SELINA Solutions for Class 9 Maths Chapter 21 - Solids [Surface Area and Volume of 3-D Solids]

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Chapter 21 - Solids [Surface Area and Volume of 3-D Solids] Exercise Ex. 21(A)

Solution 2

Let a be the one edge of a cube.

Volume

Total surface area=6

Solution 3

Volume of cinema hall

150 requires= 1 person

90000requires=persons

Therefore, 600 persons can sit in the hall.

Solution 4

Let h be height of the room.

1 person requires 16

75 person requires

 

Volume of room is 1200

Solution 5

Volume of melted single cube

Let a be the edge of the new cube.

Volume

Therefore, 6 cm is the edge of cube.

Solution 6

Solution 7

Let the side of a cube be 'a' units.

Total surface area of one cube

Total surface area of 3 cubes

After joining 3 cubes in a row, length of Cuboid =3a

Breadth and height of cuboid = a

Total surface area of cuboid equals 2 open parentheses 3 a squared plus a squared plus 3 a squared close parentheses equals 14 a squared 

Ratio of total surface area of cuboid to the total surface area of 3 cubesequals fraction numerator 14 a squared over denominator 18 a squared end fraction equals 7 over 9 

Solution 8


Solution 9

The area of the playground is 3650 m2 and the gravels are 1.2 cm deep. Therefore the total volume to be covered will be:

3650 x 0.012 =43.8 m3.

Since the cost of per cubic meter is Rs. 6.40, therefore the total cost will be:

43.8 x Rs.6.40 = Rs.280.32

Solution 10

W e space k n o w space t h a t space
1 space m m equals 1 over 10 c m
8 space m m equals 8 over 10 c m
V o l u m e equals B a s e space a r e a cross times H e i g h t
rightwards double arrow 2880 space c m cubed equals x cross times x cross times 8 over 10
rightwards double arrow 2880 cross times 10 over 8 equals x squared
rightwards double arrow x squared equals 3600
rightwards double arrow x equals 60 space c m

Solution 11

External volume of the box=

Since, external dimensions are 27 cm, 19 cm, 11 cm; thickness of the wood is 1.5 cm.

Internal dimensions

Hence, internal volume of box=

(i)

Volume of wood in the box=

(ii)

Cost of wood

(iii)

Vol. of 4 cm cube=

Number of 4 cm cubes that could be placed into the box

Solution 12

Area of sheet= Surface area of the tank

Length of the sheetits width=Area of 4 walls of the tank +Area of its base

Length of the sheet 2.5 m=

Length of the sheet= 300.8 m

Cost of the sheet = 300.8 Rs 12.50 = Rs 3760

Solution 13

Let exterior height is h cm. Then interior dimensions are 78-3=75, 19-3=16 and h-3 (subtract two thicknesses of wood). Interior volume = 75 x 16 x (h-3) which must = 15 cu dm

= 15000 cm^3

(1 dm = 10cm, 1 cu dm = 10^3 cm^3).

15000= 75 x 16 x (h-3)

h-3 = 15000/(75x16) = 12.5 cm h = 15.5 cm.

Solution 14

(i)

If the side of the cube= a cm

The length of its diagonal= cm

And,

(ii)

Total surface area of the cube=

=

Solution 15

Given that the volume of the iron in the tube 192 cm3

Let the thickness of the tube

Side of the external square=

Ext. vol. of the tube its internal vol.= volume of iron in the tube, we have,

Therefore, thickness is 1 cm.

Solution 16

Let l be the length of the edge of each cube.

The length of the resulting cuboid=

Let width (b) = l cm and its height (h)= l cm

The total surface area of the resulting cuboid

Therefore, the length of each cube is 6 cm.

Solution 1

The length, breadth and height of a rectangular solid are in the ratio 5: 4: 2.

 

Let the length, breadth and height of a cuboid be 5x, 4x and 2x.

Total surface area of a Cuboid = 1216 cm3

2(lb + bh + lh) = 1216

2(20x2 + 8x2 + 10x2) = 1216

 76x2 = 1216

 x2 = 16

 x = 4

Therefore, length = 5(4) = 20cm, breadth = 4(4) = 16cm and height = 2(4) = 8cm.

Chapter 21 - Solids [Surface Area and Volume of 3-D Solids] Exercise Ex. 21(B)

Solution 1

The given figure can be divided into two cuboids of dimensions 6 cm, 4 cm, 3 cm, and 9 cm respectively. Hence, volume of solid

Solution 2

Area of cross section of the solid

Volume of solid

Solution 3

The cross section of a tunnel is of the trapezium shaped ABCD in which and AM = BN. The height is 2.4 m and its length is 40m.

(i)

Perimeter of the cross- section of the tunnel=

Length=40 m

Internal surface area of the tunnel(except floor)

Rate of painting=Rs 5 per

Hence, total cost of painting=Rs 5408=Rs 2040

(ii)

Area of floor of tunnel

Rate of cost of paving

Total cost=

Solution 4

(i)

The rate of speed equals 5 space m over s equals 500 fraction numerator c m over denominator s end fraction

Volume of water flowing per secequals 3.2 cross times 500 space c m cubed equals 1600 space c m cubed

(ii)

Vol. of water flowing per minequals 1600 cross times 60 space c m cubed equals 96000 space c m cubed

Since 1000= 1 lt

Therefore, Vol. of water flowing per min=equals 96000 over 1000 equals 96 space l i t r e s

Solution 5

Vol. of water flowing in 1 sec=equals fraction numerator 1500 cross times 1000 over denominator 5 cross times 60 end fraction equals 5000 space c m cubed

Vol. of water flowing =area of cross sectionspeed of water

5000 space fraction numerator c m cubed over denominator s end fraction equals 2 space c m squared cross times s p e e d space o f space w a t e r
rightwards double arrow s p e e d space o f space w a t e r equals 5000 over 2 fraction numerator c m over denominator s end fraction
rightwards double arrow s p e e d space o f space w a t e r equals 2500 fraction numerator c m over denominator s end fraction
rightwards double arrow s p e e d space o f space w a t e r equals 25 space m over s

Solution 6

(i)

Area of total cross section= Area of rectangle abce+ area of

=

(ii)

The volume of the piece of metal in cubic centimeters= Area of total cross section

=

1 cubic centimetre of the metal weighs 6.6 g

of the metal weighs

The weight of the piece of metal to the nearest Kg is 352 Kg.

Solution 7

Vol. of rectangular tank

One liter= 1000

Vol. of water flowing in per sec=

 

Vol. of water flowing in 1 min=

Hence,

can be filled = 1 min

can be filled

Solution 8

 

Length of sheet=32 cm

Breadth of sheet=26 cm

Side of each square=3cm

Inner length=32-23=32-6=26 cm

Inner breadth=cm

By folding the sheet, the length of the container=26 cm

Breadth of the container= 20 cm and height of the container= 3 cm

Vol. of the container=

=

Solution 9

Length of pool= 18 m

Breadth of pool= 8 m

Height of one side= 2m

Height on second side=1.2 m

Volume of pool=

Solution 10

Consider the box 1

 

 

 

 

Thus, the dimensions of box 1 are: 60 cm, 40 cm and 30 cm.

 

 

 

 

 

 

 

Consider the box 2

 

 

 

 

Thus, the dimensions of box 2 are: 40 cm, 30 cm and 30 cm.

 

 

 

 

 

 

 

Consider the box 3

 

 

 

 

 

 

 

Thus, the dimensions of box 2 are: 40 cm, 30 cm and 20 cm.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Chapter 21 - Solids [Surface Area and Volume of 3-D Solids] Exercise Ex. 21(C)

Solution 1

The perimeter of a cube formula is, Perimeter = 4a where (a= length)

G i v e n space t h a t space p e r i m e t e r space o f space t h e space f a c e space o f space t h e space c u b e space i s space 32 space c m
rightwards double arrow 4 a equals 32 space c m
rightwards double arrow a equals 32 over 4
rightwards double arrow a equals space 8 space c m
W e space k n o w space t h a t space s u r f a c e space a r e a space o f space a space c u b e space w i t h space s i d e space apostrophe a apostrophe space equals 6 a squared
T h u s comma space S u r f a c e space a r e a equals 6 cross times 8 squared equals 6 cross times 64 equals 384 space c m squared
W e space k n o w space t h a t space t h e space v o l u m e space o f space a space c u b e space w i t h space s i d e space apostrophe a apostrophe equals a cubed
T h u s comma space v o l u m e equals 8 cubed equals 512 space c m cubed

 

 

 

 

Solution 2

G i v e n space d i m e n s i o n s space o f space t h e space a u d i t o r i u m space a r e : space 40 space m cross times 30 space m cross times 12 space m
T h e space a r e a space o f space t h e space f l o o r equals 40 cross times 30
A l s o space g i v e n space t h a t space e a c h space s t u d e n t space r e q u i r e s space 1.2 space m squared space o f space t h e space f l o o r space a r e a.
T h u s comma space M a x i m u m space n u m b e r space o f space s t u d e n t s equals fraction numerator 40 cross times 30 over denominator 1.2 end fraction equals 1000
V o l u m e space o f space t h e space a u d i t o r i u m
equals 40 cross times 30 cross times 12 space m cubed
equals V o l u m e space o f space a i r space a v a i l a b l e space f o r space 1000 space s t u d e n t s
T h e r e f o r e comma space A i r space a v a i l a b l e space f o r space e a c h space s t u d e n t equals fraction numerator 40 cross times 30 cross times 12 over denominator 1000 end fraction space m cubed equals 14.4 space m cubed

Solution 3

Length of longest rod=Length of the diagonal of the box

Solution 4

(i)

No. of cube which can be placed along length.

No. of cube along the breadth

No. of cubes along the height.

The total no. of cubes placed

(ii)

Cubes along length

Cubes along widthand cubes along height

The total no. of cubes placed

(iii)

Cubes along length

Cubes along widthand cubes along height

The total no. of cubes placed

Solution 5

Vol. of the tank= vol. of earth spread

Solution 6

 

 

 

 

Solution 7

  

 

  

Solution 8

 

 

 

 

Solution 9

 

 

 

 

Solution 10