SELINA Solution Class 9 Chapter 21 Solids [Surface area and Volume of 3-D Solids ] Exercise 21 C

Question 1

Each face of a cube has a perimeter equal to 32 cm. Find its surface area and its volume.

Sol:

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

Given that perimeter of the face of the cube is 32 cm
⇒ 4a = 32 cm
⇒ a = ${\displaystyle \frac{32}{4}}$
⇒ a = 8 cm 

We know that surface area of a cube with side 'a' = 6a² 
Thus, Surface area = 6 x 8² = 6 x 64 = 382 cm²
We know that the volume of a cube with side 'a' = a³
Thus, volume = 8³ = 512 cm³

Question 2

A school auditorium is 40 m long, 30 m broad and 12 m high. If each student requires 1.2 m² of the floor area; find the maximum number of students that can be accommodated in this auditorium. Also, find the volume of air available in the auditorium, for each student. 

Sol:

Given dimensions of the auditorium are: 40 m x 30 m x 12 m

The area of the floor = 40 x 30

Also given that each student requires 1.2 m² of the floor area. 

Thus, Maximum number of students = ${\displaystyle \frac{40\times 30}{1.2}=1000}$

Volume of the auditorium
= 40 x 30 x 12 m³ 
= Volume of air available for 1000 students

Therefore, Air available for each students
=${\displaystyle \frac{40\times 30\times 12}{1000}{\text{m}}^3=14.4{\text{m}}^3}$

Question 3

The internal dimensions of a rectangular box are 12 cm x  ${\displaystyle x}$ cm x 9 cm. If the length of the longest rod that can be placed in this box is 17 cm; find ${\displaystyle x}$.   

Sol:

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

17 = ${\displaystyle \sqrt{{12}^2+x^2+9^2}}$
17² = 12² + x² + 9²
x² = 17² - 12² -9² 
x² = 289 - 144 - 81
x² = 64
x = 8 cm  

Question 4

The internal length, breadth, and height of a box are 30 cm, 24 cm, and 15 cm. Find the largest number of cubes which can be placed inside this box if the edge of each cube is
(i) 3 cm (ii) 4 cm (iii) 5 cm

Sol:

(i) No. of the cube which can be placed along the length = ${\displaystyle \frac{30}{3}}$ =10.

No. of the cube along with the breadth = ${\displaystyle \frac{24}{3}}$ = 8

No. of cubes along with the height = ${\displaystyle \frac{15}{3}}$ =5.

∴ The total no. of cubes placed = 10 x 8 x 5 = 400

(ii) Cubes along length = ${\displaystyle \frac{30}{4}}$ = 7.5 = 7.

Cubes along width = ${\displaystyle \frac{24}{4}}$ = 6 and cubes along with height =${\displaystyle \frac{15}{4}}$ = 3.75 = 3

∴ The total no. of cubes placed = 7 x 6 x 3 = 126

(iii) Cubes along length = ${\displaystyle \frac{30}{5}}$ = 6 

Cubes along width = ${\displaystyle \frac{24}{5}}$ = 4 . 5 = 4 and cubes along with height = ${\displaystyle \frac{15}{5}}$ =3 

∴ The total no. of cubes placed = 6 x 4 x 3 = 72

Question 5

A rectangular field is 112 m long and 62 m broad. A cubical tank of edge 6 m is dug at each of the four corners of the field and the earth so removed is evenly spread on the remaining field. Find the rise in level.  

Sol:

Vol. of the tank= vol. of earth spread

4 x 6³ m³ = ( 112 x 62 - 4 x 6² ) m² x Rise in level 

Rise in level = ${\displaystyle \frac{4\times 6^3}{112\times 62-4\times 6^2}}$

                   = ${\displaystyle \frac{864}{6800}}$

                   = 0.127 m
                   = 12.7 cm

Question 6

When the length of each side of a cube is increased by 3 cm, its volume is increased by 2457 cm³. Find its side. How much will its volume decrease, if the length of each side of it is reduced by 20%?

Sol:

Let a be the side of the cube.
Side of the new cube = a + 3
Volume of the new cube = a³ + 2457
That is, ( a + 3 )³ = a³ + 2457
⇒ a³ + 3 x a x 3 ( a + 3 ) + 3³ = a³ + 2457
⇒ 9a² + 27a + 27 = 2457
⇒ 9a² + 27a - 2430 = 0
⇒ a² + 3a - 270 = 0
⇒ a ( a + 18 ) - 15 ( a + 18 ) = 0
⇒ ( a - 15 ) ( a + 18 ) = 0
⇒ a - 15 = 0 or a + 18 = 0
⇒ a = 15  or a = - 18
⇒ a = 15 cm  ...[ since side cannot be negative ] 

Volume of the cube whose side is 15 cm = 15³ = 3375 cm³ 

Suppose the length of the given cube is reduced by 20%.
Thus new side a new = a - ${\displaystyle \frac{20}{100}}$ x a

                                   = ${\displaystyle a(1-\frac{1}{5})}$

                                   = ${\displaystyle \frac{4}{5}}$ x 15

                                    =  12 cm

Volume of the new cube whose side is 12 cm = 12³ = 1728 cm³
Decrease in volume = 3375 - 1728 = 1647 cm³ 

Question 7

A rectangular tank 30 cm × 20 cm × 12 cm contains water to a depth of 6 cm. A metal cube of side 10 cm is placed in the tank with its one face resting on the bottom of the tank. Find the volume of water, in liters, that must be poured in the tank so that the metal cube is just submerged in the water. 

Sol:

The dimensions of rectangular tank : 30 cm x 20 cm x 12 cm
Side of the cube = 10 cm
Volume of the Cube = 10³  = 1000 cm³
The height of the water in the tank is 6 cm.
Volume of the cube till 6 cm = 10 x 10 x 6 cm³
Hence when the cube is placed in the tank,
then the volume of the water increases by 600 cm³.

The surface area of the water level is 30 cm x 20 cm = 600 cm²

out of this area, let us subtract the surface area of the cube.

Thus, the surface area of the Shaded part in the above figure is 500 cm²
The displaced water is spread out in 500 cm² to a height of 'h' cm.
And hence the volume of the water.
Thus, we have,
500 x h = 600 cm³
⇒ ${\displaystyle \text{h}=\left(\frac{600}{500}\right)\text{cm}}$
⇒ h = 1.2 cm

Thus, now the level of the water in the tank is = 6 + 1 . 2 = 7 . 2 cm

Remaining height of the water level,
So that the metal cube is just submerged in the water = 100 - 7 . 2 = 2 .8 cm
Thus the volume of the water that must be poured in the tank so that the metal cube is just submerged in the water = 2.8 x 500 = 1400 cm³

We know that 1000 cc = 1 liter

Thus, the required volume of water = ${\displaystyle \frac{1400}{1000}}$ = 1.4 litres

Question 8

The dimensions of a solid metallic cuboid are 72 cm × 30 cm × 75 cm. It is melted and recast into identical solid metal cubes with each edge 6 cm. Find the number of cubes formed.

Also, find the cost of polishing the surfaces of all the cubes formed at the rate Rs. 150 per sq. m.

Sol:

The dimensions of a solid are: 72 cm, 30 cm, 75 cm
The volume of the cuboid = 72 cm x 30 cm x 75 cm = 162000 cm³
Side of a cube = 6 cm 
Volume of a cube = 6³ = 216 cm³

The number of a cube = ${\displaystyle \frac{162000}{216}=750}$

The surface area  of  a cube = 6a² = 6 x 6² = 216 cm²
Total surface area of 750 cubes = 750 x 216 = 162000 cm²

Total surface area in square metres = ${\displaystyle \frac{162000}{10000}}$
                                                        = 16.2 square meters

Rates of polishing the surface per square meter = Rs. 150

The total cost of polishing the surface = 150 x 16.2 = Rs. 2430

Question 9

The dimensions of a car petrol tank are 50 cm × 32 cm × 24 cm, which is full of petrol. If a car's average consumption is 15 km per liter, find the maximum distance that can be covered by the car. 

Sol:

The dimensions of a car petrol tank are: 50 cm x 32 cm x 24 cm
Volume of the tank = 38400 cm³
We know that 1000 cm3 = 1 litre

Thus the volume of the tank = ${\displaystyle \frac{38400}{1000}}$ = 38.4 litres

The average consumption of the car = 15 km/ litre

Thus, the total distance that can be covered by the car = 38.4 x 15 = 576 km

Question 10

The dimensions of a rectangular box are in the ratio 4: 2 : 3. The difference between the cost of covering it with paper at Rs. 12 per m² and with paper at the rate of 13.50 per m² is Rs. 1,248. Find the dimensions of the box. 

SoL:

Given dimensions of a rectangular box are in the ratio 4: 2 : 3
Therefore, the total surface area of the box = 2${\displaystyle [4x\times 2x+3x+4\times 3x]}$
 = 2 ( 8x² + 6x² + 12x² ) m² 

Difference between cost of  covering the box with paper at Rs. 12 per m² and with paper at Rs. 13.50 per m² = Rs. 1,248 
⇒ 52x²[ 13 .5 - 12 ] = 1248
⇒ ${\displaystyle 52\times x^2\times 1.5}$ = 1248
⇒ 78 x x² = 1248 
⇒ x² = ${\displaystyle \frac{1248}{78}}$
⇒ x² = 16 
⇒ x = 4             ...[ length, width and height cannot be negative ]

Thus, the dimensions of the rectangular box are : 4 x 4 m,  2 x 4 m, 3 x 4 m
Thus, the dimensions are 16 m, 8 m and 12 m.   

SELINA Solution Class 9 Chapter 21 Solids [Surface area and Volume of 3-D Solids ] Exercise 21 B

Question 1

The following figure shows a solid of uniform cross-section. Find the volume of the solid. All measurements are in centimeters.
Assume that all angles in the figures are right angles.

Sol:

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

= 9 x 4 x 3 + 6 x 4 x 3
= 108 + 72 
= 180 cm³.

Question 2

A swimming pool is 40 m long and 15 m wide. Its shallow and deep ends are 1.5 m and 3 m deep respectively. If the bottom of the pool slopes uniformly, find the amount of water in liters required to fill the pool.

Sol:

Area of cross-section of the solid = ${\displaystyle \frac{1}{2}(1.5+3)\times \left(40\right)}$cm²

= ${\displaystyle \frac{1}{2}}$( 4.5 ) x ( 40 ) cm²
= 90 cm²

Volume of solid
= Area of cross section x Length
= 90 x 15 cm³
= 1350 cm³
= 1350000 liters     ....( Since 1cm³ = 1000L )

Question 3

The cross-section of a tunnel perpendicular to its length is a trapezium ABCD as shown in the following figure; also given that:

AM = BN; AB = 7 m; CD = 5 m. The height of the tunnel is 2.4 m. The tunnel is 40 m long. Calculate:

(i) The cost of painting the internal surface of the tunnel (excluding the floor) at the rate of Rs. 5 per m² (sq. meter).

(ii) The cost of paving the floor at the rate of Rs. 18 per m².

Sol:

The cross-section of a tunnel is of the trapezium-shaped ABCD in which  AB = 7 m, CD = 5 m and AM = BN. The height is 2.4 m and its length is 40 m.

(i) AM = BN =${\displaystyle \frac{7-5}{2}=\frac{2}{2}=1\text{m}}$

∴ In ΔADM,

AD² = AM² + DM²      ...[ Using pythsgor as theorem ]
       = 1² + (2 . 4)² 
       = 1  + 5.76
       = 6.67 
       = (2.6)² 
AD  = 2.6 m 

Perimeter of the cross- section of the tunnel = ( 7+ 2.6 + 2.6 + 5 ) m = 17.2 m

Length = 40 m

∴ The internal surface area of the tunnel ( except the floor ) 

= ( 17.2 x 40 - 40 x 7) m²
= ( 688 - 280 ) m²
= 408 m² 

Rate of painting = Rs. 5 per m²
Hence, total cost of painting = Rs. 5 x  5408 = Rs. 2040

(ii) Area of floor of tunnel = l x b = 40 x 7 = 280 m²

Rate of cost of paving = Rs. 18 per m² 

Total cost = 280 x 18 = Rs. 5040

Question 4

Water is discharged from a pipe of a cross-section area 3.2 cm² at the speed of 5m/s. Calculate the volume of water discharged: 
(i) In cm³ per sec.
(ii) In liters per minute.

Sol:

(i) The rate of speed  = ${\displaystyle 5\frac{\text{m}}{s}=500\frac{\text{cm}}{s}}$

The volume of water flowing per sec = 3.2 x 500cm³ = 1600 cm³ 

(ii) Vol. of water flowing per min = 1600 x 60 cm³ = 96000 cm³ 

Since 1000 cm³ = 1 litre

Therefore, Vol. of water flowing per min= ${\displaystyle \frac{96000}{1000}}$ = 96 litres

Question 5

A hose-pipe of cross-section area 2 cm² delivers 1500 liters of water in 5 minutes. What is the speed of water in m/s through the pipe? 

Sol:

Vol. of water flowing in 1 sec= ${\displaystyle \frac{1500\times 1000}{5\times 60}=5000{\text{cm}}^3}$

Vol. of water flowing =area of cross-section speed of water

${\displaystyle 5000\frac{{\text{cm}}^3}{s}=2{\text{cm}}^2\times \text{speed of water}}$

⇒ speed of water = ${\displaystyle \frac{5000}{2}\frac{\text{cm}}{s}}$

⇒ speed of water = ${\displaystyle 2500\frac{\text{cm}}{s}}$

⇒ speed of water = ${\displaystyle 25\frac{\text{m}}{s}}$

Question 6

The cross-section of a piece of metal 4 m in length is shown below. Calculate :


(i) The area of the cross-section;
(ii) The volume of the piece of metal in cubic centimeters.

If 1 cubic centimeter of the metal weighs 6.6 g, calculate the weight of the piece of metal to the nearest kg.

Sol:

(i) Area of total cross section = Area of rectangle abce + area of  Δdef

= ( 12 x 10 ) + ${\displaystyle \frac{1}{2}}$ ( 16 - 10 )( 12 - 7.5 )

= 120 + ${\displaystyle \frac{1}{2}}$ (6)( 4.5 ) cm²

= 120 + 13.5 cm²

= 133.5 cm² 

(ii) The volume of the piece of metal in cubic centimeters = Area of total cross section x length

= 133.5 cm² x 400 cm² = 53400 cm³

1 cubic centimetre of the metal weighs 6.6 g

53400 cm³ of the metal weighs 6.6 x 53400 g = ${\displaystyle \frac{6.6\times 53400}{1000}}$ kg

= 352.440 kg

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

Question 7

A rectangular water-tank measuring 80 cm x 60 cm is filled form a pipe of cross-sectional area 1.5 cm², the water emerging at 3.2 m/s. How long does it take to fill the tank?

Sol:

Vol. of the rectangular tank = 80 x 60 x 60 cm³ = 288000 cm³

One liter = 1000 cm³

Vol. of water flowing in per sec =
${\displaystyle 1.5{\text{cm}}^2\times 3.2\frac{\text{m}}{s}=1.5{\text{cm}}^2\times \frac{(3.2\times 100)\text{cm}}{\text{s}}}$

                                     = ${\displaystyle 480\frac{{\text{cm}}^3}{\text{s}}}$

Vol. of water flowing in 1 min= 480 x 60 = 28800cm³ 

Hence,

28800 cm³  can be filled = 1 min

288000cm³ can be filled =${\displaystyle (\frac{1}{28800}\times 288000)min=10min}$ 

Question 8

A rectangular cardboard sheet has length 32 cm and breadth 26 cm. Squares each of side 3 cm, are cut from the corners of the sheet and the sides are folded to make a rectangular container. Find the capacity of the container formed.

Sol:

Length of sheet = 32 cm

Breadth of sheet = 26 cm 

Side of each square = 3cm

∴  Inner length = 32 - 2 x 3 = 32 - 6 = 26 cm

Inner breadth = 26 - 2 x 3 = 26 - 6 = 20 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 = l x b x h

= 26 cm x 20 cm x 3 cm = 1560 cm³  

Question 9

A swimming pool is 18 m long and 8 m wide. Its deep and shallow ends are 2 m and 1.2 m respectively. Find the capacity of the pool, assuming that the bottom of the pool slopes uniformly. 

Sol:

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 = 18 x 8 x ${\displaystyle \frac{(2+1.2)}{2}}$ m³

= ${\displaystyle \frac{18\times 8\times 3.2}{2}}$

= 230.4 m³ 

Question 10

The following figure shows a closed victory-stand whose dimensions are given in cm.

Find the volume and the surface area of the victory stand. 

Sol:

Consider the box 1

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

Therefore, the volume of box 1 = 60 x 40 x 30 = 72000 cm³ 
Surface area of box 1 = 2 ( lb + bh + lh )
Since the box is open at the bottom and from the given figure, we have,
Surface area of box 1 = 40 x 40 + 40 x 30 + 40 x 30 + 2 ( 60 x 30 )
                                  = 1600 + 1200 + 1200 + 3600
                                 = 7600 cm²   

Consider the box 2 

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

Therefore, the volume of box 2 = 40 x 30 x 30 = 36000 cm³ 
Surface area of box 2 = 2 ( lb + bh + lh )
Since the box is open at the bottom and from the given figure, we have,
Surface area of box 2 = 40 x 30 + 40 x 30  + 2 ( 30 x 30 )
                                  = 1200 + 1200 + 1800 
                                 = 4200 cm²   

Consider the box 3

Thus, the dimensions of the box 3 are 40 cm, 30 cm, and 20 cm.

Therefore, the volume of box 3 = 40 x 30 x 20 = 24000 cm³ 
Surface area of box 3 = 2 ( lb + bh + lh )
Since the box is open at the bottom and from the given figure, we have,
Surface area of box 3 = 40 x 30 + 40 x 20  + 2 ( 30 x 20 )
                                  = 1200 + 800 + 1200 
                                 = 3200 cm²   

Total volume of the box
= volume of box 1 + volume of box 2 + volume of box 3
= 72000 + 36000 + 24000
= 132000 cm³ 

Similarly, total surface area of the box
= surface area of box 1 + surface area  of box 2 + surface area of box 3
= 7600 + 4200 + 3200
= 15000 cm² 

SELINA Solution Class 9 Chapter 21 Solids [Surface area and Volume of 3-D Solids ] Exercise 21 A

Question 1

The length, breadth, and height of a rectangular solid are in the ratio 5: 4: 2. If the total surface area is 1216 cm², find the length, the breadth, and the height of the solid.

Sol:

The rectangular solid is a cuboid.

Let the length of the cuboid = 5a, breadth = 4a, and height = 2a

Total surface area of a cuboid of length l, breadth b and height h = 2(l × b + b × h + l × h)

Given,

Total surface area of the cuboid = 1216 cm²
⇒ 2(5a × 4a + 4a × 2a + 2a × 5a) = 1216 cm²
⇒ 76a² = 1216
⇒ a = 4.

Hence, the length of the cuboid = 5a = 20 cm, breadth = 4a = 16 cm, height = 2a = 8 cm.

Question 2

The volume of a cube is 729 cm³. Find its total surface area.

Sol:

Let a be the one edge of a cube.

Volume = a³ 
729 = a³  
9³ = a³ 
9 = a
a = 9 cm

Total surface area = 6a² = 6 x 9² = 486 cm² 

Question 3

The dimensions of a Cinema Hall are 100 m, 60 m, and 15 m. How many persons can sit in the hall if each requires 150 m³ of air? 

Sol:

Volume of cinema hall = 100 x 60 x 15 = 90000 m³ 

150 m³ requires = 1 persons

90000 m³ requires = ${\displaystyle \frac{1}{150}\times 90000=600 \text{persons}}$

Therefore, 600 persons can sit in the hall. 

Question 4

75 persons can sleep in a room 25 m by 9.6 m. If each person requires 16 m³ of the air; find the height of the room. 

Sol:

Let h be the height of the room.

1 person requires 16 m³

75 person requires  75 x 16 m³ =  1200 m³

Volume of the room is 1200 m³

1200 = 25 x 9.6 x h

h = ${\displaystyle \frac{1200}{25\times 9.6}}$

h = 5 m

Question 5

The edges of three cubes of metal are 3 cm, 4 cm, and 5 cm. They are melted and formed into a single cube. Find the edge of the new cube. 

Sol:

Volume of melted single cube = 3³ + 4³ +  5³ cm³ 
= 27 + 64 + 125 cm³ 
= 216 cm³ 

Let a be the edge of the new cube.
Volume  = 216 cm³ 
a³  = 216
a³  = 6³
a = 6 cm
Therefore, 6 cm is the edge of cube.

Question 6

Three cubes, whose edges are x cm, 8 cm, and 10 cm respectively, are melted and recast into a single cube of edge 12 cm. Find 'x'.

Sol:

Volume of melted single cube x³ + 8³ + 10³ cm³

= x³ + 512 + 1000 cm³
= x³ + 1512 cm³

Given that 12 cm is edge of the single cube.

12³ = x³ + 1512 cm³
x³ = 12³ - 1512 cm³
x³ = 1728 - 1512
x³  = 216
x³ = 6³
x = 6 cm

Question 7

Three equal cubes are placed adjacently in a row. Find the ratio of the total surfaced area of the resulting cuboid to that of the sum of the total surface areas of the three cubes.

Sol:

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

The total surface area of one cube = 6a²

The total surface area of 3 cubes  = 3 x 6a² = 18a²

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

Breadth and height of cuboid = a

The total surface area of the cuboid  = 2( 3a² + a² +  3a² ) = 14a²

The ratio of total surface area of a cuboid to the total surface area of 3 cubes = ${\displaystyle \frac{14a^2}{18a^2}=\frac{7}{9}}$

Question 8

The cost of papering the four walls of a room at 75 paise per square meter Rs. 240. The height of the room is 5 meters. Find the length and the breadth of the room, if they are in the ratio 5 : 3. 

Sol:

let the length and breadth of the room is 5x and 3x respectively.

Given that the four walls of a room at 75 paise per square met Rs. 240.
Thus,

240 = Area x 0.75

Area = ${\displaystyle \frac{240}{0.75}}$

Area = ${\displaystyle \frac{24000}{75}}$

Area = 320 m

Area = 2 x Height ( Length + Breadth )
320 = 2 x 5 ( 5x + 3 x )
320 = 10 x 8x
32 = 8x
x = 4

Length = 5x
= 5 ( 4 ) m
= 20 m

breadth = 3x
= 3 ( 4 ) m
= 12 m

Question 9

The area of a playground is 3650 m². Find the cost of covering it with gravel 1.2 cm deep, if the gravel costs Rs. 6.40 per cubic metre.

Sol:

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 per cubic meter is Rs. 6.40, therefore the total cost will be:
43.8 x Rs. 6.40 = Rs. 280.32.

Question 10

A square plate of side 'x' cm is 8 mm thick. If its volume is 2880 cm³; find the value of x.

Sol:

We know that

1 mm = ${\displaystyle \frac{1}{10}}$ cm

8 mm = ${\displaystyle \frac{8}{10}}$ cm

Volume = Base area x Height

⇒ 2880 cm³ = ${\displaystyle x\times x\times \frac{8}{10}}$

⇒ 2880 x ${\displaystyle \frac{10}{8}}$ = x²

⇒ x² = 3600

⇒ x = 60 cm.

Question 11

The external dimensions of a closed wooden box are 27 cm, 19 cm, and 11 cm. If the thickness of the wood in the box is 1.5 cm; find:
(i) The volume of the wood in the box;
(ii) The cost of the box, if wood costs Rs. 1.20 per cm³;
(iii) A number of 4 cm cubes that could be placed into the box.

Sol:

The external volume of the box = 27 x 19 x 11 cm³ = 5643 cm³

Since, external dimensions are 27 cm, 19 cm, 11 cm; the thickness of the wood is 1.5 cm.
∵ Internal dimensions
= ( 27 - 2 x 1.5 )cm, ( 19 - 2 x 1.5 )cm, ( 11 - 2 x 1.5 )cm
= 24 cm, 16 cm, 8 cm

Hence, the internal volume of box = ( 24 x 16 x 8 )cm³ = 3072 cm³

(i) The volume of wood in the box = 5643 cm³ - 3072 cm³ = 2571 cm³

(ii) Cost of wood = Rs. 1.20 x 2571 = Rs. 3085.2

(iii) Vol. of 4 cm cube = 4³ = 64 cm³

Number of 4 cm cubes that could be placed into the box
= ${\displaystyle \frac{3072}{64}}$ =  48.

Question 12

A tank 20 m long, 12 m wide and 8 m deep is to be made of iron sheet. If it is open at the top. Determine the cost of iron-sheet, at the rate of Rs. 12.50 per meter, if the sheet is 2.5 m wide.

Sol:

Area of sheet = Surface area of the tank

⇒ Length of the sheets x it's width = Area of 4 walls of the tank + Area of its base

⇒ Length of the sheet x 2.5 m = 2( 20 + 12 )  x 8m² + 20 x 12m²

⇒ Length of the sheet = 300.8 m

Cost of the sheet = 300.8 x Rs. 12.50 = Rs. 3760

Question 13

A closed rectangular box is made of wood of 1.5 cm thickness. The exterior length and breadth are respectively 78 cm and 19 cm, and the capacity of the box is 15 cubic decimeters. Calculate the exterior height of the box.

Sol:

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 cm³     ......(1 dm = 10cm, 1 cu dm = 10³ cm³ ) .

15000 cm³  = 75 x 16 x ( h - 3 )

⇒ h - 3 = ${\displaystyle \frac{15000}{75\times 16}}$ = 12.5 cm
⇒ h = 15.5 cm.

Question 14

The square on the diagonal of a cube has an area of 1875 sq. cm. Calculate:
(i) The side of the cube.
(ii) The total surface area of the cube.

SoL:

(i) If the side of the cube = a cm

The length of its diagonal = a√3 cm
And,
( a√3 )² = 1875
a = 25 cm

(ii) Total surface area of the cube = 6a²
= 6( 25 )² = 3750 cm².

Question 15

A hollow square-shaped tube open at both ends is made of iron. The internal square is of 5 cm side and the length of the tube is 8 cm. There are 192 cm³ of iron in this tube. Find its thickness.

Sol:

Given that the volume of the iron in the tube 192 cm³
Let the thickness of the tube = x cm
Side of the external square = ( 5 + 2x ) cm

∵ Ext. vol. of the tube its internal vol.= volume of iron in the tube, we have,
( 5 + 2x )( 5 + 2x ) x 8 - 5 x 5 x 8 = 192
( 25 + 4x² + 20x ) x 8 - 200 = 192
200 + 32x² + 160x - 200 = 192
32x² + 160x - 192 = 0
x² + 5x - 6 = 0
x² + 6x - x - 6 = 0
x( x + 6 ) - ( x + 6 ) = 0
( x + 6 )( x - 1 ) = 0
( x - 1 ) = 0
x = 1
Therefore, thickness is 1 cm.

Question 16

Four identical cubes are joined end to end to form a cuboid. If the total surface area of the resulting cuboid as 648 m²; find the length of the edge of each cube. Also, find the ratio between the surface area of the resulting cuboid and the surface area of a cube.

Sol:

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

The length of the resulting cuboid = 4 x l = 4 l cm

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

∵ The total surface area of the resulting cuboid
       = 2( l x b + b x h + h x l )

648 = 2( 4l x l + l x l + l x 4l )

4l² + l² + 4l² = 324

 9l² = 324
l² = 36
l = 6 cm

Therefore, the length of each cube is 6 cm.

${\displaystyle \frac{\text{Surface area of the resulting cuboid}}{\text{Surface area of cube}}=\frac{648}{6l^2}}$

${\displaystyle \frac{\text{Surface area of the resulting cuboid}}{\text{Surface area of cube}}=\frac{648}{6{\left(6\right)}^2}}$

${\displaystyle \frac{\text{Surface area of the resulting cuboid}}{\text{Surface area of cube}}=\frac{648}{216}=\frac{3}{1}=3:1}$