Chapter 18 : Surface Areas and Volume of a Cuboid and Cube - Rd Sharma Solutions for Class 9 Maths CBSE

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Chapter 18 - Surface Areas and Volume of a Cuboid and Cube Excercise Ex. 18.1

Question 1
Solution 1
Question 2

Solution 2

Question 3
Solution 3
Question 4
Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with coloured paper with picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively. How many square sheets of paper of side 40 cm would she require?

Solution 4

Question 5

Solution 5
Question 6
Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Solution 6

Question 7
Solution 7
Question 8
Solution 8
Question 9
Hameed has built a cubical water tank with lid for his house, with each outer edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm. Find how much he would spend for the titles, if the cost of tiles is Rs. 360 per dozen.
Solution 9

Question 10
Solution 10
Question 11

Solution 11

Question 12

Ravish wanted to make a temporary shelter for his car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m x 3m?

Solution 12
Length of shelter = 4 m
Breadth of shelter = 3 m
Height of shelter = 2.5 m

The tarpaulin will be required for top and four sides of the shelter.
Area of Tarpaulin required = 2(lh + bh) + lb
= [2(4  2.5 + 3 2.5) + 4 3] m2
= [2(10 + 7.5) + 12] m2
= 47 m2


Question 13

An open box is made of wood 3 cm thick. Its external length, breadth and height are 1.48m, 1.16 m and 8.3 dm. Find the cost of painting the inner surface at Rs 50 per sq meter.

Solution 13

Question 14

Solution 14

Question 15
The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
Solution 15
Total surface area of one brick = 2(lb + bh + lh)
 = [2(22.5 × 10 + 10 × 7.5 + 22.5 × 7.5)]cm2
 = 2(225 + 75 + 168.75)
          = (2 × 468.75) cm2
 = 937.5 cm2   
Let n number of bricks be painted by the container.
Area of n bricks = 937.5n cm2   
Area that can be painted by the container = 9.375 m2 = 93750 cm2
 93750 = 937.5n
n = 100
Thus, 100 bricks can be painted out by the container.

 

Question 16

Solution 16

Question 17

The cost of preparing the walls of a room 12 m long at the rate of Rs 1.35 per square meter is Rs 340.20 and the cost of matting the floor at 85 paise per square meter is Rs 91.80. Find the height of the room.

Solution 17

Question 18

Solution 18

Question 19
A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm. The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm2 and the rate of painting is 10 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf.
 
                                          
Solution 19
    External length (l) of bookshelf = 85 cm
    External breadth (b) of bookshelf = 25 cm
    External height (h) of bookshelf = 110 cm
    External surface area of shelf while leaving front face of shelf
                                                  = lh + 2 (lb + bh)
                                                  = [85  110 + 2 (85 25 + 25 110)] cm2
                                                  = 19100 cm2
    Area of front face = [85 110 - 75 100 + 2 (75 5)] cm2
                                                  = 1850 + 750 cm2
                                                  = 2600 cm2
    Area to be polished = (19100 + 2600) cm2 = 21700 cm2    
    Cost of polishing 1 cm2 area = Rs 0.20
    Cost of polishing 21700 cm2 area = Rs (21700 0.20) = Rs 4340    
    
    Now, length (l), breadth (b) height (h) of each row of bookshelf is 75 cm, 20 cm, and
    30cm  respectively.
    Area to be painted in 1 row = 2 (l + h) b + lh
                                           = [2 (75 + 30) 20 + 75 30] cm2
                                           = (4200 + 2250) cm2
                                           = 6450 cm2    
    Area to be painted in 3 rows = (3 6450) cm2 = 19350 cm2
    Cost of painting 1 cm2 area = Rs 0.10
    Cost of painting 19350 cm2 area = Rs (19350 0.10) = Rs 1935    
    Total expense required for polishing and painting the surface of the bookshelf 
                                            = Rs(4340 + 1935) = Rs 6275

Chapter 18 - Surface Areas and Volume of a Cuboid and Cube Excercise Ex. 18.2

Question 1
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How, many litres of water can it holds? 
Solution 1
Volume of tank = l  b h = (6  5 4.5) m3 = 135 m3
  It is given that:
 1 m3 = 1000 litres

    
 Thus, the tank can hold 135000 litres of water.

Question 2
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?   
Solution 2
Let height of cuboidal vessel be h.
Length (l) of vessel = 10 m
Width (b) of vessel = 8 m
Volume of vessel = 380 m3
  b h = 380 
       
 10 8 h = 380
 h = 4.75
    
 Thus, the height of the vessel should be 4.75 m.    

Question 3
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs 30 per m3.

Solution 3
Length (l) of the cuboidal pit = 8 m
Width (b) of the cuboidal pit = 6 m
 Depth (h) of the cuboidal pit = 3 m
Volume of the cuboidal pit = l  b h = (8 6 3)  = 144 m3  
Cost of digging 1 m3 = Rs 30
Cost of digging 144 m3 = Rs (144 30) = Rs 4320

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16
A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?
Solution 16
Length (l) of the cuboidal tank = 20 m
Breadth (b) of the cuboidal tank = 15 m
Height (h) of the cuboidal tank = 6 m

Capacity of tank = l × b × h    = (20 × 15 × 6) m3 = 1800 m3 = 1800000 litres

Water consumed by people of village in 1 day = 4000 × 150 litres = 600000 litres

Let water of this tank lasts for n days.
Water consumed by all people of village in n days = capacity of tank
n × 600000 = 1800000
n = 3    
Thus, the water of tank will last for 3 days.

Question 17

A child playing with building blocks, which are of the shape of the cubes, has built a structure as shown in fig. If the edge of each cube is 3 cm, find the volume of the structure built by the child

Solution 17

Question 18
A godown measures 40 m  25 m  10 m. Find the maximum number of wooden crates each measuring 1.5 m  1.25 m  0.5 m that can be stored in the godown.    

Solution 18
Length  of the godown = 40 m
    Breadth  of the godown = 25 m
    Height  of the godown = 10 m

Volume of godown = l1 b1 h1 = (40  25  10)  = 10000     

    Length  of a wooden crate = 1.5 m
    Breadth  of a wooden crate = 1.25 m
    Height  of a wooden crate = 0.5 m

Volume of a wooden crate =      = (1.5  1.25  0.5) m3 = 0.9375

Let n wooden crates be stored in the godown.
Volume of n wooden crates = volume of godown
0.9375  n = 10000

 

Thus, 10666 wooden crates can be stored in godown.

Question 19

A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm x 12 cm x 8 cm, how many bricks would be required?

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Solution 22
Rate of water flow = 2 km per hour  
Depth (h) of river = 3 m
Width (b) of river = 40 m
Volume of water flowed in 1 min  
Thus, in 1 minute 4000  = 4000000 litres of water will fall into the sea.    

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

A rectangular container, whose base is a square of side 5 cm, stands on a horizontal table, and holds water upto 1 cm from the top. When a cube is placed in the water it is completely submerged, the water rises to the top and 2 cubic cm of water overflows. Calculate the volume of the cube and also the length of its edge.

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Chapter 18 - Surface Areas and Volume of a Cuboid and Cube Excercise 18.35

Question 1

begin mathsize 12px style The space length space of space the space longest space rod space that space can space be space fitted space in space straight a space cubical space vessel space of space edge space 10 space cm space long comma space is
left parenthesis straight a right parenthesis space 10 space cm
left parenthesis straight b right parenthesis space 10 square root of 2 space cm
left parenthesis straight c right parenthesis space 10 square root of 3 space cm
left parenthesis straight d right parenthesis space 20 space cm end style

Solution 1

begin mathsize 12px style The space longest space rod space that space can space be space fitted space in space straight a space cube space is space diagonal space of space straight a space cube.
Now comma space if space straight a space is space side space of space straight a space cube comma space then space diagonal space equals space square root of straight a squared plus straight a squared plus straight a squared end root equals square root of 3 straight a
rightwards double arrow Diagonal space of space straight a space cube space of space edge space 10 space cm space equals space 10 square root of 3 space cm
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 2

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

(a) 7 : 9

(b) 49 : 81

(c) 9 : 7

(d) 27 : 23

Solution 2

begin mathsize 12px style Surface space of space one space cube space equals 6 left parenthesis edge right parenthesis squared equals space 6 straight a squared
rightwards double arrow Surface space area space of space three space cubes space equals 3 space cross times 16 straight a squared equals space 18 straight a squared
Surface space Area space of space cuboid space formed
equals 2 left parenthesis lb space plus space bh space plus space hl right parenthesis
equals 2 left parenthesis 3 straight a cross times straight a space plus space straight a cross times straight a space plus space straight a space cross times space 3 straight a right parenthesis
equals 2 left parenthesis 3 straight a squared plus straight a squared plus 3 straight a squared right parenthesis
equals 2 space cross times space 7 straight a squared
equals 14 straight a squared
rightwards double arrow Required space Ratio space equals space fraction numerator Surface space area space of space resulting space cuboid over denominator Surface space area space of space three space cubes end fraction equals fraction numerator 14 straight a squared over denominator 18 straight a squared end fraction space equals 7 over 9 equals 7 space colon space 9
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 3

begin mathsize 12px style If space the space length space of space straight a space diagonal space of space straight a space cube space is space 8 square root of 3 space cm comma space then space its space surface space area space is
left parenthesis straight a right parenthesis space 512 space cm squared
left parenthesis straight b right parenthesis space 384 space cm squared
left parenthesis straight c right parenthesis space 192 space cm squared
left parenthesis straight d right parenthesis space 768 space cm squared end style

Solution 3

begin mathsize 12px style Length space of space diagonal space of space straight a space cube space of space side space straight a space equals space 8 square root of 3 space cm
rightwards double arrow square root of 3 straight a space equals space space 8 square root of 3
rightwards double arrow straight a space equals space 8 space cm
So comma space Surface space Area space of space straight a space Cube space equals space 6 straight a squared equals space 6 space cross times space 8 space cross times space 8 equals 384 space cm squared space space space
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 4

begin mathsize 12px style If space the space volumes space of space two space cubes space are space in space the space ratio space 8 space colon space 1 comma space then space the space ratio space of space their space edges space is space
left parenthesis straight a right parenthesis space 8 space colon space 1
left parenthesis straight b right parenthesis space 2 square root of 2 space colon space 1
left parenthesis straight c right parenthesis space 2 space colon space 1
left parenthesis straight d right parenthesis space none space of space these end style

Solution 4

begin mathsize 12px style If space side space of space 1 to the power of st space cube space equals space straight a space and space side space of space 2 to the power of nd space cube space equals space straight b comma space then
fraction numerator Volume space of space 1 to the power of st space cube over denominator Volume space of space 2 to the power of nd space cube end fraction space equals space straight a cubed over straight b cubed equals 8 over 1
rightwards double arrow straight a over straight b equals 2 over 1
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 5

begin mathsize 12px style The space volume space of space straight a space cube space whose space surface space area space is space 96 space cm squared comma space is
left parenthesis straight a right parenthesis space 16 square root of 2 space cm cubed
left parenthesis straight b right parenthesis space 32 space cm squared
left parenthesis straight c right parenthesis space 64 space cm cubed
left parenthesis straight d right parenthesis space 216 space cm cubed end style

Solution 5

begin mathsize 12px style Surface space Area space of space Cube space equals space 6 straight a squared space equals space 96 space cm squared
rightwards double arrow straight a squared space equals space 16 space cm squared
rightwards double arrow straight a space equals space 4 space cm
rightwards double arrow Volume space equals space straight a cubed space equals space 4 cubed space equals space 64 space cm cubed
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 6

The length, width and height of a rectangular solid are in the ratio of 3 : 2 : 1. If the volume of the box is 48 cm3, the total surface area of the box is

(a) 27 cm2

(b) 32 cm2

(c) 44 cm2

(d) 88 cm2

Solution 6

Let,

Length = 3x,

Width = 2x

Height = x

Volume = 48 cm3

L×W×H = 48 cm3

3x × 2x × x = 48 cm3

6x3 = 48 cm3

x3 = 8 cm3

x = 2 cm

Total Surface area

= 2(3x × 2x + 2x × x + 3x × x)

= 2(6x2 + 2x2 + 3x2

= 2(11x2)

= 22x2

= 22(4)

= 88 cm2

Hence, correct option is (d).

Question 7

If the areas of the adjacent faces of a rectangular block are in the ratio 2 : 3 : 4 and the volume is 9000 cm3, then the length of the shortest edge is

(a) 30 cm

(b) 20 cm

(c) 15 cm

(d) 10 cm

Solution 7

begin mathsize 12px style Three space adjacent space Faces space are space AEHD comma space DCGH comma space EHGF
Ratio space of space areas comma
bc space colon space ac space colon space ab space equals space 2 space colon space 3 space colon space 4 space space..... open parentheses 1 close parentheses
Volume comma space abc space equals space 9000 space cm cubed
From space eq. space open parentheses 1 close parentheses comma space let space bc space equals space 2 straight x comma space space ac space equals space 3 straight x comma space space ab space equals space 4 straight x
rightwards double arrow left parenthesis 2 straight x space cross times space 3 straight x space cross times space 4 straight x right parenthesis space equals space left parenthesis bc right parenthesis left parenthesis ac right parenthesis left parenthesis ab right parenthesis
rightwards double arrow 24 straight x cubed space equals space left parenthesis abc right parenthesis squared space
rightwards double arrow 24 straight x cubed space equals space left parenthesis 9000 right parenthesis squared equals 81000000
straight x cubed space equals space fraction numerator 81 space cross times space 10 to the power of 6 over denominator 24 end fraction space equals fraction numerator 27 cross times 10 to the power of 6 over denominator 8 end fraction
rightwards double arrow space straight x space equals space fraction numerator 3 space cross times space 10 squared over denominator 2 end fraction rightwards double arrow straight x space equals space 150 space cm
rightwards double arrow bc space equals space 300 space cm squared comma space space space ac space equals space 450 space cm squared comma space space space ab space equals space 600 space cm squared
Thus comma space we space have
straight a equals abc over bc equals 9000 over 300 equals space 30 space cm comma space space straight b space equals space abc over ac equals fraction numerator begin display style 9000 end style over denominator 450 end fraction equals 20 space cm comma space space straight c space equals abc over ab equals 9000 over 600 equals 15 space cm
rightwards double arrow The space length space of space the space Shortest space edge space equals space 15 space cm.
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Chapter 18 - Surface Areas and Volume of a Cuboid and Cube Excercise 18.36

Question 1

If each edge of a cube, of volume V, is doubled, then the volume of the new cube is

(a) 2 V

(b) 4 V

(c) 6 V

(d) 8 V

Solution 1

Let edge = a

Volume, V = a3

If a' = 2a, then

V' = (a')3 = (2a)3 = 8a3

V' = 8 V

Hence, correct option is (d).

Question 2

If each edge of a cuboid of surface area S is doubled, then surface area of the new cuboid is

(a) 2 S

(b) 4 S

(c) 6 S

(d) 8 S

Solution 2

begin mathsize 12px style Let space the space edges space of space Cuboid space be space straight a comma space straight b comma space straight c
straight S space equals space Surface space Area space equals space 2 left parenthesis ab space plus space bc space plus space ca right parenthesis
If space straight a apostrophe space equals space 2 straight a comma space space straight b apostrophe space equals space 2 straight b comma space space space straight c apostrophe space equals space 2 straight c comma space then space
straight S apostrophe space equals space 2 left parenthesis straight a apostrophe straight b apostrophe plus straight b apostrophe straight c apostrophe plus straight c apostrophe straight a apostrophe right parenthesis space
rightwards double arrow straight S apostrophe space equals space 2 left square bracket left parenthesis 2 straight a right parenthesis left parenthesis 2 straight b right parenthesis space plus space left parenthesis 2 straight b right parenthesis left parenthesis 2 straight c right parenthesis space plus space left parenthesis 2 straight c right parenthesis left parenthesis 2 straight a right parenthesis right square bracket
rightwards double arrow straight S apostrophe space equals space 2 left square bracket 4 ab space plus space 4 bc space plus space 4 ca right square bracket
rightwards double arrow straight S apostrophe equals space 4 space cross times space 2 open parentheses ab space plus space bc space plus space ca close parentheses
rightwards double arrow straight S apostrophe space equals space 4 space straight S
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 3

The area of the floor of a room is 15 m2. If its height is 4 m, then the volume of the air contained in the room is

(a) 60 dm3

(b) 600 dm3

(c) 6000 dm3

(d) 60000 dm3

Solution 3

begin mathsize 12px style Volume space of space room space equals space Area space of space floor space cross times space Height space of space room
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 15 space straight m squared space cross times space 4 space straight m
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1500 space dm squared space cross times space 40 space dm
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 60000 space dm cubed
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 4

The cost of constructing a wall 8 m long, 4 m high and 20 cm thick at the rate of Rs. 25 per m3 is

(a) Rs. 16

(b) Rs. 80

(c) Rs. 160

(d) Rs. 320

Solution 4

Volume space of space wall equals 8 space straight m space cross times space 4 space straight m space cross times space 20 space cm space equals 8 space straight m space cross times space 4 space straight m space cross times space 1 fifth straight m equals 32 over 5 straight m cubed
rightwards double arrow Cost space of space constructing space wall equals Rs. space open parentheses 25 space cross times space 32 over 5 close parentheses equals Rs. space 160
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. space space

Question 5

10 cubic metres clay is uniformly spread on a land of area 10 ares. The rise in the level of the ground is

(a) 1 cm

(b) 10 cm

(c) 100 cm

(d) 1000 cm

Solution 5

begin mathsize 14px style Area space of space land equals 10 space ares space equals space 10 space cross times space 100 space straight m squared space equals space 1000 space straight m squared
Volume space of space clay equals 10 space straight m cubed
rightwards double arrow Area space of space land space cross times space Rise space in space level space equals space 10 space straight m cubed
rightwards double arrow 1000 space straight m squared space cross times space Rise space in space level space equals space 10 space straight m cubed
rightwards double arrow Rise space in space level space equals space fraction numerator 10 space straight m cubed over denominator 1000 space straight m squared end fraction equals 1 over 100 straight m space equals space 1 space cm
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 6

Volume of a cuboid is 12 cm3. The volume (in cm3) of a cuboid whose sides are double of the above cuboid is

(a) 24

(b) 48

(c) 72

(d) 96

Solution 6

Let the dimensions of Cuboid be a, b, c respectively.

Volume, V = abc = 12 cm3

If a' = 2a,   b' = 2b,   c' = 2c, then 

V' = a'b'c' = 8abc = 8 × 12 = 96 cm3

Hence, correct option is (d).

Question 7

If the sum of all the edges of a cube is 36 cm, then the volume (in cm3) of that cube is

(a) 9

(b) 27

(c) 219

(d) 729

Solution 7

Let the edge of cube = a

Total no. of edge = 12

Sum of all edges = 12a

12a = 36cm

i.e. a = 3 cm

Volume = a3 = 33 = 27 cm3

Hence, correct option is (b).

Question 8

The number of cubes of sides 3 cm that can be cut from a cuboid of dimensions 10 cm × 9 cm × 6 cm, is

(a) 9

(b) 10

(c) 18

(d) 20

Solution 8

begin mathsize 12px style Volume space of space cuboid space equals space 10 space cm space cross times space 9 space cm space cross times space 6 space cm space equals space 540 space cm squared
Volume space of space Cube space equals space open parentheses 3 space cross times space 3 space cross times space 3 close parentheses space cm cubed space equals space 27 space cm cubed
No. space of space cubes space that space can space be space cut space from space straight a space cuboid space equals space 540 over 27 equals 20
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 9

On a particular day, the rain fall recorded in a terrace 6 m long and 5 m broad is 15 cm. The quantity of water collected in the terrace is

(a) 300 litres

(b) 450 litres

(c) 3000 litres

(d) 4500 litres

Solution 9

begin mathsize 12px style Volume space of space water space collected space
equals space 6 space straight m space cross times space 5 space straight m space cross times space 15 over 100 straight m
equals 450 over 100 straight m cubed
equals 4.5 space straight m
equals space 4.5 space cross times space 1000 space litres space space space space space space space space space space space space space space space space space space space space space
equals space 4500 space litres space space
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 10

begin mathsize 12px style If space straight A subscript 1 comma space straight A subscript 2 space and space straight A subscript 3 space denote space the space areas space of space three space adjacent space faces space of space straight a space cuboid comma space then space its space volume space is
left parenthesis straight a right parenthesis space straight A subscript 1 space straight A subscript 2 space straight A subscript 3
left parenthesis straight b right parenthesis space 2 straight A subscript 1 space straight A subscript 2 space straight A subscript 3
left parenthesis straight c right parenthesis space square root of straight A subscript 1 space straight A subscript 2 space straight A subscript 3 end root
left parenthesis straight d right parenthesis space cube root of straight A subscript 1 space straight A subscript 2 space straight A subscript 3 end root end style

Solution 10

begin mathsize 12px style Let space three space adjacent space faces space be space ADHE comma space DCGH comma space EHGF.
straight A subscript 1 space equals space ar left parenthesis ADHE right parenthesis space equals space bc
straight A subscript 2 space equals space ar left parenthesis DCGH right parenthesis space equals space ca
straight A subscript 3 space equals space ar left parenthesis EHGH right parenthesis space equals space ab
Volume space of space cuboid space equals space abc
Now comma space space straight A subscript 1 straight A subscript 2 straight A subscript 3 space equals space bc cross times ca cross times ab space equals space straight a squared straight b squared straight c squared
Also comma space we space can space write
abc space equals space Volume space of space Cuboid space equals space square root of straight a squared straight b squared straight c squared end root equals square root of straight A subscript 1 straight A subscript 2 straight A subscript 3 end root
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 11

begin mathsize 12px style If space straight l space is space the space length space of space straight a space diagonal space of space straight a space cube space of space volume space space straight V comma space then
left parenthesis straight a right parenthesis space 3 straight V space equals straight l cubed
left parenthesis straight b right parenthesis space square root of 3 straight V space equals space straight l cubed
left parenthesis straight c right parenthesis space 3 square root of 3 straight V space equals space 2 straight l cubed
left parenthesis straight d right parenthesis space 3 square root of 3 straight V space equals space straight l cubed end style

Solution 11

begin mathsize 12px style Length space of space diagonal space of space cube space of space side space straight a space equals space square root of 3 straight a
Volume comma space straight V space equals space straight a cubed space
rightwards double arrow left parenthesis straight V right parenthesis to the power of bevelled 1 third end exponent equals space straight a
Now comma space length space of space diagonal space comma space straight l space equals space square root of 3 space left parenthesis straight V right parenthesis to the power of 1 divided by 3 end exponent
rightwards double arrow straight l cubed space equals space 3 square root of 3 space straight V
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 12

begin mathsize 12px style If space straight V space is space the space volume space of space straight a space cuboid space of space dimensions space straight x comma space straight y comma space straight z space and space straight A space is space its space surface space area comma space then space straight A over straight V equals
left parenthesis straight a right parenthesis space straight x squared straight y squared straight z squared
left parenthesis straight b right parenthesis space 1 half open parentheses 1 over xy plus 1 over yz plus 1 over zx close parentheses
left parenthesis straight c right parenthesis space 1 half open parentheses 1 over straight x plus 1 over straight y plus 1 over straight z close parentheses
left parenthesis straight d right parenthesis space 1 over xyz end style

Solution 12

begin mathsize 12px style Sides space of space straight a space cuboid space are space straight x comma space straight y comma space straight z
Volume space equals xyz space equals space straight V
Surface space Area space equals space 2 left parenthesis xy space plus space yz space plus space zx right parenthesis space equals space straight A
straight A over straight V equals fraction numerator 2 open parentheses xy space plus space yz space plus space zx close parentheses over denominator xyz end fraction space equals space 2 open parentheses 1 over straight z plus 1 over straight x plus 1 over straight y close parentheses equals space 2 open parentheses 1 over straight x plus 1 over straight y plus 1 over straight z close parentheses
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 13

begin mathsize 12px style The space sum space of space the space length comma space breadth space and space depth space of space straight a space cuboid space is space 19 space cm space and space its space diagonal space is space 5 square root of 5 space cm. space
Its space surface space area space is space
left parenthesis straight a right parenthesis space 361 space cm squared
left parenthesis straight b right parenthesis space 125 space cm squared
left parenthesis straight c right parenthesis space 236 space cm squared
left parenthesis straight d right parenthesis space 486 space cm squared end style

Solution 13

begin mathsize 12px style If space straight a comma space straight b comma space straight c space are space the space sides space of space straight a space cuboid comma space then
straight a space plus space straight b space plus space straight c space equals space 19 space cm
Length space of space diagonal space equals space 5 square root of 5
rightwards double arrow space square root of straight a squared space plus space straight b squared space plus space straight c squared end root equals 5 square root of 5
rightwards double arrow straight a squared space plus space straight b squared space plus space straight c squared equals open parentheses 5 square root of 5 close parentheses squared equals 25 cross times 5 equals 125 space cm
Now comma space left parenthesis straight a space plus space straight b space plus space straight c right parenthesis squared space space space equals space straight a squared space plus space straight b squared space plus space straight c squared space plus space 2 left parenthesis ab space plus space bc space plus space ca right parenthesis
rightwards double arrow left parenthesis 19 right parenthesis squared space equals 125 space space plus space Surface space Area space of space Cuboid
rightwards double arrow Surface space Area space of space straight a space Cuboid space equals space 361 space minus space 125 space equals space 236 space cm squared
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 14

If each edge of a cube is increased by 50%, the percentage increase in its surface area is

(a) 50%

(b) 75%

(c) 100%

(d) 125%

Solution 14

begin mathsize 12px style Surface space Area space of space straight a space Cube space of space side space straight a space equals space 6 straight a squared
If space straight a space is space incresed space by space 50 percent sign comma
straight a apostrophe left parenthesis new space side right parenthesis space equals space straight a space plus space straight a over 2 equals fraction numerator 3 straight a over denominator 2 end fraction
So space new space Surface space Area space equals space straight S apostrophe space equals space 6 straight a apostrophe squared equals 6 open parentheses fraction numerator 3 straight a over denominator 2 end fraction close parentheses squared space equals space fraction numerator 6 cross times 9 cross times straight a squared over denominator 4 end fraction equals space 13.5 space straight a squared space
Increase space in space area space equals space 13.5 straight a squared space minus space 6 straight a squared equals space 7.5 straight a squared space
percent sign space Increase space equals space fraction numerator 7.5 up diagonal strike straight a squared end strike over denominator 6 up diagonal strike straight a squared end strike end fraction cross times 100 percent sign space equals space 125 percent sign
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Chapter 18 - Surface Areas and Volume of a Cuboid and Cube Excercise 18.37

Question 1

A cube whose volume is 1/8 cubic centimeter is placed on top of a cube whose volume is 1 cm3. The two cubes are then placed on top of a third cube whose volume is 8 cm3. The height of the stacked cubes is

(a) 3.5 cm

(b) 3 cm

(c) 7 cm

(d) none of these

Solution 1

begin mathsize 12px style Side space of space cube space whose space volume space is space space 1 over 8 cm cubed space equals space open parentheses 1 over 8 close parentheses to the power of 1 divided by 3 end exponent space equals 1 half space cm space space space open curly brackets straight V equals space open parentheses straight a close parentheses cubed close curly brackets
Side space of space Second space cube space whose space volume space is space 1 space cm cubed space equals space open parentheses 1 close parentheses to the power of 1 divided by 3 end exponent space equals space 1 space cm
Side space of space Third space cube space whose space volume space is space 8 space cm cubed equals space open parentheses 8 close parentheses to the power of 1 divided by 3 end exponent space equals space 2 space cm
1 st space Cube space is space placed space on space second space and space second space on space third comma space
then space height space of space stacked space cubes
equals Sum space of space sides space of space all space cubes
equals space open parentheses 1 half plus space 1 space plus space 2 close parentheses space cm
equals space 3.5 space cm
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

CBSE Class 9 Maths Homework Help

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