SELINA Solution Class 9 Chapter 15 Construction of Polygons (Using ruler and compass only) Exercise 15

Question 1

Construct a quadrilateral ABCD, when:
AB = 3.2 cm, BC = 5.2 cm, CD = 6.2 cm, DA = 4.2 cm and BD = 5.2 cm.

Sol:

Steps:
1. Draw AB = 3.2 cm.
2. With A as a centre draw an arc at D and with B as a centre and radius 5.2 cm draw an arc at D.
3. Join AD and DB.
4. With D and B as a centre taking radius 6.2 cm and 5.2 cm draw an arc at C. Now join BC and DC.

ABCD is the required quadrilateral.

Question 2

Construct a quadrilateral ABCD, when:
AB = 7.2 cm, BC = 5.8 cm, CD = 6.3 cm, AD = 4.3 cm and angle A = 75°.

SoL:

Steps:
1. Draw AB = 7.2cm
2. Through A draw AP such that ∠A = 75°.
3. From AP cut AD = 4.3 cm
4. With D and B as center and radii 6.2 cm and 5.8 cm respectively, draw arcs cutting each other at C.
5. Join DC and BC.

ABCD is the required quadrilateral.

Question 3

Construct a quadrilateral ABCD, when:
∠ A = 90^o, AB = 4.6 cm, BD = 6.4 cm, AC = 6.0 cm and CD = 4.2 cm.

Sol:

Steps:
1. Draw AB = 4.6 cm.
2. Through A, draw AP such that Angle A = 90°.
3. With B as a center and radii 6.4 cm draw an arc at D on AP.
4. With D and A as a center and radii, 4.2 cm and 6 cm draw an arc cutting each other at C.
5. Now join BD, AC, and CB.

ABCD is the required quadrilateral.

Question 4

Construct a quadrilateral ABCD, when:
AB = 3.8 cm, AC = 4.8 cm, AD = 2.8 cm, angle A = 105° and angle B = 60°.

Sol:

Steps:
1. Draw AD = 2.8 cm
2. Draw AB = 3.8 cm and ∠A = 105°.
3. Draw BP such that ∠B = 60°.
4. With A as a center and radii 4.8 cm draw an arc cutting BP at C.
5. Join AC, AD.

ABCD is the required quadrilateral.

Question 5

Construct a quadrilateral ABCD, when:
BC = 7.5 cm AC = 5.8 cm, AD = 3.6 cm, CD = 4.2 cm and angle A = 120°.

Sol:

Steps:
1. Draw AD = 3.6 cm
2. draw AP such that ∠A = 120°.
3. With A and D as a center and radii 5.8 cm and 4.2 cm draw arcs cutting each other at C.
4. Now join AC and CD.
5. Now with C as center and radii 7.5 cm draw an arc at B on AP.
6. Now join CB.

ABCD is the required quadrilateral.

Question 6

Construct a quadrilateral ABCD, when:
AD = AB = 4 cm, BC = 2.8 cm, CD = 2.5 cm and angle BAD = 45°.

Sol:

Steps:
1. Draw AD = 4cm.
2. Draw AP such that ∠A = 45°.
3. With A as a center with radii, 4 cm draw an arc at B on AP.
4. Now taking B and D as a center and radii 2.8 cm and 2.8 cm draw arcs cutting each other at C.
5. Now join BC and CD.

ABCD is the required quadrilateral.

Question 7

Construct a quadrilateral ABCD, when:
AB = 6.3 cm, BC = CD = 4.2 cm and ∠ABC = ∠BCD = 90°.

Sol:

Steps :
1. draw AB = 6.3 cm.
2. Draw BP such that ∠ABP = 90°.
3. With B as a center and radii, 4.2 cm draw an arc AP at C.
4. With C as a center draw a line CD with radii 4.2 cm. draw a line such that ∠BCD = 90°.
5. Now join AD.

ABCD is the required quadrilateral.

Question 8

Construct a parallelogram ABCD, when:
AB = 4.4 cm, AD = 6.2 cm and AC = 4.8 cm.

Sol:

Steps:
1. Draw AD = 6.2 cm.
2. Draw triangle ACD.
3. Then draw triangle ABC.

ABCD is the required parallelogram.

Question 9

Construct a parallelogram ABCD, when:
Diagonal AC = 6.4 cm, diagonals BD = 8.2 cm and angle between the diagonals = 60°.

Sol:

Steps:
1. Draw AC = 6.4 cm.
2. Draw line BOD such that ∠DOC = 60° and
OB = OD = ${\displaystyle \frac{1}{2}\text{BD}=\frac{1}{2}}$ x 8.2 = 4.1 cm
3. Join AB, BC, CD, and DA.

ABCD is the required parallelogram.

Question 10

Construct a parallelogram ABCD, when:
AB = 5.8 cm, diagonal AC = 8.2 cm and diagonal BD = 6.2 cm.

Sol:

Steps:
1. Since diagonal of a parallelogram bisect each other, construct OAB such that;

OA = ${\displaystyle \frac{1}{2}\text{AC}=\frac{1}{2}}$ x 8.2 cm = 4.1 cm.

OB = ${\displaystyle \frac{1}{2}\text{BD}=\frac{1}{2}}$ x 6.2 cm = 3.1 cm.
and AB = 5.8 cm.

2. Produce AO up to C, such that OC = OA = 4.1 cm and BO up to D, such that DO = OB = 3.1 cm.
3. Join AD, DC, and CB.

ABCD is the required parallelogram.

Question 11

Construct a parallelogram ABCD, when:
AB = 6.0 cm, AD = 5.0 cm and ∠A = 45°.

Sol:

Steps:
1. Draw AB = 6 cm.
2. Draw AD with radii 5 cm with an angle of 45°.
3. With D and B as a center and radii 6 cm and 5 cm draw arcs cutting each other at C.
4. Now join DC and BC.

ABCD is the required parallelogram.

Question 12

Construct a parallelogram ABCD, when:
Base AB = 6.5 cm, BC = 4 cm and the altitude corresponding to AB = 3.1 cm.

Sol:

Steps :
1. Draw AB = 6.5 cm.
2. At B, draw BP ⊥ AB.
3. From BP cut BE = 3.1 cm.
4. Through E draw a perpendicular to BP to get QR parallel to AB.
5. With B as a center and radius = AC = 4cm, draw an arc that cuts QR at C.
6. With A as a center and radius = AD = 4 cm, draw an arc that cuts QR at D.

ABCD is the required parallelogram.

Question 13

Construct a parallelogram ABCD, when:
AB = 4.5 cm, ∠B = 120° and the distance between AB and DC = 3.0 cm.

Sol:

Steps:
1. Draw AB = 4.5 cm.
2. At B, draw BP ⊥ AB.
3. From BP cut BE = 3 cm.
4. Through E draw a perpendicular to BP to get QR parallel to AB.
5. With B as a center draw an arc that cuts QR at C.
6. With A as a center draw an arc that cuts QR at D.
7. Now join AD and BC.

ABCD is the required parallelogram.

Question 14

Construct a parallelogram ABCD, when:
Base BC = 5.6 cm, diagonal BD = 6.5 cm and altitude = 3.2 cm.

Sol:

Steps:
1. Draw BC = 5.6 cm.
2. At C, draw CX perpendicular to BC.
3. with C as a centre and taking radius 3.2 cm to draw an arc to cut CX at Y.
4. Through Y draw a straight line PQ parallel to BC.
5. With B as a centre and radius, 6.5 cm draw an arc to meet PQ at D.
6. With D as a centre and radius equal to 5.6 cm, draw an arc to meet PQ at A.
7. Join BA, BD and CD.

ABCD is the required parallelogram.

Question 15

Construct a rectangle ABCD, when:
Its sides are 6.0 cm and 7.2 cm.

Sol:

Since each angle of a rectangle is 90° and the opposite sides are equal. Therefore,
Steps:
1. Draw BC = 7.2 cm.
2. With B as a centre draw a line BX taking as a 90°.
3. Now taking radius 6 cm to draw an arc at A.
4. From the point, A draw a line AY parallel to BC.
5. With A as a center taking radius 7.2 cm to draw an arc at D.
6. Now join CD.

ABCD is the required rectangle.

Question 16

Construct a rectangle ABCD, when:
One side = 4 cm and one diagonal is 5 cm. Measure the length of the other side.

Sol:

Steps:
1. Draw BC = 4 cm.
2. With C as a center and radius 5 cm draw an arc at A.
3. Now join AB and AC.
4. With A as a center draw an arc at D.
5. Now join AD and CD.

ABCD is the required rectangle.

Question 17

Construct a rectangle ABCD, when:
One diagonal = 6.0 cm and the acute angle between the diagonals = 45^o.

Sol:

Steps:
1. Draw AC = 6cm.
2. Draw the right triangle ACB.
3. Draw the right triangle ADB.
4. Join DC.

ABCD is the required rectangle.

Question 18

Construct a rectangle ABCD, when:
Area = 24 cm² and base = 4.8 cm².

Sol:

Given that the base = 4.8 cm² and Area = 24 cm²
We know that the area of the rectangle = base x Height.
Therefore,
24 = 4.8 x height
Height = 5

With base = 4.8 cm² and height = 5 cm²,
the rectangle is shown below:

Steps:
1. Draw base AB = 4.8 cm².
2. With A and B as a center draw an arc taking radius 5 cm² at D and C.
3. Now join AD, BC, and DC.

ABCD is the required rectangle.

Question 19

Construct a rectangle ABCD, when:
Area = 36 cm² and height = 4.5 cm.

Sol:

Given that the height = 4.5 cm and Area = 36 cm²
We know that the area of rectangle = base x Height
Therefore,
36 = base x 4.5
Base = 8 cm
With height = 4.5 and base = 8 cm, the rectangle is shown below:
Steps:
1. Draw base AB = 8 cm.
2. With A and B as a center draw arc taking radius = 4.5 cm at D and C.
3. Now join AD, BC, and DC.

ABCD is the required rectangle.

Question 20

Construct a trapezium ABCD, when:
AB = 4.8 cm, BC = 6.8 cm, CD = 5.4 cm, angle B = 60^o and AD // BC.

Sol:

Steps:
1. Draw BC = 6.8 cm.
2. With B as a center and radii 4.8 cm draw an arc at A such that ∠B = 60°.
3. From point draw a line AP such that AP || BC.
4. With C as a center and radii 5.4 cm draw an arc at D on the line AP.
5. Now join AB, CD.

ABCD is the required trapezium.

Question 21

Construct a trapezium ABCD, when:
AB = CD = 3.2 cm, BC = 6.0 cm, AD = 4.4 cm and AD // BC.

Sol:

Steps:
1. Draw BC = 6 cm.
2. From BC cut BE = AD = 4.1 cm.
3. Draw triangle DEC such that DE = AB = 3.2 cm and CD = 3.2cm.
4. Taking B and D as a center and radii 3.2 cm and 4.1 cm respectively, draw arcs cutting each other at A.
5. Join AB and AD.

ABCD is the required trapezium.

Question 22

Construct a rhombus ABCD, when:
Its one side = 6 cm and ∠A = 60^o.

Sol:

Steps:
1. Draw a line AB = 6 cm.
2. At A, we construct ∠BAP = 60°. 
3. From AP, we cut at D taking AD = 6 cm.
4. Through B, we draw BQ || AD.
5. Through D, we draw DC || AB to cut BQ at C.

ABCD is the required rhombus.

Question 23

Construct a rhombus ABCD, when:
One side = 5.4 cm and one diagonal is 7.0 cm.

Sol:

Steps:
1. We construct the segment AC = 7 cm.
2. With A as a centre and radius 5.4 cm, we draw an arc extending on both sides of AC.
3. With C as a centre and same radius as in step 2, we draw an arc extending on both sides of AC to cut the first arc at B and D.
4. Join AB, BC, CD and DA.

ABCD is the required rhombus.

Question 24

Construct a rhombus ABCD, when:
Diagonal AC = 6.3 cm and diagonal BD = 5.8 cm.

Sol:

Steps:
1. Draw AC = 6.3 cm.
2. Draw perpendicular bisector to AC which cuts AC at O.
3. From this perpendicular cut OD and OB such that,

OD = OB = ${\displaystyle \frac{1}{2}\text{BD}=\frac{1}{2}}$ x 5.8 = 2.9 cm.

4. Join AB, BC, CD, and DA.

ABCD is the required rhombus.

Question 25

Construct a rhombus ABCD, when:
One side = 5.0 cm and height = 2.6 cm.

Sol:

Steps:
1. Draw AB = 5 cm.
2. At B, draw BP ⊥ AB.
3. From BP, cut BE = 2.6 cm = height.
4. Through E draw a perpendicular to CP to get QR parallel to AB.

With A and B as a centre and radii 5 cm draw arcs cutting QR at D andC.

ABCD is the required rhombus.

Question 26

Construct a rhombus ABCD, when: A = 60° and height = 3.0 cm.

Sol:

Steps:
1. Draw a line AP.
2. Now draw a line AF such that ∠A = 60°.
3. At S draw a perpendicular SE of length 3 cm such that it cut at AF at D.
4. Through D draw a line QR parallel to AP.
5. Now taking the radius same as AD draws an arc at B on AP.
6. Now through and B taking radius same as AD and AB draw arcs cutting each other at C.
7. Now join BC.

ABCD is the required rhombus.

Question 27

Construct a rhombus ABCD, when: Diagonal AC = 6.0 cm and height = 3.5 cm.

SoL:

Steps:

1. draw a line AP.

2. now draw AC = 6cm and CP = 3.5cm.

3. Now draw a line BC such that AB = BC.

4. Now at C draw a line CY parallel to AP.

5. At point C and A, taking radius same as AB draw arcs cutting each other at D.

6. Now join AD.

ABCD is the required rhombus.

Question 28

Construct a square ABCD, when: One side = 4.5 cm.

Sol:

Steps:

1. Draw a line segment AB = 4.5cm

2. Draw AP ⊥ AB.

3. From AP cut off AD = 4.5cm.

4. With B as a center and radius 4.5 cm draw an arc.

5. With D as center and radius 4.5 cm draw another arc cutting the former arc at C.

6. Join BC and CD.

ABCD is the required square.

Question 29

Construct a square ABCD, when: One diagonal = 5.4 cm.

Sol:

We know that the diagonals of a square are equal and bisect each other at right angles.

Steps:

1. draw AC = 5.4cm.

2. Draw the right bisector XY of AC, meeting AC at O.

3. From O, set off OB = ${\displaystyle \frac{1}{2}}$ (5.4) = 2.7cm along OY and along OX.

4. Join AB, BC, CD, and DA.

ABCD is the required square. 

Question 30

Construct a square ABCD, when: Perimeter = 24 cm.

Sol:

The perimeter of a square
P = 4a
Where a is the length of each side
We have Perimeter = 24 cm
Therefore,
24 = 4a 
4 =6
Therefore the sides of the squares are of length 6 cm. 

Steps:

1. Draw a line segment Ab = 6cm.

2. Draw AP ⊥ AB.

3. From AP cut off AD = 6cm.

4. With B as a center and radius, 6 cm draw an arc.

5. With D as center and radius, 6 cm draw another arc cutting the former arc at C.

6. Join BC and CD.

ABCD is the required square.

Question 31

Construct a rhombus, having given one side = 4.8 cm and one angle = 75^o.

Sol:

Steps:

1.draw a line AB = 4.8cm

2. At A Draw AX such that ∠BAX = 75°.

3. With A as a center and measurement equal to AB cut off an arc at D on AX.

4. Using the same radius taking D and B as centers cut off arcs, which will intersect at C.

5. Join CD and CB.

ABCD is the required rhombus.

Question 32.1

Construct a regular hexagon of side 2.5 cm

Sol:

The length of the side of a regular hexagon is equal to the radius of its circumcircle.

Steps of construction:

1. Draw a circle of radius 2.5 cm

2. Taking any point A on the circumference of the circle as the centre, draw arcs of same radii (i.e. 2.5 cm) which cut the circumference at B and F.

3. With B and F as centres, again draw two arcs of same radii which cut the circumference at C and E respectively.

4. With C or E as the centre, draw one more arc of the same radius which cuts the circumference at point D.

In this way, the circumference of the circle is divided into six equal parts.

5. Join AB, BC, CD, DE, EF and FA.

ABCDEF is the required regular hexagon.

Question 32.2

Construct a regular hexagon of side 3.2 cm

Sol:

The length of the side of a regular hexagon is equal to the radius of its circumcircle.

Steps of construction:

1. Draw a circle of radius 3.2 cm

2. Taking any point A on the circumference of the circle as a centre, draw arcs of same radii (i.e. 3.2 cm) which cut the circumference at B and F.

3. With B and F as centres, again draw two arcs of same radii which cut the circumference at C and E respectively.

4. With C or E as a centre, draw one more arc of the same radius which cuts the circumference at point D.
In this way, the circumference of the circle is divided into six equal parts.

5. Join AB, BC, CD, DE, EF and FA.

ABCDEF is the required regular hexagon. 

Question 33

Using ruler and compasses only, construct the quadrilateral ABCD, having given AB = 5 cm, BC = 2.5 cm, CD = 6 cm. angle BAD = 90^o and the diagonal AC = 5.5 cm.

Sol:

Steps:

1. draw AB = 5cm.

2. Now draw ∠XAB such that it is 90°.

3. Taking A and B as a center and radius 2.5 cm and 5.5 cm draw arcs cut off at C.

4. Now join BC and AC.

5. Taking C as a center and radius 6 cm draw arcs at D on AX.

ABCD is the required quadrilateral.

Question 34

Using ruler and compasses only, construct a trapezium ABCD, in which the parallel sides AB and DC are 3.3 cm apart; AB = 4.5 cm, angle A = 120^o BC = 3.6 cm and angle B is obtuse.

Sol:

Steps:
1. Draw AB = 4.5cm.

2. now draw ∠BAS = 120° and draw EA ⊥ AB such that AX = 3.3cm.

3. Through X draws a line QR which is parallel to AB which cuts AS at D.
4. Through B draw an arc taking radius 3.6 cm at C on PQ.

5. Join CB.

ABCD is the required trapezium. 

Question 35

Using ruler and compasses only, construct the quadrilateral ABCD, having given AB = 5 cm, BC = 2.5 cm CD = 6 cm, ∠BAD = 90^o and diagonal BD = 5.5 cm.

Sol:

Steps:

1.Draw AB = 5cm.

2. From A draw a line AY such that ∠A =90°.

3. Taking B as a center with radius 5.5 cm draw an arc at D on AY.

4. With D and B as center and radii, 6 cm and 2.5 cm draw arcs cutting each other at C.

5. Join DC and BC.

ABCD is the required quadrilateral.

Question 36

Using ruler and compasses only, construct a parallelogram ABCD using the following data: AB = 6 cm, AD = 3 cm and ∠DAB = 45^o. If the bisector of ∠DAB meets DC at P,
prove that ∠APB is a right angle.

Sol:

Steps:

1. draw AB = 6cm.

2. With A as a center draw a line AX such that ∠BAX = 45°.

3. With A as a center and radii, 3 cm draw an arc on AD.

4. now with D and B as a center and radii 6 cm and 3 cm draw arcs cutting each other at C.

5. Join DC and BC.

ABCD is the required parallelogram.
Here
∠PAB = ∠APD                ...[ Alternate angles ]
∠CPB = ∠PBA                 ...[ Alternate angles ] 

Now,
∠DPA + ∠APB + ∠CPB = 180° …… (i)

Also, considering APB,
∠PAB + ∠PBA + ∠APB = 180° …… (ii)

Therefore, from (i) and (ii)
∠APB = 90°
Hence proved.

Question 37

The perpendicular distance between the pair of opposite sides of a parallelogram are 3 cm and 4 cm, and one of its angles measures 60^o. Using a ruler and compasses only,
construct the parallelogram.

Sol:

Steps:
1. Draw a baseline AQ.

2. A take some random distance in compass and draw one are below and above the line. Now without changing the distance in compass draw one are below and above the line. These arcs intersect each other above and below the line. Draw the line passing through these intersecting points, you will get perpendicular to the line AQ.

3. Take distance of 4 cm in compass and mark an arc on the perpendicular above the line. Draw a line parallel to line AQ passing through this arc.

4. From point A measure an angle of 60 degrees and draw the line which intersects the above-drawn line at some point label it as D.

5. Using the procedure given in step 2 again draw a perpendicular to line AD.

6. Take distance of 3 cm in compass and mark an arc on the perpendicular above the line. Draw a line parallel to line AD passing through this arc which intersects the line AQ at some point label it as B and to other lines at point C.

ABCD is the required parallelogram.  

Question 38

Draw parallelogram ABCD with the following data:
AB = 6 cm, AD = 5 cm and ∠DAB = 45^o.
Let AC and DB meet in O and let E be the mid-point of BC. Join OE.
Prove that:
(i) OE // AB
(ii) OE = ${\displaystyle \frac{1}{2}}$ AB.

Sol:

To draw the parallelogram follows the steps:
1. First, draw a line AB of measure 6cm. Then draw an angle of measure 45° at point A such that ∠DAB = 45° and AD = 5cm.

2. Now draw a line CD parallel to the line AB of measure 6cm. Then join BC to construct the parallelogram as shown below: 

3. Now it is given that E is the mid-point of BC. We join OE. Now we are to prove that OE || AB and  OE = ${\displaystyle \frac{1}{2}}$ AB.

4. Since O is the mid-point of AC and E is the midpoint of BC, therefore the line is parallel to AB and  OE = ${\displaystyle \frac{1}{2}}$ AB

Question 39

Using ruler and compasses only, construct a rectangle each of whose diagonals measure 6 cm and the diagonals interest at an angle of 45^o.

Sol:

To draw the rectangle follows the steps:
1. First draw a line AC of measure 6cm.
2. Then draw the perpendicular bisector of AC  through O.
3. At O draw an angle of measure 45°. Then produce OD  of measure 3cm and OB of measure 3cm each. 
4. Now join AD, AB, BC,  and CD to form the rectangle. 

S.chand Class 8 Maths Solution Chapter 15 Visualising Solid Shapes Exercise 15

 Exercise 15

Question 1

<fig>

1. The adjoining net is made up of two equilateral triangles and three rectangles.

(i) Name the solid it represents.

(ii) How many faces, edges and vertices does this solid have ?

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

<fig>

3. Name the solids that have :

(i) 4 faces

(ii) 1 curved surface

(iii) 6 faces

(iv) 5 faces and 5 vertices

(v) 8 triangular faces

(vi) 6 rectangular faces and 2 hexagonal faces.

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

(i) a cube

(ii) a triangular prism

(iii) a triangular pyramid

(iv) a cylinder

5.Given here is the net of an octahedron with numbers 0 to 7 on the faces.Enlarge these on a cardboard to make a model

(i) How many faces, edges and vertices does this solid have ?

(ii) Check whether Euler's rule is true for the octahedron or not.

6. Draw the 2-D representation of a

<fig>

7. On an isometric dotted paper draw the following shapes:

(i) cube

(ii) cuboid

(iii) hexagonal prism

(iv) pentagonal prism

8. Name the solid that would be formed by each net :

<fig>

Multiple Choice Questions (MCQs)

9. Which of the following solid shapes has the side

<fig>

10. Manya is building prism using straws and balls of clay. How many straws does she need to build a pentagonal prism ?

(a) 18

(b) 15

(c) 20

(d) 13

High Order Thinking Skills (HOTS)

11. How many triangles will the net of hexagonal pyramid contain ?

SChand Composite Mathematics Class 7 Chapter 15 Data Handling Exercise 15D

 Exercise 15 D

Question 1

During the festive season, the sales of the various gadgets is as under : 

$\begin{array}{|c|c|c|c|c|}\hline \text { Mixer } & \text { Microwave } & \text { Toaster } & \text { DVD Player } & \text { I Pod} \\\hline 250 & 180 & 200 & 260 & 50 \\\hline\end{array}$

Represent the above information by a bar graph.

(Diagram to be added)

Question 2

The marks obtained by a student in different subjects are:

$\begin{array}{|c|c|c|c|c|c|}\hline \text { Subject } & \text { English } & \text { Maths } & \text { Science } & \text { Hindi }&\text { Social Science } \\\hline \text { Marks } & 70 & 95 & 80 & 60 & 75 \\\hline\end{array}$

(Diagram to be added)

Question 3

The approximate population (in lac) based on 2011 census of 8 Indian States:

$\begin{array}{|c|c|}\hline \text { State } & \text { Population (in lac) } \\\hline \text { Andhra Pradesh } & 850 \\\hline \text { Bihar } & 1050 \\\hline \text { Chhatisgarh } & 260 \\\hline \text { Gujarat } & 600 \\\hline \text { Madhya Pradesh } & 730 \\\hline \text { Maharashtra } & 1130 \\\hline \text { Uttar Pradesh } & 2000 \\\hline \text { West Bengal } & 920 \\\hline\end{array}$

(Diagram to be added)

Question  4

Use the bar graph below to answer the question.

(Diagram to be added)

1. How many students scored above 75 ?

(ii) What score was made by the most students ?

(iii) If the pass score was 70 , how many did not pass ?

(iv) Form a frequency table with the help of the graph and find the mean score.

(Diagram to be added)

Question  5

The marks in a spelling test are shown in this table.

(a) Draw a bar graph to represent this set of data.

(b) What mark did most pupils score ?

(c) How many pupils scored 7 or more?

$\begin{array}{|c|c|c|c|c|c|c|c|}\hline 10 & 6 & 7 & 9 & 5 & 7 & 9 & 3 \\\hline 3 & 9 & 8 & 7 & 6 & 10 & 7 & 6 \\\hline 9 & 5 & 10 & 3 & 8 & 8 & 5 & 9 \\\hline 7  & 8 & 6 & 5 & 9 & 6 & 7 & 5 \\\hline 4 . & 9 & 8 & 7 & 6 & 8 & 10 & 7 \\\hline\end{array}$

(Diagram to be added)

Question  6

$\begin{aligned}&\text { Draw a double bar graph for the given information : }\\&\begin{array}{|l|c|c|c|c|c|c|}\hline \text { Month , } & \text { Jan. } & \text { Feb. } & \text { March } & \text { April } & \text {May } & \text { June } \\\hline \text { Actual Rainfall } & 3 \mathrm{~cm} & 5 \mathrm{~cm} & 6 \mathrm{~cm} & 7\mathrm{~cm} & 8 \mathrm{~cm} & 6 \mathrm{~cm} \\\text { Predicted Rainfall } & 2 \mathrm{~cm} & 3 \mathrm{~cm} & 3 \mathrm{~cm} & 6\mathrm{~cm} & 10 \mathrm{~cm} & 9 \mathrm{~cm} \\\hline\end{array}\end{aligned}$

(Diagram to be added)

Question  7

The table shows Arun and Sanjay's test scores. The tests were marked out of 20 .

$\begin{array}{|l|c|c|c|c|c|c|c|}\hline & \text { English } & \text { Maths } & \text { Science } & \text { History } & \text {Geography } & \text { Hindi } & \text { Art } \\\hline \text { Arun } & 20 & 7 & 12 & 14 & 16 & 18 & 11 \\\text { Sanjay } & 17 & 13 & 18 & 9 & 19 & 10 & 8 \\\hline\end{array}$

(a) Draw a dual bar chart to compare their results.

(b) If anyone scored less than 10 theN had to do the test again. Which tests did (i) Sanjay (ii) Arun have to do again ?

(Diagram to be added)

Question  8

Draw a double-bar graph to compare the data in the given table :

Enrollment in Computer Courses

$\begin{aligned}&\begin{array}{|l|c|c|c|c|}\hline \text { Year/Course } & \text { Basic } & \text { Logo } & \begin{array}{c}\text { Computer } \\\text { Science }\end{array} & \begin{array}{c}\text { Using } \\\text { Software }\end{array} \\\hline \text { Year One } & 130 & 40 & 10 & 20 \\\text { Year Two } & 260 r & 110 & 30 & 90 \\\hline\end{array}\\&\text { Which scale will you prefer to use ? }\end{aligned}$

(Diagram to be added)

Question  9

The performance of a student in 1st term and 2 nd term is given. Draw a double bar graph choosing appropriate scale and answer the following :

$\begin{array}{|l|c|c|c|c|c|}\hline \text { Subject } & \text { English } & \text { Hindi } & \text { Maths } & \text { Science } &\text { S. Science } \\\hline \text { Ist term (M.M. 100) } & 67 & 72 & 88 & 81 & 73 \\\text { 2nd term (M.M. 100) } & 70^{\circ} & 65 & 95 & 85 & 75 \\\hline\end{array}$

(a) In which subject has the child improved his Performance  the most, ?

(b) In which subject is the improvement the least?

(c) Has the performance gone down in any subject

(Diagram to be added)

Question  9

The following graph shows the points scored in five games of basketball. Study the graphs and answer the questions that follow.

(i) How many games did our team win ?

(ii) What is the highest score ? Who scored and in which game ?

(iii) By how many points did our team win the third game ?

(iv) What is the opponent's team highest score ?

(Diagram to be added)

SChand Composite Mathematics Class 7 Chapter 15 Data Handling Exercise 15C

  Exercise 15 C

Question 1

Determine the mode for each of the following sets of data : 

(a) 5,3,5,2,3,5,1,7 

Ans: Mode 5

(b) 1,7,13,19,25

Ans: No mode 

(c) 4,6,8,4,10,4,6,12,6,10

Ans: 4 and 6 

(d) 51,68,70,76,76,55,70,73,68,70,81

Ans: 70

(e) 8,4,8,4,8,4,8,4,8,4,8,4

Ans: 4 and 8 

(f) 29,29,10,10,3,3,14,10,2,2

Ans: 10 

(g) 26,32,26,21,83,26,83,85

Ans: 26 and 83

Question 2

Find the mode of the following distributions: 

(a) 

$\begin{array}{|l|l|l|l|l|l|l|}\hline \text { Size } & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { fre } & 4 & 6 & 2 & 5 & 3 & 1 \\\hline\end{array}$

Ans: Mode :3

(b) 

$\begin{array}{|l|l|l|l|l|l|}\hline \begin{array}{l}\text { Size of } \\\text { shoes }\end{array} & 6 & 7 & 8 & 9 & 10 \\\hline \begin{array}{l}\text { No of } \\\text { person }\end{array} & 12 & 20 & 40 & 15 & 7 \\\hline\end{array}$

Ans: Mode : 8 

Question 3

Determine (a) the mode (b) the median , and (c) the mean for each of the following sets of data: 

(i) $1,16,11,6,11,8,14,11,21$

Sol: mode = 11

1,6,8,11,11,14,16,21 

Median= 11

Mean $\bar{x}=\frac{Sum of observation}{No of observation}$

$=\frac{1+6+8+11+11+11+14+16+21}{9}$= $\frac{99}{9}=11$

(ii) $3,10,10,12,14,16,16,18,18,25$

Mode = 10,16 and 18 

Sol: 3,10,10,12,14,16,18,18,25 

Median = $\frac{14+16}{2}$

=$\frac{30}{2}=15$

Mean $\bar{x}=\frac{3+10+10+12+14+16+16+18+18+25}{10}$

=$\frac{142}{10}$ = 14.2

(iii) $23,2,42,6,36,11,29,9,15$

Mode = No mode

2,6,9,11,15,23,29,36,42

Median = 15

Mean

 $\begin{aligned} \bar{x} &=\frac{2+6,+9+11+15+23+29+36+42}{9} \\ &=\frac{173}{9}=19 \frac{2}{9} \end{aligned}$

(iv) $8,8,8,8,18,18,18,18$

Sol: Mode = 8 and 18 

8,8,8,8,18,18,18,18

Median = $\frac{8+18}{2}$

$=\frac{26}{2}=13$

$\operatorname{mean} \vec{x} =\frac{8,8,8,8,18,18,18,18}{8}$

$=\frac{104}{8}=13$

Question 4

Find the mode, and the mean for the set of data below.

Number of fiction books read by 28 eight grade pupils: 

$4,6,1,7,3,9,5,7,8,5,4,6,10,6,9,5,6,6,8,6,8,5,10,7,2,5,3,7$.

Sol: Mode = 6

Mean = $=\frac{4+6+1+7+3+9+5+7+8+5+4+6+10+6+9+5+6+6+8+6+8+5+10+7+2+5+3+7}{28}$

$=\frac{168}{28}$ = 6

Question 5

Find the mode and the mean of the following set showing number of hours of operating life of 25 flashlight batteries: 

$20,21,19,22,18,23,25,22,23,20,23,20,22,21,24,21,22,23,19,21,22,22,24,26,22 .$

Sol: Mode = 22 

Mean = $\frac{545}{25}=21.8$

Question 6

Tell whether the statement is true or false: 

(i) The mode is always one of the numbers in a data.- True 

(ii) The mean is one of the numbers in a data - False 

(iii) The median is always one of the numbers in a data- False 

SChand Composite Mathematics Class 7 Chapter 15 Data Handling Exercise 15B

 Exercise 15 B 

Question 1

The median of a set of numbers is the middle number when all the numbers are arranged in order of size, i.e., in descending or ascending order.

Find the median of the following. 

(1) 2,3,5,7,9 = 5

(2) 4,8,12,16,20,24, 28,32 

=16,20

median $=\frac{16+20}{2}=\frac{36}{2}=18$

(3) $60,33,63,61,44,48,51$ 

Sol: $33,44,48,51,60,61,63$

(4) $13,22,25,8,11,19,17,31,16,10$

Sol: $8,10,11,13,16,17,19,22,25,31$

median $=\frac{16+17}{2}=\frac{33}{2}=16.5$

Question 5

15 students secured the following marks in a test statistic . find the median marks 

35,2,13,17,20,30,19,29,11,10,29,23,18,25,17

Sol: Arrange in Ascending order. 

$10,11,13,17,17,18,19,20,23,25,28,29,29,30,35$

Question 6

The median of the number 85, 86,78,89 and 64 is 

Sol: $64,78,85,86,89$

Question 7

The marks scored by 10 students are 5,9,8,7,2,3,4,9, 6 and 8 . The median marks are '

Sol: $2,3,4,5,6,7,8,8,9,9$

Median $=\frac{6+7}{2}=\frac{13}{2}=6.5$

Question 8

If the median of 46, 64,88, 40 , x ,76,35,91,56,32 and 91 is 58 . find the value of x 

Sol: 32,35,40,46,56 , 64,7, 88, 91 ,91, 

x= 58 answer 

SChand Composite Mathematics Class 7 Chapter 15 Data Handling Exercise 15A

  Exercise 15 A

Question 1 

Find the mean of : 

(a) First 5 natural number 

Sol: $\bar{x}=\frac{1+2+3+4+5}{5}=\frac{15}{5}=3$

(b) First 7 whole numbers

Sol: $\bar{x}=\frac{0+1+2+3+4+5+6}{7}=\frac{21}{7}=3$

(c) First 4 prime number 

Sol: $\bar{x}=\frac{2+3+5+7}{4}=\frac{17}{4}=4.25$

(d) Rs7, Rs 19 , Rs 31, Rs 43, Rs 70

Sol: $\bar{x}=\frac{7+19+31+43+70}{5}=\frac{170}{5}=Rs 34$

Question 2

Seema obtained the following scores (Out of 100) $80,85,90,71,60,100$ . What is her mean score? 

Sol: $\bar{x}=\frac{80+85+90+71+60+100}{6}=\frac{486}{6}=81$

Question 3

Shaleen's last six batting scores were 138, 144, 155 ,142, 167,172. What was his mean score? 

Sol: $\bar{x}=\frac{138+144+155+142+167+172}{6}$

$\bar{x}=\frac{918}{6} =\bar{x}=153$

Question 4

Madhu worked $2 \frac{1}{2}$ hours on Monday, $3 \frac{1}{4}$ hrs. on Tuesday, and $2 \frac{3}{4}$ hrs. on Wednesday. What is the mean number of hours she worked on these three days?

Sol: Total number of hours worked = $2 \frac{1}{2}+3 \frac{1}{4}+2 \frac{3}{4}$

$=\frac{5}{2}+\frac{13}{4}+\frac{11}{4}$

$=\frac{10+13+11}{4}=\frac{34}{4}$

Mean $\bar{x}=\frac{\text { Sun of observation}}{\text { Number of observation }}$

$=\frac{\frac{34}{4}}{3}=\frac{34}{12}=\frac{17}{6}=2 \frac{5}{6}$ hours

Question 5

Ayushree sat for six tests and ananya sat for seven tests. Their percentage scores were: 

$\begin{array}{|l|c|c|c|c|c|c|c|}\hline \text { Ayushree } & 68 & 75 & 70 & 45 & 57 & 77 & \\\hline \text { Ananya } & 52 & 87 & 64 & 531 & 74 & 81 & 86 \\\hline\end{array}$

Sol: Mean of ayushree = $\frac{68+75+70+45+57+77}{6}$

$=\frac{392}{6}=65.33$

Mean of ananya = $\frac{52+87+64+53+74+81+86}{7}$

$=\frac{497}{7}=71$

Ananya > Ayushree

Question 6

Find the mean of first ten odd natural numbers 

Sol: $=\frac{1+3+5+7+9+11+13+15+17+19}{10}$

$=\frac{100}{10}=10$

Question 7

If the mean of 16, 14, x ,23 ,20  is 18 . find the value  of x 

Sol: $18=\frac{16+14+x+23+20}{5}$

$90=x+73 \Rightarrow x=90-73$

$x=17$ Ans 

Question 8

If the mean of x, x+ 2, x+4 ,x+6 , x+8 is 24, find x 

Sol: $24=\frac{x+x+2+x+4+x+6+x+8}{5}$

$120=5 x+20 \Rightarrow 120-20=5 x$

$\Rightarrow 100=5 x$

$\Rightarrow x=\frac{100}{5} \Rightarrow x=20$ answer

Question 9

Nisha secured 73,86,78 and 75 marks in four tests. What is the least number of marks she can secure in her next test, if she has to have a mean score of 80 marks in five test? 

Sol: Let the number =x 

$80=\frac{73+86+78+75+x}{5}$

$400=312+x=x=400-312$

$x=88$ marks

Question 10

The mean of 6 numbers is 24. If one number is excluded, the mean of remaining numbers becomes 22 find the excluded number . 

Sol: Mean = $\frac{Sum of observation}{No. of Observation }$

$24=\frac{5.0}{6} \Rightarrow 5.0 \cdot=144$

22 = $\frac{Sum of obs of 5 No}{No. of obs}$

$22=\frac{5.0}{5}$

$110=5.0$

Excluded No. = 144-110= 34 answer 

S Chand Class 10 CHAPTER 15 Three Dimensional Solids Exercise 15 C

 Exercise 15 C

Question 1

Ans: (i) Given, 

Radius = 3cm

Height = 4cm

Slant height (l) =$\sqrt{r^{2}+h^{2}}$

$=\sqrt{(3)^{2}+(4)^{2}}$

$=\sqrt{3+16}$

$=\sqrt{25}$

$=5 \mathrm{~cm}$

Curved surface = $\pi r l$

$=\pi \times 3 \times 5$

$=15 \pi \mathrm{cm}^{2}$

Area of base = $\pi r^{2}$

$\pi \times 3 \times 3$

$=9 \pi \mathrm{cm}^{2}$

Total surface area = $\pi r l+\pi r^{2}$

$=15 \pi+9 \pi$

$=24 \pi \mathrm{cm}^{2}$

Volume  =  $=\frac{1}{3} \pi r^{2} h$

$-\frac{1}{3} \pi \times 3 \times 3 \times 4 $

$=12 \pi \mathrm{cm}^{3}$

(ii) Given, 

Radius = 20cm

Slant height = 25 cm

$l^{2}=r^{2}+1^{2}$

$\left(25^{2}-r^{2}=h^{2}\right.$

$\sqrt{625}-(20)^{2}=1^{2}$

$\sqrt{225}=h$

$15=h$

$h=15 \mathrm{~cm}$

Curved surface = πrl

$=\pi \times 20 \times 25$

$=500 \pi \mathrm{cm}^{2}$

area of base =  π$r^{2}$

$=\pi \times 200 \times 20$

$=400 \pi \mathrm{cm}^{2} .$

Total surface area = $\pi r l+\pi r^{2}$

$\begin{aligned}=& 500 \pi+400 \pi \\=& 900 \pi \mathrm{cm}^{2} \end{aligned}$

Volume $=\frac{1}{3} \pi{r}^{2} h$

$=2000 \pi \mathrm{cm}^{3}$

(iii) Given, 

Height = 18cm

Slant height = 30 cm

$\therefore \quad l^{2}=h^{2}+r^{2} .$

$(30)^{2}=(18)^{2}+r^{2}$

$900-324=r^{2} .$

$\sqrt{576}=r$

$24=r$

Curved surface = πrl 

$=\pi \times 24 \times 30$

$=720 π c m^{2}$

Area of base = $\pi r^{2}$

$=\pi \times 24 \times 24$

$=576 \pi \mathrm{m}^{2}$

Total surface area= $\pi r l+\pi r^{2}$

729π + 576π

$=1296 \mathrm π{cm}^{2}$

Volume = $\frac{1}{3} \pi r^{2} h$

$=576 \times 6 \pi .$

$=3456 \pi \mathrm{cm}^{3}$

(iv) Given, 

Radius = 27cm

Height = 36cm

$\begin{aligned} \because \quad l &=\sqrt{h^{2}+r^{2}} \\ &=\sqrt{(36)^{2}+(27)^{2}} \\ &=\sqrt{1296+729} \\ &=\sqrt{2025} \\ &=45 \mathrm{~cm}\end{aligned}$

Curved surface =  $\pi$ rl

= $\pi$ $\times 27 \times 45$

$=1215 \pi \mathrm{cm}^{2} .$

Area of base =  $\pi r^{2}$ 

$=\pi \times 27 \times 27$

$=729 \pi \mathrm{cm}^{2}$

Total surface area = $\pi r l+\pi r^{2}$

$=1215 \pi+729 \pi$

$=1944 \pi \mathrm{cm}^{2}$

Volume = $\frac{1}{3}$ $\pi r l+\pi r^{2}h$

$=8748 \pi \mathrm{cm}^{3}$

(v) Given, 

Radius = 5cm,

Curved surface = 65π $\left(cm^{2}\right)$

  

∴ Curved surface =πrl 

65π = π $5\times l$

$\frac{65} π { π  \times 5}=l$

$l=13 \mathrm{~cm}$

$\because \quad l^{2}=h^{2}+r  ^{2}$

$(13)^{2}=h^{2}+(5)^{2}$

$16 y-25=h^{2}$

$144=h^{2}$

$\sqrt{144}=h$

$12=h$

$h=12 \mathrm{~cm} .$

Area of base = $πr^{2}$

$=\pi \times 5 \times 5$

$=25 \pi \mathrm{cm}^{2}$

Total surface area = $\pi r l+\pi r^{2}$

$=65 \pi+25 \pi$

$=90 \pi \mathrm{m}^{2} .$

Volume $=\frac{1}{3}\pi r^{2} h$

$=\frac{1}{3} \pi \times(5)^{2} \times 12$

$=25 \times 4 \pi$

$=100 \pi \mathrm{cm}^{3}$

(vi) Given,

Radius = 35cm

Total surface area  = $13860 \mathrm{~cm}^{2}$

Total surface area = πrl+π $r^{2}$

$13860:=\pi r(l+r) .$

$4410 \pi=\pi \times 35(l+35) .$

$4410 \pi=35 \pi(l+75)$

$126=l+35$

$126-35=l$

$91=l$

$\ell=91 \mathrm{~cm} .$

$\begin{aligned} \therefore l^{2} &=h^{2}+r^{2} \\(91)^{2} &=h^{2}+(35)^{2} \end{aligned}$

$8281-1225=h^{2}$

$\sqrt{8281-1225}=h$

$\sqrt{7056}=h$

$84=h$

$h=84 \mathrm{~cm}$

Curved surface = $\pi rl$

$=\pi \times 35 \times 91$

$=3185 \pi \mathrm{cm}^{2}$

Area of base $=\pi r^{2}$

$=\pi \times 35 \times 35$

$=1225 \pi c m^{2}$

Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \pi \times 35 \times 35 \times 84$

$=34300 \pi \mathrm{cm}^{3}$

Question 2

Ans: (i) Given, 

Height = 8m, 

Area of base = $156 \mathrm{~m}^{2}$

$\begin{aligned} \therefore \text { Area of base } &=\pi r^{2} \\ 156 &=\frac{22}{7} \times \r^{2} \\ \frac{156 \times 7}{22} &=r^{2} \end{aligned}$

$\frac{546}{11}=r^{2}$

$r^{2}=\frac{546}{11} \mathrm{~m}$

ஃ Volume = $\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{546}{11} \times 8$

$=2 \times 26 \times 8$

$=416 \mathrm{~m}^{3}$

(ii) Given,

Slant height (l)= 17cm,

Radius (r)= 8cm

$\therefore l^{2}=h^{2}+r^{2}$

$l^{2}-r^{2}=h^{2} .$

$(17)^{2}-(8)^{2}=h^{2}$

$\sqrt{289-64}=h .$

$\sqrt{225}=h$

$15=h$

$h=15 \mathrm{~cm}$

$\therefore$ Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 8 \times 8 \times 15$

$=176 \times 40$

$=\frac{7.040}{7}$

$=1005.71 \mathrm{~cm}^{3}$

(iii) Given, 

Height = 8cm,

Slant length = 10cm,

$\therefore \quad l^{2} = h^{2}+r^{2}$

$l^{2}-h^{2}=r^{2}$

$(10)^{2}-(8)^{2}=r^{2}$

$100-64=r^{2}$

$\sqrt{36}=r$

$6=r$

$r=6 \mathrm{~cm}$

$\therefore$ Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 8$

$=\frac{44 \times 48}{7}$

=301.71$cm^{3}$

(iv) Given, 

Height=5cm 

Perimeter of base = 8cm

Perimeter of base= $2 \pi r$

$8=2 \times \frac{22}{7} \times r$

$\frac{{8} \times 7}{2 \times 22}=6$

$\frac{14}{11}=r$

$r=\frac{14}{11} \mathrm{~cm}$

∴ Volume = $\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{14}{11} \times \frac{14}{11} \times 5$

$=\frac{4 \times 70}{33}$

$=\frac{280}{33}$

$=8.48 . \mathrm{cm}^{3}$

Question 3

Ans: (i) Given, 

Height = 8m,

Slant height = 10m

$\therefore l^{2}=h^{2}+r^{2}$

$l^{2}-h^{2}=r^{2} .$

$(10)^{2}-(8)^{2}=r^{2}$

$\sqrt{100-64}=r$

$\sqrt{36}=r$

$6=r$

r=6 m

∴ Curved surface area = πrl

$\begin{aligned} &=\frac{22}{7} \times 6 \times 10 \\=& \frac{1320}{7} \end{aligned}$

=199.6 $m^{2}$

(ii) Given, 

Perimeter of base = 88cm,

slant height = 2dm = $2 \times 10cm=20cm

∴ Perimeter of base $=2 \pi r$

88= $=2 \times \frac{22}{7} \times r$

$\frac{88 \times 7}{2 \times 22}=r$

r=14cm

Curved surface area $=\pi r l$

$\begin{aligned} & \frac{22}{7} \times 14 \times 20 \\=& 44 \times 20 \\=& 880 \mathrm{~cm}^{2} \end{aligned}$

(iii) Given,

Area of base= $154 \mathrm{~cm}^{2}$

Height 24cm

Area of base= $\pi r^{2}$

$154=\frac{22}{7} \times r^{2}$

$\frac{154 \times 22}{7}=r^{2}$

$49=r^{2}$

$\sqrt{49}=r$

$7=r$

$r=7 \mathrm{~cm}$

$\therefore \ell=\sqrt{h^{2}+r^{2}}$

l=$\sqrt{(24)^{2}+(7)^{2}}$

$l=\sqrt{576+49}$

$l=\sqrt{625}$

$l=25 \mathrm{~cm}$

So 

Curved surface area = π rl

$=22 \times 25$

$=550 \mathrm{~cm}^{2}$

Question 4

Ans: Given,

Radius = 5cm

Volume = 50π $\mathrm{Cm}^{3}$

Volume= $\frac{1}{3}π r^{2} h$

$50 \pi=\frac{1}{3} \pi \times 5 \times 5 \times \mathrm{h}$

$50 \pi=\frac{\pi}{3} \times 25 \mathrm{~h}$

$\frac{50 \times \pi \times 3}{\pi \times 25}=h$

$\frac{150}{25}=h$

$6=h$

$h=6 \mathrm{~cm}$

Hence, the height of the cone is 6cm

Question 5

Ans: Given,

Radius = 11.3cm

Curved surface area = $710(cm)^{2}$

Curved surface area $=\pi rl$

$710=\frac{355}{113} \times \frac{11.3}{10} \times l$

$710=\frac{355}{10} l$

$\frac{710 \times 10}{355}=l$

$\frac{7100}{355}=l$

$20=l$

$l=20 \mathrm{~cm}$

Hence , slant height is 20cm

Question 6

Ans:  Let the radius be r and height be h

Volume =$\frac{1}{3} \pi r^{2} h$

According to question 

Radius =$\frac{r}{2}$

and height= h

Volume = $\frac{1}{3} \pi\left(\frac{8}{2}\right)^{2} h$

$=\frac{1}{3} \pi \frac{r^{2}}{4} h=$ $\frac{\pi^{2} h}{12}$

∴ Ratio = $\frac{\pi r^{2} h}{12}=\frac{1}{3} \pi r^{2} h$

$=\frac{\frac{\pi r^{2} h}{12}}{\frac{1 \pi x^{2} h}{3}}$

$=\frac{1}{4}=1: 4 .$

Question 7

Ans: Given 

Curved surface area = $264 m^{2}$

Slant height = 12m

Curved surface area = πrl

$264=\frac{22}{7} \times r \times 12$

$\frac{264 \times 7}{22 \times 12}=r$

$7=r$

$r=7 \mathrm{~m}$

$\begin{aligned} \therefore l^{2} &=h^{2}+r^{2} \\ l^{2}-r^{2} &=h^{2} \end{aligned}$

Question 8

Ans: According to question, 

Height (h) = $2 \times$ diameter 

Diameter = $\frac{\text { height }}{2}$

$\therefore \quad \operatorname{Radius}=\frac{h}{2} \times \frac{1}{2}=\frac{h}{4}$.

Volume $=36 \pi \mathrm{cm}^{3}$

$\frac{1}{3} \pi r^{2} h=36 \pi$

$\frac{1}{3} \times \pi \times \frac{h}{4} \times \frac{h}{4} \times \pi=36 \pi .$

$\frac{h^{3}}{48} \pi=36 \pi$

$h^{3}=\sqrt{728}$

$h=\sqrt[3]{1728}=\sqrt{12 \times 12 \times 12}=12 . \mathrm{cm}$

Question 9

Ans: Given,

Radius and height of cone are in ratio = 3:4

Let radius be 3x 

and height be 4x 

Volume = 301.44 $(cm)^{3}$

$\because \quad$ Volume =\frac{1}{3} \pi r^{2} h$

$301.44=\frac{1}{3} \times 3.14$  $\times(3x)^{2}$  $\times$(4x)$

$301.44=3.14 \times 12 x^{3}$

$\frac{301.44}{3.14 \times 12}=x^{3}$

$\frac{30144}{3768}=x^{3}$

$8 =x^{3}$

$x^{3}=8$

$x=\sqrt[3]{8}$

$x=\sqrt{2 \times 2 \times 2}$

$x=2$

$\begin{aligned} \therefore \text { Radius } &=3 \times \\ &=3 \times 2 \\ &=6 \mathrm{~cm} . \end{aligned}$

and Hight $=4 x$

$=4 \times 2$

=8

$\begin{aligned} \therefore l^{2} &=h^{2}+r^{2} \\ l &=\sqrt{(8)^{2}+(6)^{2}} \\ l &=\sqrt{64+36} \\ l &=\sqrt{100} \\ l &=10 \mathrm{~cm} \end{aligned}$

Question 10

Ans: Given ,

The ratio of radius and slant height of cone = 4:7

Curved surface area $=792 \mathrm{~cm}^{2}$

Let radius be 4x and 

slant height be 7x

Curved surface area = πrl

$792=\frac{22}{7} \times 4 x \times 7 x$

$792=22 \times 4 x^{2}$

$792=88 x^{2}$

$\frac{792}{88}=x^{2}$

$9=x^{2}$

$x^{2}=9$

$x=\sqrt{9}$

$x=3$

Radius =4x

$=4 \times 3$

$=12 \mathrm{~cm}$

Question 11

Ans: Given, 

The ratio of radii of two cones = 3:5

Let r1 be 3x and r2 be 5x

Volume of first cone = $\frac{1}{3} \pi r_{1}^{2} h$

$=\frac{1}{3}π  (3x)^{2}\times h$

= $3 \pi x^{2} h .$

Volume of second cone =  $\frac{1}{3} \pi r_{2}^{2} h$

$\begin{aligned} &=\frac{1}{3} \pi \times (5 x)^{2} x h \\=& \frac{1}{3} \pi 25 x^{2} h \\=& \frac{25 \pi x^{2} h}{3} \end{aligned}$

ration of their volumes 

$=3 \pi x^{2}h \frac{25 \pi x^{2} 1}{3}$

$=\frac{3 \pi x^{2} h}{\frac{25 \pi x^{2} h}{3}}$

$=\frac{9}{25}$

$=9: 25$

Question 12

Ans: Given, 

Circumference = 44m

Height = 10m

Circumference = 2πr

$44=2 \times \frac{22}{7} \times r$

$\frac{44 \times 7}{44}=r$

$7=r$,

$\therefore l^{2}=h^{2}+r^{2} .$

$l=\sqrt{(10)^{2}+(7)^{2}}$

$l=\sqrt{100+49}$

$l=\sqrt{149 \mathrm{~m} .}$

∴ Curved surface area = πrl

$=\frac{22}{7} \times 7 \times \sqrt{149}$

$\begin{aligned} &=22 \sqrt{149}\\=& 22 \times 12.2 . \\=& 268.4 \mathrm{~m}^{2} . \end{aligned}$

Width of canvas(d)=  2cm= $\frac{2}{100} \mathrm{~m}$

$\therefore$ Area $=l \times b$

$268.4=l \times \frac{2}{100}$

$\frac{26840}{2}=l$

=13429=l

Hence, the length of canvas is 13420m

Question 13

Ans: Given, 

Radius =7m

Height = 24m

$\because$ Slant height $(l)=\sqrt{h^{2}+r^{2}}$

$l=\sqrt{(24)^{2}+(7)^{2}}$

$l=\sqrt{576+49}$

$l=\sqrt{625}$

$l=\sqrt{2.5 \times 25}$

$l=25 \mathrm{~m}$

Curved surface area =  πrl

$=\frac{22}{7} \times 7 \times 25$

$=22 \times 25$

$=550 \mathrm{~m}^{2}$

Width of canvas (b) = 5m

Area = L $\times$ B

$550=L \times 3$

$\frac{550}{5}=L$

$110=L$

$L=110 \mathrm{~m}$

$\therefore$ Length $=110 \mathrm{~m}$.

Question 14

Ans: Given,

Volume =  $1232 \mathrm{~m}^{3} .$

Area of base = $154 \mathrm{~m}^{2}$

Area of base = $\pi r^{2}$

$154=$ $\frac{22}{7} \times r^{2}$

$\frac{154 \times 7}{22}=r^{2}$

$49=r^{2}$

$\sqrt{49}=r$

$7=r$

$r=7 \mathrm{~m}$

Volume $=\frac{1}{3} \pi r^{2} h$

$1232=$ $\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times h$

$\frac{1232 \times 3}{22 \times 7}=h $

$8 \times 3=h$

$24=h$

$h=24 m$

$l^{2}=h^{2}+x^{2}$

$l=\sqrt{(24)^{2}+(7)^{2}}$

$l=\sqrt{576+49}$

$l=\sqrt{625}$

$l=\sqrt{25 \times 25}$

$l=25 \mathrm{~m}$

ஃ Curved surface area = πrl

$=\frac{22}{7} \times 7 \times 25$

$=22 \times 25$

$=550 \mathrm{~m}^{2}$

∴ The area of canvas $=550 \mathrm{~m}^{2}$

Question 15

Ans: Given, 

In the tent, accommodate is available for 11 persons and each person must have $4r^{2}$ of space on the ground 

So , aera of base of the tent = $11 \times{4}$

$44 m^{2}$

Air is required for each person to breadth = $20m^{3}$

Volume of air = $20 \times 11$

$=220 \mathrm{~m}^{3}$

Area of base = π$r^{2}$

$44=\frac{22}{7} \times r^{2}$

$\frac{44 \times 7}{22}=r^{2}$

$r^{2}=14 \mathrm{~m} .$

Volume = $=\frac{1}{3} \pi{r}^{2} h$

$220=\frac{1}{3}\times $\frac{22}{7}\times 14 \times h$

$220=\frac{44h}{3}$

$\frac{220 \times 3}{44}=h$

$15=h$

h=15 m

Hence , the height of the cone = 15m

Question 16

Ans: False , because the volume of a cone is one third $\left(\frac{1}{3}\right)$ of the volume of cylinder of the same radius and height 

Question 17

Ans: Given,

Height of cylinder = 9cm

and radius= $\frac{40}{2} cm=20 c m$

Height of cone = 108cm

∴ Volume of cylinder = $\pi r^{2} h$

$=\frac{22}{7} \times 20 \times 20 \times 9$

$=\frac{440 \times 180}{7}$

$=\frac{79200}{7} \mathrm{~cm}^{2} .$

ஃ Volume of cone $\frac{79200}{7}$ (Given volume of cylinder is equal to volume of cone)

Volume of cone = $=\frac{1}{3} \pi r^{2}$h

$\frac{79200}{7}=\frac{1}{3} \times \frac{22}{7} \times r^{2} \times 108$

$\frac{79200 \times 7 \times 3}{7 \times 22 \times 108}=r^{2}$

$\frac{79200 \times 21}{7 \times 22 \times 108}=r^{2}$

$\frac{79200}{792}=r^{2}$

$100=r^{2}$

$r^{2}=100$

$r=\sqrt{100}$

$r=10$

Hence, the radius of cone is 10cm

Question 18

Ans: Given, 

Radius, of cone and cylinder (r) = 7m,

Height of the cylinder (h1)=8m

And height of the conical of (h2)= 4m

(IMAGE TO BE ADDED)

∴ l =$\sqrt{h2^{2}+r^{2}}$

$=\sqrt{(4)^{2}+(7)^{2}}$ $=\sqrt{16+49}$ $=8.06M$

∴ Area of canvas = Curved surface area of cylinder + Curved surface area of cone 

= 2πrh_{1}$ + πrl

$=2 \times \frac{22}{7} \times 7 \times 8+\frac{22}{7} \times 7 \times 8.06 .$

$=352+177.32 .$

$=529.32 . \mathrm{m}^{2}$

Question 19

Ans: Given, 

Height of cylinder = 3m 

Radius of cylinder = $\frac{105}{2} \mathrm{~m}$

and slant height = 53m

Total area of the canvas = Curved surface area of cylinder +curved surface area of cone 

= 2πrh+πrl
$=22 \times \frac{22}{7} \times $3+\frac{2 x}{7}\times\frac{105}{2} \times 53$

$=22 \times 45+11 \times 795 .$

$=990+8745$

$=9735 \mathrm{~m}^{2}$

Question 20

Ans: Given, 

Edge of a cube = 9cm

Diameter of cone = 9cm (because equal to edge of cube)

Then radius = $\frac{9}{2} cm$

and height of cone = 9cm (because equal to edge of cube)

Volume of largest cone = $\frac{1}{3}πr^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{9}{2} \times \frac{9}{2} \times 9$

$=\frac{99 \times 27}{14}$

$=\frac{2673}{14}$

$=190.93 . \mathrm{cm}^{3}$

Question 21

Ans:  Given,

Height of cylinder(h1) = 3m, 

Total height of tent = 13.4m

Height of conical part (h2) = 13.5-3

=10.5m

(IMAGE TO BE ADDED)

Radius = 14 m

$l=\sqrt{h_{2}^{2}+r^{2}}$

$l=\sqrt{(10.5)^{2}+(14)^{2}}$

$l=\sqrt{110.25+196}$

$l=\sqrt{306.25}$

$l=17.5 \mathrm{~m}$

Question 22

Ans:  (IMAGE TO BE ADDED)

Given, 

Height of the cylinder (h1)= 32cm

and radius (r1) = 18cm

Height of the conical= 24 cm

Volume of sand in it = π$r_{1}^{2}$ h

$=\frac{22}{7} \times 18 \times 18 \times 32$

$=\frac{396 \times 576}{7}$

$=\frac{228096 \mathrm{cm}^{3}}{7}$

(IMAGE TO BE ADDED)

So volume of conical heap of the sand = $\frac{228096 \mathrm{~cm}^{3}}{7}$

Volume of conical heap = $\frac{1}{3} \pi r^{2} h$

$\frac{228096}{7}=$ $\frac{1}{3} \times \frac{22}{7} \times r^{2} \times 24$

$\frac{22.8096}{7}=$  $\frac{176}{7} r^{2}$

$\frac{228096 \times 7}{7 \times 176}$ $=r^{2}$

$\frac{228096}{176}=r^{2}$

$1296=r^{2} .$

$r^{2}=1296$

$r=\sqrt{1296}$

$r=36 \mathrm{~cm}$

(i) Radius of cone = 36cm

(ii) $l=\sqrt{h^{2}+r^{2}}$

$l=\sqrt{(24)^{2}+(36)^{2}}$

$l=\sqrt{576+1296}$

$l=\sqrt{1872}$

l=43.3cm

Height the slant height of heap is 43.3cm

Question 23

Ans:  (IMAGE TO BE ADDED)

Given, 

Height cylinder = 8cm

Radius of cylinder = 6cm

Volume of cylinder = $\pi r^{2} h$

$=31416 \times 6 \times 6 \times 8$

$=18.8496 \times 48$

$=904.7808 \mathrm{~cm}^{3}$

Radius of cone = 6cm 

Height of cone = 8cm

Volume of cone = $\frac{1}{3} \pi r^{2}h$

$=62832 \times 48$

$=301.5936 \mathrm{~cm}^{3}$

 

∴ Volume of remaining solid = Volume of cylinder - Volume of cone 

$=904.7808-301.5936$

$=603.1872 \mathrm{~cm}^{3}$

Question 24

Ans: (IMAGE TO BE ADDED)

Given,

Radius , of cylinder = 3cm 

and height of cylinder = 5cm 

Volume of cylinder = $\pi r^{2} h$

$=\frac{21}{7} \times 3 \times 3 \times 5$

$=\frac{990}{7} \mathrm{~cm}^{2}$

Radius of conical portion = $\frac{3}{2} c m$

and Height of conical portion = $\frac{8}{9} c h$

Volume of conical portion = $\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times \frac{8}{9}$

$=\frac{44}{21} \mathrm{~cm}^{3}$

Metal in the remaining part = Volume of cylinder - Volume of conical portion 

$=\frac{990}{7}-\frac{44}{21}$

$=\frac{2970-44}{21}$

$=\frac{2926}{21}$

According to the question 

$\frac{2926}{21}: \frac{44}{21}$

$\frac{\frac{2926}{21}}{\frac{4 4}{21}}$

$=\frac{2926 \times 21}{21 \times 44}$

$\frac{1463}{22}=$

$=133: 2$

Question 25

Ans:   (IMAGE TO BE ADDED)

Given, 

Radius of cylinder = $\frac{7}{2} \mathrm{~cm}$

Its height = 8cm

Volume of cylinder = $\pi r^{2} h$

$\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 8$

$=22 \times 14$

$=308 \mathrm{~cm}^{3} .$

Radius of cone $=\frac{7}{4} \mathrm{~cm}$

Its height = 8cm

Volume of cone = $\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7}\times \frac{7}{4} \times \frac{7}{4} \times 8$

$=\frac{77}{3} \mathrm{~cm}^{3}$

Volume of water required to fill the vessel= Volume of cylinder -Volume of cone 

$=308-\frac{77}{3}$

$=\frac{924-77}{3 .}$

$=\frac{847}{3}$

$=28 \frac{1}{3} \mathrm{~cm}^{3}$

According to question,

Height of cone = $1 \frac{3}{4}=\frac{7}{7} \mathrm{Cm} x$

its radius = 2cm

Volume of cone =  $\frac{1}{3} \pi r^{2} h$

=$\frac{22}{3} \mathrm{~cm}^{3}$

Change in volume of cones 

$=\frac{77}{3}-\frac{22}{3}$

$=\frac{77-22}{3}$

$=\frac{55}{3} \mathrm{~cm}^{3}$

Let the drop in water level be h cm

Volume = π$r^{2} h$

$\frac{55}{3}=$ $\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times h $

$\frac{55 \times 2}{3 \times 11 \times 7}=h$

$\frac{10}{21}=h $

$h=\frac{10}{21} \mathrm{~cm}$

S Chand Class 10 CHAPTER 15 Three Dimensional Solids Exercise 15 B

  Exercise 15 B

Question 1 

Ans: (i) Given,

r= 7cm, 'h=8cm

∴ Volume = $\pi r^{2} h$

$=\frac{22}{7} \times 7 \times 7+8$

$=22 \times 56$

$=1232 \mathrm{~cm}^{3}$

(ii) Given,

$\begin{aligned}&r=7 \mathrm{~cm} \\&h=12 \mathrm{~cm} .\end{aligned}$

$\therefore$ Volume $=\pi r^{2} h$

$\begin{aligned} &=\frac{22}{7} \times 7 \times 7 \times 12 \\ &=22 \times 84 \\ &=1848 \mathrm{~cm}^{3} \end{aligned}$

(iii) Given,

$r=14 \mathrm{~cm}$

$\mathrm{h}=16 \mathrm{~cm} .$

$\therefore \quad$ Volume $=\pi r^{2} h$

$-\frac{22}{7} \times 14 \times 14 \times 6$

$=44 \times 224$

$=9856 \mathrm{~cm}^{3}$

(iv) Given, 

r= 21cm,

h=40cm

Volume = $\pi r^{2} h$,

$=\frac{22}{7} \times 21 \times 21 \times 40$

$=66 \times 840$

$=\quad 55440 \mathrm{~cm}^{3}$

Question 2

Ans: (a) Given, 

Volume, = $44 \mathrm{Cm}^{3}$

Height $=3.5 \mathrm{~cm}$

ஃ Volume = $\pi r^{2}h$

$44=\frac{22}{7}\times r^{2}\times \frac{3.5}{10}$

$ 44=11 r^{2}$

$ \frac{44}{11}=r^{2}$

$4=r^{2}$

$r^{2}=4 .$

$r=\sqrt{4} .$

$r=2 \mathrm{cm} $

∴ Diameter = 2r 

$=2 \times 2$

$=4 c \mathrm{~m}$

(b) Given, 

Volume , = $385 \mathrm{~cm}^{3}$

Height $=1 \mathrm{dm}=10 \mathrm{~cm}$

$\therefore \quad$ Volume $=\pi r^{2} h$

$385=\frac{22}{7} \times r^{2} \times 10$

$\frac{49}{4}=r^{2}$

$r^{2}=\frac{49}{4}$

$r=\sqrt{\frac{49}{4}}$

$r=\frac{7}{2}$

∴ Diameter = 2r = $2 \times \frac{7}{2}=7 \mathrm{~cm}$

 

Question 3

Ans: (a) Given,

Volume = $66 \mathrm{~cm}^{3}$

Radius = 2cm

∴ Volume = π$(r^{2}$h

$\frac{66 \times 7}{22 \times 4}=h$

$\frac{21}{4}=h$

$h=\frac{21}{4}$

$h=5.25 \mathrm{~cm}$

(b) Given,

Volume = 4litres= 4000 $\mathrm{Cm}^{3}$

Radius= 5cm

$\therefore \quad$ Volume $=\pi r^{2} h$

$4000=\frac{22}{7} \times 5 \times 5 \times h$

$\frac{4000 \times 7}{22 \times 25}=h$

$\frac{80 \times 7}{11}$

$\frac{560}{11}=h$

$h=\frac{510}{11} \mathrm{~cm}$

Question 4

Ans: Given,

Height(h) = 7m

Radius(r) = $\frac{20}{2}=100 \mathrm{~cm}$ $=\frac{10}{100}$ $=\frac{1}{10}m$

$\therefore$ Volume $=\pi r^{2} h$

$=\frac{11}{50}$

According to question,

Total weight $=\frac{11}{50} \times 225$

$=\frac{99}{2}$

$=49.5 \mathrm{~kg}$

Question 5

Ans: Giver,

Internal radius. $(r)=3 \mathrm{~cm}$

Thickness of pipe= $1 \mathrm{~cm}$

$\therefore$ Outer radius $(R)=3+1$=4cm

Length = 6cm

$\begin{aligned} \therefore \quad \text { Volume } &=\pi R^{2} h-\pi r^{2} h . \\ &=\pi h\left(R^{2}-r^{2}\right) . \\ &=\frac{22}{7} \times 6\left((4)^{2}-(3)^{2}\right) \end{aligned}$

$=\frac{132}{7}(16-9)$

$=\frac{132}{7} \times 7 $

$=132 \mathrm{cm}^{3}$

Question 6

Ans: Given,

Sum of radius, of the base and the height of a cylinder (h+r) = 37cm,

Total surface area = $1628 \mathrm{~cm}^{2}$

∴ Total surface area = $2πr h+2πr^{2}$

$1628=2 \pi r(h+r)$

$1628=$ $2 \times \frac{22}{7} \times r \times 37 .$

r=7

$\therefore r=7 \mathrm{~cm} .$

So, $h+r=37$

$h+7=37$.

$h=377$

$h=30 \mathrm{~cm}$.

$\begin{aligned} \text { Volume } &=\pi r^{2} h \\ &=\frac{22}{7} \times 7 \times 7 \times 30 \\ &=22 \times 210 \\ &=4620 \mathrm{~cm}^{3} \end{aligned}$

Question 7

Ans: Given

Capacity of a cylindrical tank = $6160 \mathrm{m}^{3}$

Radius $(r)  \frac{28}{2}=14 \mathrm{~m}$

$\therefore$ Volume $=\pi r^{2} h$

$6160=\frac{22}{7} \times 14 \times 14 \times h$

$\frac{6160}{22 \times 281}=h $

h= 10m

Area of curved surface of taken inner sides = $2 \pi r h$

$=2 \times \frac{22}{7} \times 14 \times 10$

$=44 \times 20$

$=880 \mathrm{m}^{2}$

Question 8

Ans: Given,

Curved surface area of cylinder = $4400 \mathrm{~cm}^{2}$

Circumference $=110 \mathrm{~cm} $

Circumference $=2 \pi r$

$110=2 \times \frac{22}{7} \times r$

$\frac{110 \times 7}{2 \times 22}=r$

$\frac{35}{2}$=r

$r=\frac{35}{2} \mathrm{~cm}$

(i) Curved Surface area $=2$ πrh

4400= $=\frac{2}\times \frac{22}{7} \times \frac{35}{2} \times h$

$\frac{4400}{22 \times 5}=h$

$40=h$

$h=40$ cm

hence the height of cylinder is 40cm

(ii) Volume = $\pi r^{2} h$

$=110 \times 350$

$=38500 \mathrm{~cm}^{3}$

Question 9

Ans: (i) Given ,

Height of the wall = 20 meter 

Radius = $\frac{2}{2}=1$

$\because$ Volume =  πr^{2}$

$\begin{aligned} &=\frac{22}{7} \times 1 \times 1 \times 20 \\=& \frac{440}{7} \\=& 62 \frac{6}{7} \mathrm{~m}^{3} . \end{aligned}$

(ii) Curved surface area = $2 π r h$

$\begin{aligned} &=2 \times \frac{22}{7} \times 1 \times 20 \\=& \frac{44 \times 20}{7} \\=& \frac{880}{7} \end{aligned}$

Rate of plastering the inner surface = Rs 5  per m².

Total cost = 880\7 x 5

$=\frac{4400}{7}$

$=7628.57 $

Question 10

Ans: Giver

Radius of cylinder $=\frac{20}{2}=10 \mathrm{~cm}

Curved surface area = $1000 \mathrm{~cm}^{2}$

(i) Curved surface area $=2 \pi r$

$1000=2 \times \frac{22}{7} \times 10 \times h$

$\frac{175}{11}=h$

$h=\frac{175}{11}$

$h=15.9 \mathrm{~cm} .$

(ii) $\begin{aligned} \text { Volume } &=\pi r^{2}h \\ &=3.14 \times 10 \times 10 \times 15-9 . \\ &=31.4 \times 159 . \\ &=4992.6 \mathrm{~cm}^{3}\end{aligned}$

Question 11

Ans: Given, 

Radius =  $\frac{35}{2} \mathrm{~cm}$

height $=1.2 \mathrm{~m}=1.2 \times 100 \mathrm{~cm}=120 \mathrm{~cm}$.

(i) Outer lateral surface area = $2 \pi r h$

$=110 \times 120$

$=13200 \mathrm{~cm}^{2} $

(ii) Capacity = $\pi r^{2} h$

 $=55 \times 2100$

$=115500 \mathrm{~cm} .$

=115.5 liters 

Question 12

Ans: Given, 

Radius of cylindrical glass = $\frac{8}{2}=4 \mathrm{~cm}$

and its height = 15cm

Radius, of cylindrical vessel = $\frac{30}{2}=15 \mathrm{~cm}$

and its height = 80 cm

Volume of cylindrical glass= $=\pi r^{2} \mathrm{~h}$

$\pi \times 4 \times 4 \times 15$

$=240 \pi$

Volume of cylindrical vessel $=\pi r^{2} h$

$=\pi \times 15 \times 15 \times 80$

$=\pi \times 225 \times 80$

$=18000 \pi$.

$\therefore$ Number of glasses $=\frac{\text { Volume of vessel }}{\text { Volume of glass}}

$=\frac{18000 \pi}{240 \pi}$ 

$=75$ glasses

Question 13

Ans: Given, 

R= 22m

r= 20 

h=$\frac{7}{100} m$

$\pi R^{2} h-\pi{r}^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$=\frac{22}{7} \times \frac{7}{100}(R+r)(R-r)$

$=\frac{22}{7}\times{7}{100}(22) \times 18=18.48 \mathrm{~mm}$

Question 14

Ans: Given

Radius of iron cylindrical block = $=\frac{0.5m}{20}$ =$\frac{1}{4} \mathrm{~m}=$0.25m= $0.25\times $100cm  =25cm

Length = 3.5m= 3.5 $\times 100$cm = 350 cm

Volume of block =  $\pi r^{2} b$

$=\frac{22}{7} \times 25 \times 25 \times 350$

$=550 \times 1250$

$=687500 \mathrm{~cm}^{3}$

So , volume of base = $687500 \mathrm{~cm}^{3}$

Area of square base $=25 \times 25$

$=625 \mathrm{~cm}^{2}$

Height of bar= $\frac {Volume of bar}{Area of square base}$

$=\frac{687500}{625}$

$=1100 \mathrm{~cm} $

$=\frac{1100}{100}$

$=11 \mathrm{~m}$

Question 15

Ans: Given, 

Length of swimming pole (l) = 70m

Breadth (b)= 44m

and depth (h) =  3m

Volume = lbh

$=70 \times 40 \times 3$

$=924012^{3}$

Radius of pipe= $\frac{14}{2} c m=7 cm=\frac{7}{160} \mathrm{~m}$

Volume = $πr^{2}h

9240= $\frac{22}{7} \times \frac{7}{100} \times \frac{7}{100} \times h$ (Volume = 9240)

$\frac{9240 \times 100 \times 100}{22 \times 7}=h$

$\frac{92400000}{154}=h .$

$600000=h .$

$\therefore h=600000$

Let be the distance 

$\therefore \quad$ Distance $=$ speed \times $ Time.

$600000=2 \times$ Time. (speed = 2m)

$\frac{600000}{2}=$ Time.

$83 \frac{1}{3}$ hours

Question 16

Ans: Given, 

External radius of a hollow cylinder = $\frac{12}{2}=6 \mathrm{~cm}$

Internal radius of a hollow cylinder = 6_ 0.25

= 5.75cm

Length (h)= 15cm

Volume of hollow cylinder = $\pi R^{2} h-\pi r^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$=\frac{22}{7} \times 15\left((6)^{2}-\left(5.70^{2}\right)\right.$

$=\frac{330}{7}(36-33.0625) .$

$=\frac{330}{7} \times 2.9375$

$=\frac{969.375}{7 }$

Radius of solid cylinder $\frac{2}{2}=1 \mathrm{~cm}$ (given).

$\begin{aligned} \therefore \text { Volume } &=\pi r^{2} h \\ \frac{969-375}{7} &=\frac{22}{7} \times 1 \times 1 \times h \end{aligned}$

$\frac{969.375}{22}=h$ 

$\therefore h=\frac{969.375}{22}$

$h=44.0625 \mathrm{~cm} $

Question 17

Ans: Given, 

Internal radius of tube= $\frac{11.2}{20} \mathrm{~cm}=5.6 \mathrm{~cm}$

$\operatorname{Length}(h)=21 \mathrm{~cm}$.

Thickness $=0.4 \mathrm{~cm}$.

∴ Outer radius 5.6+0.4

=6cm

Volume of Metal = $\pi R^{2} h - r^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$\left.=\frac{22}{7}\times 21 \times(6)^{2}-(5-6)^{2}\right)$

$=66 \times(36-31.36)$

$=66 \times 464$

$=306.24 \mathrm{cm}^{3}$

$=306.2 \mathrm{~cm}^{3}$

Question 18

Ans: Given, 

Volume of water = 1000 lit = $1000 \times 1000 \mathrm{~cm}^{3}=1000000 \mathrm{~cm}^{3}$

Radius of pip = 0.6cm

∴ Volume = $\pi r^{2} h$

$1000000=\frac{22}{7} \times 0-6 \times 0.6 \times h .$

$\frac{1000000 \times 7}{22 \times 0-6 \times 0.6}=h .$

$\frac{700000000}{7.92}=h .$

$883838.38=h .$

$h=883838-38 \mathrm{~cm}$

Let height (h) be the distance, 

Distance = speed $\times$ times 

$883838.38=8 \times$ time.

$\frac{88383838}{800}=$ time

$110479.7975=$ Time

Time = 110479.7975sec

Time = $\frac{110479.7975}{60 \times 60}$ hours 

$=\frac{11047967975}{36000060}$ hours

$=30.69$ hours 

Question 19

Ans: Given,

Radius =$\frac{28}{2} \mathrm{~cm}=14 \mathrm{~cm}$

Height = 72cm

$\therefore$ Volume $\pi r 2 h$

$=\frac{22}{7} \times 14 \times 14 \times 72$

$=44 \times 1008$

$=44352 \mathrm{~cm}^{3}$

Length f tank = 66cm

Breadth of tank = 28cm

Volume = $=l \times b \times h$

$\begin{aligned} 44352 &=66 \times 28 \times h . \\ 44352 &=1848 h . \\ \frac{44352}{1848} &=h . \\ 24 &=h \\ \therefore h &=24 \mathrm{~cm} . \end{aligned}$

Hence, the height of the water level in the tank is 24 cm

Question 20

Ans: Given,

Radius of cylindrical vessel = $\frac{14}{2} c m=7 cm$ 

Height of water = $8 \frac{9}{14}(cm )=\frac{121}{14}$

Volume of water= $\pi r^{2} h$

$=1331 \mathrm{~cm}^{3}$

Volume of cube = $=1331 \mathrm{~cm}^{3}$

Volume of cube= $a^{3}$

$1331=a^{3}$

$\sqrt[3]{1331}=a$

$\sqrt{11 \times 11 \times 11}=a$

$a=11$

Hence , the length of the edge is 11cm

Question 21

Ans: Let the radius of the cylinder = r

And height = h

Volume = $\pi r^{2} h$

If the radius is halved 

$\therefore \quad r=\frac{r}{2}$

Height = h

$\therefore Volume=\pi\left(\frac{r}{2}\right)^{2}h $

$=\pi \frac{r^{2}}{4} h$

According to question 

$=\frac{\pi r^{2}}{4} h=\pi r^{2} h$

$=\frac{1}{4}$= 1:4

Question 22

Ans: Given,

Length of sheet = 22cm

and breadth = 12cm

If it is folded breadth wire, 

∴ Circumference = 12cm

and height = 22cm

Circumference = 2πr

$12=2 \times \frac{22}{7} \times r$

$\frac{21}{11}=r$

$y=\frac{21}{11} \mathrm{~cm}$

∴ Volume =πz $=r^{2} h$

$=\frac{22}{7} \times \frac{21}{11} \times \frac{21}{11} \times 22$

$=\frac{462 \times 42}{77}$

$=\frac{19404}{72}$

$=269.5 \mathrm{~cm}^{3}$

IF it is folded length wire 

$\therefore$ circumference  $=22 \mathrm{~cm}$. 

and Height: $12 \mathrm{~cm}$.

So circumference =$2 \pi r$

$22=2 \times \frac{22}{7} \times r$

$\frac{22 \times 7}{2 \times 22}=r$

$\frac{7}{2}=r$

$r=\frac{7}{2} cm$

Volume =$\left.\pi r^{2}\right)h$

$\frac{27}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12$

$=22 \times 21$

$=482cm^{3}$

According to question, 

$=462-269-5$

$=192.5 \mathrm{~cm}^{3}$

Question 23

Ans: Given, 

Depth of a wall = 20m

Radius = $\frac{7}{2} \mathrm{~m}$

Volume of earth = $\pi r^{2} h$

$=22 \times 35$

$=770 \mathrm{m}^{3}$

Length of platform= 22m,

Breadth = 14m

Volume of platform = $770 \mathrm{~m} 3$

Volume = lbh 

770= $22 \times 14 \times h$

$\frac{770}{22 \times 14}=h$

$\frac{770}{308}=h$

$2-5=h$

$h=2.5 \mathrm{~m}$

Hence , the height of the platform is 2.5m

Question 24

Ans: Given,

Height of cylindrical barrel of pen = 7cm

Radius = $\frac{5}{2} \cdot mm =$ $\frac{5}{2} \times \frac{1}{10} \mathrm{~cm}$=1\4cm

Volume of ink in it =  $\pi r^{2} h$

$-\frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times 7$

$=\frac{11}{8} \mathrm{~cm}^{3}$

Volume of link in bottle =  $\frac{1}{5}l=\frac{1}{5} \times 1000 \mathrm{~cm}=200 \mathrm{cm}^{3}$

$\therefore$ Total number of barrels $=200 \div \frac{11}{8}$

$=200 \times \frac{8}{11}$

$=\frac{1600}{11}$

Word written in one barrel = 310 words 

Total number of words = $\frac{1600}{11} \times 320$

$=\frac{496000}{11 .}$

$=45080.90 .$

$=45090$ words 

Question 25

 

Ans: Given , 

Length of a rectangular box= 40cm,

Breadth = 30cm and 

Height = 25cm

Volume = lbh 

$=40 \times 30 \times 25$

$=1200 \times 25$

$=30000 \mathrm{~cm}^{3}$

 Radius of cylindrical tin= 17.5cm

Volume = $30000 \mathrm{~cm}^{3}$

Volume $=\pi r^{2} h$

$30000=3-14 \times 17.5 \times 17.5 \times h $

$30000=3-14 \times 306.25 \dot{x h}$

$30000=961.625h $

$\frac{30000000}{961625}=h$

$31 \cdot 2$

$h=31-2 \mathrm{~cm}$

Hence the height of the cylindrical tin is 31.2cm

Question 26

 

Ans: Given, 

Diameter of a circular tank = 17.5m

So, the radius = $\frac{175}{20}=\frac{35}{4}=$ 8.75m

Outer radius = 8.75+4

= 12.75m

Height= 2m 

Volume of the embankment = $\pi R^{2} h-\pi r^{2} h$

$\begin{aligned} & \pi h\left(R^{2}-r^{2}\right) \\=& \frac{22}{7} \times 2\left((12-75)^{2}-(8.75)^{2}\right) \\=& \frac{44}{7}(162.5625-76.5625) \\=& \frac{44}{7} \times 86 . \\=& \frac{3784}{7} \\ 540.57 \mathrm{~m}^{3} . \end{aligned}$

 

∴ Volume of earth of the tank = $540.57 \mathrm{~m}^{3}$

$\begin{aligned} \therefore \text { Volyme } &=\pi r^{2} h \\ 540.57 &=\frac{22}{7} \times 8.75 \times 8.75 \times h . \end{aligned}$

$540.57=\frac{1684.375 \mathrm{~h}}{7}$

$2.25=h$

$h=2.25 \mathrm{~m} .$

Hence the depth of the circular tank is 2.25m

Question 27

 

Ans: Given, 

Speed of water = $7 m=700 \mathrm{~cm}$

Internal radius= $\frac{2}{2} c m=1 c m$

Radius of tank = 40cn

Time =$\frac{1}{2}$ hour $=$ $\frac{1}{2} \times 60 \times 60=\frac{3600}{2} \mathrm{sec}=$

=1800sec 

∴ Distance  (h) = Times into speed 

$=1800 \times 700$

$=1260000 \mathrm{~cm}$

Volume $=\pi r^{2} \mathrm{~h}$. 

=$\frac{22}{7}1 \times1 \times \times 1260000$

$=3960000 \mathrm{~cm}^{3}$

∴ Volume of water in the tank =$3960000 \mathrm{cm}^{3} $

Volume $=\pi r^{2} \mathrm{h}$

$3960000=\frac{22}{7} \times 40 \times 40 \times h$.

$\frac{17325}{22}=h$

$787.5=h$

$h=787.5 \mathrm{cm} $

Question 28

 

Ans: Given, 

Radius of pipe = $\frac{7}{2} cm=\frac{7}{200}$ m

Speed = 36 km\hr = $36 \times \frac{1000}{60}$ = 600m

Radius of tank = $35 \mathrm{~cm}=\frac{35}{100} \mathrm{~m}$

Height = 1m

∴ Volume of tank =π$r^{2}h$

$\frac{22}{7} \times \frac{35}{100} \times \frac{35}{100} \times 1$

$=\frac{77}{200} \mathrm{~m}^{3}$

$\because$ Volume $=\pi r^{2} h .$

$\frac{77}{200}=\frac{22}{7} \times \frac{7}{200} \times \frac{7}{200} \times h .$

$\frac{2200}{221}=h$

$100=h .$

$h=100$

Let height (h) be the distance 

Distance = speed into Time

100= $600 \times Time$

Time $=\frac{100}{600}$

Time $=0.167 \mathrm{~min} $

Question 29

 

Ans: Given, 

Radius of coin =$\frac{1.5}{20} \mathrm{~cm}=\frac{3}{4} \mathrm{~cm}$

and its thickness = 0.2 cm

Volume of one coin = $\pi r^{2} h$

$=\frac{22}{7} \times \frac{3}{4} \times \frac{3}{4} \times 0.2 .$

$=\frac{39.6}{112}$

$=0.35 \mathrm{~cm}^{3}$

Radius of cylinder = $\frac{4.5}{20}=\frac{9}{4} \mathrm{~cm}$

Height = 10cm

Volume of cylinder = $\pi r^{2} h$

$\begin{aligned} &=\frac{22}{7} \times \frac{9}{4} \times \frac{9}{4} \times 10 \\=& \frac{17820}{112} \\=& 159.11 \mathrm{~cm}^{3} . \end{aligned}$

Volume of cylinder = volume of one coin X Number of coin

159.11 = $0.35 \times$ Number of coin.

$\frac{159 \cdot 11}{0.35}$ =Number of coin.

$454.6=$ Number of coin

Number of coin $=454.6$

Question 30

 

Ans: Given

Weight of  $1cm^{3}=21 g$

Length of pipe = $1 \mathrm{~m}=100 \mathrm{~cm}$

Internal radius = $\frac{3}{2} \mathrm{~cm}=1.5 \mathrm{~cm}$

External radius = 1.5+1= 2.5cm

∴ Volume = Volume of outer - Volume of inner surface 

$π R^{2} h$-$π r^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$=\frac{22}{7} \times 100\left((2.5)^{2}-(1.5)^{2}\right)$

$=\frac{2200}{7} \times(6.25-2.25) .$

$=\frac{2200}{7} \times 4 $

$=\frac{8800}{7} \mathrm{~cm}^{3} $

∴ Total Weight of metal = $\frac{8800}{7} \times {21}$

$=26400 \mathrm{~g}$