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.

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

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

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