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SELINA Solution Class 9 Chapter 18 Statistics Exercise 18B

Question 1

Construct a frequency polygon for the following distribution:

Class-intervals	0-4	4 - 8	8 - 12	12 - 16	16 - 20	20 - 24
Frequency	4	7	10	15	11	6

Sol:

The frequency polygon is shown in the following figure

Steps:

(i) Drawing a histogram for the given data.

(ii) Marking the mid-point at the top of each rectangle of the histogram drawn.

(iii) Also, marking mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.

(iv) Joining the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

Question 2

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

Class-Intervals	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60
Frequency	3	5	6	4	2

Sol:

Steps:

1. Draw a histogram for the given data.

2. Mark the mid-point at the top of each rectangle of the histogram drawn.

3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.

4. Join the consecutive mid-point marked by straight lines to obtain the required frequency polygon.

5. The require combined histogram and frequency polygon are shown in the following figure:

Question 3

Construct a frequency polygon for the following data:

Class-Intervals	10 - 14	15 - 19	20 - 24	25 - 29	30 - 34
Frequency	5	8	12	9	4

Sol:

The class intervals are inclusive. We will first convert them into the exclusive form.

Class-Interval	Frequency
9.5 - 14.5	5
14.5 - 19.5	8
19.5 - 24.5	12
24.5 - 29.5	9
29.5 - 34.5	4

Steps:

Draw a histogram for the given data.
Mark the mid-point at the top of each rectangle of the histogram drawn.
Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

The required frequency polygon is as follows:

Question 4

The daily wages in a factory are distributed as follows:

Daily wages (in Rs.)	125 - 175	175 - 225	225 - 275	275 - 325	325 - 375
Number of workers	4	20	22	10	6

Draw a frequency polygon for this distribution.

Sol:

Steps:

Draw a histogram for the given data.
Mark the mid-point at the top of each rectangle of the histogram drawn.
Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

The required frequency polygon is as follows:

Question 5.1

Draw frequency polygons for each of the following frequency distribution:

(a) using histogram

(b) without using histogram

C.I	10 - 30	30 - 50	50 - 70	70 - 90	90 - 110	110 - 130	130 - 150
ƒ	4	7	5	9	5	6	4

Sol:

using histogram

C.I	ƒ
10 - 30	4
30 - 50	7
50 - 70	5
70 - 90	9
90 - 110	5
110 - 130	6
130 - 150	4

Steps:

1. Draw a histogram for the given data.
2. Mark the mid-point at the top of each rectangle of the histogram drawn.
3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

Without using Histogram:

Steps:

Find the class-mark (mid-value) of each given class-interval.
class mark = mid-value = $\frac{Upper limit + Lower limit}{2}$
On a graph paper, mark class-marks along X-axis and frequencies along Y-axis.
On this graph paper, mark points taking values of class-marks along the X-axis and the values of their corresponding frequencies along Y-axis.
Draw line segments joining the consecutive points marked in step (3) above.

C.I. Class-mark f

-10 - 10 0 0

10 - 30 20 4

30 - 50 40 7

50 - 70 60 5

70 - 90 80 9

90 - 110 100 5

110 - 130 120 6

130 - 150 140 4

150 - 170 160 0

Question 5.2

Draw frequency polygons for each of the following frequency distribution:
(a) using histogram
(b) without using histogram

C.I	5 -15	15 -25	25 -35	35 - 45	45-55	55-65
ƒ	8	16	18	14	8	2

Sol:

Using Histogram:

C.I.	f
5 - 15	8
15 - 25	16
25 - 35	18
35 - 45	14
45 - 55	8
55 - 65	2

Steps:

Draw a histogram for the given data.
Mark the mid-point at the top of each rectangle of the histogram drawn.
Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

Without using Histogram:
Steps:

Find the class-mark (mid-value) of each given class-interval.
On a graph paper, mark class-marks along X-axis and frequencies along Y-axis.
On this graph paper, mark points taking values of class-marks along the X-axis and the values of their corresponding frequencies along the Y-axis.
Draw line segments joining the consecutive points marked in step (3) above.

C.I.

Class-mark

f

-5 - 5

0

0

5 - 15

10

8

15 - 25

20

16

25 - 35

30

18

35 - 45

40

14

45 - 55

50

8

55 - 65

60

2

65 - 75

70

0

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

Exercise 18B

Question 1

Copy each net into your book and add the face that is needed to make the net of named solid . Shade the added face .

Prime:- Two parallel congruent Polygonal faced connected

Pyramid : - Polygon base and triangular faces meet at a common vertex.

(diagram to be added)

Question 2

Label each net as of :

(a) cuboid

(b) cube

(d) rectangular pyramid

(e) cone

(f) cylinder

(g) square pyramid

(h) triangular prism

Question 3

Draw the net of a cube. Put the number I to 6 on the faces so that the number on opposite faces add to 7. Can this be done in more than one. If so, give all ways.

One has been done for you :

(diagram to be added)

Question 4

Name the different plane shapes needed to draw the net of :

(a) a cube

(b) a triangular prism

(d) a cylinder

(diagram to be added)

Question 5

Which of these nets will not fold up to give a cube?

[Hint. Make the cardboard cutouts of the following nets and fold them to sce whether a cube is formed or not]

(diagram to be added)

Question 6

Only one of these nets could form a pyramid. Which one ?

S.chand class 6 Mathematics Chapter 18 Exercise 18B

Exercise 18B

Question 1

1. Line $l$ is a transversal to lines $n$ and $m$. Name the following.

<img>

(i) A pair of interior angles on the same side of the transversal.

(ii) A pair of exterior angles on the same side of the transversal.

(iii) Two pairs of alternate interior angles.

(iv) Four pairs of corresponding angles.

2. In the given figure, write the name of

<img>

(i) alternate interior angles.

(ii) corresponding angles.

(iii) vertical angles.

(iv) co-interior angles with respect to both the transversals $P Q$ and $P R$.

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

Exercise 18B

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

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Question 1:

In the given figure, ABCD is a quadrilateral in which AC = 24 cm, BL ⊥ AC and DM ⊥ AC such that BL = 8 cm and DM = 7 cm. Find the area of quad. ABCD.

Answer 1:

$Area of quadrilateral ABCD = (Area of ∆ ADC) + (Area of ∆ ACB)$
$= (\frac{1}{2} \times AC \times DM) + (\frac{1}{2} \times AC \times BL) = [(\frac{1}{2} \times 24 \times 7) + (\frac{1}{2} \times 24 \times 8)] {cm}^{2} = (84 + 96) {cm}^{2} = 180 {cm}^{2}$

$Hence, the area of the quadrilateral is 180 {cm}^{2} .$

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

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Question 2:

In the given figure, ABCD is a quadrilateral-shaped field in which diagonal BD is 36 m, AL ⊥ BD and CM ⊥ BD such that AL = 19 m and CM = 11 m. Find the area of the field.

Answer 2:

$Area of quadrilateral ABCD = (Area of ∆ ABD) + (Area of ∆ BCD)$
$= (\frac{1}{2} \times BD \times AL) + (\frac{1}{2} \times BD \times CM)$
$= [(\frac{1}{2} \times 36 \times 19) + (\frac{1}{2} \times 36 \times 11)] m^{2} = (342 + 198) m^{2} = 540 m^{2}$

$Hence, the area of the field is 540 m^{2} .$

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

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Question 3:

Find the area of pentagon ABCDE in which BL ⊥ AC, DM ⊥ AC and EN ⊥ AC such that AC = 18 cm, AM = 14 cm, AN = 6 cm, BL = 4 cm, DM = 12 cm and EN = 9 cm.

Answer 3:

$Area of pentagon ABCDE = (Area of ∆ AEN) + (Area of trapezium EDMN) + (Area of ∆ DMC) + (Area of ∆ ACB)$
$= (\frac{1}{2} \times AN \times EN) + (\frac{1}{2} \times (EN + DM) \times NM) + (\frac{1}{2} \times MC \times DM) + (\frac{1}{2} \times AC \times BL) = (\frac{1}{2} \times AN \times EN) + (\frac{1}{2} \times (EN + DM) \times (AM - AN)) + (\frac{1}{2} \times (AC - AM) \times DM) + (\frac{1}{2} \times AC \times BL) = [(\frac{1}{2} \times 6 \times 9) + (\frac{1}{2} \times (9 + 12) \times (14 - 6)) + (\frac{1}{2} \times (18 - 14) \times 12) + (\frac{1}{2} \times 18 \times 4)] {cm}^{2} = (27 + 84 + 24 + 36) {cm}^{2} = 171 {cm}^{2}$

$Hence, the area of the given pentagon is 171 {cm}^{2} .$

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

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Question 4:

Find the area of hexagon ABCDEF in which BL ⊥ AD, CM ⊥ AD, EN ⊥ AD and FP ⊥ AD such that AP = 6 cm, PL = 2 cm, LN = 8 cm, NM = 2 cm, MD = 3 cm, FP = 8 cm, EN = 12 cm, BL = 8 cm and CM = 6 cm.

Answer 4:

$Area of hexagon ABCDEF = (Area of ∆ AFP) + (Area of trapezium FENP) + (Area of ∆ ALB)$
$= (\frac{1}{2} \times AP \times FP) + (\frac{1}{2} \times (FP + EN) \times PN) + (\frac{1}{2} \times ND \times EN) + (\frac{1}{2} \times MD \times CM) + (\frac{1}{2} \times (CM + BL) \times LM) + (\frac{1}{2} \times AL \times BL) = (\frac{1}{2} \times AP \times FP) + (\frac{1}{2} \times (FP + EN) \times (PL + LN)) + (\frac{1}{2} \times (NM + MD) \times CM) + (\frac{1}{2} \times MD \times CM) + (\frac{1}{2} \times (CM + BL) \times (LN + NM)) + (\frac{1}{2} \times (AP + PL) \times BL) = [(\frac{1}{2} \times 6 \times 8) + (\frac{1}{2} \times (8 + 12) \times (2 + 8)) + (\frac{1}{2} \times (2 + 3) \times 12) + (\frac{1}{2} \times 3 \times 6) + (\frac{1}{2} \times (6 + 8) \times (8 + 2)) + (\frac{1}{2} \times (6 + 2) \times 8)] {cm}^{2} = (24 + 100 + 30 + 9 + 70 + 32) {cm}^{2} = 265 {cm}^{2}$

$Hence, the area of the hexagon is 265 {cm}^{2} .$

Q5 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Question 5:

Find the area of pentagon ABCDE in which BL ⊥ AC, CM ⊥ AD and EN ⊥ AD such that AC = 10 cm, AD = 12 cm, BL = 3 cm, CM = 7 cm and EN = 5 cm.

Answer 5:

$Area of pentagon ABCDE = (Area of ∆ ABC) + (Area of ∆ ACD) + (Area of ∆ ADE)$
$= (\frac{1}{2} \times AC \times BL) + (\frac{1}{2} \times AD \times CM) + (\frac{1}{2} \times AD \times EM) = [(\frac{1}{2} \times 10 \times 3) + (\frac{1}{2} \times 12 \times 7) + (\frac{1}{2} \times 12 \times 5)] {cm}^{2} = (15 + 42 + 30) {cm}^{2} = 87 {cm}^{2}$

$Hence, the area of the pentagon is 87 {cm}^{2} .$

Q6 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Question 6:

Find the area enclosed by the given figure ABCDEF as per dimensions given herewith.

Answer 6:

$Area enclosed by the given figure = (Area of trapezium FEDC) + (Area of square ABCF)$
$= [\{\frac{1}{2} \times (6 + 20) \times 8\} + (20 \times 20)] {cm}^{2} = (104 + 400) {cm}^{2} = 504 {cm}^{2}$

$Hence, the area enclosed by the figure is 504 {cm}^{2} .$

Page-211

Q7 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Question 7:

Find the area of given figure ABCDEFGH as per dimensions given in it.

Answer 7:

$We will find the length of AC . From the right triangles ABC and HGF, we have : {AC}^{2} = {HF}^{2} = \{{(5)}^{2} - {(4)}^{2}\} cm$
$= (25 - 16) c m = 9 c m$

$AC = HF = \sqrt{9} c m = 3 c m$
$Area of the given figure ABCDEFGH = (Area of rectangle ADEH) + (Area of ∆ ABC) + (Area of ∆ HGF)$
$= (Area of rectangle ADEH) + 2 (Area of ∆ ABC) = (AD \times DE) + 2 (Area of ∆ ABC) = \{(AC + CD) \times DE\} + 2 (\frac{1}{2} \times BC \times AC) = \{(3 + 4) \times 8\} + 2 (\frac{1}{2} \times 4 \times 3) {cm}^{2} = (56 + 12) cm = 68 {cm}^{2}$
$Hence, the area of the given figure is 68 {cm}^{2}$ .

Q8 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Question 8:

Find the area of a regular hexagon ABCDEF in which each side measures 13 cm and whose height is 23 cm, as shown in the given figure.

Answer 8:

$Let AL = DM = x cm LM = BC = 13 cm ∴ x + 13 + x = 23 \Rightarrow 2 x + 13 = 23 \Rightarrow 2 x = (23 - 13) \Rightarrow 2 x = 10 \Rightarrow x = 5 ∴ AL = 5 cm From the right ∆ AFL, we have : F L^{2} = A F^{2} - A L^{2} \Rightarrow F L^{2} = \{(13^{2}) - {(5)}^{2}\} \Rightarrow F L^{2} = (169 - 25) \Rightarrow F L^{2} = 144 \Rightarrow F L = \sqrt{144} \Rightarrow F L = 12 c m ∴ FL = BL = 12 cm Area of a regular hexagon = (A rea of t h e trapezium ADEF) + (A rea of t h e trapezium ABCD)$
$= 2 (A r e a o f t r a p e z i u m A D E F) = 2 \{\frac{1}{2} \times (A D + E F) \times F L\} = 2 \{\frac{1}{2} \times (23 + 13) \times 12\} c m^{2} = 2 (\frac{1}{2} \times 36 \times 12) c m^{2} = 432 c m^{2}$
$Hence, the area of the given regular hexagon is 432 {cm}^{2} .$

C.I.	Class-mark	f
-10 - 10	0	0
10 - 30	20	4
30 - 50	40	7
50 - 70	60	5
70 - 90	80	9
90 - 110	100	5
110 - 130	120	6
130 - 150	140	4
150 - 170	160	0

C.I.	Class-mark	f
-5 - 5	0	0
5 - 15	10	8
15 - 25	20	16
25 - 35	30	18
35 - 45	40	14
45 - 55	50	8
55 - 65	60	2
65 - 75	70	0

SELINA Solution Class 9 Chapter 18 Statistics Exercise 18B

Question 1

Question 2

Question 3

Question 4

Question 5.1

Question 5.2

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

Exercise 18B

S.chand class 6 Mathematics Chapter 18 Exercise 18B

Exercise 18B

Question 1

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

Exercise 18B

Page-210

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

Question 1:

Answer 1:

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

Question 2:

Answer 2:

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

Question 3:

Answer 3:

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

Question 4:

Answer 4:

Q5 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

Question 5:

Answer 5:

Q6 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

Question 6:

Answer 6:

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Q7 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

Question 7:

Answer 7:

Q8 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

Question 8:

Answer 8:

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