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SELINA Solution Class 9 Chapter 18 Statistics Exercise 18A

Question 1.1

State, the following variable is continuous or discrete:
number of children in your class.

Ans: Discrete Variable

Question 1.2

State, the following variable is continuous or discrete:
Distance travelled by car.

Ans: Continuous Variable

Question 1.3

State, the following variable is continuous or discrete:
Sizes of shoes.

Ans: Discrete Variable

Question 1.4

State, the following variable is continuous or discrete: Time.

Ans: Continuous Variable

Question 1.5

State, the following variable is continuous or discrete: Number of patients in a hospital.

Ans: Discrete Variable

Question 2

Given below are the marks obtained by 30 students in an examination:

08	17	33	41	47	23	20	34
09	18	42	14	30	19	29	11
36	48	40	24	22	02	16	21
15	32	47	44	33	01

Taking class intervals 1 - 10, 11 - 20, ....., 41 - 50;
make a frequency table for the above distribution.

Sol:

The frequency table for the given distribution is

Marks	Tally Marks	Frequency
1 - 10		4
11 - 20		8
21 - 30		6
31 - 40		6
41 - 50		6

Question 3

The marks of 24 candidates in the subject mathematics are given below:

45	48	15	23	30	35	40	11
29	0	3	12	48	50	18	30
15	30	11	42	23	2	3	44

The maximum marks are 50. Make a frequency distribution taking class intervals 0 - 10, 10-20, .......

Sol:

The frequency table for the given distribution is

Marks	Tally Marks	Frequency
0 - 10	\|\|\|\|	4
10 - 20		6
20 - 30	\|\|\|	3
30 - 40	\|\|\|\|	4
40 - 50		7

In this frequency distribution, the marks 30 are in the class of interval 30 - 40 and not in 20 - 30. Similarly, marks 40 are in the class of interval 40 - 50 and not in 30 - 40.

Question 4.1

Fill in the blank :
A quantity which can very from one individual to another is called a .............

Sol:

A quantity which can very from one individual to another is called a Variable.

Question 4.2

Fill in the blank :
Sizes of shoes are ........... variables.

Sol:

Sizes of shoes are Discrete variables.

Question 4.3

Fill in the blank :
Daily temperatures is ........... variable.

Daily temperatures is Continuous variable.

Sol:

Daily temperatures is Continuous variable.

Question 4.4

Fill in the blank :
The range of data 7, 13, 6, 25, 18, 20, 16 is ............

Sol:

The range of the data 7, 13, 6, 25, 18, 20, 16 is The range is 25 - 6 = 19.

Question 4.5

Fill in the blank :
In the class interval 35 - 46; the lower limit is .......... and the upper limit is .........

Sol:

In the class interval 35 - 46; the lower limit is 35 and the upper limit is 46.

Question 4.6

Fill in the blank :
The class mark of class interval 22 - 29 is ...........

Sol:

The class mark of class interval 22 - 29 is
The classmark is 22 - 29 = $\frac{22 + 29}{2} = \frac{51}{2}$ = 25.5.

Question 5

Find the actual lower class limits, upper-class limits and the mid-values of the classes:
10 - 19, 20 - 29, 30 - 39 and 40 - 49.

Sol:

In case of frequency 10 - 19 the lower class limit is 10, the upper class limit is 19 and mid-value is
$\frac{10 + 19}{2}$ = 14.5

In case of frequency 20 - 29 the lower class limit is 20, the upper class limit is 29 and mid-value is
$\frac{20 + 29}{2}$ = 24.5

In case of frequency 30 - 39 the lower class limit is 30, the upper class limit is 39 and mid-value is
$\frac{30 + 29}{2}$ = 34.5

In case of frequency 40 - 49 the lower class limit is 40, the upper class limit is 49 and mid-value is
$\frac{40 + 49}{2}$ = 44.5

Question 6

Find the actual lower and upper-class limits and also the class marks of the classes:
1.1 - 2.0, 2.1 -3.0 and 3.1 - 4.0.

Sol:

In the case of frequency 1.1 - 2.0 the lower class limit is 1.1, upper-class limit is 2.0 and class mark

is $\frac{1.1 + 2.0}{2}$ = 1.55

In the case of frequency 2.1 - 3.0 the lower class limit is 2.1, upper-class limit is 3.0 and class mark

is $\frac{2.1 + 3.0}{2}$ = 2.55

In the case of frequency 3.1 - 4.0 the lower class limit is 3.1, the upper class limit is 4.0 and class mark

is $\frac{3.1 + 4.0}{2}$ = 3.55

Question 7

Use the table given below to find:
(a) The actual class limits of the fourth class.
(b) The class boundaries of the sixth class.
(c) The class mark of the third class.
(d) The upper and lower limits of the fifth class.
(e) The size of the third class.

Class Interval	Frequency
30 - 34	7
35 - 39	10
40 - 44	12
45 - 49	13
50 - 54	8
55 - 59	4

Sol:

(a) The actual class limit of the fourth class will be 44.5 - 49.5.

(b) The class boundaries of the sixth class will be 54.5 - 59.5

(c) The class mark of the third class will be the average of the lower bound and the upper bound of the interval. Therefore the class mark will be: $\frac{40 + 44}{2}$ = 42

(d) The upper and lower limit of the fifth class is 54 and 50 respectively.

(e) The size of the third class will be 44 - 40 + 1 = 5.

Question 8.1

Construct a cumulative frequency distribution table from the frequency table given below:

Class Interval	Frequency
0 -8	9
8 - 16	13
16 - 24	12
24 - 32	7
32 - 40	15

Sol:

The cumulative frequency distribution table is

C.I	c.f
0 -8	9
8 - 16	22
16 - 24	34
24 - 32	41
32 - 40	56

Question 8.2

Construct a cumulative frequency distribution table from the frequency table given below:

Class Interval	Frequency
1 - 10	12
11 - 20	18
21 - 30	23
31 - 40	15
41 - 50	10

Sol:

The cumulative frequency distribution table is

C.I	c.f
1 - 10	12
11 - 20	30
21 - 30	53
31 - 40	68
41 - 50	78

Question 9.1

Construct a frequency distribution table from the following cumulative frequency distribution:

Class Interval	Cumulative Frequency
10 - 19	8
20 - 29	19
30- 39	23
40- 49	30

Sol:

The frequency distribution table is

C. I	c.f
10 - 19	8
20 - 29	11
30 - 39	4
40 - 49	7

Question 9.2

Construct a frequency distribution table from the following cumulative frequency distribution:

C.I	C.F
5 - 10	18
10 - 15	30
15 - 20	46
20 - 25	73
25 - 30	90

Sol:

The frequency distribution table is

C .I	c.f
5 - 10	18
10 - 15	12
15 - 20	16
20 - 25	27
25 - 30	17

Question 10

Construct a frequency table from the following data:

Marks	No. of students
less than 10	6
less than 20	15
less than 30	30
less than 40	39
less than 50	53
less than 60	70

Sol:

The frequency table is

C. I	c.f
0 - 10	6
10 - 20	9
20 - 30	15
30 - 40	9
40 - 50	14
50 - 60	17

Question 11

Construct the frequency distribution table from the following cumulative frequency table:

Ages	No. of students
Below 4	0
Below 7	85
Below 10	140
Below 13	243
Below 16	300

(i) State the number of students in the age group 10 - 13.
(ii) State the age-group which has the least number of students.

Sol:

The frequency distribution table is

C. I	c.f
4 - 7	85
7 - 10	55
10 - 13	103
13 - 16	57

(i)The number of students in the age group is 10 -13 is 103

(ii)The age group which has the least number of students is 7 - 10

Question 12

Fill in the blank in the following table:

Class interval	Frequency	Cumulative Frequency
25 - 34	......	15
35 - 44	......	28
45 - 54	21	......
55 - 64	16	......
65 - 74	......	73
75 - 84	12	......

Sol: (To be added)

Question 13

Fill in the blank in the following table:

Class interval	Frequency	Cumulative Frequency
25 - 34	......	15
35 - 44	......	28
45 - 54	21	......
55 - 64	16	......
65 - 74	......	73
75 - 84	12	......

Sol:

Fill in the blank in the following table:

Class interval	Frequency	Cumulative Frequency
25 - 34	......	15
35 - 44	......	28
45 - 54	21	......
55 - 64	16	......
65 - 74	......	73
75 - 84	12	......

SChand Composite Mathematics Class 7 Chapter 18 Visualising Solid Shapes Exercise 18A

Exercise 18 A

Question 1

Leonhard Euler (1707-83) was a famous mathematician who discovered a rule about solids.-He observed the number of faces, vertices and edges of many solids before discovering this rule.

(a) Let us see if you can derive the Euler's rule $(F+V-E=2$ ) by completing the following table.

(Diagram to be added)

Question 2

For each solid, count the number of faces, vertices and edges. Check the Euler's rule for each one.

(Diagram to be added)

(a) (Diagram to be added)

Sol: F= 7

V=10

E=15

F+V-F=2

17-15=2

(b) (Diagram to be added)

Sol: $F=6$

$V=8$

$E=12$

$F+V-E$

$6+6-12$

$14-12$ = 2

Sol:

$\begin{aligned} F &=7 \\ V &=7 \\ E &=12 \\ 14-12 \\ &=2 \end{aligned}$

(d) (Diagram to be added)

Sol: $F=9$

$V=9$

$E=16$

$18-16$

$=2$

Question 3

Fill in the blanks. Name a solid that has :

(i) 4 faces

(ii) 4 vertices

(iii) 2 edges

(iv) 6 vertices

(v) No edges.

(vi) 6 faces.

(Diagram to be added)

Multiple choice question

Question 4

Which of these solids has the maximum number of vertices ?

(a) Cons

(b) Cylinder

(c) Cuboid

(d) Pyramid

Question 5

If $\mathrm{F}=6$ and $\mathrm{V}=4$, then the value of $\mathrm{E}$ using Euler's formula is

(a) 10

(b) 8

(d) 4

S.chand class 6 Mathematics Chapter 18 Exercise 18A

Exercise 18A

Question 1

Identify parallel line segments in each of the figures given below:

<img>

2. Classify the following as vertical, horizontal or oblique.

(i) The bottom edge of a door.

(ii) A soldier standing in the attention position.

(iii) The hand rail of a staircase.

(iv) A ladder leaning against a wall.

(v) The hinged edge of a door.

(vi) Horizon at sea.

3. In which plane do the wheels of a moving bike revolve?

4. Based on the adjacent figure, name the following.

<img>

(i) All horizontal lines

(ii) All vertical lines

(iii) All vertical planes

(iv) All horizontal planes

(v) All lines perpendicular to DC

RS Aggarwal solution class 8 chapter 18 Area of Trapezium and a Polygon Exercise 18A

Exercise 18A

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Q1 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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Question 1:

Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.

Answer 1:

$Area of a trapezium = \frac{1}{2} \times (Sum of parallel sides) \times (Distance between them)$

$= \{\frac{1}{2} \times (24 + 20) \times 15\} {cm}^{2} = (\frac{1}{2} \times 44 \times 15) {cm}^{2} = (22 \times 15) {cm}^{2} = 330 {cm}^{2}$

$Hence, the area of the trapezium is 330 {cm}^{2} .$

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Question 2:

Find the area of a trapezium whose parallel sides are 38.7 cm and 22.3 cm, and the distance between them is 16 cm.

Answer 2:

$Area of a trapezium = \frac{1}{2} \times (Sum of parallel sides) \times (Distance between them)$

$= \{\frac{1}{2} \times (38.7 + 22.3) \times 16\} {cm}^{2} = (\frac{1}{2} \times 61 \times 16) {cm}^{2} = (61 \times 8) {cm}^{2} = 488 {cm}^{2}$

$Hence, the area of the trapezium is 488 {cm}^{2} .$

Q3 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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Question 3:

The shape of the top surface of a table is trapezium. Its parallel sides are 1 m and 1.4 m and the perpendicular distance between them is 0.9 m. Find its area.

Answer 3:

$Area of a trapezium = \frac{1}{2} \times (Sum of parallel sides) \times (Distance between them)$
$= \{\frac{1}{2} \times (1 + 1.4) \times 0.9\} m^{2} = (\frac{1}{2} \times 2.4 \times 0.9) m^{2} = (1.2 \times 0.9) m^{2} = 1.08 m^{2}$
$Hence, the area of the top surface of the table is 1.08 m^{2} .$

Q4 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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Question 4:

The area of a trapezium is 1080 cm². If the lengths of its parallel sides be 55 cm and 35 cm, find the distance between them.

Answer 4:

$Let the distance between the parallel sides be x . Now, Area of trapezium = \{\frac{1}{2} \times (55 + 35) \times x\} {cm}^{2}$

$= (\frac{1}{2} \times 90 \times x) c m^{2} = 45 x c m^{2}$
$Area of the trapezium = 1080 {cm}^{2} (Given) ∴ 45 x = 1080 \Rightarrow x = \frac{1080}{45} \Rightarrow x = 24 cm Hence, the distance between the parallel sides is 24 cm .$

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Question 5:

A field is in the form of a trapezium. Its area is 1586 m² and the distance between its parallel sides is 26 m. If one of the parallel sides is 84 m, find the other.

Answer 5:

$Let the length of the required side be x cm . Now, Area of trapezium = \{\frac{1}{2} \times (84 + x) \times 26\} m^{2}$

$= (1092 + 13 x) m^{2}$

$Area of trapezium = 1586 m^{2} (Given) ∴ 1092 + 13 x = 1586 \Rightarrow 13 x = (1586 - 1092) \Rightarrow 13 x = 494 \Rightarrow x = \frac{494}{13} \Rightarrow x = 38 m Hence, the length of the other side is 38 m .$

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Question 6:

The area of a trapezium is 405 cm². Its parallel sides are in the ratio 4 : 5 and the distance between them is 18 cm. Find the length of each of the parallel sides.

Answer 6:

$Let the lengths of the parallel sides of the trapezium be 4 x cm and 5 x cm, respectively . Now, Area of trapezium = \{\frac{1}{2} \times (4 x + 5 x) \times 18\} {cm}^{2}$
$= (\frac{1}{2} \times 9 x \times 18) c m^{2} = 81 x c m^{2}$
$Area of trapezium = 405 {cm}^{2} (Given) ∴ 81 x = 405 \Rightarrow x = \frac{405}{81} \Rightarrow x = 5 cm Length of one side = (4 \times 5) cm = 20 cm Length of the other side = (5 \times 5) cm = 25 cm$

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Question 7:

The area of a trapezium is 180 cm² and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, find the two parallel sides.

Answer 7:

$Let the lengths of the parallel sides be x cm and (x + 6) cm . Now, Area of trapezium = \{\frac{1}{2} \times (x + x + 6) \times 9\} {cm}^{2}$
$= (\frac{1}{2} \times (2 x + 6) \times 9) {cm}^{2} = 4.5 (2 x + 6) {cm}^{2} = (9 x + 27) {cm}^{2}$

$Area of trapezium = 180 {cm}^{2} (Given) ∴ 9 x + 27 = 180 \Rightarrow 9 x = (180 - 27) \Rightarrow 9 x = 153 \Rightarrow x = \frac{153}{9} \Rightarrow x = 17 Hence, the lengths of the parallel sides are 17 cm and 23 cm, that is, (17 + 6) cm .$

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Question 8:

In a trapezium-shaped field, one of the parallel sides is twice the other. If the area of the field is 9450 m² and the perpendicular distance between the two parallel sides is 84 m, find the length of the longer of the parallel sides.

Answer 8:

$Let the lengths of the parallel sides be x cm and 2 x cm . Area of trapezium = \{\frac{1}{2} \times (x + 2 x) \times 84\} m^{2}$
$= (\frac{1}{2} \times 3 x \times 84) m^{2} = (42 \times 3 x) m^{2} = 126 x m^{2}$

$Area of the trapezium = 9450 m^{2}$

$(Given) ∴ 126 x = 9450 \Rightarrow x = \frac{9450}{126} \Rightarrow x = 75$

Thus, the length of the parallel sides are 75 m and 150 m, that is, 2×75 m, and the length of the longer side is 150 m.

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Question 9:

The length of the fence of a trapezium-shaped field ABCD is 130 m and side AB is perpendicular to each of the parallel sides AD and BC. If BC = 54 m, CD = 19 m and AD = 42 m, find the area of the field.

Answer 9:

$Length of the side AB = (130 - (54 + 19 + 42)) m$
$= 15 m$
$Area of the trapezium - shaped field = \{\frac{1}{2} \times (AD + BC) \times AB\}$
$= \{\frac{1}{2} \times (42 + 54) \times 15\} m^{2} = (\frac{1}{2} \times 96 \times 15) m^{2} = (48 \times 15) m^{2} = 720 m^{2}$

$Hence, the area of the field is 720 m^{2} .$

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Question 10:

In the given figure, ABCD is a trapezium in which AD||BC, ∠ABC = 90°, AD = 16 cm, AC = 41 cm and BC = 40 cm. Find the area of the trapezium.

Answer 10:

$∠ ABC = 90 ° From the right ∆ ABC, we have : {AB}^{2} = ({AC}^{2} - {BC}^{2}) \Rightarrow {AB}^{2} = \{(41^{2}) - (40^{2})\} \Rightarrow {AB}^{2} = (1681 - 1600) \Rightarrow {AB}^{2} = 81 \Rightarrow AB = \sqrt{81} \Rightarrow AB = 9 cm ∴ Length AB = 9 cm Now, Area of the trapezium = \{\frac{1}{2} \times (AD + BC) \times AB\}$
$= (\frac{1}{2} \times (16 + 40) \times 9) {cm}^{2} = (\frac{1}{2} \times 56 \times 9) {cm}^{2} = (28 \times 9) {cm}^{2} = 252 {cm}^{2}$
$Hence, the area of the trapezium is 252 {cm}^{2} .$

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Question 11:

The parallel sides of a trapezium are 20 cm and 10 cm. Its nonparallel sides are both equal, each being 13 cm. Find the area of the trapezium.

Answer 11:

$Let ABCD be the given trapezium in which AB ∥ DC, AB = 20 cm, DC = 10 cm and AD = BC = 13 cm .$

$Draw CL ⊥ AB and CM ∥ DA meeting AB at L and M, respectively .$

$Clearly, AMCD is a parallelogram .$

$Now, AM = DC = 10 cm MB = (AB - AM) = (20 - 10) cm = 10 cm$

$Also, CM = DA = 13 cm$

$Therefore, ∆ CMB is an isosceles triangle and CL ⊥ MB . L is the midpoint of B .$

$\Rightarrow ML = LB = (\frac{1}{2} \times MB)$
$= (\frac{1}{2} \times 10) cm = 5 cm$

$From right ∆ CLM, we have : {CL}^{2} = ({CM}^{2} - {ML}^{2}) {cm}^{2}$

$\Rightarrow {CL}^{2} = \{{(13)}^{2} - {(5)}^{2}\} {cm}^{2}$

$\Rightarrow {CL}^{2} = (109 - 25) {cm}^{2}$

$\Rightarrow {CL}^{2} = 144 {cm}^{2}$

$\Rightarrow CL = \sqrt{144} cm$

$\Rightarrow CL = 12 cm$

$∴ Length of CL = 12 cm$

$Area of the trapezium = \{\frac{1}{2} \times (AB + DC) \times CL\}$
$= \{\frac{1}{2} \times (20 + 10) \times 12\} {cm}^{2} = (\frac{1}{2} \times 30 \times 12) {cm}^{2} = (15 \times 12) {cm}^{2} = 180 {cm}^{2}$
$Hence, the area of the trapezium is 180 {cm}^{2} .$

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Question 12:

The parallel sides of a trapezium are 25 cm and 11 cm, while its nonparallel sides are 15 cm and 13 cm. Find the area of the trapezium.

Answer 12:

$Let ABCD be the trapezium in which AB ∥ DC, AB = 25 cm, CD = 11 cm, AD = 13 cm and BC = 15 cm .$

$Draw CL ⊥ AB and CM ∥ DA meeting AB at L and M, respectively . Clearly,$

$AMCD is a parallelogram .$

$Now, MC = AD = 13 cm AM = DC = 11 cm$

$\Rightarrow MB = (AB - AM)$
$= (25 - 11) cm = 14 cm$

$Thus, in ∆ CMB, we have : CM = 13 cm MB = 14 cm BC = 15 cm$

$∴ s = \frac{1}{2} (13 + 14 + 15) cm = \frac{1}{2} 42 cm = 21 cm (s - a) = (21 - 13) cm = 8 cm (s - b) = (21 - 14) cm = 7 cm (s - c) = (21 - 15) cm = 6 cm$

$∴ Area of ∆ CMB = \sqrt{s (s - a) (s - b) (s - c)}$
$= \sqrt{21 \times 8 \times 7 \times 6} {cm}^{2} = 84 {cm}^{2}$

$∴ \frac{1}{2} \times MB \times CL = 84 {cm}^{2} \Rightarrow \frac{1}{2} \times 14 \times CL = 84 {cm}^{2}$
$\Rightarrow CL = \frac{84}{7} \Rightarrow CL = 12 cm$
$Area of the trapezium = \{\frac{1}{2} \times (AB + DC) \times CL\}$
$= \{\frac{1}{2} \times (25 + 11) \times 12\} {cm}^{2} = (\frac{1}{2} \times 36 \times 12) {cm}^{2} = (18 \times 12) {cm}^{2} = 216 {cm}^{2}$

$Hence, the area of the trapezium is 216 {cm}^{2} .$

08	17	33	41	47	23	20	34
09	18	42	14	30	19	29	11
36	48	40	24	22	02	16	21
15	32	47	44	33	01

08	17	33	41	47	23	20	34
09	18	42	14	30	19	29	11
36	48	40	24	22	02	16	21
15	32	47	44	33	01

Question 1.1

Question 1.2

Question 1.3

Question 1.4

Question 1.5

Question 2

Question 3

Question 4.1

Question 4.2

Question 4.3

Question 4.4

Question 4.5

Question 4.6

Question 5

Question 6

Question 7

Question 8.1

Question 8.2

Question 9.1

Question 9.2

Question 10

Question 11

Question 12

Question 13

Exercise 18 A

Exercise 18A

Question 1

Exercise 18A

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Q1 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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Answer 1:

Q2 | Ex-18A | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon | myhelper

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Question 11:

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Question 12:

Answer 12:

Contact Form

08	17	33	41	47	23	20	34
09	18	42	14	30	19	29	11
36	48	40	24	22	02	16	21
15	32	47	44	33	01