Category: Class 8

Students can Download Maths Chapter 2 Measurements Ex 2.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.4

Miscellaneous Practice Problems

Question 1.
Two gates are fitted at the entrance of a library. To open the gates easily, a wheel is fixed at 6 feet istance from the wall ito which the gate is fixed. If one of the gates is opened to 90°, find the distance moved by the wheel (π = 3.14).

Solution:
Let A be the position of the wall AC be the gate in initial position and AB be position when it is moved 90°.
Now the arc length BC gives the distance moved by the wheel.
Length of the arc
= \(\frac{\theta}{360^{\circ}}\) × 2πr units
= \(\frac{90^{\circ}}{360^{\circ}}\) × 2 × 3.14 × 6 feets
= 3.14 × 3 feets
= 9.42 feets
∴ Distance moved by the wheel = 9.42 feets.

Question 2.
With his usual speed, if a person covers a circular track of radius 150 ra in 9 minutes, find the distance that he covers in 3 minutes (π = 3.14).
Solution:
Radius of the circular track = 150m
Distance covers in 9 minutes = Perimeter of the circle = 2 × π × r units
Distance covered in 9 min = 2 × 3.14 × 150 m

Distance he covers in 3 min = 314 m

Question 3.
Find the area of the house drawing given in the figure.

Solution:
Area of the house = Area of a square of side 6 cm + Area of a rectangle with l = 8cm, h = 6 cm + Area of a ∆ with b = 6 cm and h = 4 cm + Area of a parallelogram with b = 8 cm, h = 4 cm
= (side × side) + (l × b) + (\(\frac{1}{2}\) × b × h) + 6h cm²
= (6 × 6) + (8 × 6) + (\(\frac{1}{2}\) × 6 × 4) + (8 × 4) cm²
= 36 + 48 + 12+ 32 cm²
Required Area = 128 cm²

Question 4.
Draw the top, front and side view of the following solid shapes.

Solution:

Question 5.
Draw the net for the cube of side 4 cm in a graph sheet.
Solution:

Challenging Problems

Question 6.
Guna has fixed a single door of 3 feet wide in his room whereas Nathan has fixed a double door, each 1 \(\frac{1}{2}\) feet wide in his room. From the closed state, if each of the single and double doors can open up to 120°, whose door requires a minimum area?

Solution:
(a) Width of the door that Guna fixed = 3 feet.
When the door is open the radius of the sector = 3 feet
Angle covered = 120°
∴ Area required to open the door = \(\frac{120^{\circ}}{360^{\circ}}\) × πr² = \(\frac{120^{\circ}}{360^{\circ}}\) × π × 3 × 3 = 37π feet²

(b) Width of the double doors that Nathan fixed = 1\(\frac{1}{2}\) feet.
Angle described to open = 120°
Area required to open = 2 × Area of the sector
= 2 × \(\frac{120^{\circ}}{360^{\circ}} \times \pi \times \frac{3}{2} \times \frac{3}{2} \text { feets }^{2}=\frac{3 \pi}{2}\) feet²
= \(\frac{1}{2}\) (3π) feet²
∴ The double door requires the minimum area.

Question 7.
In a rectangular field which measures 15 m × 8m, cows are tied with a rope of length 3m at four corners of the field and also at the centre. Find the area of the field where none of the cow can graze. (π = 3.14).
Solution:
Area of the field where none of the cow can graze = Area of the rectangle – [Area of 4 quadrant circles] – Area of a circle

Area of the rectangle = l × b units²
= 15 × 8 m² = 120 m²
Area of 4 quadrant circles = 4 × \(\frac{1}{4}\) πr² units
Radius of the circle = 3 m
Area of 4 quadrant circles = 4 × \(\frac{1}{4}\) × 3.14 × 3 × 3 = 28.26m²
Area of the circle at the middle = πr² units
= 3.14 × 3 × 3m² = 28.26m²
∴ Area where none of the cows can graze
= [120 – 28.26 – 28.26]m² = 120 – 56.52 m² = 63.48m²

Question 8.
Three identical coins, each of diameter 6 cm are placed as shown. Find the area of the shaded region between the coins, (π = 3.14) ( \(\sqrt{3}\) = 1.732)

Solution:
Given diameter of the coins = 6 cm
∴ Radius of the coins = \(\frac{6}{2}\) = 3 cm
Area of the shaded region = Area of equilateral triangle – Area of 3 sectors of angle 60°
Area of the equilateral triangle

∴ Area of the shaded region = 15.588 – 14.13 cm² = 1.458 cm²
Required area 1.458 cm² (approximately)

Question 9.
Using graph sheet, draw the net for the cuboid whose length is 5cm, breadth is 4cm and height is 3cm and also find its area.
Solution:
Net for the cuboid is:

One of the possible nets for a cuboid of length = 5 cm, breadth = 4 cm, height = 3 cm is given above
Area of the cuboid
= 20 cm² + 15 cm² + 20 cm² + 15 cm² + 12 cm² + 12 cm² = 94 cm²
Using formula,
Surface area of a cuboid
= 2 (lb + bh + lh) unit²
= 2(5 × 4 + 4 × 3 + 5 × 3) cm²
= 2(20 + 12 + 15) cm²
= 94 cm²

Question 10.
Using Euler’s formula, find the unknowns.

Solution:
Euler’s formula is given by F + V- E = 2
(i) V = 6, E = 14
By Euler’s formula
= F + 6 – 14 = 2
F = 2 + 14 – 6
F = 10

(ii) F = 8, E = 10
By Euler’s formula
= 8 + V – 10 = 2
V = 2 – 8 + 10
V = 4

(iii) F = 20, V = 10
By Euler’s formula
= 20 + 10 – E = 2
30 – E = 2
E = 30 – 2
E = 28
Tabulating the required unknowns

Students can Download Maths Chapter 2 Measurements Ex 2.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.3

Question 1.
Fill in the blanks:
(i) The three dimensions of a cuboid are _____.
(ii) The meeting point of more than two edges- is called as ______.
(iii) A cube has _____ faces.
(iv) The cross section of a solid cylinder is ______.
(v) If a net of a 3-D shape has six plane squares, then it is called ______.
Solution:
(i) length, breadth and height
(ii) vertex
(iii) six
(iv) circle
(v) cube

Question 2.
Match the following:

Solution:
(i) b
(ii) a
(iii) d
(iv) c

Question 3.
Which 3-D shapes do the following nets represent? Draw them.

Solution:
(i) The net represents cube, because it has 6 squares

(ii) The net represents cuboid

(iii) The net represents Triangular prism

(iv) The net represents square pyramid

(v) The net represents cylinder

Question 4.
For each solid, three views are given. Identify for each solid, the corresponding top, front and side (T, F & S) views.

Solution:

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

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

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

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

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

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

Question 6.
Find the area of the given nets.

Solution:
(i) Area = Area of 6 squares of side 4 cm
= 6 × a² sq. units
= 6 × 4 × 4 cm²
= 96 cm²
(ii) Area = Area of 2 rectangles of
l = 10, b = 6 + Area of 2 rectangles of l = 6, b = 4 + Area of 2 rectangles of l= 10,b = 4
= (10 × 6) + (6 × 4)+ (10 × 4) cm²
= 60 + 24 + 40 cm²
= 124 cm²

Question 7.
Can a polyhedron have 12 faces, 22 edges and 17 vertices?
Solution:
By Euler’s formula F + V- E = 2 fora polyhedron.
Here F = 12, V = 17, E = 22
F + V – E = 12 + 17 – 22
= 29 – 22
= 7 ≠ 2
∴ The polyhedron cannot have 12 faces 22 edges and 17 vertices.

Students can Download Maths Chapter 5 Information Processing Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 5 Information Processing Intext Questions

Try these (TextBook Page No 122)

Question 1.
Find the number of all possible triangles that can be formed from the triangle given below.
Solution:

Question 2.
Use the given figure to form a 3 × 3 magic square.

Solution:

Question 3.
Convert the tree diagram into a numeric expression

Solution:
(10 × 5) + (9 × 4)

Question 4.
(i) Find out the total time taken by the bus to reach from A to E via B , C and D.
(ii) Find which is the shortest route from A to E.
Solution:

(i) Route ⇒ 2A ➝ B ➝ C ➝ D ➝ E
Time taken ⇒ (7 + 5 + 3 + 6) hrs = 21 hrs

Try this (TextBook Page No 127)

Question 1.
Kumar has four different hats. He always wears a hat. Sometimes he wears the same hat more than once in a week as in the figure.

In how many different ways might he decide to wear his hats in one week?
Solution:

H1, H2, H3, H4 be Hat 1, Hat 2, Hat 3, Hat 4 respectively.
∴ In 7 × 4 = 28 different ways kumar may decide to wear his hat in one week.

Exercise 5.1

Try these (TextBook Page No 134)

Question 1.
Find any four SETs among these set of cards.

Solution:

Question 2.
This is an example for a magic square in SETs, can you make another one?

Solution:

Try this (TextBook PageNo 137)

Question 1.
Think why this graph colouring is invalid?

Solution:
The objective of graph colouring is to assign minimum number of colours to the verties do not have the same colour.
Here adjacent edges have the same colour.
∴ It is invalid.

Students can Download Maths Chapter 1 Rational Numbers Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 1 Rational Numbers Intext Questions

Exercise 1.1
Try This (Textbook Page No. 6)
Question 1.
Is the number \(\frac{7}{-5}\) a fraction or a rational number ? Why?
Solution:
\(\frac{7}{-5}\) is a rational number
Because of rational number is a rational number which is of the form \(\frac{p}{q}\) , q ≠ 0 and p and q are integers.
But fraction is part of a whole.

Question 2.
Write any 6 rational numbers of your choice.
Solution:
0, \(\frac{-1}{2}\), \(\frac{1}{2}\), \(\frac{3}{4}\), \(\frac{6}{7}\), -5, 6

Try This (Textbook Page No. 7)

Question 1.
Explain why the following statements are true?
(i) 0.8 = \(\frac{4}{5}\)
(ii) 1.4 > \(\frac{1}{4}\)
(iii) 0.74 < \(\frac{3}{4}\) (iv) 0.4 > 0.386
(v) 0.096 < 0.24
(vi) 1.128 = 0.1280
Solution:


(vi) 0.128 = \(\frac{128}{1000}=\frac{1280}{10000}\)
0.1280 = \(\frac{1280}{10000}\)
∴ Both are equal. i.e., 0.128 = 0.1280

Try This (Textbook Page No. 9)

Question 1.
Which of the pairs are equivalent rational numbers?

Solution:

Question 2.
Find the standard form of
(i) \(\frac{36}{-96}\)
(ii) \(\frac{-56}{-72}\)
(iii) \(\frac{27}{18}\)
Solution:

Question 3.
Mark the following rational numbers on a number line.
(i) \(\frac{-2}{3}\)
(ii) \(\frac{-8}{-5}\)
(iii) \(\frac{5}{-4}\)
Solution:
(i) \(\frac{-2}{3}\) lies between 0 and -1.
The unit part between 0 and -1 is divided into 3 equal parts and second part is taken.

(ii) \(\frac{-8}{-5}=1 \frac{3}{5}\)
\(1 \frac{3}{5}\) lies between 1 and 2. The unit part between 1 and 2 is divided into 5 equal parts and the third part is taken.

(iii) \(\frac{5}{-4}=-\frac{5}{4}=-1 \frac{1}{4}\)
\(-1 \frac{1}{4}\) lies between -1 and The unit part between -1 and -2 is divided into four equal parts and the first part is taken.

Try This (Textbook Page No. 15)

Question 1.
Are there any rational numbers between \(\frac{-7}{11}\) and \(\frac{6}{-11}\) ?
Solution:

Try This (Textbook Page No. 19)

Question 1.
Divide: (i) 5 by \(\frac{-7}{3}\)
(ii) \(\frac{-7}{3}\) by 5
Solution:

Exercise 1.2
Try This (Textbook Page No. 25)

Question 1.
The closure property on integers holds for subtraction and not for division. What about rational numbers? Verify.
Solution:
Let 0 and \(\frac{1}{2}\) be two rational numbers 0 – \(\frac{1}{2}\) = \(\frac{1}{2}\) is a rational number
∴ Closure property for subtraction holds for rational numbers.
But consider the two rational number \(\frac{5}{2}\) and 0.
\(\frac{5}{2} \div 0=\frac{5}{2 \times 0}=\frac{5}{0}\)
Here denominator = 0 and it is not a rational number.
∴ Closure property is not true for division of rational numbers.

Question 2.
Check whether \(\frac{3}{5}-\frac{7}{8}=\frac{7}{8}-\frac{3}{5}\)
Solution:

∴ Subtraction of rational numbers is not commutative

Question 3.
Is \(\frac{3}{5} \div \frac{7}{8}=\frac{7}{8} \div \frac{5}{3}\)? So, what do you conclude?
Solution:

∴ Commutative property not hold good for division of rational numbers.

Try This (Textbook Page No. 26)

Question 1.
Check whether the associative property holds for subtraction and division.
Solution:
Consider the rational numbers \(\frac{2}{3}\), \(\frac{1}{2}\) and \(\frac{3}{4}\)


∴ Associative property does not hold for division of rational numbers

Question 2.
Observe that

Use your reasoning skills, to find the sum of the first 7 numbers in the pattern given above.
Solution:

Students can Download Maths Chapter 5 Information Processing Ex 5.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 5 Information Processing Ex 5.1

I. Choose the best answer.

Question 1.
How many outcomes can you get when you toss three coins once?
(A) 6
(B) 8
(C) 3
(D) 2
Solution:
(B) 8

Question 2.
In how many ways can you answer 3 multiple choice questions, with the choices A, B, C and D ?
(1) 4
(2) 3
(3) 12
(4) 64
Solution:
(4) 64

Question 3.
How many 2 digit numbers contain the number 7 ?
(A) 10
(B) 18
(C) 19
(D) 20
Solution:
(B) 18

II. Answer the following.

Question 1.
You are going to have an ice cream or a cake. There are three flavours (chocolate, strawberry , vanilla) in ice creams, and two flavours (orange or red velvet) in the cake. In how many possible ways can you choose an ice cream or the cake?

Solution:
We are going to have either a ice cream or a cake.
Ice cream can be selected from 3 flavors and cake from two flavors. Both the events cannot occur simultaneously selecting ice cream and cake.
∴ Number of possible ways = 3 + 2
= 5 ways

Question 2.
In how many ways, can the teacher choose 3 students, in all one each from 10 students in VI std, 15 students in VII std and 20 students in VIII std to go to an excursion?
Solution:
The teacher is going to selected one student from class VI out of 10 students in 10 ways from
class VII out of 15 students in 15 ways and from class VIII out of 20 students in 20 ways.
∴ Number of ways 3 students can be selected
= 10 + 15 + 20 = 45 ways

Question 3.
If you have 2 school bags and 3 water bottles then, in how many different ways can you carry a school bag and a water bottle, while going to school?
Solution:
We can select one school bag from 2 and one bottle from 3 as follows.

∴ A bag and a water bottle can be selected in 2 × 3 = 6 ways.

Question 4.
Roll numbers are created with a letter followed by 3 digits in it. From the letter A, B, C, D, E and any 3 digits from 0 to 9. Then, in how many possible ways can the roll numbers be generated?
Solution:
We have a letter followed by 3 digits in the roll number.
The letter is selected from A, B, C, D, E.
For these 5 letters we have to select a 3 digit number using the digits 0 to 9.
Once place can be formed using any one of the 10 number 0 to 9 in 10 ways.
Tens place can be formed in 10 ways.
∴ A two digit number can be formed in 10 × 10 = 100 ways.
Thousands place can be formed in 10 ways
∴ A 3 digit number can be formed in 10 × 10 × 10 = 1000 ways.
∴ 5 letters can be attached in 5 × 1000 = 5000 ways.
∴ The roll number can be formed in 5000 ways.

Question 5.
A safety locker in a jewel shop requires a 4 digit unique code. The code has the digits from 0 to 9. How many unique codes are possible ?

Solution:
The unique code has 4 digits.
Each digit is formed using any of the 10 numbers from 0 to 9.
∴ Single digit number can be formed in 10 ways.
A double digit number can be formed in 10 × 10 ways.
A three digit number can be formed in 10 × 10 × 10 ways.
A four digit number can be formed in 10 × 10 × 10 × 10 ways. = 10,000 ways

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