Bhagya

Students can download Maths Chapter 1 Numbers Ex 1.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 1 Numbers Ex 1.2

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

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

Solution:

1. <
2. >
3. =

Question 2.
Say True or False.

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

Solution:

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

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

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

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

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

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

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

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

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

Objective Type Questions

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

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

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

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

Students can download Maths Chapter 1 Numbers Ex 1.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 1 Numbers Ex 1.1

Question 1.
Fill in the blanks.

1. The smallest 7 digit number is _______
2. The largest 8 digit number is _______
3. The place value of 5 in 7005380 is ________
4. The expanded form of the number 76,70,905 is _______

Solution:

1. 10,00,000
2. 9,99,99,999
3. 5 × 1000 = 5000
4. 7 × 10,00,000 + 6 × 1,00,000 + 7 × 10,000 + 0 + 9 × 100 + 0 + 5 × 1
   (or)
   70,00,000 + 6,00,000 + 70,000 + 900 + 5

Question 2.
Say True or False.

1. In the Indian System of Numeration, the number 67999037 is written as 6,79,99,037
2. The successor of a one-digit number is always a one-digit number 9 + 1 = 10
3. The predecessor of a 3-digit number is always a 3 or 4 digit number 100 – 1 = 99
4. 88888 = 8 × 10000 + 8 × 100 + 8 × 10 + 8 × 1

Solution:

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

Question 3.
Complete the given order.
Ten crores, crore, ten lakh, ………, ……….., ……….., ………, …….., ……..
Solution:
Ten crores, Crore, Ten lakh, Lakh, Ten Thousand, Thousand, Hundred, Ten, One

Question 4.
How many ten thousands are there in the smallest 6 digit number?
Solution:
Smallest 6 digit number = 1,00,000
= \(\frac{100000}{10000}\) = 10
1 Lakh = 10 Ten Thousands

Question 5.
Using the digits 5, 2, 0, 7, 3 forms the largest 5 digit number and the smallest 5 digit number.
Solution:
Given digits = 5, 2, 0, 7, 3
Largest 5 digit number – 75320
Smallest 5 digit number – 20357

Question 6.
Observe the commas and write down the place value of 7.
(i) 56,74,56,345
(ii) 567,456,345
Solution:
(i) 70,00,000
(ii) 7,000,000

Question 7.
Write the following numbers in the International System by using commas.

1. 347056
2. 7345671
3. 634567105
4. 1234567890

Solution:

1. 347,056
2. 7,345,671
3. 634,567,105
4. 1,234,567,890

Question 8.
Write the largest six-digit number and put commas in the Indian and the International Systems.
Solution:
Largest 6 digit number = 9,99,999
Indian System: 9,99,999 (Nine Lakh Ninety-Nine Thousand Nine Hundred Ninety-Nine)
International System: 999,999 (Nine Hundred Ninety-Nine Thousand, Nine Hundred Ninety-Nine)

Question 9.
Write the number names of the following numerals in the Indian System.
(i) 75,32,105
(ii) 9,75,63,453
Solution:
(i) Seventy-five lakh thirty-two thousand one hundred five.
(ii) Nine crore seventy-five lakh sixty-three thousand four hundred fifty-three.

Question 10.
Write the number of names in words using the International System.

1. 345,678
2. 8,343,710
3. 103,456,789

Solution:

1. Three hundred forty-five thousand six hundred seventy-eight.
2. Eight million three hundred forty-three thousand seven hundred ten.
3. One hundred three million four hundred fifty-six thousand seven hundred eighty-nine.

Question 11.
Write the number name in numerals.

1. Two crores thirty lakhs fifty-one thousand nine hundred eighty.
2. Sixty-six million three hundred forty-five thousand twenty-seven.
3. Seven hundred eighty-nine million, two hundred thirteen thousand four hundred fifty-six.

Solution:

1. 2,30,51,980
2. 66,345,027
3. 789,213,456

Question 12.
Tamil Nadu has about twenty-six thousand three hundred forty-five square kilometres of Forest land. Write the number mentioned in the statement in the Indian System.
Solution:
26,345

Question 13.
The number of employees in the Indian Railways is about 10 lakh. Write this in the International System of numeration.
Solution:
1,000,000 (One Million)

Objective Type Questions

Question 14.
1 billion is equal to
(a) 100 crore
(b) 100 million
(c) 100 lakh
(d) 10000 lakh
Solution:
(a) 100 crore

Question 15.
The successor of 10 million is
(a) 1000001
(b) 10000001
(c) 9999999
(d) 100001
Solution:
(b) 10000001

Question 16.
The difference between successor and predecessor of 99999 is
(a) 90000
(b) 1
(c) 2
(d) 99001
Solution:
(c) 2

Question 17.
The expanded form of the number 6,70,905 is
(a) 6 × 10000 + 7 × 1000 + 9 × 100 + 5 × 1
(b) 6 × 10000 + 7 × 1000 + 0 × 100 + 9 × 100 + 0 × 10 + 5 × 1
(c) 6 × 1000000 + 7 × 10000 + 0 × 1000 + 9 × 100 + 0 × 10 + 5 × 1
(d) 6 × 100000 + 7 × 10000 + 0 × 1000 + 9 × 100 + 0 × 10 + 5 × 1
Solution:
(d) 6 × 100000 + 7 × 10000 + 0 × 1000 + 9 × 100 + 0 × 10 + 5 × 1

Students can download Maths Chapter 9 Probability Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 9 Probability Additional Questions

I. Choose the Correct Answer

Question 1.
Which of the following cannot be taken as probability of an event?
(a) 0
(b) 0.35
(c) \(\frac{7}{20}\)
(d) –\(\frac{7}{20}\)
Solution:
(d) –\(\frac{7}{20}\)

Question 2.
A letter is chosen at random from the word “MATHEMATICS” the probability of getting a vowel is ……..
(a) \(\frac{2}{11}\)
(b) \(\frac{3}{11}\)
(c) \(\frac{4}{11}\)
(d) \(\frac{5}{11}\)
Solution:
(c) \(\frac{4}{11}\)

Question 3.
If P(A) =\(\frac{1}{3}\) then P(A)’ is ………
(a) \(\frac{1}{3}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{3}{2}\)
(d) 1
Solution:
(b) \(\frac{2}{3}\)

Question 4.
An integer is chosen from the first twenty natural number, the probability that it is a prime number is ……..
(a) \(\frac{1}{5}\)
(b) \(\frac{2}{5}\)
(c) \(\frac{3}{5}\)
(d) \(\frac{4}{5}\)
Solution:
(b) \(\frac{2}{5}\)

Question 5.
From a well shuffled pack of 52 cards one card is drawn at random. The probability of getting not a king is ………
(a) \(\frac{12}{13}\)
(b) \(\frac{1}{13}\)
(c) \(\frac{4}{13}\)
(d) \(\frac{2}{13}\)
Solution:
(a) \(\frac{12}{13}\)

II. Answer the Following Questions

Question 6.
1500 families with 2 children were selected randomly and the following data were recorded.

Compute the probability of a family chosen at random having one girl.
Solution:
Total number of families = 1500
∴ n(S) = 1500
Let E be the event of getting one girl
n(E) = 814
P(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{814}{1500}\)
= \(\frac{407}{750}\)

Question 7.
The record of weather station shows that out of the part 250 consecutive days its whether forecast were correct 175 time. What is the probability that it was not correct on a given data?
Solution:
n(S) = 250
Let E be the event of getting whether forecast were correct
n(E) = 175
P(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{175}{250}\)
= \(\frac{7}{10}\)
Forecast was not correct on a given day = 1 – P(E)
= 1 – \(\frac{7}{10}\)
= \(\frac{10-7}{10}\)
= \(\frac{3}{10}\)

Question 8.
If A coin is tossed 200 times and is found that a tail comes up for 120 times. Find the probability of getting a tail.
Solution:
Number of trials = 200
n(S) = 200
Let E be the event of getting a tail
n(E) = 120
p(E) = \(\frac{n(E)}{n(S)}\)
= \(\frac{120}{200}\)
= \(\frac{12}{20}\)
= \(\frac{3}{5}\)

Question 9.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of drawing a red ball, then find the number of blue balls in the bag.
Solution:
Let the number of blue balls be “x”
Total number of balls = 5 + x
∴ n(S) = 5 + x
Let B be the event of drawing a blue ball and R be the event of drawing a red ball
Given P(B) = 3P(R)

∴ x = 15
Number of blue balls = 15

Question 10.
Find the probability that a non leap year selected at random will have 53 fridays.
Solution:
No. of days in a non leap year = 365 days
This year contain 52 weeks and one day
Sample space = {Sun, Mon, Tue, Wed, Thu, Fri, Sat}
n(S) = 7
Let A be the event of getting a friday
n(A) = 1
P(A) = \(\frac{n(A)}{n(S)}\)
= \(\frac{1}{7}\)

Students can download Maths Chapter 3 Ratio and Proportion Ex 3.3 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 3 Ratio and Proportion Ex 3.3

Question 1.
Fill in the blanks.
(i) 3 : 5 :: …….. : 20
(ii) ……… : 24 :: 3 : 8
(iii) 5 : …….. :: 10 : 8 :: 15 : ………
(iv) 12 : ……… :: ………. :: 4 = 8 : 16
Solution:
(i) 12
Hint:
5x = 3 × 20 ⇒ x = 12
(ii) 9
8x = 24 × 3 ⇒ x = 9
(iii) 4, 12
Hint:
10x = 8 × 5 = 40 ⇒ x = 4
10y = 8 × 15 = 120 ⇒ y = 12
(iv) 24, 2
Hint:
16y = 8 × 4 ⇒ y = 2
12 × 4 = 2x ⇒ x = 24

Question 2.
Say True or False.
(i) 2 : 7 : : 14 : 4
(ii) 7 Persons are to 49 Persons as 11 kg is to 88 kg
(iii) 10 books are to 15 books as 3 books are to 15 books.
Solution:
(i) False
Hint:
7 × 14 ≠ 4 × 2
98 ≠ 8
(ii) False
Hint:
7 : 49 :: 11 : 48
49 × 11 ≠ 7 × 48
539 ≠ 336
(iii) False
Hint:
10 : 15 :: 3 : 5
\(\frac{10}{15}=\frac{5 \times 2}{5 \times 3}=\frac{2}{3} \Rightarrow \frac{3}{5}=\frac{3}{5}\)

Question 3.
Using the numbers 3, 9, 4, 12 write two ratios that are in proportion.
Solution:
(i) 3, 9, 4, 12
Here product of extremes = 3 × 12 = 36
Product of means = 9 × 4 = 36
3 : 9 : : 4 : 12

(ii) Also if we take 9, 3, 12, 4
Product of extremes = 9 × 4 = 36
Product of means = 3 × 12 = 36
9 : 3 : : 12 : 4

Question 4.
Find whether 12, 24,18, 36 are in order that can be expressed as two ratios that are in proportion.
Solution:
Yes, 12 : 24 : : 18 : 36
Because product of extremes 12 × 36 = 432
Product of means = 24 × 18 = 432
12 : 24 :: 18 : 36.

Question 5.
Write the mean and extreme terms in the following ratios and check whether they are in proportion.
(i) 78 liters is to 130 liters and 12 bottles are to 20 bottles
(ii) 400 gm is to 50 gm and 25 rupees is to 625 rupees
Solution:
(i) 78 : 130 :: 12 : 20
Extreme terms are 78 and 20.
Mean terms are 130 and 12.
Product of Extremes = 78 × 20 = 1560
Product of Means = 130 × 12 = 1560
Product of Extremes = Product of means
It is in proportion.

(ii) 400 : 50 : : 25 : 625
Product of extremes = 400 × 625 = 2,50,000
Product of means = 50 × 25 = 1250
Here product of extremes ≠ product of means
400 : 50 and 25 : 625 are not in proportion.

Question 6.
America’s famous Golden Gate bridge is 6480 ft long with 756 ft tall towers. A model of this bridge exhibited in a fair is 60 ft long with 7 ft tall towers. Is the model in proportion to the original bridge?

Solution:
6480 : 756, 60 : 7
Product of the means = 756 × 60 = 45360
Product of the extremes = 6480 × 7 = 45360
ad = bc
∴ They are in proportion

Objective Type Questions

Question 7.
Which of the following ratios are in proportion?
(a) 3 : 5, 6 : 11
(b) 2 : 3, 9 : 6
(c) 2 : 5, 10 : 25
(d) 3 : 1, 1 : 3
Solution:
(c) 2 : 5, 10 : 25
Hint:
2 × 25 = 5 × 10
⇒ 50 = 50

Question 8.
If the ratios formed using the numbers 2, 5, x, 20 in the same order are in proportion, then ‘x’ is
(a) 50
(b) 4
(c) 10
(d) 8
Solution:
(d) 8
5x = 2 × 20 ⇒ x = 8

Question 9.
If 7 : 5 is in proportion to x : 25, then ‘x’ is
(a) 27
(b) 49
(c) 35
(d) 14
Solution:
(c) 35
Hint:
5x = 7 × 25 ⇒ x = 35

Students can download Maths Chapter 3 Ratio and Proportion Ex 3.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 3 Ratio and Proportion Ex 3.1

Question 1.
Fill in the blanks.
(i) Ratio of ₹ 3 to ₹ 5 = ____
(ii) Ratio of 3 m to 200 cm = ______
(iii) Ratio of 5 km 400 m to 6 km = ____
(iv) Ratio of 75 paise to ₹ 2 = ____
Solution:
(i) 3 : 5
(ii) 3 : 2
Hint: 3m = 300 cm
(iii) 9 : 10
Hint: 5km 400 m = 5400m and 6 km = 6000 m
(iv) 3 : 8
Hint: ₹ 2 = 200 paise

Question 2.
Say whether the following statements are True or False.
(i) The ratio of 130 cm to 1 m is 13 : 10
(ii) One of the terms in a ratio cannot be 1
Solution:
(i) True
Hint: 1m = 100 cm
(ii) False

Question 3.
Find the simplified form of the following ratios.
(i) 15 : 20
(ii) 32 : 24
(iii) 7 : 15
(iv) 12 : 27
(v) 75 : 100
Solution:
(i) 15 : 20
= \(\frac{15}{20}\)
= \(\frac{3}{4}\)
= 3 : 4

(ii) 32 : 24
= \(\frac{32}{24}\)
= \(\frac{4}{3}\)
= 4 : 3

(iii) 7 : 15

(iv) 12 : 27
= \(\frac{12}{27}\)
= \(\frac{4}{9}\)
= 4 : 9

(v) 75 : 100
= \(\frac{75}{100}\)
= \(\frac{3}{4}\)
= 3 : 4

Question 4.
Akilan walks 10 km in an hour while Selvi walks 6 km in an hour. Find the simplest ratio of the distance covered by Akilan to that of Selvi.
Solution:
Ratio of the distance covered by Akilan to that of Selvi = 10 km : 6 km
= \(\frac{10}{6}\)
= \(\frac{5}{3}\)
= 5 : 3

Question 5.
The cost of parking a bicycle is Rs 5 and the cost of parking a scooter is Rs 15. Find the simplest ratio of the parking cost of a bicycle to that of a scooter.
Solution:
Ratio of the parking cost of a bicycle to that of a scooter = Rs 5 : Rs 15
= \(\frac{5}{15}\)
= \(\frac{1}{3}\)
= 1 : 3

Question 6.
Out of 50 students in a class, 30 are boys. Find the ratio of
(i) number of boys to the number of girls.
(ii) the number of girls to the total number of students.
(iii) the number of boys to the total number of students.
Solution:
Total no of students = 50
No of boys = 30
No of girls = 50 – 30 = 20
(i) Ratio of boys to girls = 30 : 20
= \(\frac{30}{20}\)
= \(\frac{3}{2}\)
= 3 : 2

(ii) Ratio of girls to the total number of students = 20 : 50
= \(\frac{20}{50}\)
= \(\frac{2}{5}\)
= 2 : 5

(iii) Ratio of boys to the total no of students = 30 : 50
= \(\frac{30}{50}\)
= \(\frac{3}{5}\)
= 3 : 5

Objective Type Questions

Question 7.
The ratio of ₹ 1 to 20 paise is _____
(a) 1 : 5
(b) 1 : 2
(c) 2 : 1
(d) 5 : 1
Solution:
(d) 5 : 1
Hint: ₹ 1 = 100 paise

Question 8.
The ratio of 1 m to 50 cm is
(a) 1 : 50
(b) 50 : 1
(c) 2 : 1
(d) 1 : 2
Solution:
(c) 2 : 1

Question 9.
The length and breadth of a window are 1 m and 70 cm respectively. The ratio of the length to the breadth is
a) 1 : 7
(b) 7 : 1
(c) 7 : 10
(d) 10 : 7
Solution:
(d) 10 : 7

Question 10.
The ratio of the number of sides of a triangle to the number of sides of a rectangle is
(a) 4 : 3
(b) 3 : 4
(c) 3 : 5
(d) 3 : 2
Solution:
(b) 3 : 4

Question 11.
If Azhagan is 50 years old and his son is 10 years old then the simplest ratio between the age of Azhagan to his son is
(a) 10 : 50
(b) 50 : 10
(c) 5 : 1
(d) 1 : 5
Solution:
(c) 5 : 1

Students can download Maths Chapter 2 Introduction to Algebra Ex 2.3 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 2 Introduction to Algebra Ex 2.3

Miscellaneous Practice Problems

Question 1.
Complete the following pattern.
9 – 1 =
98 – 21 =
987 – 321 =
9876 – 4321 =
98765 – 54321 =
What comes next?
Solution:
9 – 1 = 8
98 – 21 = 77
987 – 321 = 666
9876 – 4321 = 5555
98765 – 54321 = 44444
Next will be 987654 – 654321 = 333333

Question 2.
A piece of wire is ‘12s’ cm long. What will be the length of the side if it is formed as
(i) an equilateral triangle.
(ii) a square?
Solution:
(i) An equilateral triangle has 3 equal sides.
Length of the wire = ‘12s’ cm
Length of each side = \(\frac{12 s}{3}\) cm.
Length of each side of the triangle = 4s cm
(ii) A square has four equal sides.
Length of each side \(\frac{12 s}{4}\) cm
Length of each side of the square 3s cm

Question 3.
Identify the value of the shapes and figures in the table given below and verify their addition horizontally and vertically.

Solution:

Question 4.
The table given below shows the results of the matches played by 8 teams in a kabaddi championship tournament.

Find the value of all the variables in the table given above.
Solution:
k = 3, m = 1, n = 10, a = 9, b = 6, c = 4, x = 4, y = 9

Challenging Problems

Question 5.
Gopal is 8 years younger than Karnan. If the sum of their ages is 30, how old is Karnan?
Solution:
Let the age of Kaman be x years
Gopal’s age = x – 8
Ace to the problem, x + x – 8 = 30
2x – 8 = 30
2x = 30 + 8
2x = 38
x = \(\frac{38}{2}\)
x = 19
Age of Kaman = 19 years

Question 6.
The rectangles made of identical square blocks with varying lengths but having only two square blocks as width are give below

(i) How many small size squares are there in each of the rectangles P,Q, R and S?
(ii) Fill in the boxes.

Solution:
(i) P = 2; Q = 8; R = 6; S = 10

Question 7.
Find the variables from the clues given below and solve the crossword puzzle.


Solution:

Students can download Maths Chapter 5 Statistics Ex 5.3 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Ex 5.3

Question 1.
Read the given Bar Graph which shows the percentage of marks obtained by Brinda in different subjects in an assessment test.

Observe the Bar Graph and answer the following questions.
(i) 1 Unit = ………… % of marks on vertical line.
(ii) Brinda has scored maximum marks in …….. subject.
(iii) Brinda has scored minimum marks in …….. subject.
(iv) The percentage of marks scored by Brinda in Science is ……..
(v) Brinda scored 60 % marks in the subject ……..
(vi) Brinda scored 20% more in …….. subject than ……… subject.
Solution:
(i) 10
(ii) Mathematics
(iii) Language
(iv) 65%
(v) English
(vi) Mathematics, Science

Question 2.
Chitra has to buv Laddus in order to distribute to her friends as follows:

Draw a Bar Graph for this data.
Solution:
Distribution of Laddus by Chitra to her friends
Scale : 1 unit = 10 Laddus

Question 3.
The fruits liked by the students of a class are as follows:

Solution:
Draw a bar graph for data this data
Scale 1 unit = 2 fruits

Question 4.
The pictograph below gives the number of absentees on different days of the week in class six. Draw the Bar graph for the same.

Solution:
No of absentees on different days of the week in a class
Scale 1 unit = 2 students

Objective Type Questions

Question 5.
A bar graph can be drawn using
(a) Horizontal bars only
(b) Vertical bars only
(c) Both horizontal bars and Vertical bars
(d) Either horizontal bars or vertical bars
Solution:
(d) Either horizontal bars or vertical bars

Question 6.
The spaces between any two bars in a bar graph
(a) can be different
(b) are the same
(c) are not the same
(d) all of these
Solution:
(b) are the same

Students can download Maths Chapter 5 Statistics Ex 5.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Ex 5.2

Question 1.
Fill in the blanks.

(iii) Representation of data by using pictures is known as ………
Solution:
(i) 150
(ii)
(iii) Pictograph

Question 2.
Draw a pictograph for the given data.

(Choose your own suitable scale)
Solution:
Number of computers sold
Scale : 1 Unit = 100 computers

Question 3.
The following table shows the number of tourists who visited the places in the month of May. Draw a pictograph

(Choose your own suitable scale)
Solution:
Number of Tourists who visited the places in the month of May.

Question 4.
The following pictograph shows the number of students playing different games in a school.

Answer the following questions.
(i) Which is the most popular game among the students?
(ii) Find the number of students playing Kabaddi.
(iii) Which two games are played by an equal number of students?
(iv) What is the difference between the number of students playing Kho-Kho and Hockey?
(v) Which is the least popular game among the students?
Solution:
(i) Kabaddi is the most popular game among students.
(ii) There are 11 × 10 = 110 students playing kabaddi.
(iii) Kho-Kho and Hockey are played by an equal number of students.
(iv) Difference is 90 – 90 = 0.
(v) Basketball is the least popular game among students.

Objective Type Questions

Question 5.
The pictorial representation for a phrase is a ____
(a) Picto
(b) Tally mark
(c) Frequency
(d) Data
Solution:
(d) Data

Question 6.
A pictograph is also known as
a) Pictoword
(b) Pictogram
(c) Pictophrase
(d) Pictograph
Solution:
(b) Pictogram

Students can download Maths Chapter 5 Statistics Ex 5.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Ex 5.1

Question 1.
Fill in the blanks.

1. The collected information is called ……..
2. An example of Primary data is ………
3. An example of Secondary Data is ……….
4. The tally marks for number 8 in standard form is ………

Solution:

1. Data
2. List of absentees in a class
3. Cricket scores gathered from a website
4. 

Question 2.
Viji threw a die 30 times and noted down the result each time as follows. Prepare a table on the numbers shown using Tally Marks.
1, 4, 3, 5, 5, 6, 6, 4, 3, 5, 4, 5, 6, 5, 2, 4, 2, 6, 5, 5, 6, 6, 4, 5, 6, 6, 5, 4, 1, 1
Solution:

Question 3.
The following list tells colours liked by 25 students. Prepare a table using Tally Marks.

Solution:

Question 4.
The following are the marks obtained by 30 students in a class test out of 20 in Mathematics subject.

Prepare a table using Tally Marks.

Question 5.
The table shows the number of calls recorded by a Fire Service Station in one year.

Complete the table and answer the following questions.
(i) Which type of call was recorded the most?
(ii) Which type of call was recorded the least?
(iii) How many calls were recorded in all?
(iv How many calls were recorded as False Alarms?
Solution:

(i) The call for “Other Fires” was recorded the most
(ii) The call for “Rescues” was recorded the least
(iii) The total of 35 calls was recorded
(iv) There are 7 calls were recorded as False alarm.

Objective Type Questions

Question 6.
The tally marks for the number 7 in standard form is ………

Solution:
(b)

Question 7.

(a) 5
(b) 8
(c) 9
(d) 10
Solution:
(c) 9

Question 8.
The plural form of ‘datum’ is ____
(a) datum
(b) datums
(c) data
(d) dates
Solution:
(c) data

Students can download Maths Chapter 4 Geometry Additional Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Additional Questions

I. Multiple choice questions

Question 1.
If a straight line intersects the sides AB and AC of a ∆ABC at D and E respectively and is parallel to BC, then \(\frac { AE }{ AC } \) = …………
(1) \(\frac { AD }{ DB } \)
(2) \(\frac { AD }{ DB } \)
(3) \(\frac { DE }{ BC } \)
(4) \(\frac { AD }{ EC } \)
Answer:
(2) \(\frac { AD }{ DB } \)

Hint:
By BPT theorem,
\(\frac { AE }{ AC } \) = \(\frac { AD }{ AB } \)

Question 2.
In ∆ABC, DE is || to BC, meeting AB and AC at D and E. If AD = 3 cm, DB = 2 cm and AE = 2.7 cm, then AC is equal to ……..
(1) 6.5 cm
(2) 4.5 cm
(3) 3.5 cm
(4) 3.5 cm
Answer:
(2) 4.5 cm
Hint:
By BPT theorem,

\(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \)
\(\frac { 3 }{ 2 } \) = \(\frac { 2.7 }{ EC } \)
∴ EC = \(\frac{2.7 \times 2}{3}\) = 1.8 CM
AC = AE + EC
= 2.7 + 1.8 = 4.5 cm

Question 3.
In ∆PQR, RS is the bisector of ∠R. If PQ = 6 cm, QR = 8 cm, RP = 4 cm then PS is equal to …………..
(1) 2 cm
(2) 4 cm
(3) 3 cm
(4) 6 cm
Answer:
(2) 2 cm
Hint:

By ABT theorem,
\(\frac { PS }{ SQ } \) = \(\frac { PR }{ QR } \) ⇒ \(\frac { x }{ 6-x } \) = \(\frac { 4 }{ 8 } \)
24 – 4x = 8x ⇒ 24 = 12x
x = 2
PS = 2 cm

Question 4.
In figure, if \(\frac { AB }{ AC } \) = \(\frac { BD }{ DC } \), ∠B = 40° and ∠C = 60°, then ∠BAD = ……………
(1) 30°
(2) 50°
(3) 80°
(4) 40°
Answer:
(4) 40°
Hint:
\(\frac { AB }{ AC } \) = \(\frac { BD }{ DC } \)

AD is the internal bisector of ∠BAC
∠A + ∠B + ∠C = 180°
∠A + 40° + 60° = 180° ⇒ ∠A = 80°
∴ ∠BAC = \(\frac { 80 }{ 2 } \) = 40°

Question 5.
In the figure, the value x is equal to …………
(1) 4.2
(2) 3.2
(3) 0.8
(4) 0.4

Answer:
(2) 3.2
Hint:
By Thales theorem
(DE || BC)
\(\frac { AD }{ BD } \) = \(\frac { AE }{ EC } \)
\(\frac { x }{ 8 } \) = \(\frac { 4 }{ 10 } \) ⇒ x = \(\frac{8 \times 4}{10}\) = 3.2

Question 6.
In triangles ABC and DEF, ∠B = ∠E, ∠C = ∠F, then ………….
(1) \(\frac { AB }{ DE } \) = \(\frac { CA }{ EF } \)
(2) \(\frac { BC }{ EF } \) = \(\frac { AB }{ FD } \)
(3) \(\frac { AB }{ DE } \) = \(\frac { BC }{ EF } \)
(4) \(\frac { CA }{ FD } \) = \(\frac { AB }{ EF } \)
Answer:
(3) \(\frac { AB }{ DE } \) = \(\frac { BC }{ EF } \)
Hint:

Question 7.
From the given figure, identify the wrong statement.

(1) ∆ADB ~ ∆ABC
(2) ∆ABD ~ ∆ABC
(3) ∆BDC ~ ∆ABC
(4) ∆ADB ~ ∆BDC
Answer:
(2) ∆ ABD ~ ∆ ABC

Question 8.
If a vertical stick 12 m long casts a shadow 8 m long on the ground and at the same time a tower casts a shadow 40 m long on the ground, then the height of the tower is …………..
(1) 40 m
(2) 50 m
(3) 75 m
(4) 60 m
Answer:
(4) 60 m
Hint:

\(\frac { AB }{ DE } \) = \(\frac { BC }{ EF } \)
\(\frac { 12 }{ h } \) = \(\frac { 8 }{ 40 } \); h = \(\frac{40 \times 12}{8}\) = 60

Question 9.
The sides of two similar triangles are in the ratio 2:3, then their areas are in the ratio ………….
(1) 9 : 4
(2) 4 : 9
(3) 2 : 3
(4) 3 : 2
Answer:
(2) 4 : 9
Hint:
Ratio of the Area of two similar triangle
= 2² : 3² = 4 : 9
[squares of their corresponding sides]

Question 10.
Triangles ABC and DEF are similar. If their areas are 100 cm² and 49 cm² respectively and BC is 8.2 cm then EF = …………….
(1) 5.47 cm
(2) 5.74 cm
(3) 6.47 cm
(4) 6.74 cm
Answer:
(2) 5.74 cm
Hint:

Question 11.
The perimeters of two similar triangles are 24 cm and 18 cm respectively.
If one side of the first triangle is 8 cm, then the corresponding side of the other triangle is ………
(1) 4 cm
(2) 3 cm
(3) 9 cm
(4) 6 cm
Answer:
(4) 6 cm
Hint:

Question 12.
A point P is 26 cm away from the centre O of a circle and PT is the tangent drawn from P to the circle 10 cm, then OT is equal to …………
(1) 36 cm
(2) 20 cm
(3) 18 cm
(4) 24 cm
Answer:
(4) 24 cm
Hint:

OT² = OP² – PT²
= 26² – 10²
= (26 + 10) (26 – 10)
OT² = 36 × 16
OT = 6 × 4 = 24 cm

Question 13.
In the figure, if ∠PAB = 120° then ∠BPT = ……….

(1) 120°
(2) 30°
(3) 40°
(4) 60°
Answer:
(4) 60°
Hint:
∠BCP + ∠PAB = 180°
(sum of the opposite angles of a cyclic quadrilateral)
∠BCP = 180° – 120° = 60°
∠BPT = 60°
(By tangent chord theorem)

Question 14.
If the tangents PA and PB from an external point P to circle with centre O are inclined to each other at an angle of 40°, then ∠POA = ………………
(1) 70°
(2) 80°
(3) 50°
(4) 60°
Answer:
(1) 70°
Hint:
∠OPA = \(\frac { 40 }{ 2 } \) = 20°
In ∆OAP.
∠POA + ∠OAP + ∠APO = 180°
(sum of the angles of a triangle)

∠POA + 90° + 20° = 180°
∠POA = 180° – 110° = 70°

Question 15.
In the figure, PA and PB are tangents to the circle drawn from an external point P. Also CD is a tangent to the circle at Q. If PA = 8 cm and CQ = 3 cm, then PC is equal to ……….

(1) 11 cm
(2) 5 cm
(3) 24 cm
(4) 38 cm
Answer:
(2) 5 cm
Hint:
PA = PB (tangent of a circle)
PB = 8 cm
PC + BC = 8
PA + QC = (BC = QC tangent)
PC + 3 = 8
∴ PC = 8 cm – 3 cm = 5 cm

Question 16.
∆ABC is a right angled triangle where ∠B = 90° and BD ⊥ AC. If BD = 8 cm, AD = 4 cm, then CD is …………
(1) 24 cm
(2) 16 cm
(3) 32 cm
(4) 8 cm
Answer:
(2) 16 cm
Hint:
∆DCB ~ ∆DBA
\(\frac { DC }{ DB } \) = \(\frac { DB }{ DA } \)

DB² = DC × DA
8² = DC × 4
DC = \(\frac { 64 }{ 4 } \) = 16 cm

Question 17.
The areas of two similar triangles are 16 cm2 and 36 cm2 respectively. If the altitude of the first triangle is 3 cm, then the corresponding altitude of the other triangle is
(1) 6.5 cm
(2) 6 cm
(3) 4 cm
(4) 4.5 cm
Answer:
(4) 4.5 cm
Hint:

Question 18.
The perimeter of two similar triangles ∆ABC and ∆DEF are 36 cm and 24 cm respectively. If DE =10 cm, then AB is …………
(1) 12 cm
(2) 20 cm
(3) 15 cm
(4) 18 cm
Answer:
(3) 15 cm
Hint:
Perimeter of

Question 19.
In the given diagram θ is ………….
(1) 15°
(2) 30°
(3) 45°
(4) 60°

Answer:
(2) 30°
Hint:
∠BAD = 180° – 150° = 30°
= 180° – 150° = 30°
∠DAC = θ = 30°

Question 20.
If AD is the bisector of ∠A then AC is …………

(1) 12
(2) 16
(3) 18
(4) 20
Answer:
(4) 20
Hint. In ∆ABC, AD is the internal bisector of ∠A
\(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \)
\(\frac { 4 }{ 10 } \) = \(\frac { 8 }{ x } \)
4x = 10 × 8
x = \(\frac{10 \times 8}{4}\) = 20 cm

Question 21.
In ∆ABC and ∆DEF, ∠A = ∠E and ∠B = ∠F. Then AB : AC is ………….
(1) DE : DF
(2) DE : EF
(3) EF : ED
(4) DF : EF
Answer:
(3) EF : ED
Hint:

\(\frac { AB }{ EF } \) = \(\frac { BC }{ FD } \) = \(\frac { AC }{ ED } \)

Question 22.
Two circles of radius 8.2 cm and 3.6 cm touch each other externally, the distance between their centres is ………….
(1) 1.8 cm
(2) 4.1 cm
(3) 4.6 cm
(4) 11.8 cm
Answer:
(4) 11.8 cm
Hint:
Distance between the two centres

= r₁ + r₂
= 8.2 + 3.6
= 11.8 cm

Question 23.
In the given diagram PA and PB are tangents drawn from P to a circle with centre O. ∠OPA = 35° then a and b is …………

(1) a = 30°, b = 60°
(2) a = 35°, b = 55°
(3) a = 40°, b = 50°
(4) a = 45°, b = 45°
Answer:
(2) a = 35°, b = 55°
Hint:
∠OAP = 90° (tangent of the circle)
∠AOP + ∠OPA + ∠PAO = 180°
b + 35° + 90° = 180°
b = 180° – 125°
= 55°
OP is the angle bisector of ∠P
∴ a = 35°

II. Answer the following questions

Question 1.
The image of a man of height 1.8 m, is of length 1.5 cm on the film of a camera. If the film is 3 cm from the lens of the camera, how far is the man from the camera?
Solution:
Let AB be the height of the man, CD be the height of the image of the man of height 1.8 m (180 cm). LM be the distance between man and lens. LN be the distance between Lens and the film.

Given, AB = 1.8 m (180 cm)
CD = 1.5 cm and LN = 3 cm
Consider ∆ LAB and ∆ LCD
∠ALB = ∠DLC (vertically opposite angles)
∠LAB = ∠LDC (alternate angles) (AB || CD)
∴ ∆ LAB ~ ∆ LDC (AA similarity)
∴ \(\frac { AB }{ CD } \) = \(\frac { LM }{ LN } \) ⇒ \(\frac { 180 }{ 1.5 } \) = \(\frac { x }{ 3 } \) ⇒ 1.5x = 180 × 3
x = \(\frac{180 \times 3}{1.5}\) = \(\frac{180 \times 3 \times 10}{15}\)
x = 360 cm (or) 3.6 m
∴ The distance between the man and the camera = 3.6 cm

Question 2.
A girl of height 120 cm is walking away from the base of a lamp-post at a speed of 0. 6 m/sec. If the lamp is 3.6 m above the ground level, then find the length of her shadow after 4 seconds.
Solution:
Let AB be the height of the lamp-post above the ground level.
AB = 3.6 m = 360 cm
Let CD be the height of the girl.
CD = 1.2 m = 120 cm
The distance travelled by the girl in 4 seconds (AC)
= 4 × 0.6 = 2.4m = 240 cm
Consider ∆ECD and ∆EAB
Given (CD || AB)
∠EAB = ∠ECD = 90°
∠E is common

∴ ∆ EAB = ∆ ECD = 90°
\(\frac { AB }{ CD } \) = \(\frac { AE }{ CE } \)
= \(\frac { x+240 }{ x } \) ⇒ 3 = \(\frac { x+240 }{ x } \)
3x = x + 240
2x = 240 ⇒ x = \(\frac { 240 }{ 2 } \) = 120 cm
∴ Lenght of girls shadow after 4 seconds = 120 cm

Question 3.
In ∆ ABC, AB = AC and BC = 6 cm. D is a point on the side AC such that AD = 5 cm and CD = 4 cm. Show that ∆BCD ~ ∆ACB and hence find BD.
Solution:
Given, In ∆ ABC, AB = AC = 9 cm and BC = 6 cm, AD = 5 cm and CD = 4 cm
\(\frac { BC }{ AC } \) = \(\frac { 6 }{ 9 } \) = \(\frac { 2 }{ 3 } \) …..(1)
\(\frac { CD }{ CB } \) = \(\frac { 4 }{ 6 } \) = \(\frac { 2 }{ 3 } \) ……..(2)

From (1) and (2) we get,
\(\frac { BC }{ AC } \) = \(\frac { CD }{ CB } \)
In ∆ BCD and ∆ ACB
∠C = ∠C (common angle)
\(\frac { BC }{ AC } \) = \(\frac { CD }{ CB } \)
∴ ∆ BCD ~ ∆ ACB
\(\frac { BD }{ AB } \) = \(\frac { BC }{ AC } \) ⇒ \(\frac { BD }{ 9 } \) = \(\frac { 6 }{ 9 } \) ⇒ ∴ BD = 6 cm

Question 4.
The lengths of three sides of a triangle ABC are 6 cm, 4 cm and 9 cm. ∆PQR ~ ∆ABC. One of the lengths of sides of APQR is 35 cm. What is the greatest perimeter possible for ∆PQR?
Solution:
Given, ∆ PQR ~ ∆ ABC

Perimeter of ∆ ABC = 6 + 4 + 9 = 19 cm
When the perimeter of ∆ PQR is the greatest only the corresponding side QR must be equal to 35 cm

Question 5.
A man sees the top of a tower in a mirror which is at a distance of 87.6 m from the tower. The mirror is on the ground, facing upward. The man is 0.4 m away from the mirror, and the distance of his eye level from the ground is 1.5 m. How tall is the tower? (The foot of man, the mirror and the foot of the tower lie along a straight line).
Solution:
Let AB and ED be the heights of the man and the tower respectively. Let C be the point of incidence of the tower in the mirror.
In ∆ ABC and ∆ EDC, we have

∠ABC = ∠EDC = 90°
∠BCA = ∠DCE
(angular elevation is same at the same instant, i.e., the angle of incidence and the angle of reflection are same.)
∴ ∆ ABC ~ ∆ EDC (AA similarity criterion)
Thus,
\(\frac { ED }{ AB } \) = \(\frac { DC }{ BC } \) (corresponding sides are proportional)
ED = \(\frac { DC }{ BC } \) × AB = \(\frac { 87.6 }{ 0.4 } \) × 1.5 = 328.5
Hence, the height of the tower is 328.5 m.

Question 6.
In ∆ PQR, given that S is a point on PQ such that ST || QR and \(\frac { PS }{ SQ } \) = \(\frac { 3 }{ 5 } \). If PR = 5.6 cm, then find PT.
Solution:
In ∆ PQR, we have ST || QR and by Thales theorem,
\(\frac { PS }{ SQ } \) = \(\frac { PT }{ TR } \) …….(1)
Let PT = x
Thus, TR = PR – PT = 5.6 – x

From (1), we get PT = TR (\(\frac { PS }{ SQ } \))
x = (5.6 – x) (\(\frac { 3 }{ 5 } \))
5x = 16.8 – 3x
8x = 16.8
x = \(\frac { 16.8 }{ 8 } \) = 2.1
That is PT = 2.1 cm

Question 7.
In a ∆ ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 4x – 3, BD = 3x – 1, AE = 8x – 7 and EC = 5x – 3, then find the value of x.
Solution:
In ∆ ABC, DE || BC. By BPT theorem. (Thales theorem)
We get \(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \) ⇒ \(\frac { 4x-3 }{ 3x-1 } \) = \(\frac { 8x-7 }{ 5x-3 } \)

(8x – 7) (3x – 1) = (4x – 3) (5x – 3)
24x² – 8x – 21x + 7 = 20x² – 12x – 15x + 9
24x² – 8x – 21x + 7 = 20x² – 27x + 9
24x² – 29x + 7 – 20x² + 27x – 9 = 0
∴ 4x² – 2x – 2 = 0
÷ 2 ⇒ 2x² – x – 1 = 0
(x – 1) (2x + 1) = 0

x – 1 = 0 (or) 2x + 1 = 0
x = 1 (or) 2x = – 1
x = – \(\frac { 1 }{ 2 } \) ⇒ since, x ≠ – \(\frac { 1 }{ 2 } \)
∴ The value of x = 1

Question 8.
In the figure AC || BD and CE || DF. if OA = 12 cm, AB = 9cm, OC = 8 cm and EF = 4.5 cm, then find FO.
Solution:

In OBD, AC || BD
∴ \(\frac { OA }{ AB } \) = \(\frac { OC }{ CD } \) (By Thales theorem)
\(\frac { 12 }{ 9 } \) = \(\frac { 8 }{ CD } \)
∴ CD = \(\frac{9 \times 8}{12}\) = 6 CM
In ODF, CE || DF
\(\frac { OC }{ CD } \) = \(\frac { OE }{ EF } \) (By Thales theorem)
\(\frac { 8 }{ 6 } \) = \(\frac { OE }{ 4.5 } \) ⇒ OE = \(=\frac{8 \times 4.5}{6}\) = 6 cm
FO = FE + EO = 4.5 + 6 = 10.5 cm
∴ The value of FO = 10.5 cm

Question 9.
Check whether AD is the bisector of ∠A of ∆ABC in each of the following.
(i) AB = 4 cm, AC = 6 cm, BD 1.6 cm, and CD = 2.4 cm.
Solution:
\(\frac { BD }{ DC } \) = \(\frac { 1.6 }{ 2.4 } \) = \(\frac { 16 }{ 24 } \) = \(\frac { 2 }{ 3 } \) …..(1)
\(\frac { AB }{ AC } \) = \(\frac { 4 }{ 6 } \) = \(\frac { 2 }{ 3 } \) ……..(2)
From (1) and (2) we get,

\(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \)
By the converse of angle bisector theorem we have,
∴ AD is the internal bisector of ∠A

(ii) AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 3 cm
Solution:
\(\frac { BD }{ DC } \) = \(\frac { 1.5 }{ 3 } \) = 0.5 …….(1)
\(\frac { AB }{ AC } \) = \(\frac { 6 }{ 8 } \) = \(\frac { 3 }{ 4 } \) …….(2)

From (1) and (2) we get,
\(\frac { BD }{ DC } \) ≠ \(\frac { AB }{ AC } \)
Hence AD is not the bisector of ∠A.

Question 10.
In a ∆ABC, AD is the internal bisector of ∠A, meeting BC at D. If AB 5.6 cm, AC = 6 cm and DC = 3 cm, find BC.
Solution:
Given, AB = 5.6 cm, AC = 6 cm, DC = 3 cm
Let BD be x
In ∆ ABC, AD is the internal bisector of ∠A.
By Angle bisector theorem, we have,

\(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \) ⇒ \(\frac { x }{ 3 } \) = \(\frac { 5.6 }{ 6 } \)
x = \(\frac{3 \times 5.6}{6}\) = 2.8 cm
∴ BC = BD + DC = 2.8 + 3 = 5.8 cm
Length of BC = 5.8 cm

Question 11.
In the figure, tangents PA and PB are drawn to a circle with centre O from an external point P. If CD is a tangent to the circle at E and AP = 15 cm, find the perimeter of ∆PCD.
Solution:
We know that the lengths of the two tangents from an exterior point to a circle are equal.
∴ CA = CE, DB = DE and PA = PB
Now, the perimeter of ∆PCD

= PC + CD + DP
= PC + CE + ED + DP
= PC + CA + DB + DP
= PA + PB = 2 PA (PB = PA)
Thus, the perimeter of APCD = 2 × 15 = 30 cm

Question 12.
ABCD is a quadrilateral such that all of its sides touch a circle. If AB = 6 cm, BC = 6.5 cm and CD = 7 cm, then find the length of AD.
Solution:
Let P, Q, R and S be the points where the circle touches the quadrilateral.
We know that the lengths of the two tangents drawn from an exterior point to a circle are equal. Thus, we have,

AP = AS, BP = BQ, CR = CQ and DR = DS
Hence, AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ AB + CD = AD + BC
⇒ AD = AB + CD – BC
= 6 + 7 – 6.5 = 6.5
Thus, AD = 6.5 cm

Question 13.
A man goes 10 m due east and then 24 m due north. Find the distance from the starting point.
Solution:
Let the initial position of the man be “O” and the final position be B.
In the ∆AOB,

OB² = OA² + AB²
OB² = 10² + 24²
= 100 + 576 = 676
OB = \(\sqrt { 676 }\) = 26 m
The man is at a distance of 26 m from the starting point.

Question 14.
Suppose AB, AC and BC have lengths 13, 16 and 20 respectively. If \(\frac { AF }{ FB } \) = \(\frac { 4 }{ 5 } \) and \(\frac { CE }{ EA } \) = \(\frac { 5 }{ 12 } \) Find BD and DC.
Solution:
Given that AB = 13, AC = 16 and BC = 20. Let BD = JC and DC = y.
Using Ceva’s theorem we have

\(\frac { BD }{ DC } \) × \(\frac { CE }{ EA } \) × \(\frac { AF }{ FB } \) = 1 ………(1)
Substitute the given values
\(\frac { x }{ y } \) × \(\frac { 5 }{ 12 } \) × \(\frac { 4 }{ 5 } \) = 1 ⇒ \(\frac { x }{ y } \) × \(\frac { 1 }{ 3 } \) = 1
\(\frac { x }{ y } \) = 3 ⇒ x = 3y
x – 3y = 0 ……(2)
Given BC = 20
x + y = 20 …..(3)
Subtract (2) and (3)

Question 15.
ABC is a right – angled triangle at B. Let D and E be any two points on AB and BC respectively. prove that AE² + CD² = AC² + DE².
Solution:
In the right ∆ ABE right- angled at B.
AE² = AB² + BE² …..(1)
In the right ∆ DBC,CD² = BD² + BC² ………(2)
Adding (1) and (2) we get
AE² + CD² = AB² + BE² + BD² + BC²
= (AB² + BC²) + (BC² + BD²)
AE² + CD² = AC² + DE²
Hence it is proved

[AC² = AB² + BC²]
[DE² = BE² + BD²]

III. Answer the following questions

Question 1.
A point O in the interior of a rectangle ABCD is joined to each of the vertices A, B, C and D. Prove that OA² + OC² = OB² + OD².
Solution:
Given: O is any point inside the rectangle ABCD.
To prove: OA² + OC² = OB² + OD²
Construction: Through “O” draw EF || AB.
Proof: Using Pythagoras theorem

In the right ∆OEA,
∴ OA²= OE² + AE² …(1)
(By Pythagoras theorem)
In the right ∆OFC,
OC² = OF² + FC² …(2)
(By Pythagoras theorem)
OB² = OF² + FB² … (3)
(By Pythagoras theorem)
In the right ∆OED,
OD² = OE² + ED² … (4)
(By Pythagoras theorem)
By adding (3) and (4) we get
OB² + OD² = OF² + FB² + OE² + ED²
= (OE² + FB²) + (OF² + ED²)
= (OE² + EA²) + (OF² + FC²)
[FB = EA and ED = FC]
OB² + OD² = OA² + OC² using (1) and (2)

Question 2.
A lotus is 20 cm above the water surface in a pond and its stem is partly below the water surface. As the wind blew, the stem is pushed aside so that the lotus touched the water 40 cm away from the original position of the stem. How much of the stem was below the water surface originally?
Solution:
Let O be the bottom of the stem immersed in water.
Let B be the lotus, AB be the length of the stem above the water surface
AB = 20 cm
Let OA be the length of the stem below the water surface
Let OA = x cm
Let C be the point where the lotus touches the water surface when the wind blow.
OC = OA + AB
OC = x + 20 cm
In ∆ AOC, OC² = OA² + AC²
(x + 20)² = x² + 40²
x² + 400 + 40x = x² + 1600

40x = 1600 – 400
40x = 1200
x = \(\frac { 1200 }{ 40 } \) = 30 cm
The stem is 30 cm below the water surface.

Question 3.
In the figure, DE || BC and \(\frac { AD }{ BD } \) = \(\frac { 3 }{ 5 } \), calculate the value of

Solution:
(i) Given, DE || BC and \(\frac { AD }{ BD } \) = \(\frac { 3 }{ 5 } \)
Let AD = 3k and BD = 5k; AB = 3k + 5k = 8k
In ∆^le ABC and ∆ADE
∠A = ∠A (common angle)
∠ABC = ∠ADE (corresponding angle)
Since DE || BC
∴ ∆ ABC ~ ∆ ADE

(ii) Let Area of ∆ ADE be 9k and Area of ∆ ABC be 64 k
Area of ∆ BCED = Area of ∆ ABC – Area of ∆ ADE
= 64 k – 9 k
= 55 k

Question 4.
A boy is designing a diamond shaped kite, as shown in the figure where AE = 16 cm, EC = 81 cm. He wants to use a straight cross bar BD. How long should it be?
Solution:
Let in AB be “x”, BE be “z” and BC be “y”
In the right ∆ AEB,
AB² = AE² + BE²
x² = 16² + Z² …….(1)
In the right ∆ BEC,
BC² = EC² + BE²
y² = 81² + z² ……(2)

In the right ∆ ACD,
AC² = AD² + DC²
97² = x² + y² ….(3)
Add (1) and (2) ⇒ x² + y² = 162 + z² + 81² + z²
x² + y² = 2z² + 16² + 81²
97² = 2z² + 16² + 81² (from 3)
9409 = 2z² + 256 + 6561
= 2z² + 6817
2z² = 9409 – 6817 = 2592
z² = \(\frac { 2592 }{ 2 } \) = 1296
z = \(\sqrt { 1296 }\) = 36
∴ Length of cross bar BD = 2 × BE = 2 × 36 = 72 cm

Question 5.
Find the unknown values in each of the following figures. All lengths are given in centimetres (All measures are not in scale)

Solution:
(i) In ∆ ABC and ∆ ADE,
∠A = ∠A (common angle)
∠ABC = ∠ADE (corresponding angle) [BC || DE]
∆ ABC ~ ∆ ADE (By AAA similarity)

In ∆ EAG and ∆ ECF,
∠E || ∠E (common angle)
∠ECF = ∠EAG (corresponding angle)
Given CF || AG
∆ EAG ~ ∆ ECF
\(\frac { EC }{ EA } \) = \(\frac { CF }{ AG } \) ⇒ \(\frac { 8 }{ x+8 } \) = \(\frac { 6 }{ y } \) ⇒ \(\frac { 8 }{ 12 } \) = \(\frac { 6 }{ y } \) (x = 4)
8y = 6 × 12
y = \(\frac{12 \times 6}{8}\) = 9 cm
∴ The value of x = 4 cm and y = 9 cm.

(ii) In ∆ HBC and ∆ HFG,
∠H = ∠H (common angle)
∠HFG = ∠HBC (corresponding angle)
Given FG || BC
\(\frac { HF }{ HB } \) = \(\frac { FG }{ BC } \) ⇒ \(\frac { 4 }{ 10 } \) = \(\frac { x }{ 9 } \) ⇒ 10x = 36
x = 3.6 cm
In ∆ FBD and ∆ FHD,
∠BFD = ∠HFG (vertically opposite angle)
∠FBD = ∠FHG (Alternate angles)
By AA similarity
∆ FBD ~ ∆ FHG
\(\frac { FG }{ FD } \) = \(\frac { FH }{ FB } \) ⇒ \(\frac { x }{ 3+y } \) = \(\frac { 4 }{ 6 } \)
4 (3 + y) = 3.6 × 6
3 + y = \(\frac{3.6 \times 6}{4}\) = 5.4 ⇒ y = 5.4 – 3 = 2.4 cm
In ∆ AEG and ∆ ABC,
∠A = ∠A (common angle)
∠AEG = ∠ABC (corresponding angles)
Given EG || BC
\(\frac { AE }{ AB } \) = \(\frac { EG }{ BC } \)
\(\frac { z }{ z+5 } \) = \(\frac { x+y }{ 9 } \) ⇒ \(\frac { z }{ z+5 } \) = \(\frac { 3.6+2.4 }{ 9 } \)
\(\frac { z }{ z+5 } \) = \(\frac { 6 }{ 9 } \) ⇒ \(\frac { z }{ z+5 } \) = \(\frac { 2 }{ 3 } \)
3z = 2z + 10 ⇒ 3z – 2z = 10 ⇒ z = 10
∴ The values of x = 3.6 cm, y = 2.4 cm and z = 10 cm.

Question 6.
The internal bisector of ∠A of AABC meets BC at D and the external bisector of ∠A meets BC produced at E. Prove that \(\frac { BD }{ BE } \) = \(\frac { CD}{ CE } \)
Solution:
Given: In ∆ ABC, AD is the internal bisector of ∠A meets BC at D. AE is the external bisector of ∠A meets BC produced to E.
To Proof: \(\frac { BD }{ BE } \) = \(\frac { CD }{ CE } \)

Proof: In ∆ ABC, AD is the internal bisector of ∠A.
By ABT theorem we have,
∴ \(\frac { BE }{ CD } \) = \(\frac { AB }{ AC } \) ……….(1)
In ∆ABC, AE is the external bisector of ∠A.
By ABT theorem we have,
∴ \(\frac { BE }{ CE } \) = \(\frac { AB }{ AC } \) ……….(2)
From (1) and (2) we get
\(\frac { BD }{ CD } \) = \(\frac { BE }{ CE } \) ⇒ ∴ \(\frac { BD }{ BE } \) = \(\frac { CD }{ CE } \)

Question 7.
In a quadrilateral ABCD, the bisectors of ∠B and ∠D intersect on AC at E. Prove that \(\frac { AB }{ BC } \) = \(\frac { AD }{ DC } \)
Solution:
Given: ABCD is a quadrilateral. BE is the bisector of ∠B intersecting AC at E, DE is the bisector of ∠D intersecting AC at E.
To proove: \(\frac { AB }{ BC } \) = \(\frac { AD }{ DC } \)
Proof: In ∆ABC, BE is the internal bisector of ∠D.
By Angle bisector theorem we have,
\(\frac { AE }{ EC } \) = \(\frac { AB }{ BC } \) ……..(1)

In ∆ ACD, DE is the internal bisector of ∠C.
By Angle bisector theorem we have,
∴ \(\frac { AE }{ EC } \) = \(\frac { AD }{ DC } \) ……….(2)
From (1) and (2) we get,
\(\frac { AB }{ BC } \) = \(\frac { AD }{ DC } \)

Question 8.
ABCD is a quadrilateral with AB parallel to DC. A line drawn parallel to AB meets AD at P and BC at Q. Prove that \(\frac { AP }{ PD } \) = \(\frac { BQ }{ QC } \)
Solution:
Given: ABCD is a quadrilateral. AB || DC.
The line PQ intersect AD at P and BC at Q

To prove: \(\frac { AP }{ PD } \) = \(\frac { BQ }{ QC } \)
Proof: In the ∆ABC, OQ || AB
By BPT, theorem we have, \(\frac { AO }{ OC } \) = \(\frac { BQ}{ QC } \) ……………(1)
In the ∆ACD, PO || DC
By BPT, theorem we have, \(\frac { AO }{ OC } \) = \(\frac { AP }{ PD } \) ……..(2)
From (1) and (2) we get, \(\frac { AP }{ PD } \) = \(\frac { BQ }{ QC } \)

Question 9.
D is the midpoint of the side BC of AABC. If P and Q are points on AB and on AC such that DP bisects ∠BDA and DQ bisects ∠ADC, then prove that PQ || BC.
Solution:
In ∆ABD,DP is the angle bisector of ∠BDA.
∴ \(\frac { AP }{ PB } \) = \(\frac { AD }{ BD } \) (angle bisector theorem) ……..(1)
In ∆ADC, DQ is the bisector of ∠ADC
∴ \(\frac { AQ }{ QC } \) = \(\frac { AD }{ DC } \) (angle bisector theorem) ……….(2)

But, BD = DC (D is the midpoint of BC)
Now (2) ⇒ ∴ \(\frac { AQ }{ QC } \) = \(\frac { AD }{ BD } \)
From (1) and (3) We get,
∴ \(\frac { AP }{ PB } \) = \(\frac { AQ }{ QC } \)
Thus PQ || BC (converse of thales theorem)

Question 10.
ABCD is a trapezium with AB || DC. The diagonal AC and BD intersect at E. If ∆AED ~ ∆BEC. Prove that AD = BC.
Solution:
By given data ABCD is a trapezium with AB || DC.
In ∆ ECD and ∆ ABE

∠EDC = ∠EBA
∠ECD = ∠EAB
∴ ∆ DEC ~ ∆ BEA by AA – similarity
\(\frac { DE }{ BE } \) = \(\frac { EC }{ EA } \) = \(\frac { DC }{ BA } \)
\(\frac { DE }{ BE } \) = \(\frac { EC }{ EA } \)
\(\frac { DE }{ EC } \) = \(\frac { BE }{ EA } \) ……….(1)

Also given ∆ DEA ~ ∆ CEB
\(\frac { DE }{ CE } \) = \(\frac { EA }{ EB } \) = \(\frac { DA }{ CB } \)
\(\frac { DE }{ CE } \) = \(\frac { EA }{ EB } \) …………(2)
From (1) and (2) we get
\(\frac { BE }{ EA } \) = \(\frac { EA }{ EB } \) ⇒ EB² = EA²
∴ EB = EA
Substitute in (2) we get
\(\frac { EA }{ EA } \) = \(\frac { DA }{ CB } \)
1 = \(\frac { DA }{ CB } \) ⇒ ∴ AD = DC Hence it is proved