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Chapter-Test Statistics ML Aggarwal Class-9 ICSE Solutions Chapter-20. Statistics is a collection of mathematical techniques that help to analyze and present data. Therefore we provide Statistics of ML Aggarwal Class-9 ICSE Maths Solutions of Chapter-20 . Hence Step by step answer of Exercise-20.1 , Exercise-20.2, Exercise-20.3, MCQ and Chapter-Test of Exercise-20 Questions. So Statistics Solved Questions for Class-9 of ML Aggarwal Understanding APC Mathematics are available here. Visit […]

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]]>**Chapter-Test Statistics ML Aggarwal Class-9 ICSE Solutions Chapter-20. Statistics** is a collection of mathematical techniques that help to analyze and present data. Therefore we provide **Statistics **of **ML Aggarwal Class-9** ICSE Maths Solutions of Chapter-20 . Hence Step by step answer of Exercise-20.1 , Exercise-20.2, Exercise-20.3, MCQ and Chapter-Test of Exercise-20 Questions. So **Statistics** Solved Questions for **Class-9** of **ML** **Aggarwal** Understanding APC Mathematics are available here. Visit official website CISCE for detail information about ICSE Board Class-9.

Board | ICSE |

Publications | Avichal Publishig Company (APC) |

Subject | Maths |

Class | 9 th |

Chapter-20 | Statistics (Chapter-test) |

Writer | ML Aggarwal |

Book Name | Understanding |

Topics | Solution of Exercise-20.1 , Exercise-20.2, Exercise-20.3, MCQ and Chapter-Test Questions. |

Academic Session | 2021-2022 |

**–: Select Topics :–**

**Chapter-Test , **

Note:- Before viewing Solutions of Chapter -20 **Statistics Class-9 **APC Understanding** ML** **Aggarwal** Solutions . Read the Chapter Carefully . Then solve all example given in Exercise-20.1 , Exercise-20.2, Exercise-20.3.Statistics ML Aggarwal Class-9 ICSE is important chapter.

**Page 500**

Find the mean and the median of the following set of numbers:

8, 0, 5, 3, 2, 9, 1, 5, 4, 7, 2, 5.

In descending order

0, 1, 2, 2, 3, 4, 5, 5, 5, 7, 8, 9

n = 12 which is even

Mean (x̄) = Ʃ x_{i}/n

= (0 + 1 + 2 + 2 + 3 + 4 + 5 + 5 + 5 + 7 + 8 + 9)/12

= 51/12

= 17/4

= 4.25

Median = ½ [12/2th + (12/2 + 1)th terms]

= ½ [6^{th} + 7^{th} terms]

= ½ (4 + 5)

= 9/2

= 4.5

Find the mean and the median of all the (positive) factors of 48.

Positive factors of 48 are

1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Here N = 10 which is even

Mean (x̄) = Ʃ x_{i}/ n

= (1 + 2 + 3 + 4 + 6 + 8 + 12 + 16 + 24 + 48)/10

= 124/10

= 12.4

Median = ½ [10/2 th + (10/2 + 1)th terms]

= ½ [5^{th} + 6^{th} terms]

= ½ (6 + 8)

= 14/2

= 7

The mean weight of 60 students of a class is 52.75 kg. If the mean weight of 35 of them is 54 kg, find the mean weight of the remaining students.

Mean weight of 60 students of a class=52.75 kg

So the total weight of 60 students = 52.75 × 60

= 3165 kg

Mean weight of 35 students among them = 54 kg

So the total weight of 35 students = 54 × 35

= 1890 kg

Remaining students = 60 – 35

= 25

Total weight of 25 students = 3165 – 1890

= 1275 kg

So the mean weight of 25 students = 1275 ÷ 25

= 51 kg

Hence, the mean weight of the remaining students is 51 kg.

The mean age of 18 students of a class is 14.5 years. Two more students of age 15 years and 16 years join the class. What is the new mean age?

Mean age of 18 students = 14.5 years

Total age = 14.5 × 18

= 261 years

Total age of 2 more students = 15 + 16

= 31 years

Total age of 18 + 2 = 20 students = 261 + 31

= 292 years

Mean age = 292/20 =

14.6 years

Hence, the new mean age is 14.6 years.

If the mean of the five observations x + 1, x + 3, x + 5, 2x + 2, 3x + 3 is 14, find the mean of first three observations.

Mean of the five observations

x + 1, x + 3, x + 5, 2x + 2, 3x + 3 is 14

Mean = (x + 1 + x + 3 + x + 5 + 2x + 2 + 3x + 3)/5

= (8x + 14)/5

(8x + 14)/5 =14

8x + 14 = 70

8x = 70 – 14 = 56

x = 56/8 = 7

Mean of x + 1 + x + 3 + x + 5

Mean = (x + 1 + x + 3 + x + 5)/3

= (3x + 9)/3

= x + 3

Substituting the value of x

= 7 + 3

= 10

Therfore, the mean of first three observations is 10.

The mean height of 36 students of a class is 150.5 cm. Later on, it was detected that the height of one student was wrongly copied as 165 cm instead of 156 cm. Find the correct mean height.

Mean height of 36 students of a class=150.5 cm

Total height = 150.5 × 36

= 5418 cm

Difference in height which was wrongly copied = 165 + 56

= 9 cm

Actual height = 5418 – 9

= 5409 cm

Actual mean height = 5409/36

= 150.25 cm

Therfore, the correct mean height is 150.25 cm.

The mean of 40 items is 35. Later on, it was discovered that two items were misread as 36 and 29 instead of 63 and 22. Find the correct mean.

Mean of 40 items = 35

Total of 40 items = 35 × 40 = 1400

Difference between two items which were wrongly read = (63 + 22) – (36 + 29)

= 85 – 65

= 20

Here,

Actual total = 1400 + 20 = 1420

Correct mean = 1420/40 = 35.5

Therefore, the correct mean is 35.5.

The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.

29, 32, 48, 50, x, x + 2, 72, 75, 87, 91.

N = 10 which is even

Median = ½ [n/2 th + (n/2 + 1)th term]

63 = ½ [10/2th + (10/2 + 1)th term]

63 = ½ (5^{th} + 6^{th} terms)

63 = ½ (x + x + 2)

63 × 2 = 2x + 2

2x = 126 – 2 = 124

By division

x = 124/2 = 62

Hence, the value of x is 62.

Draw a histogram showing marks obtained by the students of a school in a Mathematics paper carrying 60 marks.

Marks |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |

Students |
4 |
5 |
10 |
8 |
30 |
40 |

**I**n a class of 60 students, the marks obtained in a surprise test were as under:

Marks | 14-20 | 20-26 | 26-32 | 32-38 | 38-44 | 44-50 | 50-56 | 56-62 |

No. of students | 4 | 10 | 9 | 15 | 12 | 5 | 3 | 2 |

Construct a combined histogram and frequency polygon for the following distribution:

Classes | 91-100 | 101-110 | 111-120 | 121-130 | 131-140 | 141-150 | 151-160 |

Frequency | 16 | 28 | 44 | 20 | 32 | 12 | 4 |

the classes in continuous frequency classes:

Classes | Classes after adjustment | Frequency |

91-100 | 90.5-100.5 | 16 |

101-110 | 100.5-110.5 | 28 |

111-120 | 110.5-120.5 | 44 |

121-130 | 120.5-130.5 | 20 |

131-140 | 130.5-140.5 | 32 |

141-150 | 140.5-150.5 | 12 |

151-160 | 150.5-160.5 | 4 |

The electricity bills (in rupees) of 40 houses in a locality are given below:

78 87 81 52 59 65 101 108 115 95

98 65 62 121 128 63 76 84 89 91

65 101 95 81 87 105 129 92 75 105

78 72 107 116 127 100 80 82 61 118

Form a frequency distribution table with a class size of 10. Also represent the above data with a histogram and frequency polygon.

Least term = 52

Greatest term = 129

Range = 129 – 52 = 77

Class Interval |
Tally Numbers |
Frequency |

50-60 | II | 2 |

60-70 | 6 | |

70-80 | 5 | |

80-90 | 8 | |

90-100 | 5 | |

100-110 | 7 | |

110-120 | III | 3 |

120-130 | IIII | 4 |

Total | 40 |

The data given below represent the marks obtained by 35 students:

21 26 21 20 23 24 22 19 24

26 25 23 26 29 21 24 19 25

26 25 22 23 23 27 26 24 25

30 25 23 28 28 24 28 28

Taking class intervals 19-20, 21-22 etc., make a frequency distribution for the above data.

Construct a combined histogram and frequency polygon for the distribution.

Least mark = 19

Greatest marks = 30

Range = 30 – 19 = 11

Class Interval |
Actual Intervals |
Frequency |

19-20 | 18.5-20.5 | 3 |

21-22 | 20.5-22.5 | 5 |

23-24 | 22.5-24.5 | 10 |

25-26 | 24.5-26.5 | 10 |

27-28 | 26.5-28.5 | 5 |

29-30 | 28.5-30.5 | 2 |

Total | 35 |

The given histogram and frequency polygon shows the ages of teachers in a school. Answer the following:

(i) What is the class size of each class?

(ii) What is the class whose class mark is 48?

(iii) What is the class whose frequency is maximum?

(iv) Construct a frequency table for the given distribution.

**(i)** The class size of each class is 6.

**(ii)** The class whose class mark is 48 is 45 – 51

= (45 + 51)/2

= 96/2

= 48

**(iii)** The class 51-57 has the maximum frequency i.e., 20.

**(iv)** Frequency table for the given distribution:

Classes | 27-33 | 33-39 | 39-45 | 45-51 | 51-57 | 57-63 |

Frequency | 4 | 12 | 18 | 6 | 20 | 8 |

–: end of Statistics solutions (chapter-test) :–

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]]>Exe-20.3 Statistics ML Aggarwal Class-9 ICSE Solutions Chapter-20. Statistics is a collection of mathematical techniques that help to analyze and present data. Therefore we provide Statistics of ML Aggarwal Class-9 ICSE Maths Solutions of Chapter-20 . Hence Step by step answer of Exercise-20.1 , Exercise-20.2, Exercise-20.3, MCQ and Chapter-Test of Exercise-20 Questions. So Statistics Solved Questions for Class-9 of ML Aggarwal Understanding APC Mathematics are available here. Visit […]

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]]>**Exe-20.3 Statistics ML Aggarwal Class-9 ICSE Solutions Chapter-20. Statistics** is a collection of mathematical techniques that help to analyze and present data. Therefore we provide **Statistics **of **ML Aggarwal Class-9** ICSE Maths Solutions of Chapter-20 . Hence Step by step answer of Exercise-20.1 , Exercise-20.2, Exercise-20.3, MCQ and Chapter-Test of Exercise-20 Questions. So **Statistics** Solved Questions for **Class-9** of **ML** **Aggarwal** Understanding APC Mathematics are available here. Visit official website CISCE for detail information about ICSE Board Class-9.

Board | ICSE |

Publications | Avichal Publishig Company (APC) |

Subject | Maths |

Class | 9 th |

Chapter-20 | Statistics (Exercise 20.3) |

Writer | ML Aggarwal |

Book Name | Understanding |

Topics | Solution of Exercise-20.1 , Exercise-20.2, Exercise-20.3, MCQ and Chapter-Test Questions. |

Academic Session | 2021-2022 |

**–: Select Topics :–**

**Exercise-20.3, **

Note:- Before viewing Solutions of Chapter -20 **Statistics Class-9 **APC Understanding** ML** **Aggarwal** Solutions . Read the Chapter Carefully . Then solve all example given in Exercise-20.1 , Exercise-20.2, Exercise-20.3.Statistics ML Aggarwal Class-9 ICSE is important chapter.

**Page 494**

The area under wheat cultivation last year in the following states, correct to the nearest lacs hectares was:

State | Punjab | Haryana | U.P. | M.P. | Maharashtra | Rajasthan |

Cultivated area | 220 | 120 | 100 | 40 | 80 | 30 |

**The number of books sold by a shopkeeper in a certain week was as follows:**

Day | Monday | Tuesday | Wednesday | Thursday | Friday | Saturday |

No. of books | 420 | 180 | 230 | 340 | 160 | 120 |

**Page 495**

Given below is the data of percentage of passes of a certain school in the ICSE for consecutive years:

Year | 2000 | 2001 | 2002 | 2003 | 2004 | 2005 | 2006 |

% of passes | 92 | 80 | 70 | 86 | 54 | 78 | 94 |

Birth rate per thousand of different countries over a certain period is:

Country | India | Pakistan | China | U.S.A. | France |

Birth rate | 36 | 45 | 12 | 18 | 20 |

Given below is the data of number of students (boys and girls) in class IX of a certain school:

Class | IX A | IX B | IX C | IX D |

Boys | 28 | 22 | 40 | 15 |

Girls | 18 | 34 | 12 | 25 |

Draw a bar graph to represent the above data.

Draw a histogram to represent the following data:

Marks obtained | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |

No. of students | 4 | 10 | 6 | 8 | 5 | 9 |

Draw a histogram to represent the following frequency distribution of monthly wages of 255 workers of a factory.

Monthly wages (in rupees) | 850-950 | 950-1050 | 1050-1150 | 1150-1250 | 1250-1350 |

No. of workers | 35 | 45 | 75 | 60 | 40 |

Draw a histogram for the following data:

Class marks | 12.5 | 17.5 | 22.5 | 27.5 | 32.5 | 37.5 |

Frequency | 7 | 12 | 20 | 28 | 8 | 11 |

Draw a histogram for the following frequency distribution:

Age (in years) | Below 2 | Below 4 | Below 6 | Below 8 | Below 10 | Below 12 |

No. of children | 12 | 15 | 36 | 45 | 72 | 90 |

First convert the given cumulative frequency into frequency distribution table:

Age (in years) | c.f. | f. |

0-2 | 12 | 12 |

2-4 | 15 | 3 |

4-6 | 36 | 21 |

6-8 | 45 | 9 |

8-10 | 72 | 27 |

10-12 | 90 | 18 |

represent age on x-axis and number of children on y-axis :

Draw a histogram for the following data:

Classes | 59-65 | 66-72 | 73-79 | 80-86 | 87-93 | 94-100 |

Frequency | 10 | 5 | 25 | 15 | 30 | 10 |

Classes | Classes after adjustment | Frequency |

59-56 | 58.5-65.5 | 10 |

66-72 | 65.5-72.5 | 5 |

73-79 | 72.5-79.5 | 25 |

80-86 | 79.5-86.5 | 15 |

87-93 | 86.5-93.5 | 30 |

94-100 | 93.5-100.5 | 10 |

Represent classes on x-axis and frequency on y-axis :

**Page 496**

Draw a frequency polygon for the following data:

Class intervals | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |

Frequency | 15 | 28 | 45 | 32 | 41 | 18 |

In a class of 60 students, the marks obtained in a monthly test were as under:

Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |

Students | 10 | 25 | 12 | 08 | 05 |

In a class of 90 students, the marks obtained in a weekly test were as under:

Marks | 16-20 | 21-25 | 26-30 | 31-25 | 36-40 | 41-45 | 46-50 |

No. of students | 4 | 12 | 18 | 26 | 14 | 10 | 6 |

Marks | Classes after adjustment | Class Mark | No. of students (f) |

16-20 | 15.5-20.5 | 18.0 | 4 |

21-25 | 20.5-25.5 | 23.0 | 12 |

26-30 | 25.5-30.5 | 28.0 | 18 |

31-35 | 30.5-35.5 | 33.0 | 26 |

36-40 | 35.5-40.5 | 38.0 | 14 |

41-45 | 40.5-45.5 | 43.0 | 10 |

46-50 | 45.5-50.5 | 48.0 | 6 |

Represent marks on x-axis and frequency on y-axis :

In a city, the weekly observations made in a study on the cost of living index are given in the following table:

Cost of living index | 140-150 | 150-160 | 160-170 | 170-180 | 180-190 | 190-200 |

Number of weeks | 5 | 10 | 20 | 9 | 6 | 2 |

Cost of living index | No. of weeks | Mid-points |

140-150 | 5 | 145 |

150-160 | 10 | 155 |

160-170 | 20 | 165 |

170-180 | 9 | 175 |

180-190 | 6 | 185 |

190-200 | 2 | 195 |

Mark the mid-points of the cost of living index on x-axis and number of weeks on the y-axis:

Construct a combined histogram and frequency polygon for the following data:

Weekly earnings (in rupees) | 150-165 | 165-180 | 180-195 | 195-210 | 210-225 | 225-240 |

No. of workers | 8 | 14 | 22 | 12 | 15 | 6 |

In a study of diabetic patients, the following data was obtained:

Age (in years) | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |

No. of patients | 3 | 8 | 30 | 36 | 27 | 15 | 6 |

Represent the above data by a histogram and a frequency polygon.

The water bills (in rupees) of 32 houses in a locality are given below:

30, 48, 52, 78, 103, 85, 37, 94, 72, 73, 66, 52, 92, 65, 78, 81, 64, 60, 75, 78, 108, 63, 71, 54, 59, 75, 100, 103, 35, 89, 95, 73.

Taking class intervals 30-40, 40-50, 50-60, ….. ,form frequency distribution table. Construct a combined histogram and frequency polygon.

Class Intervals | Tally Marks | Frequency |

30-40 | III | 3 |

40-50 | I | 1 |

50-60 | IIII | 4 |

60-70 | 5 | |

70-80 | 9 | |

80-90 | III | 3 |

90-100 | III | 3 |

100-110 | IIII | 4 |

Take class intervals on x-axis and frequency on y-axis:

The number of matchsticks in 40 boxes on counting was found as given below:

44, 41, 42, 43, 47, 50, 51, 49, 43, 42, 40, 42, 44, 45, 49, 42, 46, 49, 45, 49, 45, 47, 48, 43, 43, 44, 48, 43, 46, 50, 43, 52, 46, 49, 52, 51, 47, 43, 43, 45.

Taking classes 40-42, 42-44 ….., construct the frequency distribution table for the above data. Also draw a combined histogram and frequency polygon to represent the distribution.

Class Intervals | Tally Marks | Frequency |

40-42 | II | 2 |

42-44 | 12 | |

44-46 | 7 | |

46-48 | 6 | |

48-50 | 7 | |

50-52 | IIII | 4 |

52-54 | II | 2 |

Total | 40 |

Take class on x-axis and frequency on y-axis :

**Page 497**

The histogram showing the weekly wages (in rupees) of workers in a factory is given alongside.

Answer the following about the frequency distribution:

(i) What is the frequency of the class 400-425?

(ii) What is the class having minimum frequency?

(iii) What is the cumulative frequency of the class 425 – 450?

(iv) Construct a frequency and cumulative frequency table for the given distribution.

(i) The frequency of the class 400-425 is 18.

(ii) The class having minimum frequency is 475-500.

(iii) The cumulative frequency of the class 425-450 is (6 + 18 + 10) = 34.

(iv) The frequency table for the given distribution is:

Classes | Frequency | Cumulative Frequency |

375-400 | 6 | 6 |

400-425 | 18 | 24 |

425-450 | 10 | 34 |

450-475 | 20 | 54 |

475-500 | 4 | 58 |

The runs scored by two teams A and B on the first 42 balls in a cricket match are given below:

No. of balls | 1-6 | 7-12 | 13-18 | 19-24 | 25-30 | 31-36 | 37-42 |

Runs scored by Team A | 2 | 1 | 8 | 9 | 4 | 5 | 6 |

Runs scored by Team B | 5 | 6 | 2 | 10 | 5 | 6 | 3 |

No. of balls | Class marks | Team A | Team B |

1-6 | 3.5 | 2 | 5 |

7-12 | 9.5 | 1 | 6 |

13-18 | 15.5 | 8 | 2 |

19-24 | 21.5 | 9 | 10 |

25-30 | 27.5 | 4 | 5 |

31-36 | 33.5 | 5 | 6 |

37-42 | 39.5 | 6 | 3 |

–: End of Statistics ML Aggarwal Solutions for ICSE Class-9 :–

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]]>OP Malhotra Class-9 ICSE Maths S Chand SK Gupta Anubhuti Gangal Solutions . Solutions of Exercise ,MCQ , Chapter Test of Class-9 ICSE S Chand Publication Maths of OP Malhotra. Visit official website CISCE for detail information about ICSE Board Class-9. We are providing valuable and important information for ICSE Maths Class 9th. OP Malhotra Class-9 ICSE […]

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]]>OP Malhotra Class-9 ICSE Maths S Chand SK Gupta Anubhuti Gangal Solutions . Solutions of Exercise ,MCQ , Chapter Test of Class-9 ICSE S Chand Publication Maths of OP Malhotra. Visit official website CISCE for detail information about ICSE Board Class-9. We are providing valuable and important information for** **ICSE Maths Class 9th.

**Name the Author:** O.P. Malhotra | S.K. Gupta | Anubhuti Gangal.

**Class :** 9 (Nine).

**Subject:** Mathematics.

**Board:** ICSE.

**Year **– 2020-21 (on reduced Syllabus)

**Publication** – S Chand

- Plan a self made time table of Chapters which you can follow
- Keep yourself away from obstacle such as Phone Call, Game , Over Browsing on Net
- Read the Concept of Certain Chapter carefully
- Focus on Formulas used
- Attention on when and which formulas is suitable for certain questions
- Practice Example given of your textbook
- Also Practice example of other famous publications
- Now try to solve Exercise
- If feel any problems then view our Solutions given in Sequence of textbook
- Solving model paper of ML Aggarwal is also best
- Keep a self written Formula on your study table
- Visit Contact Us Menu of icsehelp to get Mobile number of Maths Teacher without hesitation

**Unit 1 : Pure Arithmetic**

Chapter 1. Rational and Irrational Numbers

**Unit 2 : Commercial Mathematics**

**Unit 3 : Algebra**

Chapter 5 : Simultaneous Linear Equations in two variables

Chapter 6 : Indices / Exponents

**Unit 4 : Geometry**

Chapter 9 : Mid-Point and Intercept Theorems

Chapter 10 : Pythagoras Theorem

Chapter 11 : Rectilinear Figures

**Unit 5 : Statistics**

Chapter 14 : Statistics, Introduction, Data and Frequency Distribution

Chapter 15 : Mean, Median and Frequency Polygon

**Unit 6 : Mensuration**

Chapter 16 : Area of Plane Figures

Chapter 17 : Circle – Circumference and Area

Chapter 18 : Surface Area and Volume of 3D Solids (Cuboid and Cube)

**Unit 7 : Trigonometry**

Chapter 19 : Trigonometrical Ratios

**Unit 8 : Coordinate Geometry**

Chapter 20 : Coordinate and Graphs of Simultaneous Linear Equations

Return to – ICSE Class-9 Textbook Solutions

Thanks

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]]>Distance Formula Class-9 Concise Selina ICSE Maths Solutions Chapter-28 . We provide step by step Solutions of Exercise / lesson-28 Distance Formula for ICSE Class-9 Concise Selina Mathematics by R. K. Bansal. Our Solutions contain all type Questions with Exe-28 to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics. Distance Formula Class-9 […]

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]]>**Distance Formula Class-9 Concise** Selina ICSE Maths Solutions Chapter-28 . We provide step by step Solutions of Exercise / lesson-28 **Distance Formula** for ICSE **Class-9 Concise** Selina Mathematics by R. K. Bansal.

Our Solutions contain all type Questions with Exe-28 to develop skill and confidence. Visit official Website **CISCE** for detail information about ICSE Board Class-9 Mathematics.

–: Select Topics :–

To calculate the **distance between two points** in a plane, we have to use distance formula as per described in coordinate geometry. With the help of this formula, we can find the distance between any two points marked in an x-y coordinate

Distance between two points P(x_{1},y_{1}) and Q(x_{2},y_{2}) is given by:

PQ = √[(x_{2}– x_{1})^{2}+ (y_{2}– y_{1})^{2}]

It is known as **distance formula.**

Find the distance between the following pairs of points:

(i) (-3, 6) and (2, -6 )

(ii) (-a, -b) and (a, b)

(iii)(3/5 ,2) and (-1/5 , 1 ²/5)

(iv) (√3+1, 1) and (0, √3)

**Answer 1:**

Find the distance between the origin and the point:

(i) (-8, 6)

(ii) (-5, -12)

(iii) (8, -15)

**Answer 2:**

The distance between the points (3, 1) and (0, x) is 5. Find x.

**Answer 3:**

Find the co-ordinates of points on the x-axis which are at a distance of 17 units from the point (11, -8).

**Answer 4:**

Find the coordinates of the points on the y-axis, which are at a distance of 10 units from the point (-8, 4).

**Answer 5:**

A point A is at a distance of √10 unit from the point (4, 3). Find the co-ordinates of point A, if its ordinate is twice its abscissa.

**Answer 6:**

A point P (2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.

**Answer 7:**

What point on the x-axis is equidistant from the points (7, 6) and (-3, 4)?

**Answer 8:**

Find a point on the y-axis which is equidistant from the points (5, 2) and (-4, 3).

**Answer 9:**

A point P lies on the x-axis and another point Q lies on the y-axis.

(i) Write the ordinate of point P.

(ii) Write the abscissa of point Q.

(iii) If the abscissa of point P is -12 and the ordinate of point Q is -16; calculate the length of line segment PQ.

**Answer 10:**

Show that the points P (0, 5), Q (5, 10) and R (6, 3) are the vertices of an isosceles triangle.

**Answer 11:**

Prove that the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of a rectangle PQRS

**Answer 12:**

Prove that the points A (1, -3), B (-3, 0) and C (4, 1) are the vertices of an isosceles right-angled triangle. Find the area of the triangle.

**Answer 13:**

Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square ABCD.

**Answer 14:**

Show that (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.

**Answer 15:**

Points A (-3, -2), B (-6, a), C (-3, -4) and D (0, -1) are the vertices of quadrilateral ABCD; find a if ‘a’ is negative and AB = CD.

**Answer 16:**

The vertices of a triangle are (5, 1), (11, 1) and (11, 9). Find the co-ordinates of the circumcentre of the triangle.

**Answer 17:**

Given A = (3, 1) and B = (0, y – 1). Find y if AB = 5.

**Answer 18:**

Given A = (x + 2, -2) and B (11, 6). Find x if AB = 17.

**Answer 19:**

The centre of a circle is (2x – 1, 3x + 1). Find x if the circle passes through (-3, -1) and the length of its diameter is 20 unit.

**Answer 20:**

The length of line PQ is 10 units and the co-ordinates of P are (2, -3); calculate the co-ordinates of point Q, if its abscissa is 10.

**Answer 21:**

Point P (2, -7) is the center of a circle with radius 13 unit, PT is perpendicular to chord AB and T = (-2, -4); calculate the length of:

(i). AT

(ii). AB

**Answer 22:**

Calculate the distance between the points P (2, 2) and Q (5, 4) correct to three significant figures.

**Answer 23:**

Calculate the distance between A (7, 3) and B on the x-axis whose abscissa is 11.

**Answer 24:**

Calculate the distance between A (5, -3) and B on the y-axis whose ordinate is 9.

**Answer 25:**

Find the point on y-axis whose distances from the points A (6, 7) and B (4, -3) are in the ratio 1: 2.

**Answer 26:**

The distances of point P (x, y) from the points A (1, – 3) and B (- 2, 2) are in the ratio 2: 3.

Show that: 5x^{2} + 5y^{2} – 34x + 70y + 58 = 0.

**Answer 27:**

The points A (3, 0), B (a, -2) and C (4, -1) are the vertices of triangle ABC right angled at vertex A. Find the value of a.

**Answer 28:**

— End of **Distance Formula Class-9 Concise** Selina Solutions :–

Return to **– Concise Selina Maths Solutions for ICSE Class -9**

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