Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Students can download Maths Chapter 8 Statistics 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 8 Statistics Additional Questions

I. Choose the Best Answer

Question 1.
The Arithmetic mean of all the factors of 10 is ………
(a) 4.5
(b) 5.5
(c) 10
(d) 55
Solution:
(a) 4.5

Question 2.
The mean of five numbers is 27, if one number is excluded, then mean is 25. Then the excluded number is ……..
(a) 0
(b) 15
(c) 25
(d) 35
Solution:
(d) 35

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 3.
The mean of 8 numbers is 15. If each number is multiplied by 2, then the new mean will be ……..
(a) 7.5
(b) 30
(c) 10
(d) 25
Solution:
(b) 30

Question 4.
The median of 11, 8, 4, 9, 7, 5, 2, 4, 10 is ……..
(a) 1
(b) 8
(c) 4
(d) 11
Solution:
(a) 1

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 5.
Median is …….
(a) the most frequent value
(b) the least frequent value
(c) middle most value
(d) mean of first and last values
Solution:
(c) middle most value

Question 6.
The mode of the distribution is …….
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 1
(a) 3
(b) 4
(c) 6
(d) 14
Solution:
(b) 4

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 7.
Mode is …….
(a) the middle value
(b) extreme value
(c) minimum value
(d) the most repeated value
Solution:
(d) the most repeated value

Question 8.
The mode of the data 72, 33, 44, 72, 81, 72, 15 is ……..
(a) 12
(b) 33
(c) 81
(d) 15
Solution:
(a) 12

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 9.
The Arithmetic mean of 10 number is -7. If 5 is added to every number, then the new arithmetic mean is ……..
(a) 17
(b) 12
(c) -2
(d) -7
Solution:
(c) -2

Question 10.
The Arithmetic mean of integers from -5 to 5 is ……..
(a) 25
(b) 10
(c) 3
(d) 0
Solution:
(d) 0

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

II. Answer the Following Questions

Question 1.
Find the Arithmetic mean for the following data.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 2
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 3
Arithmetic mean (\(\bar { X }\)) = \(\frac{Σfx}{Σf}\)
= \(\frac{5540}{96}\)
= 57.7
∴ Arithmetic mean = 57.7

Question 2.
Calculate the Arithmetic mean of the following data using step deviation method.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 4
Solution:
Assumed mean = 35
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 5
Arithmetic mean (\(\bar { X }\)) = A + \(\frac{Σfd}{Σf}\) × c
= 35 + \(\frac{(-20)}{100}\) × 10
= 35 – 2
= 33
∴ Arithmetic mean = 33

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 3.
Find the median for the following data.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 6
Solution:
The given class intervals is inclusive type we convert it into exclusive type.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 7
\(\frac{N}{2}\) = \(\frac{70}{2}\)
= 35
Median class is 25.5 – 30.5
Here l = 25.5, f = 26, m = 30, c = 5
Median = l + \(\frac{\frac{N}{2}-m}{f}\) × c
= 25.5 + \(\frac{35-30}{26}\) × 5
= 25.5 + \(\frac{5×5}{26}\)
= 25.5 + 0.96
= 26.46
∴ Median = 26.46

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 4.
Calculate the mode of the following data.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 8
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 9
The highest frequency is 30. Corresponding class interval is the modal class.
Here l = 25, f = 30, f1 = 18, f2 = 20 and c = 5
Mode
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 10
∴ Arithmetic mean = 25 + 2.727
= 25 + 2.73
= 27.73
mode = 27.73

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Question 5.
Find the mean, median and mode of marks obtained by 20 students in an examination. The marks are given below.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 11
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 12
Arithmetic mean:
Here Σfx = 560, Σf = 20
\(\bar { X }\) = \(\frac{Σfx}{Σf}\)
= \(\frac{560}{20}\)

Median:
Median class is 20 – 30
Here l = 20, \(\frac{N}{2}\) = 10, m = 5, c = 10, f = 5
Median
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 13
= 20 + 10
= 30

Mode:
The highest frequency is 8
Modal class is 30 – 40
Here, l = 30, f = 8, f1 = 5, f2 = 2, c = 10
Mode
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions 14
= 33.33
Mean = 28, median = 30, mode = 33.33

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Additional Questions

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Students can download Maths Chapter 8 Statistics Ex 8.4 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 8 Statistics Ex 8.4

Question 1.
Let m be the mid point and b be the upper limit of a class in a continuous frequency distribution. The lower limit of the class is …….
(a) 2m – b
(b) 2m + b
(c) m – b
(d) m – 2b
Solution:
(a) 2m – b

Question 2.
The mean of a set of seven numbers is 81. If one of the numbers is discarded, the mean of the remaining numbers is 78. The value of discarded number is ……..
(a) 101
(b) 100
(c) 99
(d) 98
Solution:
(c) 99
Hint:
Total of 8 numbers = 81 × 7 = 567
Total of 7 numbers = 78 × 6 = 468
The number is = 567 – 468
= 99

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 3.
A particular observation which occurs maximum number of times in a given data is called its ………
(a) frequency
(b) range
(c) mode
(d) median
Solution:
(c) mode

Question 4.
For which set of numbers do the mean, median and mode all have the same values?
(a) 2, 2, 2, 4
(b) 1, 3, 3, 3, 5
(c) 1, 1, 2, 5, 6
(d) 1, 1, 2, 1, 5
Solution:
(b) 1, 3, 3, 3, 5

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 5.
The algebraic sum of the deviations of a set of n values from their mean is ……..
(a) 0
(b) n – 1
(c) n
(d) n + 1
Solution:
(a) 0

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 6.
The mean of a, b, c, d and e is 28. If the mean of a, c and e is 24, then mean of b and d is ……..
(a) 24
(b) 36
(c) 26
(d) 34
Solution:
(d) 34
Hint:
Mean = 28
a + b + c + d + e = 28 × 5 = 140
= 140 …… (1)
But \(\frac{a+c+e}{3}\) = 24
a + c + e = 72
a + b + c + d + e = 140
b + d + 72 = 140
b + d = 140 – 72
= 68
Mean = \(\frac{68}{2}\)
= 34

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 7.
If the mean of five observations x, x + 2, x + 4, x + 6, x + 8, is 11, then the mean of first three observations is …….
(a) 9
(b) 11
(c) 13
(d) 15
Solution:
(a) 9
Hint:
Mean = \(\frac{x+x+2+x+4+x+6+x+8}{5}\)
11 = \(\frac{5x+20}{5}\)
5x + 20 = 55
5x = 55 – 20
= 35
x = \(\frac{35}{5}\)
= 7
Mean of first 3 observation is = \(\frac{7+9+11}{3}\)
= 9

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 8.
The mean of 5, 9, x, 17 and 21 is 13, then find the value of x.
(a) 9
(b) 13
(c) 17
(d) 21
Solution:
(b) 13
Hint:
Mean = \(\frac{5+9+x+17+21}{5}\)
13 = \(\frac{52+x}{5}\)
65 = 52 + x
x = 65 – 52
= 13

Question 9.
The mean of the square of first 11 natural numbers is
(a) 26
(b) 46
(c) 48
(d) 52
Solution:
(b) 46
Hint:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4 1
= 46

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.4

Question 10.
The mean of a set of numbers is \(\bar { X }\). If each number is multiplied by z, the mean is ……..
(a) \(\bar { X }\) + z
(b) \(\bar { X }\) – z
(c) z\(\bar { X }\)
(d) \(\bar { X }\)
Solution:
(c) z\(\bar { X }\)
Hint:
If each observation is multiplied by k, k ≠ 0 then the arithmetic mean is also multiplied by the same quantity.

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Students can download Maths Chapter 8 Statistics Ex 8.3 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 8 Statistics Ex 8.3

Question 1.
The monthly salary of 10 employees in a factory are given below:
Rs 5000, Rs 7000, Rs 5000, Rs 7000, Rs 8000, Rs 7000, Rs 7000, Rs 8000, Rs 7000, Rs 5000
Find the mean, median and mode.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 1
Mean = 6600
Median:
Arrange in ascending order we get.
5000, 5000, 5000, 7000, 7000, 7000, 7000, 7000, 8000, 8000
The number of values = 10
Median = Average of (\(\frac{10}{2})^{th}\) and (\(\frac{10}{2}+1)^{th}\)
= Average of 5th value and 6th value
= \(\frac{7000+7000}{2}\)
∴ Median = 7000
Mode: 7000 repeated 5 times
∴ Mode = 7000

Question 2.
Find the mode of the given data: 3.1, 3.2, 3.3, 2.1, 1.3, 3.3, 3.1
Solution:
3.1 occuring two times
3.3 occuring two times
∴ 3.1 and 3.3 are the mode (bimodal)

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Question 3.
For the data 11, 15, 17, x + 1, 19, x – 2, 3 if the mean is 14, find the value of x. Also find the mode of the data.
Solution:
Arithmetic mean
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 2
∴ 2x + 64 = 14 × 7
2x = 98 – 64
2x = 34
x = \(\frac{34}{2}\)
= 17
The given numbers are 11, 15, 17, 18, 19, 15 and 3
15 occuring two times
∴ Mode = 15
The value of x = 17 and mode = 15

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Question 4.
The demand of track suit of different sizes as obtained by a survey is given below:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 3
Which size is demanded more?
Solution:
The highest frequency is 37
The corresponding value is the mode
∴ Mode = 40
Size 40 is demanded more.

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Question 5.
Find the mode of the following data:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 4
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 5
The highest frequency is 46
20 – 30 is the modal class
Here l = 20, f = 46, f1 = 38, f2 = 34 and c = 10
Mode
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 6
= 20 + 4
= 24
∴ Mode = 24

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Question 6.
Find the mode of the following distribution
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 7
Solution:
In the given table the class intervals are in inclusive form; convert them into exclusive form.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 8
The highest frequency is 14
Modal class is 54.5 – 64.5
Here l = 54.5, f = 14, f1 = 10, f2 = 8 and c = 10
mode
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3 9
= 58.5
∴ Mode = 58.5

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.3

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Students can download Maths Chapter 8 Statistics Ex 8.2 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 8 Statistics Ex 8.2

Question 1.
Find the median of the given values: 47, 53, 62, 71, 83, 21, 43, 47, 41
Solution:
Arrange the values in ascending order we get
21, 41, 43, 47, 47, 53, 62, 71, 83
The number of values = 9 which is odd
Median = (\(\frac{9+1}{2})^{th}\) variable
= 5th variable
∴ Median = 47

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Question 2.
Find the median of the given data: 36, 44, 86, 31, 37, 44, 86, 35, 60, 51
Solution:
Arrange the values in ascending order we get
31, 35, 36, 37, 44, 44, 51, 60, 86, 86
The number of values = 10 which is even
Median = Average of (\(\frac{10}{2})^{th}\) and (\(\frac{10}{2}+1)^{th}\) value
= Average of 5th and 6th value
= \(\frac{44+44}{2}\)
= \(\frac{88}{2}\)
= 44
∴ Median = 44

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Question 3.
The median of observation 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 arranged in ascending order is 24. Find the values of x.
Solution:
The given observation is 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 (is ascending order)
The number of values =10
Median = Average of (\(\frac{10}{2})^{th}\) and (\(\frac{10}{2}+1)^{th}\) value
= Average of 5th and 6th value
24 = \(\frac{x+2.+x+4}{2}\)
24 = \(\frac{2x+6}{2}\)
2x + 6 = 48
2x = 48 – 6
2x = 42
x = \(\frac{42}{2}\)
= 21
The value of x = 21

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Question 4.
A researcher studying the behavior of mice has recorded the time (in seconds) taken by each mouse to locate its food by considering 13 different mice as 31, 33, 63, 33, 28, 29, 33, 27, 27, 34, 35, 28, 32. Find the median time that mice spent in searching its food.
Solution:
Arrange the value in ascending order we get
27, 27, 28, 28, 29, 31, 32, 33, 33, 33, 34, 35, 63
The number of values = 13 which is odd
Median = (\(\frac{13+1}{2})^{th}\) value
= (\(\frac{14}{2})^{th}\)
= 7th value
7th value is = 32
∴ Median = 32

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Question 5.
The following are the marks scored by the students in the Summative Assessment exam.
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2 1
Calculate the median.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2 2
\(\frac{N}{2}\) = \(\frac{50}{2}\)
= 25
Here l = 30, f = 10; m = 24 and c = 10
Median = l + \(\frac{(\frac{N}{2}-m)×c}{f}\)
= 30 + \(\frac{(25-24)10}{10}\)
= 30 + 1
= 31
∴ Median = 31

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Question 6.
The mean of five positive integers is twice their median. If four of the integers are 3, 4, 6, 9 and median is 6, then find the fifth integer.
Solution:
Let the 5th positive integer be x
\(\bar { x }\) = \(\frac{3+4+6+9+x}{5}\)
= \(\frac{22+x}{5}\)
Median = 6
Mean = 2 × median
\(\frac{22+x}{5}\) = 2 × 6
22 + x = 60
x = 60 – 22
= 38
The fifth integer is 38.

Samacheer Kalvi 9th Maths Guide Chapter 8 Statistics Ex 8.2

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions

Students can download Maths Chapter 5 Coordinate Geometry 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 5 Coordinate Geometry Additional Questions

I. Multiple Choice Questions.

Question 1.
On which quadrant does the point (- 4, 3) lie?
(a) I
(b) II
(c) III
(d) IV
Solution:
(b) II

Question 2.
The point whose abscissa is 5 and lies on the x-axis is …….
(a) (-5, 0)
(b) (5, 5)
(c) (0, 5)
(d) (5, 0)
Solution:
(d) (5, 0)

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions

Question 3.
A point which lies in the III quadrant is ……..
(a) (5, 4)
(b) (5, -4)
(c) (-5, -4)
(d) (-5, 4)
Solution:
(c) (-5, -4)

Question 4.
A point on the y-axis is ……..
(a) (1, 1)
(b) (6, 0)
(c) (0, 6)
(d) (-1, -1)
Solution:
(c) (0, 6)

Question 5.
The distance between the points (4, -1) and the origin is ……..
(a) \(\sqrt{24}\)
(b) \(\sqrt{37}\)
(c) \(\sqrt{26}\)
(d) \(\sqrt{17}\)
Solution:
(d) \(\sqrt{17}\)

Question 6.
The distance between the points (-1, 2) and (3, 2) is ……..
(a) \(\sqrt{14}\)
(b) \(\sqrt{15}\)
(c) 4
(d) 0
Solution:
(c) 4

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions

Question 7.
The centre of a circle is (0, 0). One end point of a diameter is (5, -1), then the radius is …….
(a) \(\sqrt{24}\)
(b) \(\sqrt{37}\)
(c) \(\sqrt{26}\)
(d) \(\sqrt{17}\)
Solution:
(c) \(\sqrt{26}\)

Question 8.
The point (0, -3) lies on
(a) + ve x-axis
(b) + ve y-axis
(c) – ve x-axis
(d) – ve y-axis
Solution:
(d) – ve y-axis

Question 9.
The point which is on y-axis with ordinate -5 is ……..
(a) (0, -5)
(b) (-5, 0)
(c) (5, 0)
(d) (0, 5)
Solution:
(a) (0, -5)

Question 10.
The diagonal of a square formed by the points (1, 0), (0, 1), (-1, 0) and (0, -1) is …….
(a) 2
(b) 4
(c) √2
(d) 8
Solution:
(a) 2

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions

Question 11.
The distance between the points (-2, 2) and (3, 2) is ……..
(a) 10 units
(b) 5 units
(c) 5√3 units
(d) 20 units
Solution:
(b) 5 units

Question 12.
The midpoint of the line joining the points (1, -1) and (-5, 3) is ……..
(a) (2, 1)
(b) (2, -1)
(c) (-2, -1)
(d) (-2, 1)
Solution:
(d) (-2, 1)

Question 13.
If the centroid of a triangle is at (1, 3) and two of its vertices are (-7, 6) and (8, 5) then the third vertex is ……..
(a) (-2, 2)
(b) (2, -2)
(c) (-2, -2)
(d) (2, 2)
Solution:
(b) (2, -2)

Question 14.
The ratio in which the X-axis divides the line segment joining the points (6, 4) and (1, -7) is ……..
(a) 1 : 2
(b) 2 : 3
(c) 4 : 7
(d) 7 : 4
Solution:
(c) 4 : 7

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions

Question 15.
The centroid of a triangle (3, -5), (-7, 4) and (10, -2) is …….
(a) (2, -1)
(b) (2, 1)
(c) (-2, 1)
(d) (1, -2)
Solution:
(a) (2, -1)

II. Answer the Following Questions.

Question 1.
Show that the given points (1, 1), (5, 4), (-2, 5) are the vertices of an isosceles right angled triangle.
Solution:
Let A (1, 1), B (5, 4) and G (-2, 5)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 1
AB = 5, AC = 5
∴ ABC is an isosceles triangle …….. (1)
BC² = AB² + AC²
50 = 25 + 25 ⇒ 50 = 50
∴ ∠A = 90° ……… (2)
From (1) and (2) we get ABC is an isosceles right angle triangle.

Question 2.
Show that the point (3, -2), (3, 2), (-1, 2) and (-1, -2) taken in order are the vertices of a square.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 2
= \(\sqrt{16}\)
= 4
AB = BC = CD = DA = 4. All the four sides are equal.
∴ ABCD is a Rhombus ……..(1)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 3
Diagonal AC = Diagonal BD = \(\sqrt{32}\) ……..(2)
From (1) and (2) we get ABCD is a square.

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions

Question 3.
Show that the point A (3, 7) B (6, 5) and C (15, -1) are collinear.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 4
AB + BC = AC ⇒ \(\sqrt{13}\) + 3\(\sqrt{13}\) = 4\(\sqrt{13}\)
∴ The points A, B, C are collinear.

Question 4.
Find the type of triangle formed by (-1, -1), (1, 1) and (-√, √3)
Solution:
Let the point A (-1, -1), B (1, 1) and C (-√3, √3)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 5
AB = BC = AC = √8
∴ ABC is an equilateral triangle.

Question 5.
Find x such that PQ = QR where P(6, -1) Q(1, 3) and R(x, 8) respectively.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 6
But PQ = QR
\(\sqrt{(x-1)^{2}+25}\) = \(\sqrt{41}\)
Squaring on both sides
(x – 1)² + 25 = 41
(x – 1)² = 41 – 25 = 16
x – 1 = \(\sqrt{16}\) = ± 4
x – 1 = 4 (or) x – 1 = – 4
x = 5 (or) x = -4 + 1 = -3
The value of x = 5 or – 3

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions

Question 6.
Find the coordinate of the point of trisection of the line segment joining (4, -1) and
Solution:
Let A (4, -1) and B (-2, -3) are the given points
Let P (a, b) and Q (c, d) be the points of trisection of AB.
∴ AP = PQ = QB
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 7
The required coordinate P is (2, –\(\frac{5}{3}\)) and Q is (0, –\(\frac{7}{3}\))

Question 7.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
Given points are A(-3, 10), B(6, -8) and P(-1, 6)
divides AB internally in the ratio m : n
By section formula.
A line divides internally in the ratio m : n the point P =
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 8
∴ \(\frac{6m-3n}{m+n}\) = -1
6m – 3n = -m – n
6m + m = 3n – n
7m = 2n ⇒ \(\frac{m}{n}\) = \(\frac{2}{7}\)
∴ m : n = 2 : 7
and
\(\frac{-8m+10n}{m+n}\) = 6
-8m + 10n = 6m + 6n
-8m – 6m = 6n – 10n
14m = 4n
∴ \(\frac{m}{n}\) = \(\frac{14}{4}\) = \(\frac{2}{7}\)
Hence P divides AB internally in the ratio 2 : 7

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions

Question 8.
If (1, 2) (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find “x” and “y”.
Solution:
Let A(1, 2), B(4, y), C(x, 6) and D(3, 5)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions 9
Since ABCD is a parallelogram the diagonal bisect each other
Mid point of AC = Mid point of BD
(\(\frac{1+x}{2}\), 4) = (\(\frac{7}{2}\), \(\frac{y+5}{2}\))
\(\frac{1+x}{2}\) = \(\frac{7}{2}\)
1 + x = 7
x = 7 – 1
= 6
and
\(\frac{y+5}{2}\) = 4
y + 5 = 8
y = 8 – 5
= 3
∴ The value of x = 6 and y = 3

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.3

Students can download Maths Chapter 6 Trigonometry Ex 6.3 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 6 Trigonometry Ex 6.3

Question 1.
Find the value of the following:
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.3 1
(iii) tan 15° tan 30° tan 45° tan 60° tan 75°
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.3 2
Solution:
(i) cos 45° = \(\frac{1}{√2}\)
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.3 3
= 1² + 1² – 2(\(\frac{1}{√2}\))²
= 1 + 1 – 2(\(\frac{1}{2}\))
= 2 – 1
= 1

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.3

(ii) cos 60° = \(\frac{1}{2}\)
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.3 4
= 1 + 1 + 1 – 8(\(\frac{1}{2}\))²
= 3 – 8 × \(\frac{1}{4}\)
= 3 – 2
= 1

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.3

(iii) tan 30° = \(\frac{1}{√3}\), tan 45° = 1, tan 60° = √3
tan 15° . tan 30°. tan 45° . tan 60°. tan 75° = tan 15° . \(\frac{1}{√3}\) . 1 . √3 tan 75°
= tan 15° × tan 75° × \(\frac{1}{√3}\) × 1 × √3
= tan(90° – 75°) × \(\frac{1}{cot 75°}\) × 1 [tan 90° – θ = cot θ]
= cot 75° × \(\frac{1}{cot 75°}\) × 1
= 1

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.3

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.3 5
= 1 + 1
= 2

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.3

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.2

Students can download Maths Chapter 6 Trigonometry Ex 6.2 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 6 Trigonometry Ex 6.2

Question 1.
Verify the following equalities:
(i) sin² 60° + cos² 60° = 1
Solution:
sin 60° = \(\frac{√3}{2}\); cos 60° = \(\frac{1}{2}\)
L.H.S = sin² 60° + cos² 60°
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.2 1
L.H.S = R.H.S
Hence it is proved.

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.2

(ii) 1 + tan² 30° = sec² 30°
Solution:
tan 30° = \(\frac{1}{√3}\); sec 30° = \(\frac{2}{√3}\)
L.H.S = 1 + tan² 30°
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.2 2
∴ L.H.S = R.H.S
Hence it is proved.

(iii) cos 90° = 1 – 2sin² 45° = 2cos² 45° – 1
Solution:
cos 90° = 0, sin 45° = \(\frac{1}{√2}\), cos 45° = \(\frac{1}{√2}\)
cos 90° = 0 ……. (1)
1 – 2 sin² 45° = 1 – 2 (\(\frac{1}{√2}\))²
= 1 – 2 × \(\frac{1}{2}\)
= 1 – 1 = 0 → (2)
2 cos² 45° – 1 = 2(\(\frac{1}{√2}\))² – 1
= \(\frac{2}{2}\) – 1
= \(\frac{2 – 2}{2}\) = 0 → (3)
From (1), (2) and (3) we get
cos 90° = 1 – 2 sin² 45° = 2 cos² 45° – 1
Hence it is proved.

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.2

(iv) sin 30° cos 60° + cos 30° sin 60° = sin 90°
Solution:
sin 30° = \(\frac{1}{2}\); cos 60° = \(\frac{1}{2}\); cos 30° = \(\frac{√3}{2}\); sin 60° = \(\frac{√3}{2}\); sin 90° = 1
L.H.S = sin 30° cos 60° + cos 30° sin 60°
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.2 3
= 1
R.H.S = sin 90° = 1
L.H.S = R.H.S
Hence it is proved.

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.2

Question 2.
Find the value of the following:
(i) \(\frac{tan 45°}{cosec 30°}\) + \(\frac{sec 60°}{cot 45°}\) – \(\frac{5 sin 90°}{2 cos 0°}\)
(ii) (sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° – cos 45°)
(iii) sin²30° – 2 cos³ 60° + 3 tan4 45°
Solution:
(i) tan 45° = 1, cosec 30° = 2; sec 60° = 2; cot 45° = 1; tan 45°, sin 90° = 1; cos 0° = 1
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.2 4
= 0

(ii) sin 90° = 1; cos 60° = \(\frac{1}{2}\); cos 45° = \(\frac{1}{√2}\); sin 30° = \(\frac{1}{2}\); cos 0° = 1
(sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° – cos 45°)
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.2 5
= \(\frac{7}{4}\)

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.2

(iii) sin 30° = \(\frac{1}{2}\); cos 60° = \(\frac{1}{2}\); tan 45° = 1
sin² 30° – 2 cos³ 60° + 3 tan4 45°
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.2 6
= 3

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.2

Question 3.
Verify cos 3 A = 4 cos³ A – 3 cos A, when A = 30°
Solution:
L.H.S = cos 3 A
= cos 3 (30°)
= cos 90°
= 0
R.H.S = 4 cos³ A – 3 cos A
= 4 cos³ 30° – 3 cos 30°
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.2 7
= 0
∴ L.H.S = R.H.S
Hence it is proved.

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.2

Question 4.
Find the value of 8 sin2x, cos 4x, sin 6x, when x = 15°.
Solution:
8 sin 2x cos 4x sin 6x = 8 sin 2 (15°) × cos 4 (15°) × sin (6 × 15°)
= 8 sin 30° × cos 60° × sin 90°
= 8 × \(\frac{1}{2}\) × \(\frac{1}{2}\) × 1
= 2

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.2

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Students can download Maths Chapter 6 Trigonometry Ex 6.1 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 6 Trigonometry Ex 6.1

Question 1.
From the given figure, find all the trigonometric ratios of angle B.
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 1
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 2

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 2.
From the given figure, find the values of
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 3
(i) sin B
(ii) sec B
(iii) cot B
(v) tan C
(vi) cosec C
Solution:
In the right ΔABD,
AD² = AB² – BD²
= 13² – 5²
= 169 – 25
= 144
AD = \(\sqrt{144}\)
= 12
In the right ΔADC,
AC² = AD² + DC²
= 12² + 16²
= 144 + 256
= 400
AC = \(\sqrt{400}\)
= 20
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 4

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 3.
If 2 cos θ = √3, then find all the trigonometric ratios of angle θ.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 5
2 cos θ = √3 ⇒ cos θ = \(\frac{√3}{2}\)
AB² = AC² – BC²
= 2² – (√3)² ⇒ = 4 – 3 = 1
AB = √1 = 1
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 6

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 4.
If cos A =\(\frac{3}{5}\), then find the value of \(\frac{sin A-cos A}{2 tan A}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 7
cos A = \(\frac{3}{5}\)
ΔABC
BC² = AC² – AB²
= 5² – 3²
= 25 – 9
= 16
BC = \(\sqrt{16}\) = 4
sin A = \(\frac{4}{5}\); tan A = \(\frac{4}{3}\)
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 8
∴ The value of \(\frac{sin A-cos A}{2 tan A}\) = \(\frac{3}{40}\)

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 5.
If cos A = \(\frac{2x}{1+x_{2}}\) then find the values of sin A and tan A in terms of x.
Solution:
cos A = \(\frac{2x}{1+x_{2}}\)
In the triangle ABC
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 9
BC² = AC² – AB²
= (1 + x²)² – (2x)²
= 1 + x4 + 2x² – 4x²
= x4 – 2x² + 1
= (x² – 1)² (or) (1 – x²)² (using (a – b)²)
BC = \(\sqrt{(x^{2}-1)^{2}}\) (or) \(\sqrt{(1-x^{2})^{2}}\)
BC = x² – 1
The value of sin A = \(\frac{BC}{AC}\) = \(\frac{x²-1}{x²+1}\)
tan A = \(\frac{BC}{AB}\) = \(\frac{x²-1}{2x}\)
and
BC = 1 – x²
The value of sin A = \(\frac{1-x²}{1+x²}\)
tan A = \(\frac{1-x²}{2x}\)

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 6.
If sin θ = \(\frac{a}{\sqrt{a²+b²}}\) then show that b sin θ = a cos θ.
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 10
sin θ = \(\frac{a}{\sqrt{a²+b²}}\)
In the triangle ΔABC
BC² = AC² – AB²
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 11
L. H. S = R. H. S

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 7.
If 3 cot A = 2, then find the value of \(\frac{4sin A-3cos A}{2 sin A+3 Cos A}\)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 12
3 cot A = 2 ⇒ cot A = \(\frac{2}{3}\)
AC² = AB² + BC²
= 3² + 2²
= 9 + 4
AC = \(\sqrt{13}\)
cos A = \(\frac{AB}{AC}\) = \(\frac{3}{\sqrt{13}}\)
sin A = \(\frac{BC}{AC}\) = \(\frac{2}{\sqrt{13}}\)
The value of \(\frac{4sin A-3cos A}{2 sin A+3 Cos A}\)
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 13
The value is (\(\frac{-1}{13}\))

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 8.
If cos θ : sin θ = 1 : 2, then find the value of \(\frac{8cos θ-2cos θ}{4 cos θ+2 sin θ}\)
Solution:
cos θ : sin θ = 1 : 2
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 14
Aliter:
cos θ : sin θ = 1 : 2
2 cos θ = sin θ ⇒ 2 = \(\frac{sin θ}{cos θ}\) ⇒ 2 = tan θ
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 15
∴ The value is \(\frac{1}{2}\)

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 9.
From the given figure, prove that θ + ∅ = 90°. Also prove that there are two other right angled triangles. Find sin α, cos β and tan ∅.
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 16
Solution:
In the ΔABC,
AB = 9 + 16 = 25
AC = 15; BC = 20
AB² = 25²
= 625 ……. (1)
AC² + BC² = 15² + 20²
= 225 + 400
= 625 …….. (2)
From (1) and (2) we get
AB² = AC² + BC²
ABC is a right angle triangle at C (Pythagoras theorem)
∴ ∠C = 90°
θ + ∅ = 90°
Also ADC is a right angle triangle ∠ADC = 90° (Given)
BDC is also a right angle triangle ∠BDC = 90° (since ADB is a straight line sum of the two angle is 180°)
From the given diagram
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 17

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1

Question 10.
A boy standing at a point O finds his kite flying at a point P with distance OP = 25 m. It is at a height of 5 m from the ground. When the thread is extended by 10 m from P, it reaches a point Q. What will be the height QN of the kite from the ground? (use trigonometric ratios).
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 18
Solution:
Let the angle O be “θ”
In ΔONQ
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.1 19
In ΔOMP
sin θ = \(\frac{PM}{OP}\) ⇒ sin θ = \(\frac{5}{25}\)
sin θ = \(\frac{1}{5}\) ……… (2)
From (1) and (2) we get
\(\frac{h}{35}\) = \(\frac{1}{5}\)
5 h = 35 ⇒ h = \(\frac{35}{5}\) = 7
The height of the kite from the ground is 7m.

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.5

Students can download Maths Chapter 6 Trigonometry Ex 6.5 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 6 Trigonometry Ex 6.5

Question 1.
If sin 30° = x and cos 60° = y, then x² + y² is …….
(a) \(\frac{1}{2}\)
(b) 0
(c) sin 90°
(d) cos 90°
Solution:
(a) \(\frac{1}{2}\)
Hint:
sin 30° = x = \(\frac{1}{2}\)
cos 60° = y = \(\frac{1}{2}\)
x² + y²
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.5 1

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.5

Question 2.
If tan θ = cot 37°, then the value of θ is ………
(a) 37°
(b) 53°
(c) 90°
(d) 1°
Solution:
(b) 53°
Hint:
tan θ = cot 37°
= cot (90° – 53°)
= tan 53°
The value of θ is 53°

Question 3.
The value of tan 72°. tan 18° is ………
(a) 0
(b) 1
(c) 18°
(d) 72°
Solution:
(b) 1
Hint:
tan 72° . tan 18° = tan 72° . tan (90° – 72°)
= tan 72° . cot 72°
= tan 72° × \(\frac{1}{tan 72°}\)
= 1

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.5

Question 4.
The value of \(\frac{2 tan 30°}{1 – tan^{2} 30°}\) is equal to ………
(a) cos 60°
(b) sin 60°
(c) tan 60°
(d) sin 30°
Solution:
(c) tan 60°
Hint:
Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.5 2
= √3 = tan 60°

Question 5.
If 2 sin 2θ = √3 then the value of θ is ………
(a) 90°
(b) 30°
(c) 45°
(d) 60°
Solution:
(b) 30°
Hint:
2 sin 2θ = √3 ⇒ sin 2θ = \(\frac{√3}{2}\)
sin 2θ = sin 60° ⇒ 2θ = 60°
θ = 30°

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.5

Question 6.
The value of 3 sin 70° sec 20° + 2 sin 39° sec 51° is ………
(a) 2
(b) 3
(c) 5
(d) 6
Solution:
(c) 5
Hint:
3 sin 70° sec 20° + 2 sin 39° sec 51°
= 3. sin 70° . sec (90° – 70°) + 2 sin 39° . sec (90° – 39°)
= 3. sin 70° . cosec 70° + 2 sin 39° . cosec 39°
= 3. sin70° × \(\frac{1}{sin 70°}\) + 2 sin 39° × \(\frac{1}{sin 39°}\)
= 3 + 2
= 5

Question 7.
The value of \(\frac{1-tan^{2}45°}{1+tan^{2}45°}\) is ……..
(a) 2
(b) 1
(c) 0
(d) \(\frac{1}{2}\)
Solution:
(c) 0
Hint:
\(\frac{1-tan^{2}45°}{1+tan^{2}45°}\) = \(\frac{1-1}{1+1}\)
= \(\frac{0}{2}\) = 0

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.5

Question 8.
The value of cosec (70° + θ) – sec (20° + θ) + tan (65° + θ) – cot (25° + θ) is ……..
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0
Hint:
cosec (70° + θ) – sec (20° – θ) + tan (65° + θ) – cot (25° – θ)
= sec [90° – (70° + θ)] – sec (20° – θ) + tan (65° + θ) – tan [90° – (25° – θ)]
= sec (20° – θ) – sec (20° – θ) + tan (65° + θ) – tan (65° + θ)
= 0

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.5

Question 9.
The value of tan 1° . tan 2° . tan 3°…. tan 89° is ………
(a) 0
(b) 1
(c) 2
(d) \(\frac{√3}{2}\)
Solution:
(b) 1
Hint:
tan 1° . tan 2° . tan 3° …….. tan 89°
= tan (90° – 89°). tan (90° – 88°) .tan (90° – 87°) …….. tan 45° . tan (89°)
= cot 89° . cot 88°. cot 87°. ……. tan 45° …….. tan 87°. tan 88°. tan 89°
= 1

Samacheer Kalvi 9th Maths Guide Chapter 6 Trigonometry Ex 6.5

Question 10.
Given that sin α = \(\frac{1}{2}\) and cos β = \(\frac{1}{2}\), then the value of α + β is ………
(a) 0°
(b) 90°
(c) 30°
(d) 60°
Solution:
Hint:
sin α = \(\frac{1}{2}\)
sin 30° = \(\frac{1}{2}\) ∴ α = 30°
cos β = \(\frac{1}{2}\) ∴ β = 60°
∴ α + β = 30° + 60°
= 90°

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.6 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 5 Coordinate Geometry Ex 5.6

Question 1.
If the y-coordinate of a point is zero, then the point always lies ……..
(a) in the I quadrant
(b) in the II quadrant
(c) on x-axis
(d) on y-axis
Solution:
(c) on x-axis

Question 2.
The points (-5, 2) and (2, -5) lie in the ……….
(a) same quadrant
(b) II and III quadrant respectively
(c) II and IV quadrant respectively
(d) IV and II quadrant respectively
Solution:
(c) II and IV quadrant respectively

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6

Question 3.
On plotting the points O (0, 0), A (3, -4), B (3, 4) and C (0, 4) and joining OA, AB, BC and CO, which of the following figure is obtained?
(a) Square
(b) Rectangle
(c) Trapezium
(d) Rhombus
Solution:
(c) Trapezium

Question 4.
If P (-1, 1), Q ( 3, -4), R (1, -1), S (-2, -3) and T (- 4, 4) are plotted on a graph paper, then the points in the fourth quadrant are ………
(a) P and T
(b) Q and R
(c) only S
(d) P and Q
Solution:
(b) Q and R

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6

Question 5.
The point whose ordinate is 4 and which lies on the v-axis is ……….
(a) (4, 0)
(b) (0, 4)
(c) (1, 4)
(d) (4, 2)
Solution:
(b) (0, 4)

Question 6.
The distance between the two points (2, 3) and (1, 4) is ……..
(a) 2
(b) \(\sqrt{56}\)
(c) \(\sqrt{10}\)
(d) √2
Solution:
(d) √2
Hint:
\(\sqrt{(1-2)^{2}+(4+3)^{2}}\)
= \(\sqrt{(-1)^{2}+1^{2}}\)
= \(\sqrt{1+1}\)
= √2

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6

Question 7.
If the points A (2, 0), B (-6, 0), C (3, a – 3) lie on the x-axis then the value of a is ……..
(a) 0
(b) 2
(c) 3
(d) -6
Solution:
(c) 3
Hint:
a – 3 = 0 ⇒ a = 3

Question 8.
If (x + 2, 4) = (5, y – 2), then the coordinates (x, y) are ………
(a) (7, 12)
(b) (6, 3)
(c) (3, 6)
(d) (2, 1)
Solution:
(c) (3, 6)
Hint:
x + 2 = 5
∴ x = 5 – 2 = 3
and
4 = y – 2
4 + 2 = y
∴ y = 6

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6

Question 9.
If Q1, Q2, Q3, Q4 are the quadrants in a Cartesian plane then Q2 ∩ Q3 is ……..
(a) Q1 U Q2
(b) Q2 U Q3
(C) Null set
(d) Negative x-axis
Solution:
(c) Null set

Question 10.
The distance between the point (5, -1 ) and the origin is ………
(a) \(\sqrt{24}\)
(b) \(\sqrt{37}\)
(c) \(\sqrt{26}\)
(d) \(\sqrt{17}\)
Solution:
(c) \(\sqrt{26}\)
Hint:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6 1

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6

Question 11.
The coordinates of the point C dividing the line segment joining the points P (2, 4) and Q (5, 7) internally in the ratio 2 : 1 is ……..
(a) (\(\frac{7}{2}, \frac{11}{2}\))
(b) (3, 5)
(c) (4, 4)
(d) (4, 6)
Solution:
(d) (4, 6)
Hint:
A line divides internally in the ratio m : n
m = 2, n = 1
x1 = 2, x2 = 5
y1 = 4, y2 = 7
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6 2
= (4, 6)

Question 12.
If p (\(\frac{a}{3}, \frac{b}{2}\)) is the mid-point of the line segment joining A (-4, 3) and B (-2, 4) then (a, b) is ………
(a) (-9, 7)
(b) (-3, \(\frac{7}{2}\))
(c) (9, -7)
(d) (3, –\(\frac{7}{2}\))
Solution:
(a) (-9, 7)
Hint:
Mid point of a line
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6 3
\(\frac{a}{3}\) = -3 ⇒ a = -9
\(\frac{b}{2}\) = \(\frac{7}{2}\) ⇒ b = 7
(a, b) is (-9, 7)

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6

Question 13.
In what ratio does the point Q (1, 6) divide the line segment joining the points P (2, 7) and R(-2, 3) ………
(a) 1 : 2
(6) 2 : 1
(c) 1 : 3
(d) 3 : 1
Solution:
(c) 1 : 3
Hint:
A line divides internally in the ratio m : n the point P =
(\(\frac{mx_{2}+nx_{1}}{m+n}\), \(\frac{my_{2}+ny_{1}}{m+n}\))
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6 4
(1, 6) = (\(\frac{-2m+2n}{m+n}\), \(\frac{3m+7n}{m+n}\))
\(\frac{-2m+2n}{m+n}\) = 1
-2m + 2n = m + nx
-3m = n – 2n
3m = n
\(\frac{m}{n}\) = \(\frac{1}{3}\)
∴ m : n = 1 : 3
and
\(\frac{3m+7n}{m+n}\) = 6
3m + 7n = 6m + 6n
6m – 3n = 7n – 6n
3m = n
\(\frac{m}{n}\) = \(\frac{1}{3}\)

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6

Question 14.
If the coordinates of one end of a diameter of a circle is (3, 4) and the coordinates of its centre is (-3, 2), then the coordinate of the other end of the diameter is ……..
(a) (0, -3)
(b) (0, 9)
(c) (3, 0)
(d) (-9, 0)
Solution:
(d) (-9, 0)
Hint:
Let the other end of the diameter be (a, b)
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6 5
Mid point of a line =
(\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
(-3, 2) = \(\frac{3+a}{2}, \frac{4+b}{2}\)
\(\frac{3+a}{2}\) = -3
3 + a = -6
a = -6 – 3 = -9
and
\(\frac{4+b}{2}\) = 2
4 + b = 4
b = 4 – 4 = 0
The other end is (-9, 0)

Question 15.
The ratio in which the x-axis divides the line segment joining the points A (a1, b1) and B (a2, b2) is ……..
(a) b1 : b2
(b) -b1 : b2
(c) a1 : a2
(d) -a1 : a2
Solution:
(b) -b1 : b2
Hint:
A line divides internally in the ratio m : n the point P is,
(\(\frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}+ny_{1}}{m+n}\))
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6 6
The point P is (a, 0) = (\(\frac{ma_{2}+na_{1}}{m+n}, \frac{mb_{2}+nb_{1}}{m+n}\))
∴ \(\frac{mb_{2}+nb_{1}}{m+n}\) = 0
mb2 + nb1 = 0 ⇒ mb2 = -nb1
\(\frac{m}{n}\) = \(\frac{b_{1}}{b_{2}}\)
∴ m : n = -b1 : b2

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6

Question 16.
The ratio in which the x-axis divides the line segment joining the points (6, 4) and (1, -7) is ………
(a) 2 : 3
(b) 3 : 4
(c) 4 : 7
(d) 4 : 3
Solution:
(c) 4 : 7
Hint:
A line divides internally in the ratio m : n the point P
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6 7
-7m + 4n = 0
4n = 7m
\(\frac{m}{n}\) = \(\frac{4}{7}\)
The ratio is 4 : 7

Question 17.
If the coordinates of the mid-points of the sides AB, BC and CA of a triangle are (3, 4), (1, 1) and (2, -3) respectively, then the vertices A and B of the triangle are ……….
(a) (3, 2), (2, 4)
(b) (4, 0), (2, 8)
(c) (3, 4), (2, 0)
(d) (4, 3), (2, 4)
Solution:
(b) (4, 0), (2, 8)
Hint:
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6 8
x1 + x2 = 6 → (1)
y1 + y2 = 8 → (2)
x2 + x3 = 2 → (3)
y2 + y3 = 2 → (4)
x1 + x3 = 4 → (5)
y1 + y3 = -6 → (6)
Adding (1) + (3) + (5) we get,
2(x1 + x2 + x3) = 3
x1 + x2 + x3 = 6
x1 + x3 = 4
∴ x2 = 6 – 4 = 2
x2 + x3 = 2
x1 = 6 – 2 = 4
Adding (2) + (4) + (6) we get,
2 (y1 + y2 + y3) = 4
y1 + y2 + y3 = 2
y2 + y3 = 2
∴ y1 = 2 – 2 = 0
y1 + y3 = -6
y2 = 2 + 6 = 8
∴ The vertices A is (4, 0) and B is (2, 8).

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6

Question 18.
The mid-point of the line joining (-a, 2b) and (-3a, -4b) is ……..
(a) (2a, 3b)
(b) (-2a, -b)
(c) (2a, b)
(d) (-2a, -3b)
Solution:
(b) (-2a, -b)
Mid points of line
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6 9
= (-2a, -b)

Question 19.
In what ratio does the y-axis divides the line joining the points (-5, 1) and (2, 3) internally ………
(a) 1 : 3
(b) 2 : 5
(c) 3 : 1
(d) 5 : 2
Solution:
(d) 5 : 2
Hint:
When it cut the y-axis the point P is (0, a)
A line divides internally in the ratio m : n the point
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6 10
2m – 5n = 0 ⇒ 2m = 5n
\(\frac{m}{n}\) = \(\frac{5}{2}\) ⇒ m : n = 5 : 2

Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6

Question 20.
If (1, -2), (3, 6), (x, 10) and (3, 2) are the vertices of the parallelogram taken in order, then the value of x is ………
(a) 6
(b) 5
(c) 4
(d) 3
Solution:
(b) 5
Hint:
Since ABCD is a parallelogram
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6 11
Mid-point of AC = Mid-point of BD
Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Ex 5.6 12
\(\frac{1+x}{2}\) = \(\frac{6}{2}\) ⇒ 1 + x = 6 ⇒ x = 6 – 1 = 5
The value of x = 5