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.