Category: Class 8

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 7 Information processing Ex 7.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 7 Information processing Ex 7.4

Question 1.
Find the best buy of the following purchases:
(i) A pack of 5 chocolate bars for ₹ 175 or 3 chocolate bars for ₹ 114?
Answer:

Best buy is a packet of 5 chocolate bars for ₹ 175

(ii) Basker buy 1\(\frac { 1 }{ 2 }\) dozen of eggs for 81 and Aruna buy 15 eggs for ₹ 64.50?
Answer:

Best buy is 15 eggs for ₹ 64.5

Question 2.
Using the given picture find the total special offer price of fresh sweets and bakery products to buy \(\frac { 1 }{ 2 }\) kg laddu, 1 kg cake, 6 pockets of bread.

Answer:

Question 3.
Using the given picture prepare a price list.
Suppose you planto buy 1\(\frac { 1 }{ 2 }\) kg of apple, 2 kg of pomegranate, 2 kg of banana, 3 kg of mango, \(\frac { 1 }{ 2 }\) kg of papaya, 3 kg of onion, \(\frac { 1 }{ 2 }\) kg of tomato, and 1 kg of carrot in shop 1, how much will you save compared to shop 2.
Shop 1

Shop 2

Answer:


Price in shop 2 – Shop 1 = 715 – 675.75
= ₹ 39.25
We can save ₹ 39.25 in shop 1 compared to shop 2.

Question 4.
You want to buy some grocery items as per your shopping list that are given in the picture with their price. Also you have a bag that capacity of carrying 7 kg. Using weight ratio approach tabulate to find the total price and how much can
you buy more grocery items within your budget of ₹ 1000 and not exceeding 7 kg.
Shopping list
1. 2kg of red chili
2. 2 kg of coriander
3. 1 kg of garlic
4. 1 kg of tamarind
5. 2 kg of toor dal

Red chilli

₹ 141/1 kg

Coriander

₹ 130/1 kg

Garlic

₹ 82/1 kg

Tamarind

₹ 99/1 kg

Toor dal

₹ 78/1 kg
Answer:

Objective Type Questions

Question 5.
Online or television advertisements influence people on spending decisions by
(a) using special music
(b) making them think that they need the item
(c) using attractive pictures
(d) all the above
Answer:
(d) all the above

Question 6.
When I go shopping, I will buy
(a) something that looks attractive
(b) something my friend has
(c) something that I need to purchase
(d) the first thing I see in the store
Answer:
(c) something that I need to purchase

Question 7.
The best shopping choice is to
(a) shop at brand name stores always buy
(b) compare the choices before buying
(c) the same thing my friends bought
(d) buy at a regular shop always
Answer:
(b) compare the choices before buying

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 7 Information processing Ex 7.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 7 Information processing Ex 7.3

Question 1.
Fill in the blanks (Use Atbash Cipher that is given in code 3)
Hint:
For this question, we need to use Atbash cipher. For Atbash cipher, first we write the alphabets from A to Z and then in reverse from Z to A below that.
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Z Y X W V U T S R Q P O N M L K J I H G F E D C B A

(i) G Z N R O = _________
Answer:
TAMIL
Hint:
Now to solve, we look up the corresponding letter from the table to replace in code to get the actual word.
So, for G Z N R O, from table,
for G, it is T
for Z, it i sA
for N, it is M
for R, it is i
for 0, it is L
So, the actual word is TAMIL

(ii) V M T O R H S = ________
Answer:
ENGLISH

Hint:
VMTORHS
To solve, we look up the corresponding letter from table to replace in code to get the actual word.
For V, it is E
for M, it is N
for T, it is G
for O, it is L
for R, it is I,
for H, it is S
for S, it is H
Therefore we get E N G L I S H = ENGLISH

(iii) N Z G S V N Z G R X H _________
Answer:
MATHEMATICS
Hint:
Similarly as above for

(iv) H X R V M X V __________
Answer:
SCIENCE
Hint:
For

(v) H L X R Z O H X R V M X V = __________
Answer:
SOCIAL SCIENCE
Hint:

Question 2.
Match the following (a = 00 ………… Z = 25).

Answer:
(i) – c
(ii) – d
(iii) – a
(iv) – e
(v) – b
Hint:
(i) Mathematics is

So matching option is c

(ii) addition is

Matching option is d

(iii) Subtraction is

multiplication is a

(iv) multiplication is

Matching option is e
however instead of 25, it should be 08

(v) division is

Matching option is b

Question 3.
Frame Additive cipher table (key = 4).
Answer:
Step 1 : write all alphabets
Step 2 : Assign numbers to each alphabet starting from 00 till 25.
Step 3 : add key value (here it is 4) to the numbers assigned in step 2 to form cipher table

Question 4.
A message like “Good Morning” written in reverse would instead be “Doog Gninrom” In the same way decode the sentence given below:
“Ot dnatsrednu taht scitamehtam nac eb decneirepxe erehwreve ni erutan dna laer efil”
Answer:
Given that good morning written in reverse is doog gninrom.
We have to decode the below by reversing, so,
Ot dnatsrednu taht scitamehtam nac eh decneirepxe erehwreve ni erutan dna laer efil.

Ans: to understand that mathematics can be experienced everywhere in nature and real life.

Question 5.
Decode the given Pigpen Cipher text and compare your answer to get result.


I. The room number in which the treasure took place

Answer:
The room in which the treasure took place = 28

II. Place of the treasure

Answer:
The place of treasure = Chair

III. The name of the treasure

Answer:
Identity of treasure = Gift voucher.

Question 6.
Praveen recently got the registration number for his new two-wheeler. Here, the number is given in the form of mirror-image. Encode the image and find the correct registration number of praveen’s two-wheeler.
T N 1 2 H 2 5 8 9
(a)

(b)

(c)

(d)

Answer:
The mirror image is

When we place an imaginary mirror & visualize the image seen in the mirror, we will get the below.

∴ The answer option c

Objective Type Questions

Question 7.
In questions (i) and (ii), there are four groups of letters in each set. Three of these sets are alike in some way while one is different. Find the one which is different.
(i).(A) C R D T
(B) A P B Q
(C) E U F V
(D) G W H X
Answer:
(i) (A) C R D T
Hint:
The four groups of letters are
CRDT APBQ EUFV GWHX
The above can be written as

We find that when we take 1^st & 3^rd letter & 2^nd & 4^th letter as 2 pairs, the 3 letter is the next letter alphabetically to the 1st letter.
Similarly the 4^th letter is alphabetically the next letter of the 2^nd letter.
i.e CD, AB, EF, GH & PQ, UV, WX
Only in CRDT, we have T instead of ‘S’
So, Ans: in CRDT ⇒ Option (a)

(ii). (A) H K N Q
(B) I L O R
(C) J M P S
(D) A D G J
Answer:
(D) A D G J
Hint:
The four groups of letters are
HKNQ ILOR IMPS ADGI
If we notice, we find that 2 letters are missing in the sequence. le.
H_IJ K_LI N_OP Q
I_JK L_MN O_PQ R
J_KL M_NO P_QR S
A_BC D_EF G_H I
We find that only in ADGI, the difference is only one letter between G & I.
Hence it is the odd one out.

Question 8.
A group of letters are given. A numerical code has been given to each letter. These letters have to be unscrambled into a meaningful word. Find out the code for the word so formed from the 4 answers given.

(A) 2 3 4 1 5 6
(B) 5 6 3 4 2 1
(C) 6 1 3 5 2 4
(D) 4 2 1 3 5 6
Answer:
(B) 5 6 3 4 2 1

Hint:
Given code is

Option (a) is 234156. When we substitute number for each letter from code, we get,

Option (b) is 563421, similarly, we get

Option (c) is

Option d) is

So, only in option b, we get a meaningful word, i.e PENCIL.
Hence, answer is Option b.

Question 9.
Questions (iii) and (iv) are based on code language. Find the correct answer from the four alternatives given.
(iii)Inacertain code, ‘M E D I C I N E’ is coded as ‘E O J D J E F M’,then how is ‘C O M P U T E R’ written in the same code?
(A) C N P R V U F Q
(B) C M N Q T U D R
(C) R F I J V Q N P C
(D) R N V F T U D Q
Answer:
(C) R F I J V Q N P C
Hint:
lt is given that in a certain code MEDlClNE is coded as E O J D J E F M
When we observe the word & the code, we find that, there is a pattern.

[to understand, see the matching shapes]
To get the code from the word, we follow the below steps

1.

E [E D I C I N] M

2. For the middle letters, replace the letters with their alphabetically next letters, so we get

3. Now we have to reverse the order of the middle letters in the bracket, so we get
E [O J D J E F] M = E O J D J E F M
Thus we get the code
So similarly, we have to follow the 3 steps to get code, therefore:

Step 1: Swap 1st & last letters, so we get
R [O M P U T E] C

Step 2: For the middle letters, replace the letters with alphabetically next letters, so we get
R [P N Q V L F] C

Step 3: Reverse the word of letters in the bracket to we get
R [F U V Q N P] C
∴ Ans is R F U V Q N P C ⇒ option c

(iv) lf the word ’P H O N E’ is coded as ’S K R Q H’, how will ’R A D I O’ be coded?
(A) S C G N H
(B) V R G N G
(C) U D G L R
(D) S D H K Q
Answer:
(C) U D G L R
Hint:
If PHONE is coded as S K R Q H
Find that code for R A D I O

We find that we get the code, by 3 letter alphabetically so. from P. skipping Q & R. we get S

Similarly, from H, skipping I & J, we get K
Like wise for R A D I O. skipping the 2 alphabets,
From R skip S & T → U
From A skip B & C → D
From D skip E & F → G
From l skip J & K → L
From O skip P & Q → R

∴ Ans is UDGLR ⇒ option c

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 7 Information processing Ex 7.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 7 Information processing Ex 7.2

Question 1.
Using repeated division method, find the HCF of the following:
(i) 455 and 26
Answer:

Step 1: The larger number should be dividend 455 & smaller number should be divisor = 26
Step 2: After 1st division, the remainder becomes new divisor & the previous divisor becomes next dividend.
Step 3: This is done till remainder is zero.
Step 4: The last divisor is the HCF L.
∴ Ans: HCF is 13.

(ii) 392 and 256
Answer:
256 is smaller, so it is the 1st divisor

∴ HCF = 8

(iii) 6765 and 610
Answer:

∴ HCF = 5

(iv) 184, 230 and 276
Answer:
First let us take 184 & 230

∴ 46 is the HCF of 184 and 230
Now the HCF of the first two numbers is the dividend for the third number.

∴ Ans: HCF of 184, 230 & 276 is 46

Question 2.
Using repeated subtraction method, find the HCF of the following:
(i) 42 and 70
Answer:
Let number be m & n m > n
We do, m – n & the result of subtraction becomes new ‘m’. if m becomes less than n,
we do n – m and then assign the result as n. We should do this till m n. When m = n then ‘m’ is the HCF.
42 and 70
m = 70 n = 42
70 – 42 = 28

now m = 42, n = 28
42 – 28 = 14.

now m = 28, n = 14
28 – 14 = 14.

now m = 14. n = 14;
we stop here as m = n
∴ HCF of 42 & 70 is 14

(ii) 36 and 80
Answer:

28 – 8 = 20
20 – 8 = 12
12 – 8 = 4
8 = 4 = 4
now m = n = 4
∴ HCF is 4

(iii) 280 and 420
Answer:
Let m = 420, n = 280
m – n = 420 – 280 = 140
now m = 280, n = 140
m – n = 280 – 140 = 140
now m = n = 140
∴ HCF is 140

(iv) 1014 and 654
Answer:
Let m = 1014, n = 654
m – n = 1014 – 654 = 360
now m = 654, n = 360
m – n = 654 – 360 = 294
now m = 360, n = 294
m – n = 360 – 294 = 66
now m = 294 n = 66
m – n = 294 – 66 = 228
now m = 66, n = 228
n – m = 228 – 66 = 162
now m = 162, n = 66
= 162 – 66 = 96
n – m = 96 – 66 = 30
Similarly, 66 – 30 = 36
36 – 30 = 6
30 – 6 = 24
24 – 6 = 18
18 – 6 = 12
12 – 6 = 6 now m = n
∴ HCF of 1014 and 654 is 6

Question 3.
Do the given problems by repeated subtraction method and verify the result.
(i) 56 and 12

Answer:
Let n = 56 & n = 12
m – n = 56 – 12 = 44
now m = 44, n = 12
m – n = 44 – 12 = 32
m – n = 32 – 12 = 20
m – n = 20 – 12 = 8
n – m = 12 – 84
m – n = 8 – 4 = 4. now m = n
∴ HCF of 56 & 12 is 4

(ii) 320, 120 and 95

Answer:
Let us take 320 & 120 first m = 320, n = 120
m – n = 320 – 120 = 200
m = 200, n = 120
∴ m – n = 200 – 120 = 80
120 – 80 = 40
80 – 40 = 40
∴ m = n = 40 → HCF of 320, 120
Now let us find HCF of 40 & 95
m = 95, n = 40
∴ m – n = 95 – 40 = 55
55 – 40 = 15
40 – 15 = 25
25 – 15 = 10
15 – 10 = 5
HCF of 40 & 95 is 5
10 – 5 = 5
∴ HCF of 320 120 & 95 is 5

Question 4.
Kalai wants to cut identical squares as big as she can, from a piece of paper measuring 168mm and by 196mm. What is the length of the side of the biggest square? (To find HCF using repeated subtraction method)
Answer:
Sides are 168 & 196
To find HCF of 168 & 196, we are to use repeated subtraction method.
∴ m = 196, n = 168
m – n = 196 – 168 = 28

now n = 28, m = 168
m – n = 168 – 28 = 140

now m = 140, n = 28
m – n = 140 – 28 = 112

now m = 112, n = 28
m – n = 112 – 28 = 84

now m = 84, n = 28
m – n = 84 – 28 = 56

now m = 56, n = 28
m – n = 56 – 28 = 28
∴ HCF is 28
∴ Length of biggest square is 28

Objective Type Questions

Question 5.
What is the eleventh Fibonacci number?
(a) 55
(b) 77
(c) 89
(d) 144
Answer:
(c) 89
Hint:

∴ 11^th Fibonacci number is 89

Question 6.
If F(n) is a Fibonacci number and n = 8, which of the following is true?
(a) F(8) = F(9) + F( 10)
(b) F(8) = F(7) + F(6)
(c) F(8) = F(10) × F(9)
(d) F(8) = F(7) – F(6)
Answer:
(b) F(8) = F(7) + F(6)
Hint:
Given F(n) is a Fibonacci number & n = 8
∴ F(8) = F(7) + F (6) as any term in Fibonacci series is the sum of preceding 2 terms

Question 7.
Every 3^rd number of the Fibonacci sequence is a multiple of_______
(a) 2
(b) 3
(c) 5
(d) 8
Answer:
(a) 2
Hint:
Every 3^rd number in Fibonacci sequence is a multiple of 2

Question 8.
Every _____ number of the Fibonacci sequence is a multiple of 8
(a) 2^th
(b) 4^th
(c) 6^th
(d) 8^th
Answer:
(c) 6^th

Question 9.
The difference between the 18^th and 17^th Fibonacci number is
(a) 233
(b) 377
(c) 610
(d) 987
Answer:
(d) 987
Hint:
F(18) = F(17) + F(16)
F(18) – F(17) = F(16) = F(15) + F(14)
= 610 + 377 = 987

Question 10.
Common prime factors of 30 and 250 are
(a) 2 × 5
(b) 3 × 5
(c) 2 × 3 × 5
(d) 5 × 5
Answer:
(a) 2 × 5
Prime factors of 30 are 2 × 3 × 5
Prime factors of 250 are 5 × 5 × 5 × 2
∴ Common prime factors are 2 × 5

Question 11.
Common prime factors of 36,60 and 72 are
(a) 2 × 2
(b) 2 × 3
(c) 3 × 3
(d) 3 × 2 × 2
Answer:
(d) 3 × 2 × 2
Hint:
Prime factors of 36 are 2 × 2 × 3 × 3
Prime factors of 60 are 2 × 2 × 3 × 5
Prime factors of 72 are 2 × 2 × 2 × 3 × 3
∴ Common prime factors are 2 × 2 × 3

Question 12.
Two numbers are said to be co-prime numbers if their HCF is
(a) 2
(b) 3
(c) 0
(d) 1
Answer:
(d) 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 7 Information processing Ex 7.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 7 Information processing Ex 7.1

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

Answer:
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.
Shanthi has 5 chudithar sets and 4 frocks. In how many possible ways, can she wear either a chudithar or a frock?
Answer:
Shanthi her 5 chudidhar sets and 4 frocks.
She wear either chudidhar or a frock.
∴ Total possible ways = 5 + 4 = 9 ways

Question 3.
In a Higher Secondary School, the following groups are available in XI standard
I. Science Group:
(i) Physics, Chemistry, Biology and Mathematics
(ii) Physics, Chemistry, Mathematics and Computer Science.
(iii) Physics, Chemistry, Biology and Home Science

Il. Arts Group:
(i) Accountancy, Commerce, Economics and Business Maths
(ii) Accountancy, Commerce, Economics and Computer Science
(iii) History, Geography, Economics and Commerce

III. Vocational Group:
(i) Biology, Nursing Theory, Nursing Practical I and Nursing Practical II
(ii) Home Science, Textiles and Dress Designing Theory, Textiles and Dress
Designing Practical I and Textiles and Dress Designing Practical Il
In how many possible ways, can a student choose a group?
Answer:
The student either select any one of science group in 3 ways or any of the arts group in 3
ways or any of the vocational group in 2 ways.
∴ Total possible ways = 3 + 3 + 2 = 8 ways

Question 4.
If you have 2 school bags and 3 water bottles then, in how many different ways can you choose each one of them, while going to school?
Answer:
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 5.
Roll numbers are created with a letter followed by 3 digits in it, from the letters A, B, C, D and E and any 3 digits from 0 to 9. In how many possible ways can the roll numbers be generated? (except A000, B000, C000, D000 and E000)
Answer:
We have a letter followed by 3 digits in the roll number.
The letter is selected from the five letters A, B, C, D, E.
For these 5 letters we have to select a 3 digit number using the digits O to 9.
Ones place can be formed using any one of the 10 number 0 to 9 in 10 ways.
Tens place can be formed in I O ways.
∴ A two digit number can be formed in 10 × 10 = 100 ways.
Thousands place can be formed in lo 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 – 5 = 4995 ways.

Question 6.
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?

Answer:
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

Question 7.
An examination paper has 3 sections, each with five questions and students are instructed to answer one question from each section. In how many different ways of can the questions be answered?
Answer:
The tree diagram for this may be

∴ Number of possible ways to select one questions from each of 3 sections is 3 × 5 = 15 ways

Question 8.
The given spinner is spun twice and the two numbers got are used to form a 2 digit number. How many different 2 digits numbers are possible?

Answer:
On the first spin we get any of the five numbers to form ones place then insecond spin the number got will fill 10’s place.
∴ Number of ways = 5 × 5 = 25 ways.
Removing the repetitions (11, 22, 33, 44, 55) once we get 25 – 5 = 20 ways.
20 different two digit numbers are possible

Question 9.
Ramya wants to paint a pattern in her living room wall with minimum budget. Help her to colour the pattern with 2 colours but make sure that no two q adjacent boxes are the same colour. The pattern is shown in the picture.

Answer:

Question 10.
Colour the regions in the maps with few colours as possible but make sure that no two adjacent countries are of the same colour.

Answer:

Objective Type Questions

Question 11.
In a class there are 26 boys and 15 girls. The teacher wants to select a boy or a girl to represent a quiz competition. In how many ways can the teacher make this selection?
(A) 41
(B) 26
(C) 15
(D) 390
Answer:
41

Question 12.
How many outcomes can you get when you toss three coins once?
(A) 6
(B) 8
(C) 3
(D) 2
Answer:
8

Question 13.
In how many ways can you answer 3 multiple choice questions, with the choices A,B,C and D?
(A) 4
(B) 3
(C) 12
(D) 64
Answer:
64

Question 14.
How many 2 digit numbers contain the number 7?
(A) 10
(B) 18
(C) 19
(D) 20
Answer:
18

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 6 Statistics InText Questions Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 6 Statistics InText Questions

Try These (Text Book page No. 212)

Question 1.
Arrange the given data in ascending and descending order: 9, 34, 4, 13, 42, 10, 25, 7, 31, 4, 40
Answer:
Ascending order: 4, 4, 7, 9, 10, 13, 25, 31, 34, 40, 42.
Descending order: 42, 40, 34, 31, 25, 13, 10, 9, 7, 4, 4

Question 2.
Find the_range of the given data: 53, 42,61, 9, 39, 63, 14, 20,06, 26, 31, 4, 57
Answer:
Ascending order of the given data:
4, 6, 9, 14, 20, 26, 31, 39, 42, 53, 57, 61, 63
Here largest value = 63
Smallest value = 4
∴ Range = Largest value – smallest value
= 63 – 4 = 59

Try These (Text Book page No. 217)

Question 1.
Prepare a frequency table for the data:
3, 4, 2, 4, 5, 6, 1, 3, 2, 1, 5, 3, 6, 2, 1, 3, 2, 4
Answer:
Ascending order of the given data.
1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 6
The distribution table:

Data	Tally marks	Frequency
1	|||	3
2	||||	4
3	||||	4
4	|||	3
5	||	2
6	||	2

∴ Frequency Table:

Question 2.
Prepare a grouped frequency table for the data:
10, 9, 3, 29, 17, 34, 23, 20, 39, 42, 5, 12, 19, 47, 18, 19, 27, 7, 13, 40, 38, 24,34,15,40
Answer:
Largest value = 47
Smallest value = 3
Range = Largest value – Smallest value
= 47 – 3 = 44
Suppose we take class size as 10, then Number of class intervals possible

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Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 6 Statistics Ex 6.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 6 Statistics Ex 6.3

Miscellaneous Practice Problems

Question 1.
Draw a pie chart for the given table.

Answer:
Converting the area in percentage into components parts of 360°. we have.

Continental Area.

Question 2.
The data on modes of transport used by the students to come to school are given below. Draw a pie chart for the data.

Answer:
Converting the percentage into components parts of 360°. we have

Mode of Transport by students.

Question 3.
Draw a histogram for the given frequency distribution.

Answer:
The given distribution is discontinuous.
Lower boundary = lower limit – \(\frac { 1 }{ 2 }\) (gap between the adjacent class interval)
= 41 – \(\frac { 1 }{ 2 }\)(1) = 40.5
Upper boundary = Upper limit + \(\frac { 1 }{ 2 }\) (gap between the adjacent class interval)
= 45 + \(\frac { 1 }{ 2 }\)(1) = 45.5
Now continuous frequency table is as below

Question 4.
Draw a histogram and the frequency polygon in the same diagram to represent the following data.

Answer:
The given distribution is discontinuous.
Lower boundary = lower limit – \(\frac { 1 }{ 2 }\) (gap between the adjacent class interval)
= 50 – \(\frac { 1 }{ 2 }\)(1) = 49.5
Upper boundary = Upper limit + \(\frac { 1 }{ 2 }\)(gap between the adjacent class interval)
= 55 + \(\frac { 1 }{ 2 }\)(1) = 55.5
∴ The continuous frequency table is as below.

Challenging problems

Question 5.
Form a continuous frequency distribution table and draw histogram from the following data.

Age (in years)	No of persons
Under 5	1
Under 10	12
Under 15	19
Under 20	26
Under 25	27
Under 30	35
Under 35	38
Under 40	45
Under 45	48
Under 50	53

Answer:
Converting into continuous distribution we have

Class interval	No. of persons
0 – 5	1
5 – 10	11
10 – 15	7
15 – 20	7
20 – 25	1
25 – 30	8
30 – 35	3
35 – 40	7
40 – 45	3
45 – 50	5
Total	53

Question 6.
A rupee spent in a cloth manufacturing company is distributed as follows. Represent this in a pie chart.

Particulars	Paise
Farmer	20
Spinner	35
Dyer	15
Weaver	15
Printer	05
Salary	10

Answer:
1 Rupee = 100 paise.

Expenditure of a cloth manufacturing company.

Question 7.
Draw a histogram for the following data.

Answer:
Since mid values are given, the given distributors is discontinuous.
Lower boundary lower limit – \(\frac { 1 }{ 2 }\)(gap between the adjacent class interval)
= 15 – \(\frac { 1 }{ 2 }\)(10) = 10
Upper boundary = Upper limit + (gap between the adjacent class interval)
= 15 + \(\frac { 1 }{ 2 }\)(10) = 20
The continuous distribution will he as folIows.

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 6 Statistics Ex 6.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 6 Statistics Ex 6.2

Question 1.
Which of the following data can be represented in a histogram?
(i) The number of mountain climbers in the age group 20 to 60 in TamilNadu.
Answer:
Yes

(ii) Production of cycles in different years.
Answer:
No

(iii) The number of students in each class of a school.
Answer:
No

(iv) The number votes polled from 7 am to 6 pm in an election.
Answer:
Yes

(v) The wickets fallen from 1 over to 50th over in a one day cricket match.
Answer:
Yes

Question 2.
Fill in the blanks:
(i) The total area of the histogram is _________ to the total frequency of the given data.
Answer:
proportional

(ii) A graph that displays data that changes continuously over the periods of time is _________ .
Answer:
Histogram

(iii) Histogram is a graphical representation of_________ data.
Answer:
grouped

Question 3.
In a village, there are 570 people who have cell phones. An NGO survey their cell phone usage. Based on this survey a histogram is drawn. Answer the following questions.

(i) How many people use the cell phone for less than 3 hours?
Answer:
330 people (110 + 220)

(ii) How many of them use the cell phone for more than 5 hours?
Answer:
150 of them (100 + 50)

(iii) Are people using cell phone for less than 1 hour?
Answer:
No

Question 4.
Draw a histogram for the following data.

Answer:
The given (tata IS continuous frequency distribution taking class intervals on X axis and No. of students on Y-axis, the histogram is given below.

Question 5.
Construct a histogram from the following distribution of total marks of 40 students in a class.

Answer:
The given distribution is continuous taking marks on X axis and No. of students on Y-axis the histogram is constructed.

Question 6.
The distribution of heights (in cm ) of loo people is given below. Construct a histogram and the frequency polygon imposed on it.

Answer:
The given distribution is discontinuous.
Converting into continuous distribution we have
Lower boundary = lower limit – \(\frac { 1 }{ 2 }\) (gap between adjacent class interval)
= 125 – \(\frac { 1 }{ 2 }\) (1) = 124.5
Upper boundary = Upper limit + \(\frac { 1 }{ 2 }\) (gap between the adjacent class interval)
= 135 + \(\frac { 1 }{ 2 }\) = 135.5
∴ The new frequency table is

Question 7.
In a study of dental problem, the following data were obtained.

Represent the above data by a frequency polygon.
Answer:
Finding the midpoints of the class interval we get.

Ages	Mid point (x)	No. of patients
0 – 10	5	5
10 – 20	15	13
20 – 30	25	25
30 – 40	35	14
40 – 50	45	30
50 – 60	55	35
60 – 70	65	43
70 – 80	75	50

The points tobe plotted are A(5,5), B(15, 13), C(25, 25), D(35, 14), E(45, 30), G(65, 43.), H(75, 50) to obtain the frequency polygon ZABCDEFGHI.
Where I imagined class between 80 and 90.

Question 8.
The marks obtained by 50 students in Mathematics are given below (j) Make a frequency distribution table taking a class size of 10 marks (ii) Draw a histogram and a frequency polygon.

Answer:
Maximum marks obtained = 89
Minimum marks obtained = 08
Range = Maximum marks – Minimim marks
= 89 – 08
= 81
Taking the class size = 10, then

= \(\frac{81}{10}\) = 8.1 = 9

Now we have the continuous frequency table.

We will draw the histogram taking class interval in x-axis and frequency in y-axis as follows.

Objective Type Questions

Question 9.
Data is a collection of _________
(A) numbers
(B) words
(C) measurements
(D) all the three
Answer:
(D) all the three

Question 10.
The number of times an observation occurs in the given data is called _________
(A) tally marks
(B) data
(C) frequency
(D) none of these
Answer:
(C) frequency

Question 11.
The difference between the largest value and the smallest value of the given data is ________
(A) range
(B) frequency
(C) variable
(D) none of these
Answer:
(A) range

Question 12.
The data that can take values between a certain range is called_________
(A) ungrouped
(B) grouped
(C) frequency
(D) none of these
Answer:
(B) grouped

Question 13.
Inclusive series is a _________series.
(A) continuous
(B) discontinuous
(C) both
(D) none of these
Answer:
(B) discontinuous

Question 14.
In a class interval the upper limit of one class is the lower limit of the other class. This is _________ series.
(A) Inclusive
(B) exclusive
(C) ungrouped
(D) none of these
Answer:
(B) exclusive

Question 15.
The graphical representation of ungrouped data is ________
(A) histogram
(B) frequency polygon
(C) pie chart
(D) all the three
Answer:
(C) pie chart

Question 16.
Histogram is a graph of a ________ frequency distribution.
(A) continuous
(B) discontinuous
(C) discrete
(D) none of these
Answer:
(A) continuous

Question 17.
A _________ is a line graph for the graphical representation of the continuous frequency
distribution.
(A) frequency polygon
(B) histogram
(C) pie chart
(D) bar graph
Answer:
(A) frequency polygon

Question 18.
The graphical representation of grouped data is _________
(A) bar graph
(B) pictograph
(C) pie chart
(D) histogram
Answer:
(D) histogram

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 6 Statistics Ex 6.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 6 Statistics Ex 6.1

Question 1.
Fill in the blanks:
(i) Data has already been collected by some other person is ________ data.
Answer:
Secondary

(ii) The upper limit of the class interval (25-35) is ________ .
Answer:
35

(iii) The range of the data 200, 15, 20, 103, 3, 196, is ________ .
Answer:
197

(iv) If a class size is 10 and range is 80 then the number of classes are ________ .
Answer:
8

(v) Pie chart is a ________ graph.
Answer:
circular

Question 2.
Say True or False:
(i) Inclusive series is a continuous series.
Answer:
False

(ii) Comparison of parts of a whole may be done by a pie chart.
Answer:
True

(iii) Media and business people use pie charts.
Answer:
True

(iv) A pie diagram is a circle broken down into component sectors.
Answer:
True

Question 3.
Represent the following data in ungrouped frequency table which gives the number of children in 25 families.
1, 3, 0, 2, 5, 2, 3, 4, 1, 0, 5, 4, 3, 1, 3, 2, 5, 2, 1, 1, 2, 6, 2, 1, 4
Answer:
The data given is raw data.
Ascending order : 0, 1, 2, 3, 4, 5, 6

∴ Tabulating in frequency distribution table we get

Question 4.
Form a continuous frequency distribution table for the marks obtained by 30 students in a X std public examination.
328, 470, 405, 375, 298, 326, 276, 362, 410, 255, 391, 370, 455, 229, 300, 183, 283, 366, 400, 495, 215, 157, 374, 306, 280, 409, 321, 269, 398, 200.
Answer:
Maximum mark obtained = 495
Minimum marks obtained = 157
Range = Maximum value – Minimum value
Range = 495 – 157
= 338

If we take the class size as 50 then the number of class intervals possible

= \(\frac{338}{50}\) = 6.76
≅ 7

Question 5.
A paint company asked a group of students about their favourite colours and made a pie chart of their findings. Use the information to answer the following questions.
(i) What percentage of the students like red colour?
(ii) How many students liked green colour?
(iii) What fraction of the students liked blue?
(iv) How many students did not like red colour?
(v) How many students liked pink or blue?
(vi) How many students were asked about their favourite colours?

Answer:
Total percentage of students = 100 %
∴ 50students = 100% – (30% + 20% + 25% + 15%)
= 100% – 90%
50 students = 10%
10% of total students = 50
∴ \(\frac { 10 }{ 100 }\) (Total students) = 50
Total students = \(\frac{50 \times 100}{10}\) = 500.
Total students = 500.

(i) 20% of the students like red colour.

(ii) 15% of the students liked green colour.
\(\frac{15}{100}\) × 500 = 75 students liked green colour.

(iii) 25% students liked blue students liked blue.
⇒ \(\frac{25}{100}\) students liked blue.
⇒ \(\frac{1}{4}\) students liked blue.

(iv) Percentage of students liked red colour
= 20%
Percentage of students did not like red colour
= 100% – 20%
= 80%
∴ Number of students did not like red colour
= 80% of 500
= \(\frac{80}{100}\) × 500 = 400
400 students did not like red colour.

(v) Students liked pink or blue = students liked pink + students liked blue.
= 30% of 500 + 25% of 500
= \(\frac{30}{100}\) × 500 + \(\frac{25}{100}\) × 500
= 150 + 125
= 275

(vi) Total number of students = 500
500 students were asked about their favourite colour.

Question 6.
A survey gives the following information of food items preferred by people. Draw a Pie chart.

Answer:
Total number of people = 160 + 90 + 80 + 50 + 30 + 40 = 450
Converting the number of people prefer various food items into components part of 360°

Food items are preferred by people.

Question 7.
Income from various sources for Government of India from a rupee is given below. raw a pie chart.

Answer:

Income from various sources for Government of India in a rupee.

Question 8.
Monthly expenditure of Kumaran’s family is given below. Draw a suitable Pie chart.

Also
1. Find the amount spent for education if Kumaran spends ₹ 6000 for Rent.
2. What is the total salary of Kumaran?
3. How much did he spend more for food than education?
Answer:

Monthly expenditure of kumaran’s family.

1. Given Kumaran spends ₹ 6000 for Rent.
∴ 15% of’ total expenditure = 6000
\(\frac{15}{100}\) (Total Expenditure) = 6000
Total Expenditure = \(\frac{6000 \times 100}{15}\)
Total Expenditure = ₹ 40,000
Amount spend l’or education = 20% of total expenditure.
\(\frac{20}{100}\) × 40,000
= ₹ 8000

2. Total salary of Kumaran = ₹ 40,000

3. Amount spend for food = 50% of (40,000)
Amount spend for the food than education
= 20,000 – 8,000
= ₹ 12,000

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 5 Geometry InText Questions Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 5 Geometry InText Questions

Answer the following questions by recalling the properties of triangles:

Question 1.
The sum of the three angles of a triangle is _________ .
Answer:
180°

Question 2.
The exterior angle of a triangle is equal to the sum of the _______ angles to it.
Answer:
interior

Question 3.
In a triangle, the sum of any two sides is ________ than the third side.
Answer:
greater

Question 4.
Angles opposite to equal sides are ________ and vice – versa.
Answer:
Equal

Question 5.
What is ∠A in the triangle ABC?

Answer:
The exterior angle = sum of interior opposite angles.
∴ ∠A + ∠C = 1500 in ∆ABC
But ∠C = 40°
[∵Vertically opposite angler are equal]
∴ ∠A + ∠C = 150°
⇒ ∠A + ∠40° = 150°
∠A = 150° – 40°
∠A = 1100

Try These (Text Book page No. 157)

Identify the pairs of figures which are similar and congruent and write the letter pairs.

Answer:
Similar shapes:
(i) Wand L
(ii) B and J
(iii) A and G
(iv) B and J
(v) B and Y
(vi) E and N
(vii) H and Q
(viii) R and T
(ix) S and T

Congruent shapes:
(i) Z and I
(ii) J and Y
(iii) C and P
(iv) B and K
(v) R and S
(vi) I and Z
You can find more.

Try These (Text Book page No. 158)

Question 1.
Match the following by their congruence property

Answer:
1. – (iv),
2. – (iii),
3. – (i),
4. – (ii)

Think (Text Book page No. 160)

In the figure, DA = DC and BA = BC. Are the triangles DBA and DBC congruent? Why?

Answer:
Here AD = CD
AB = CB
DB = DB (common)
∆DB ≡ ∆DBC
[∵ By SSS congruency]
Also RHS rule also bind here to say their congruency.

Activity (Text Book page No. 169)

Question 1.
We can construct sets of Pythagorean triplets as follows.
Let m and n be any two positive integers (m > n):
(a, b, c) is a Pythagorean triplet if a = m² – n², b = 2mn and c = m² + n² (Think, why?)
Complete the table.

Answer:

Question 2.
Find all integer-sided right angled triangles with hypotenuse 85.
Answer:
(x + y)² – 2xy = 85²

Think (Text Book page No. 173)

Question 1.
In any acute angled triangle, all three altitudes are inside the triangle. Where will be the orthocentre? In the interior of the triangle or in its exterior?

Answer:
Interior of the triangle

Question 2.
In any right angled triangle, the altitude perpendicular to the hypotenuse is inside the triangle; the other two altitudes are the legs of the triangle. Can you identify the orthocentre in this case?

Answer:
Vertex containing 90°

Question 3.
In any obtuse angled triangle, the altitude connected to the obtuse vertex is inside the triangle, and the two altitudes connected to the acute vertices are outside the triangle. Can you identify the orthocentre in this case?

Answer:
Exterior of the triangle.

Try These (Text Book page No. 177)

Identify the type of segment required in each triangle:
(median, altitude, perpendicular bisector, angle bisector)

(i)

AD = _______
Answer:
AD = Altitude

(ii)

l₁ = _______
Answer:
l₁ = perpendicular bisector

(iii)

BD = _______
Answer:
BD = Median

(iv)

CD = _______
Answer:
CD = Angular bisector

Think (Text Book page No. 187)

Is it possible to construct a quadrilateral PQRS with PQ = 5 cm, QR = 3cm, RS = 6cm, PS = 7cm and PR = 10cm. If not,why?
Answer:

The lower triangle cannot be constructed as the sum of two sides 5 + 3 = 8 < 10 cm. So this quadrilateral cannot be constructed,

Try These (Text Book page No. 187)

Question 1.
The area of the trapezium is ________ .
Answer:
\(\frac { 1 }{ 2 }\) × h × (a + b) sq. units

Question 2.
The distance between the parallel sides of a trapezium is called as ________ .
Answer:
its height

Question 3.
If the height and parallel sides of a trapezium are 5cm, 7cm and 5cm respectively, then
its area is ________ .
Answer:
30

Hints:
= \(\frac { 1 }{ 2 }\) × h × (a + b) sq. units
= \(\frac { 1 }{ 2 }\) × 5 × (7 + 5) = \(\frac { 1 }{ 2 }\) × 5 × 12 = 30 sq. cm

Question 4.
In an isosceles trapezium, the non-parallel sides are _________ in length.
Answer:
equal

Question 5.
To construct a trapezium, _________ measurements are enough.
Answer:
Four

Question 6.
If the area and sum of the parallel sides are 60 cm² and 12 cm, its height is ________ .
Answer:
10 cm
Hint:
Area of the trapezium = \(\frac { 1 }{ 2 }\) × h (a + b)
60 = \(\frac { 1 }{ 2 }\) × h × (12)
∴ h = \(\frac{60 \times 2}{12}\) = 10cm

Activity (Text Book page No. 193 & 194)

Question 1.
A pair of identical 30° – 60° – 90° set-squares are needed for this activity. Place them as shown in the figure.

(i) What is the shape we get? It is a parallelogram.
Answer:

(ii) Are the opposite sides parallel?
Answer:
Yes

(iii) Are the opposite sides equal?
Ans:
Yes

(iv) Are the diagonals equal?
Answer:
No

(v) Can you get this shape by using any other pair of identical set-squares?
Answer:
Yes

Question 2.
We need a pair of 30° – 60° – 90° set- squares for this activity. Place them as shown in the figure.

(i) What is the shape we get’?
Ans:
Rectangle

(ii) Is it a parallelogram?
Answer:
Yes

(iii) It is a quadrilateral; infact it is a rectangle. (How?)
Ans:
Opposits sides are equal.
All angles = 90°

(iii) What can we say about its lengths of sides, angles and diagonals?
Discuss and list them out.
Answer:
Opposite sides are equal
All angles are equal and are = 90°
Diagonals arc equal

Question 3.
Repeat the above activity, this time with a pair of 45° – 45° – 90° set-squares.

(i) How does the figure change now? Is it a parallelogram? It becomes a square! (How did it happen?)
Ans:
All sides are equal

(ii) What can we say about its lengths of sides, angles and diagonals? Discuss and list them out.
Answer:
All sides are equal
All angles = 90°
Diagonals equal

(iii) How does it differ from the list we prepared for the rectangle?
Answer:
All sides are equal.
Diagonals bisects each other.

Question 4.
We again use four Identical 30° – 60° – 90° set- squares for this activity.
Note carefully how they are placed touching one another.

(j) Do we get a parallelogram now?
Answer:
Yes

(ii) What can we say about its lengths of sides, angles and diagonals’?
Answer:
All sides equal.
opposite sides arc equal and parallel.

(iii) What is special about their diagonals?
Answer:
Diagonals bisects perpendicularly.

Try These (Text Book Page No. 195 & 196)

Question 1.
Say True or False:
(a) A square is a special rectangle.
Answer:
True

(b) A square is a parallelogram.
Answer:
True

(c) A square is a special rhombus.
Answer:
True

(d) A rectangle is a parallelogram
Answer:
True

Question 2.
Name the quadrilaterals

(a) Which have diagonals bisecting each other.
Answer:
Square, rectangle, parallelogram, rhombus.

(b) In which the diagonals are perpendicular bisectors of each other.
Answer:
Rhombus and square.

(c) Which have diagonals of different lengths.
Answer:
Parallelogram and Rhombus

(d) Which have equal diagonals.
Answer:
Rectangle, square.

(e) Which have parallel opposite sides.
Answer:
Square, Rectangle. Rhombus, parallelogram.

(f) In which opposite angles are equal.
Answer:
Square, rectangle. rhombus, parallelogram

Question 3.
Two sticks are placed on a ruled sheet as shown. What figure is formed if the four corners of the sticks are joined?

(a)

Two unequal sticks. Placed such that their midpoints coincide.
Answer:
parallelogram

(b)

Two equal sticks. Placed such that their midpoints coincide.
Answer:
Rectangle

(c)

Two unequal sticks. Placed intersecting at mid points perpendicularly.
Answer:
Rhombus

(d)

Two equal sticks. Placed intersecting at mid points perpendicularly.
Answer:
Square

(e)

Two unequal sticks. Tops are not on the same ruling. Bottoms on the same ruling. Not cutting at the mid point of either.
Answer:
Quadrilateral

(f)

Two unequal sticks. Tops on the same ruling. Bottoms on the same ruling. Not necessarily cutting at
the mid point of either.
Answer:
Trapezium

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 5 Geometry Ex 5.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 5 Geometry Ex 5.4

I. Construct the following quadrilaterals with the given measurements and also find their area.

Question 1.
ABCD, AB = 5 cm, BC = 4.5 cm, CD = 3.8 cm, DA = 4.4 cm and AC = 6.2 cm.
Answer:
Given ABCD, AB = 5 cm, BC = 4.5 cm, CD = 3.8 cm, DA = 4.4 cm and AC = 6.2 cm.

Steps:

• Draw a line segment AB = 5 cm
• With A and B as centers drawn arcs of radii 6.2 cm and 4.5cm respectively and let them cut at C.
• Joined AC and BC.
• With A and C as centrers drawn arcs of radii 4.4cm and 3.8 cm respectively and let them at D.
• Joined AD and CD.
• ABCD is the required quadrilateral.

Calculation of Area:
Area of the quadrilateral ABCD = \(\frac { 1 }{ 2 }\) × d × (h₁1 + h₂) sq. units
= \(\frac { 1 }{ 2 }\) × 6.2 × (2.6 + 3.6)cm² = 3.1 × 6.2 = 19.22 cm²

Question 2.
PLAY, PL = 7cm, LA = 6cm, AY = 6cm, PA = 8cm and LY = 7cm.
Answer:
Given PLAY, PL = 7cm, LA = 6cm, AY = 6cm, PA = 8cm and LY = 7cm

Steps:

• Drawn a line segment PL = 7 cm
• With P and L as centers, drawn arcs of radii 8 cm and 6 cm respectively, let them cut at A.
• Joined PA and LA.
• With L and A as centers, drawn arcs of radii 7 cm and 6 cm respectively and let them cut at Y.
• Joined LY, PY and AY.
• PLAY is the required quadrilateral.

Calculation of Area:
Area of the quadrilateral PLAY = \(\frac { 1 }{ 2 }\) × d × (h₁ + h₂) sq. units = \(\frac { 1 }{ 2 }\) × 8 × (5.1 + 1.4) cm²
\(\frac { 1 }{ 2 }\) × 8 × 6.5 cm² = 26 cm²
Area of the quadrilateral = 26 cm²

Question 3.
PQRS, PQ = QR = 3.5 cm, RS = 5.2 cm, SP = 5.3 cm and ∠Q = 120°.
Answer:
Given PQ = QR = 3.5 cm, RS = 5.2 cm, SP = 5.3 cm and ∠Q = 120°

Steps:

• Draw a line segment PQ = 3.5 cm
• Made ∠Q = 120°. Drawn the ray QX.
• With Q as centre drawn an arc of radius 3.5 cm. Let it cut the ray QX at R.
• With R and P as centres drawn arcs of radii 5.2cm and 5.5 cm respectively and let them cut at S.
• Joined PS and RS.
• PQRS is the required quadrilateral.

Calculation of Area:
Area of the quadrilateral PQRS = \(\frac { 1 }{ 2 }\) × d × (h₁ + h₂) sq. units
= \(\frac { 1 }{ 2 }\) × 6 × (4.3 + 17)cm²
= 3 × 6cm²
= 18 cm²
Area of the quadrilateral PQRS = 18 cm²

Question 4.
MIND, MI = 3.6 cm, ND = 4 cm, MD = 4 cm, ∠M = 50° and ∠D = 100°.
Answer:
Given MI = 3.6 cm, ND = 4 cm, MD = 4 cm, ∠M = 50° and ∠D = 100°

Steps:

• Drawn a line segment MI = 3.6 cm
• At M on MI made an angle ∠IMX = 500
• Drawn an arc with center M and radius 4 cm let it cut MX it D
• At D on DM made an angle ∠MDY = 100°
• With I as center drawn an arc of radius 4 cm, let it cut DY at N.
• Joined DN and IN.
• MIND is the required quadrilateral.

Calculation of Area:
Area of the quadrilateral MIND = \(\frac { 1 }{ 2 }\) × d × (h₁ + h₂) sq. units = \(\frac { 1 }{ 2 }\) × 3.2 × (2.7 + 33) cm²
= \(\frac { 1 }{ 2 }\) × 3.2 × 6 cm² = 9.6 cm²
Area of the quadrilateral = 9.6 cm²

Question 5.
AGRI, AG = 4.5 cm, GR = 3.8 cm, ∠A = 60°, ∠G = 110° and ∠R = 90°.
Answer:
AG = 4.5 cm, GR = 3.8 cm, ∠A = 60°, ∠G = 110° and ∠R = 90°.

Steps:

• Draw a line segment AG = 4.5 cm
• At G on AG made ∠AGX = 110°
• With G as centre drawn an arc of radius 3.8 cm let it cut GX at R.
• At R on GR made ∠GRZ = 90°
• At A on AG made ∠GAY = 90°
• AY and RZ meet at I.
• AGRI is the required quadrilateral.

Calculation of Area:
Area of the quadrilateral AGRI = \(\frac { 1 }{ 2 }\) × d × (h₁ + h₂) sq. units = × 6.8 × (2.9 + 2.4) cm²
= \(\frac { 1 }{ 2 }\) × 6.8 × 5.3 × cm²
Area of the quadrilateral = 18.02 cm²

II. Construct the following trapeziums with the given measures and also find their area.

Question 1.
AIMS with \(\overline{\mathrm{AI}}\) || \(\overline{\mathrm{SM}}\), AI = 6cm, IM = 5cm, AM = 9cm and MS = 6.5cm.
Answer:
Given AI = 6 cm, IM = 5 cm
AM = 9 cm, and \(\overline{\mathrm{AI}}\) || \(\overline{\mathrm{SM}}\)
MS = 6.5 cm

Rough Diagram

Construction:
Steps:

• Draw a line segment AI = 6cm.
• With A and I as centres, draw arcs of radii 9 cm and 5 cm respectively and let them cut at M
• Join AM and IM.
• Draw MX parallel to AI
• With M as centre, draw an arc of radius 6.5 cm cutting MX at S.
• Join AS AIMS is the required trapezium.

Calculation of Area
Area of the trapezium AIMS = \(\frac { 1 }{ 2 }\) × h × (a + b) sq.units
= \(\frac { 1 }{ 2 }\) × 4.6 × (6 + 6.5)
= \(\frac { 1 }{ 2 }\) × 4.6 × 12.5
= 28.75 Sq.cm

Question 2.
CUTE with \(\overline{\mathrm{CD}}\) || \(\overline{\mathrm{ET}}\), CU = 7cm, ∠ UCE = 80° CE = 6cm and TE = 5cm.
Answer:
Given: In the trapezium CUTE,
CU = 7cm, ∠UCE = 80°,
CE = 6cm,TE = 5cm and \(\overline{\mathrm{CD}}\) || \(\overline{\mathrm{ET}}\)

Rough Diagram

Construction:
Steps:

• Draw a line segment CU = 7cm.
• Construct an angle ∠UCE = 80° at C
• With C as centre, draw an arc of radius 6 cm cutting CY at E
• Draw EX parallel to CU
• With E as centre, draw an arc 7 cm of radius 5 cm cutting EX at T
• Join UT. CUTE is the required trapezium.

Calculation of area:
Area of the trapezium CUTE = \(\frac { 1 }{ 2 }\) × h × (a + b) sq. units
= \(\frac { 1 }{ 2 }\) × 5.9 × (7 + 5)sq. units
= 35.4 sq.cm

Question 3.
ARMY with \(\overline{\mathrm{AR}}\) || \(\overline{\mathrm{YM}}\), AR = 7cm, RM = 6.5 cm ∠RAY = 100° and ∠ARM = 60°
Answer:
Given: In the trapezium ARMY
AR = 7cm,RM = 6.5cm,
∠RAY = 100° and ARM 60°, \(\overline{\mathrm{AR}}\) || \(\overline{\mathrm{YM}}\)

Rough Diagram

Construction:
Steps:

• Draw a line segment AR = 7 cm.
• Construct an angle ∠RAX = 100° at A
• Construct an angle ∠ARN = 60° at R
• With R as centre, draw an arc of radius 6.5 cm cutting RN at M
• Draw MY parallel to AR
• ARMY is the required trapezium.

Calcualtion of Area:
Area of the trapezium ARMY = \(\frac { 1 }{ 2 }\) × h × (a + b) sq. units
= \(\frac { 1 }{ 2 }\) × 5.6 × (7 + 4.8) sq. units
= \(\frac { 1 }{ 2 }\) × 5.6 × 1.18
= 33.04 sq.cm

Question 4.
CITY with \(\overline{\mathrm{CI}}\) || \(\overline{\mathrm{YT}}\), CI = 7cm, IT = 5 .5 cm, TY = 4cm and YC = 6cm.
Answer:
Given: In the trapezium CITY,
CI = 7cm,IT = 5.5cm, TY = 4cm
YC = 6cm, and \(\overline{\mathrm{CI}}\) || \(\overline{\mathrm{YT}}\)

Rough Diagram

Construction:
Steps:

• Draw a line segment CI = 7 cm.
• Mark a point D on CI such that CD = 4cm
• With D and I as centres, draw arcs of radii 6 cm and 5.5 cm respectively.
  Let them cut at T. Join DT and IT.
• With C as centre, draw an arc of radius 6 cm.
• Draw TY parallel to Cl. Let the line cut the previous arc at Y.
• Join CY. CITY is the required trapezium.

Construction of area:
Area of the trapezium CITY = \(\frac { 1 }{ 2 }\) × h × (a + b) sq. units
= \(\frac { 1 }{ 2 }\) × 5.5 × (7 + 4)sq.units
= \(\frac { 1 }{ 2 }\) × 5.5 × 11
= 30.25 sq.cm