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

Ex. 21

Test Yourself

Solids [Surface Area and Volume of 3-D Solids] Exercise Ex. 21

Solution 1(a)

Correct option: (ii) 4 cm

Let the edge of the cube = 'a' cm

Then, length of the diagonal

Therefore,

Area of a square on the diagonal of a cube

Solution 1(b)

Correct option: (iii) 144 cm²

Therefore, volume of a cuboid =

= 4 × 3 × 12

= 144 cm³

* Given length, breadth and diagonal of a cuboid, volume of cube cannot be determined.

Solution 1(c)

Correct option: (i) 6 cm

Let the edge of a cube = 'a' cm

Now,

Volume of a cube = Volume of a cuboid

Solution 1(d)

Correct option: (iv) 18 cm, 12 cm and 12 cm

Let the dimensions of the cuboid be as follows:

Then,

Therefore,

Solution 1(e)

Correct option: (iii) 75 cm

Now,

Volume of the soil spread = Area of the field × Rise in level of field

⇒ 960 m³ = 1280 × Rise in level of field

⇒ Rise in level of field = 0.75 m = 75 cm

Solution 1(f)

Correct option: (i) 0.0144 m³

Volume of water flowing in 1 second = 12 × 20 cm³ = 240 cm³

Therefore,

Volume of water flowing in 1 minute = 240 × 60 cm³ = 14,400 cm³ = 0.0144 m³

Solution 1(g)

Correct option: (iv)

Volume of given solid = Area of cross-section × length

= Area of trapezium × length

Solution 2

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.

Solution 3

Let a be the one edge of a cube.

Volume

Total surface area=6

Solution 4

Volume of cinema hall

150 requires= 1 person

90000requires=persons

Therefore, 600 persons can sit in the hall.

Solution 5

Let h be height of the room.

1 person requires 16

75 person requires

Volume of room is 1200

Solution 6

Volume of melted single cube

Let a be the edge of the new cube.

Volume

Therefore, 6 cm is the edge of cube.

Solution 7

Solution 8

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 cubes $equals fraction numerator 14 a squared over denominator 18 a squared end fraction equals 7 over 9$

Solution 9

Solution 10

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 11

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

Solution 12

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 13

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 14

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 15

(i)

If the side of the cube= a cm

The length of its diagonal= cm

And,

(ii)

Total surface area of the cube=

Solution 16

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 17

Area of cross section of the solid

Volume of solid

Solution 18

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 19

(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 sec $equals 3.2 cross times 500 space c m cubed equals 1600 space c m cubed$

(ii)

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

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 20

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 21

(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 22

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

Solids [Surface Area and Volume of 3-D Solids] Exercise Test Yourself

Solution 1

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 2

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 3

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 4

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 5

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.

Solution 6

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 7

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