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

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

Let *a* be the one edge of a cube.

Volume

Total surface area=6

Volume of cinema hall

150 requires= 1 person

90000requires=persons

Therefore, 600 persons can sit in the hall.

Let *h* be height of the room.

1 person requires 16

75 person requires

Volume of room is 1200

Volume of melted single cube

Let *a* be the edge of the new cube.

Volume

Therefore, 6 cm is the edge of cube.

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

Ratio of total surface area of cuboid to the total surface area of 3 cubes

The area of the playground is 3650 m^{2} and the gravels are 1.2 cm deep. Therefore the total volume to be covered will be:

3650 x 0.012 =43.8 m^{3}.

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

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

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

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.

(i)

If the side of the cube= *a* cm

The length of its diagonal= cm

And,

(ii)

Total surface area of the cube=

=

Given that the volume of the iron in the tube 192 cm^{3}

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.

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.

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 cm^{3}

⟹ 2(lb + bh + lh) = 1216

⟹ 2(20x^{2} + 8x^{2}
+ 10x^{2}) = 1216

⟹ 76x^{2} = 1216

⟹ x^{2} = 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)

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

Area of cross section of the solid

Volume of solid

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=

(i)

The rate of speed

Volume of water flowing per sec

(ii)

Vol. of water flowing per min

Since 1000= 1 lt

Therefore, Vol. of water flowing per min=

Vol. of water flowing in 1 sec=

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

(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.

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

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=

=

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=

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)

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

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

(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

Vol. of the tank= vol. of earth spread

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