intended properties. Colored or opaque bottles block out light, preventing this decomposition and preserving the integrity of the silver salts.

c) Some specific types of medicines are stored in refrigerators because their chemical stability and efficacy are temperature-dependent. Many ingredients in these medicines are sensitive to heat. Higher temperatures can cause them to degrade, lose their potency, or even become toxic. Refrigeration slows down these chemical degradation processes, maintaining the medicine’s effectiveness and ensuring its safety for use throughout its shelf life.

Question 26.
Define the following
a) Thermochemical Reactions
b) Photochemical Reactions
c) Electrochemical Reactions
Answer:
a) The reactions in which heat energy is liberated or absorbed are known as thermochemical reactions.
b) The reactions in which light energy is absorbed or liberated are known as photochemical reactions
c) If electrical energy is absorbed or produced during a chemical reaction, it is known as an electrochemical reaction.

Question 27.
a) What is electrolysis?
b) Write any practical application of electrolysis.
Answer:
a) The process of dissociation of a substance by absorbing electrical energy is known as electrolysis.
b) Electroplating

Question 28.
Draw a diagram of the setup for electroplating gold onto a copper bangle and label its parts
Answer:
Class 8 Basic Science Chapter 4 Important Questions Kerala Syllabus Chemistry of Changes 2

Class 8 Basic Science Chapter 4 Important Questions Kerala Syllabus Chemistry of Changes

Question 29.
A picture of the arrangement of copper plating on an iron bangle is drawn. What are the mistakes in this?
Class 8 Basic Science Chapter 4 Important Questions Kerala Syllabus Chemistry of Changes 3
Answer:
Copper should be connected to the positive terminal of the battery.
The iron bangle should be connected to the negative terminal of the battery. Copper sulphate solution should be used as the electrolyte (or solution).

Question 30.
Match the following

A B C
Sublimation Chemical energy is converted into electrical energy. Naphthalene
Firefly Change of state Photochemical reaction
Dry cell Bioluminescence Electrochemical reaction

Answer:

A B C
Sublimation Change of state Naphthalene
Firefly Bioluminescence Photochemical reaction
Dry cell Chemical energy is converted into electrical energy. Electrochemical reaction

Class 8 Basic Science Chapter 3 Important Questions Kerala Syllabus Pressure

The comprehensive approach in SCERT Class 8 Basic Science Textbook Solutions Chapter 3 Pressure Important Questions ensure conceptual clarity.

Pressure Extra Questions and Answers Class 8 Basic Science Chapter 3 Kerala Syllabus

Pressure Class 8 Important Questions

Question 1.
Which of the following is the unit of atmospheric pressure?
(N, parsec, kg/m3, bar)
Answer:
bar

Question 2.
It is possible to drink juice using a straw because of ……………………
Answer:
Atmospheric pressure

Question 3.
Correct the following. As the number of particles increases, gas pressure decreases.
Answer:
As the number of particles increases, gas pressure increases.

Question 4.
The instrument used to measure atmospheric pressure is a ………………………
Answer:
Barometer.

Question 5.
The pressure exerted on an area of 10 m2 by a force of 50 N is ………………….. .
(5 N, 5 N/m2, 0.2 N/m2, 0.2 N)
Answer:
5 N/m2
P = F/A, F = 50 N, A =10 m2
P = 50 /10 = 5 N/m2

Class 8 Basic Science Chapter 3 Important Questions Kerala Syllabus Pressure

Question 6.
The instrument used to measure liquid pressure is ……………………. .
Answer:
Manometer

Question 7.
Choose the correct equation of pressure.
Answer:
(P = F × A, P = A/F, P = F/A)
Answer:
P = F/A

Question 8.
Observe the diagram
Class 8 Basic Science Chapter 3 Important Questions Kerala Syllabus Pressure 1
Water flowing down through the holes P, Q and R on a bottle is depicted.
a) Which is the correct figure? i/ii/iii
b) What is the reason for this inference?
Answer:
a) Figure iii is correct

b) As depth of liquid increases pressure increases. So the water flowing from the hole R will have the greatest force and covers more distance.

Question 9.
Observe the picture below. Both beakers contain the same volume of water.
Class 8 Basic Science Chapter 3 Important Questions Kerala Syllabus Pressure 2
a) Which beaker experiences maximum pressure at the bottom?
b) Justify your answer.
Answer:
a) Beaker A experience maximum pressure at the bottom.

b) The height of liquid column (or depth) is more in beaker A than beaker B. So more pressure is experienced at the bottom of beaker A.

Question 10.
Liquids also exert pressure. Write down any two factors that influence liquid pressure. (Term II, 2022)
Answer:

  1. Height of the liquid column
  2. Density of the liquid

Class 8 Basic Science Chapter 3 Important Questions Kerala Syllabus Pressure

Question 11.
The weight of the air column over unit area of earth’s surface is atmospheric pressure.
a) What is standard atmospheric pressure?
b) Name the instrument used to measure atmospheric pressure. (Annual 2022)
Answer:
a) Atmospheric pressure at sea level is known as standard atmospheric pressure. It is defined as the weight of a mercury column that is 0.76 m high, with a unit cross sectional area, exerting a pressure of 1 atm.

b) Barometer.

Question 12.
Two identical balloons one filled with water and other with kerosene of equal quantity.
Class 8 Basic Science Chapter 3 Important Questions Kerala Syllabus Pressure 3
a) Identify the balloon filled with water.
b) Why is the expansion of balloons appear different in figures (a) and (b)? (Annual 2022)
Answer:
a) Figure (b)

b) Water has higher density than kerosene. As density increases, liquid pressure increases. So the balloon which is expanded has water filled in it and the other balloon which is small in size has kerosene filled in it.

Question 13.
Check whether the given statement is correct or wrong. Correct the wrong statement.
(i) Atmospheric pressure increases as height from the Earth’s surface increases.
(ii) The porters in railway stations place a piece of cloth before carrying luggage on head to decrease the area of contact between luggage and head to reduce the pressure felt on their head.
Answer:
(i) Statement is wrong
Corrected statement: Atmospheric pressure decreases as height from the Earth’s surface increases.

(ii) Statement is wrong
Corrected statement: The porters in railway stations place a piece of cloth to increase the area of contact between luggage and head to reduce the pressure felt on their head.

Question 14.
Calculate the force exerted by atmospheric pressure on a table surface of area 3 m2. Atmospheric pressure is 101325 Pascal.
Answer:
Force = Pressure × Area
= 1 Pa × 3 m2 = 101325 × 3
= 303975 N

Question 15.
The thrust acting per unit area by a liquid is the liquid pressure.
Class 8 Basic Science Chapter 3 Important Questions Kerala Syllabus Pressure 4
a) Three points A, B and C are marked on a cylindrical glass vessel filled with water as shown in figure. If three identical inflated balloons are placed on these points inside the cylinder, which one will be the largest in size? Justify your answer.
b) Do you expect any change in size to the balloons, if the water is replaced with kerosene? Explain.
Answer:
a) The balloon placed in the position A will be the largest in size. As depth decreases liquid pressure also decreases. The pressure experienced at position A is the least compared to positions B and C and so the balloon with largest size occupies position A.

b) Yes, when kerosene is replaced with water the size of the balloons become comparatively larger than in water because kerosene has low density than water. When density decreases liquid pressure also decreases. So when the pressure experienced is less, the size of balloons (volume of balloons) is comparitively larger.

Class 8 Basic Science Chapter 3 Important Questions Kerala Syllabus Pressure

Question 16.
When area increases, pressure decreases. Find three situations where this principle is applied.
Answer:
a) We can walk easily through a muddy place by placing a plank over it.
b) Ducks walk easily over mud.
c) Constructing dams with wider basements.

Question 17.
Write down reasons for the following statements.
a) Astronauts wear specially designed suits.
b) Sewing needles are made with sharp tips.
c) Connecting the wheels of tanks and other such vehicles using wide chains.
Answer:
a) The suit helps us to maintain the pressure between the dress and the body equal to the atmospheric pressure.

b) Sharp tips have less surface area of contact. As area of contact decreases, pressure increases so that sewing needles can pierce into the materials to be stitched easily.

c) Connecting wheels of tanks and other such vehicles using wide chains helps in making more surface area of contact with the ground to exert less pressure over it. This ensures its smooth motion.