Rd Sharma 2020 2021 Solutions for Class 9 Maths Chapter 18 Surface Area And Volume Of A Cuboid And Cube are provided here with simple step-by-step explanations. These solutions for Surface Area And Volume Of A Cuboid And Cube are extremely popular among Class 9 students for Maths Surface Area And Volume Of A Cuboid And Cube Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2020 2021 Book of Class 9 Maths Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2020 2021 Solutions. All Rd Sharma 2020 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 18.14:

#### Question 1:

Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm.

#### Answer:

Dimensions are given as

Length

Breadth

Height

We have to find lateral surface area and total area

Hence its, lateral surface area,

Total surface area,

The lateral surface area of the cuboids is and total surface area of it is.

#### Page No 18.14:

#### Question 2:

Find the lateral surface area and total surface area of a cube of edge 10 cm.

#### Answer:

Edge of the given cube,

We have to find lateral and total surface area

Lateral surface area,

Total surface area,

The lateral surface area of the cube is and its total surface area is.

#### Page No 18.14:

#### Question 3:

Find the ratio of the total surface area and lateral surface area of a cube.

#### Answer:

Let the length of the edge of the cube be

We have to find the ratio of total surface area and lateral surface area

Total surface area of the cube,

Lateral surface area of the cube,

The desired ratio,

The ratio of the total surface area and the lateral surface area of a cube is.

#### Page No 18.14:

#### Question 4:

Marry 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?

#### Answer:

The dimensions of the cubical block are,

We are asked to find the number of square sheet paper whose side is 40 cm

Let the total surface area of the block be.

So, Mary would require in total of colored paper.

But the paper is available in square sheets of side,

Area of a single square sheet,

The number of square sheets required=

Mary would require square sheets of paper.

#### Page No 18.14:

#### Question 5:

The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs. 7.50 m^{2}.

#### Answer:

Dimensions of the room are,

Let,

*S* The total surface area to whitewash

*A*_{1}* *The lateral surface area of the room

*A*_{2 } The surface area of ceiling

*R *The rate of whitewashing per

We know that,

We are asked to find the cost of whitewashing

Now, the total surface area to whitewash,

Total cost of whitewashing,

Hence the cost of whitewashing the room and the ceiling is .

#### Page No 18.14:

#### 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 surfaces areas of the three cubes.

#### Answer:

Let,

Side of each cube

Surface area of each cube

So,

Hence,

Sum of surface areas of three cubes,

The length (say *l*) of the newly formed cuboids is;

Its breadth (say *b*) and height (say *h*) will be the same as that of each cube.

Total surface area of the new cuboids is;

Required Ratio,

The total surface area of the new cuboids to that of the sum of the surface area of the three cubes is

#### Page No 18.14:

#### Question 7:

A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes.

#### Answer:

We can define the following notations as follows

Side of cube

Volume of cube

Side of cube

Volume of cube

Let;

Number of cubes formed

Surface area of a single small cube

We know,

And;

The number of cubes formed,

Total surface area of all the small cubes formed

The total surface area of all the small cubes is .

#### Page No 18.14:

#### Question 8:

The length of a hall is 18 m and the width 12 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the hall.

#### Answer:

The hall is cubical.

Let,

Length of the cuboids

Breadth of the cuboids

Height of the cuboids

We have,

We need to find the Height of the hall

It is given that,

The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls

Using abbreviations, we can write the same as,

Height of the wall is .

#### Page No 18.14:

#### Question 9:

Hameed has built a cubical water tank with lid for his house, with each other 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 his tiles, if the cost of tiles is Rs 360 per dozen.

#### Answer:

The water tank is cubical.

So let,

Side of the cube

Total surface area covered by tiles

Side of each square tile

Area of each square tile

Number of tiles required

Cost of each tile

We are asked to find the total cost of the tiles

We have,

.So,

We have,

So;

Now,

The cost of tiles is per dozen.

Hence,

Total cost for the tiles

Hameed would spend for the tiles.

#### Page No 18.14:

#### Question 10:

Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.

#### Answer:

Let,

Initial edge of the cube

Initial surface area of the cube

Increased edge of the cube

Increased surface area of the cube

We need to find the percentage increase in the total surface area of the cube

We know that,

And

Now,

Percentage increase in

Percentage increase in the surface area of the cube is .

#### Page No 18.14:

#### Question 11:

A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of Rs 5 per metre sheet, sheet being 2 m wide.

#### Answer:

We know,

Length of the iron tank

Width of the iron tank

Depth of the iron tank

Width of the iron sheet

Rate of the iron sheet

We need to find the cost of iron sheet used

Total surface area of the iron tank,

Length of the iron sheet required,

Cost of the required iron sheet,

The total cost of iron sheet used is .

#### Page No 18.14:

#### Question 12:

Ravish wanted to make a temporary shelter for his car by making a box-like structure with tarpaulin that covers al the four sides and the top of the car (with the front face as 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 witt base dimensions 4 m ✕ 3 m?

#### Answer:

The shelter is a box-like structure and hence cubical.

We have,

Length of the cuboid

Breadth of the cuboid

Height of the cuboid

The total surface area of the cuboid,

The area of the base,

The quantity of tarpaulin required,

So the quantity of tarpaulin required is.

#### Page No 18.14:

#### Question 13:

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

#### Answer:

The external dimensions of the wooden box,

Length

Breadthand height is

Thickness of the wood

We are asked to find the cost of painting

So, the internal dimensions of the box are,

Length

Breadth

Height

The internal surface area of the box,

We are given the rate of painting per square meter is

So the total cost of painting is,

The total cost of painting is .

#### Page No 18.14:

#### Question 14:

The dimensions of a room are 12.5 m by 9 m by 7 m . There are 2 doors and 4 windows in the room; each door measures 2.5 m by 1.2 m and each window 1.5 m by 1 m . Find the cost of painting the walls at Rs 3.50 per square metre.

#### Answer:

We are given the dimensions of the room as *l = *12.5 m, *b *= 9 m, *h *= 7 m

The lateral surface area of the room,

Surface area of each door,

Surface area of each window,

There are 2 doors and 4 windows in the room.

Hence, total area to be painted,

Rate of painting the wall at the rate of,

So, total cost of painting,

The total cost of painting is .

#### Page No 18.14:

#### Question 15:

The paint in a certain container is sufficient to paint on area equal to 9.375 m^{2}. How many bricks of dimension 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?

#### Answer:

The paint in the container can paint the area,

Dimensions of a single brick,

Length

Breadth

Height

We need to find the number of bricks that can be painted

Surface area of a brick,

Number of bricks that can be painted

Hencebricks can be painted out of the container.

#### Page No 18.15:

#### Question 16:

The dimensions of a rectangular box are in the ratio of 2 : 4 and the difference between the cost of covering it with sheet of paper at the rates of Rs 8 and Rs 9.50 per m^{2} is Rs 1248. Find the dimensions of the box.

#### Answer:

The dimensions of the rectangular box are in the ratio.

So let the dimensions be,

Length

Breadth

Height

We are asked to find the dimensions of the box

The total surface area of the box,

The cost of covering it at the rate of per

The cost of covering it at the rate of per

We know that, the difference between above two costs is.

So,

So the dimensions of the box are;

Hence the dimensions of the box are.

#### Page No 18.15:

#### Question 17:

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

#### Answer:

We have,

Cost of matting the floor

Rate of matting per square meter

Length of the floor

Let,

Area of the floor

Width of the room

So,

Now, we have,

The cost of preparing the walls

The rate of preparing the walls

Let,

Lateral surface area of the room

Height of the room

So,

Hence, height of the room is .

#### Page No 18.15:

#### Question 18:

The length and breadth of a hall are in the ratio 4 : 3 and its height is 5.5 metres. The cost of decorating its walls (including doors and windows) at Rs 6.60 per square metre is Rs 5082. Find the length and breadth of the room.

#### Answer:

The length and breadth of the hall are in the ratio 4 : 3.

Hence *l *= 4*x*, *b *= 3*x*, *h = *5.5 m

Rate of decorating the wall, *R *= 6.6 per square meter

Total cost of decoration *C = Rs.*5082

We have to find the length and breadth of the room

Surface area of the walls,

Cost of decoration = *A* × *R*

Hence,

Length

Breadth

The length and breadth of the hall are and respectively.

#### Page No 18.15:

#### Question 19:

A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (See Figure). The thickness of the plank is 5 cm every where. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm^{2} and the rate of painting is 10 paise per cm^{2}. Find the total expenses required for polishing and painting the surface of the bookshelf.

#### Answer:

External dimensions of the bookshelf are,

Length,

Breadth,

Height,

External surface area of the bookshelf excluding the front face,

Area of the front face,

Area to be polished,

Rate of polishing

Total cost of polishing,

Now, above diagram will make it clear that for each row of bookshelf,

Length

Breadth

Height

Hence, area to be painted in one row,

Area to be painted in three rows,

Rate of painting

Total cost of painting,

Total Expense

Therefore, the total expenses are.

#### Page No 18.29:

#### 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 hold?

#### Answer:

Dimensions of the water tank, *l *= 6 m, *b *= 5 m, *h *= 4.5 m

We need to find the capacity of the tank

Capacity of the tank,

The tank can hold of water.

#### Page No 18.29:

#### Question 2:

A cubical vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?

#### Answer:

We have,

Length of the vessel

Width of the vessel

Capacity of the vessel

Let: Minimum required height of the vessel

So,

Thus, to hold of liquid, the vessel must be minimum high.

#### Page No 18.29:

#### 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 m^{3}.

#### Answer:

We have, the dimensions of the cubical pit are,

Length

Breadth

Depth

Rate of digging

Volume of the pit,

The cost of digging,

The cost of digging the pit is.

#### Page No 18.29:

#### Question 4:

If the areas of three adjacent faces of a cuboid are 8 cm^{2}, 18 cm^{3}^{ }and 25 cm^{3}. Find the volume of the cuboid.

#### Answer:

We know that, areas of three adjacent faces of the cuboid are respectively.

Where,

Length of the cuboid

Breadth of the cuboid

Height of the cuboid

Let,

Volume of the cuboid

We have, areas of three adjacent faces of the cuboid are respectively,

So their product,

The volume of the cuboid is.

#### Page No 18.29:

#### Question 5:

The breadth of a room is twice its height, one half of its length and the volume of the room is 512 cu. dm. Find its dimensions.

#### Answer:

Let,

Length of the room

Breadth of the room

Height of the room

Volume of the room

We have, *b = *2*h, l *= 2*b* and volume of room is 512 dm^{3}

We have to find the dimensions

We know that

We have,

Therefore,

Hence, the dimensions of the cuboid are,

Length, Breadth, Height

Therefore, length is equal to 16 dm, breadth is 8 m and Height is 4 dm.

#### Page No 18.29:

#### Question 6:

Three metal cubes with edges 6 cm, 8 cm and 10 cm respectively are melted together and formed into a single cube. Find the volume, surface area and diagonal of the new cube.

#### Answer:

Let,

Sides of the three small cubes

Volumes of the three small cubes

Side of the new cube formed

Volume of the new cube formed

Surface area of the new cube formed

Diagonal of the new cube formed

We have,

We need to find the volume, surface area and diagonal of the new cube

Now,

We know,

So, the volume, surface area and the diagonal of the new cube will be respectively.

#### Page No 18.30:

#### Question 7:

Two cubes, each of volume 512 cm^{3}^{ }are joined end to end. Find the surface area of the resulting cuboid.

#### Answer:

We have volume of each cube

Let,

Side of each cube

The two cubes are joined together and we are asked to find the surface area of new cuboid

We know that,

When the two cubes are joined end to end,

Dimensions of the resulting cuboid are,

Length

Breadth

Height

Hence, its surface area

The surface area of the resulting cuboid will be .

#### Page No 18.30:

#### Question 8:

A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edge of the two smaller cubes are 6 cm and 8 cm, find the edge of the third smaller cube.

#### Answer:

Let,

Side of the cube

Volume of the cube

Sides of the three smaller cubes

Volumes of the three smaller cubes

We have,

,

We know that,

Edge of the third smaller cube is .

#### Page No 18.30:

#### Question 9:

The dimensions of a cinema hall are 100 m, 50 m and 18 m. How many persons can sit in the hall, if each person requires 150 m^{3} of air?

#### Answer:

Dimensions of the cinema hall are,

Length

Breath

Height

Each person requires of air (say, *v*)

We are asked to find the number of persons who can sit in the cinema hall

Let,

Volume of the hall, then

The number of people that can sit in the hall,

Maximum people can sit in the hall.

#### Page No 18.30:

#### Question 10:

Given that 1 cubic cm of marble weighs 0.25 kg, the weight of marble block 28 cm in with and 5 cm thick is 112 kg. Find the length of the block.

#### Answer:

We are given that 1 cubic cm of marble weighs 0.25 kg

Let,

Volume of the block

Length of the block

We have,

Width of the block

Thickness of the block

Weight of the block

We need to find the length of the block

We have, of marble occupies of volume.

So, of marble will occupy the volume,

The length of the block is.

#### Page No 18.30:

#### Question 11:

A box with lid is made of 2 cm thick wood. Its external length, breadth and height are 25 cm, 18 cm and 15 cm respectively. How much cubic cm of a liquid can be placed in it? Also, find the volume of the wood used in it.

#### Answer:

External dimensions of the box are,

Length

Breadth

Height

Thickness of the wood

We need to find the volume used

So, internal dimensions of the box are,

Length

Breadth

Height

Capacity of the box,

Volume of the wood,

Maximum of liquid can be placed in the box.

Volume of the wood used in the box is .

#### Page No 18.30:

#### Question 12:

The external dimensions of a closed wooden box are 48 cm, 36 cm, 30 cm. The box is made of 1.5 cm thick wood. How many bricks of size 6 cm × 3 cm × 0.75 cm can be put in this box?

#### Answer:

External dimensions of the closed wooden box,

Length

Breath

Height

Thickness of the wood

We need to find number of bricks that can be put inside the box of dimension

Internal dimensions of the box,

Length

Breadth

Height

Capacity of the box,

Volume of each brick,

Number of bricks that can be put in the box,

The box can contain maximum bricks.

#### Page No 18.30:

#### Question 13:

A cube of 9 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base are 15 cm and 12 cm, find the rise in water level in the vessel.

#### Answer:

“When an object is completely immersed in a liquid, the volume of the liquid displaced is equal to the volume of the object”

Using this principle, we shall now solve this problem.

We have,

Edge of the immersed cube

Length of the rectangular container

Breadth of the rectangular container

Let,

The rise in water level

Volume of the immersed cube

As per the above mentioned principle,

The rise in water level is .

#### Page No 18.30:

#### Question 14:

A field is 200 m long and 150 m broad. There is a plot, 50 m long and 40 m broad, near the field. The plot is dug 7 m deep and the earth taken out is spread evenly on the field. By how many metres is the level of the field raised? Give the answer to the second place of decimal.

#### Answer:

We have, dimensions of the plot that is dug,

Length

Breadth

Depth

Length of the field

Breadth of the field

We need to find the level of field raised

Here, the volume of earth taken out,

So the rise in the level of the field

The level of the field is raised by .

#### Page No 18.30:

#### Question 15:

A field is in the form of a rectangle of length 18 m and width 15 m. A pit, 7.5 m long, 6 m broad and 0.8 m deep, is dug in a corner of the field and the earth taken out is spread over the remaining area of the field. Find out the extent to which the level of the field has been raised.

#### Answer:

We have,

Length of the field (*L*) = 18 m

Width of the field (*B*) = 15 m

Length of the pit (*l*) = 7.5 m

Breadth of the pit (*b*) = 6 m

Depth of the pit (*h*) = 0.8 m

We have to find the level of field raised

Volume of the earth dug out

The area on which the earth has to be spread,

The rise in the level of the field

The level of the field has been raised to .

#### Page No 18.30:

#### 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?

#### Answer:

Let,

Number of days the water will last for

One villager requires of water per day.

Population of the village is 4000

Measurement of the tank is

We need to find for how many days the water of the tank will last

The requirement of water for all 4000 villagers for days,

But we are given;

The water of this tank will last for.

#### Page No 18.30:

#### Question 17:

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

#### Answer:

We have,

Number of boxes

In the above structure we need to find the total volume

Edge of each cube

Volume of each cube

Hence, total volume of the structure,

The volume of the structure built by the child is.

#### Page No 18.31:

#### Question 18:

A godown measure 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.

#### Answer:

We have,

Volume of the godown

Volume of each crate

We need to find the maximum number of crates in the godown that can be placed

Hence, the number of crates that can be stored,

But, we can not place this amount of crates in the godown, as this is not an integer.

So, we can place maximumcrates in the godown.

#### Page No 18.31:

#### 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 × 12cm × 8cm, how many bricks would be required.

#### Answer:

We have,

Length of the wall

Height of the wall

Thickness of the wall

Dimension of the brick is

We need to find the number of bricks

Here,

Volume of the wall,

Dimensions of the brick are,

So, number of bricks in the wall,

As this is not an integer, we should take least integer greater than.

So, we need bricks to build the wall.

#### Page No 18.31:

#### Question 20:

If *V* is the volume of a cuboid of dimensions *a*, *b*, *c* and *S* is its surface area, then prove that

$\frac{1}{V}=\frac{2}{S}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

#### Answer:

We have,

Volume of the cuboid

Surface area of the cuboid

Dimensions of the cuboid

We need to prove,

We know that,

And

Hence,

#### Page No 18.31:

#### Question 21:

The areas of three adjacent faces of a cuboid are x, y and z. If the volume is *V*, prove that *V*^{2} = *xyz*.

#### Answer:

Let,

Length of the cuboid

Breadth of the cuboid

Height of the cuboid

Volume of the cuboid

Areas of three adjacent faces of the cuboid

We know that, areas of three adjacent faces of the cuboid are *lb*, *bh*, and *hl* respectively

Hence,

Hence,

#### Page No 18.31:

#### 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?

#### Answer:

River is flowing at the speed of,

Width of the river

Depth of the river

We need to find the water flowed in one minute

So, in one minute, the river covers the distance (say, *l*) of .

Quantity of water that will fall into the sea in one minute,

So, in one minute of water will fall into the sea.

#### Page No 18.31:

#### Question 23:

Water in a canal 30 dm wide and 12 dm deep, is flowing with a velocity of 100 km per hour. How much area will it irrigate in 30 minutes if 8 cm of standing water is desired?

#### Answer:

We are given;

Velocity of the water

So, in 30 min, it will go the distance (say, *l*) 5,00,000 dm.

Width of the canal

Depth of the canal

In 30 min, quantity of water flown,

If 8 cm of standing water is desired, then the area that will be irrigated,

In 30 min, it will irrigate the area of .

#### Page No 18.31:

#### Question 24:

Half cubic metre of gold-sheet is extended by hammering so as to cover an area of 1 hectare. Find the thickness of the gold-sheet.

#### Answer:

Volume of the gold sheet,

Area it covers,

Let,

Thickness of the sheet

We know that,

Thickness of the sheet is .

#### Page No 18.31:

#### Question 25:

How many cubic centimetres of iron are there in an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cubic cm of iron weighs 15 g, find the weight of the empty box in kg.

#### Answer:

We have,

External dimensions of the iron box are,

Length

Breadth

Height

Thickness of ironand of iron weighs

We are asked to find the volume of the metal used in the box and weight of the empty box

Internal dimensions of the box,

Length

Breadth

Height

Let,

External volume of the box

Internal volume of the box

Volume of the iron

So,

We have, of iron weighs,

So, weight of of iron,

In that open box, there areof iron, and weight of the empty box is .

#### Page No 18.31:

#### 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.

#### Answer:

“When an object is completely immersed in a liquid, the volume of the liquid displaced is equal to the volume of the object”

Using this principle, we shall now solve this problem.

We have,

Length of the container

Breadth of the container

Height to which the water raised

Volume of the water displaced,

= volume of the water raised + volume of the water over flown

We need to calculate the volume and edge of the cube

Let,

Volume of the cube submerged

Edge of the cube submerged

According to the principle mentioned above,

Volume of the cube is and edge of the cube is .

#### Page No 18.31:

#### Question 27:

A rectangular tank is 80 m long and 25 m broad. Water flows into it through a pipe whose cross-section is 25 cm^{2}, at the rate of 16 km per hour. How much the level of the water rises in the tank in 45 minutes.

#### Answer:

It is given that

Length of the tank (*l*) = 80 m

Breadth of the tank (*b*) = 25 m

= 2500 cm

Area of the cross section of the pipe is 25 cm^{2}

The water flows at the rate of 16 km/hr.

We are asked to find the level of tank raised in 45 minutes

In 45 minutes, the water through the pipe will go,

Area of the cross-section of the pipe is 25 cm^{2}.

So, the quantity of the water poured in 45 minutes,

Let,

Height to which the water is raised

So,

In 45 minutes, the water level rises by .

#### Page No 18.31:

#### Question 28:

Water in a rectangular reservoir having base 80 m by 60 m is 6.5 deep. In what time can the water be emptied by a pipe of which the cross-section is a square of side 20 cm, if the water runs through the pipe at the rate of 15 km/hr.

#### Answer:

Dimensions of the reservoir are,

Length

Breadth

Depth

Side of the cross-section of the pipe

The flow of water through the water is;

We are asked to find the time in which the reservoir can be emptied

Here, volume of the water in the reservoir,

Since the side of the cross-section of the pipe

So, the area of the cross-section of the pipe,

Velocity of the water,

Let,

Time required emptying the reservoir

So,

Using that pipe, the water is emptied in.

#### Page No 18.35:

#### Question 1:

The length of the longest rod that can be fitted in a cubical vessel of edge 10 cm long, is

(a) 10 cm

(b) 10$\sqrt{2}$ cm

(c) 10$\sqrt{3}$ cm

(d) 20 cm

#### Answer:

The longest rod that can be fitted in the cubical vessel is its diagonal.

Side of the cube

So, the diagonal of the cube,

So, the length of the longest rod that can be fitted in the cubical box is.

Hence, the correct choice is (c).

#### Page No 18.35:

#### 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

#### Answer:

Let, Side of each cube

So, the dimensions of the resulting cuboid are,

Length

Breadth

Height

Total surface area of the cuboid,

Sum of the surface areas of the three cubes,

Required ratio,

Thus, the required ratio is.

Hence the correct choice is (a).

#### Page No 18.35:

#### Question 3:

If the length of a diagonal of a cube is $8\sqrt{3}$ cm, then its surface area is

(a) 512 cm^{2}

(b) 384 cm^{2}

(c) 192 cm^{2}

(d) 768 cm^{2}

#### Answer:

Let,

Side of the cube

Length of the diagonal

We have to find the surface area of the cube

Surface area of the cube,

Thus, surface area of the cube is.

Hence, the correct choice is (b).

#### Page No 18.35:

#### Question 4:

If the volumes of two cubes are in the ratio 8: 1, then the ratio of their edges is

(a) 8 : 1

(b) $2\sqrt{2}:1$

(c) 2 : 1

(d) none of these

#### Answer:

Let,

Volumes of the two cubes

Edges of the two cubes

We know that,

So,

Ratio of their edges is.

So, the correct choice is (c).

#### Page No 18.35:

#### Question 5:

The volume of a cube whose surface area is 96 cm^{2}, is

(a) $16\sqrt{2}c{m}^{3}$

(b) 32 cm^{3}

(c) 64 cm^{3}

(d) 216 cm^{3}

#### Answer:

Let,

Side of the cube

Volume of the cube

Surface area of the cube

We have,

So,

Thus, volume of the cube is.

Hence the correct choice is (c).

#### Page No 18.35:

#### 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 48cm^{3}, the total surface area of the box is

(a) 27 cm^{2}

(b) 32 cm^{2}

(c) 44 cm^{2}

(d) 88 cm^{2}

#### Answer:

Length (*l*), width (*b*) and height (*h*) of the rectangular solid are in the ratio 3 : 2 : 1.

So we can take,

We need to find the total surface area of the box

Volume of the box,

Thus,

Surface area of the box,

Thus total surface area of the box is.

Hence, the correct option is (d).

#### Page No 18.35:

#### Question 7:

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

(a) 30 cm

(b) 20 cm

(c) 15 cm

(d) 10 cm

#### Answer:

Let, the edges of the cuboid be *a* cm, *b* cm and *c* cm.

And, *a* < *b* < *c*

The areas of the three adjacent faces are in the ratio 2 : 3 : 4.

So,

*ab* : *ca* : *bc* = 2 : 3 : 4, and its volume is 9000 cm^{3}

We have to find the shortest edge of the cuboid

Since;

Similarly,

Volume of the cuboid,

As and

Thus, length of the shortest edge is.

Hence; the correct choice is (c).

#### Page No 18.35:

#### Question 8:

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

#### Answer:

Let, Initial edge of the cube

So,

In the new cube, let,

Edge of new cube

Volume of the new cube,

Volume of the new cube is.

Hence, the correct choice is (d).

#### Page No 18.35:

#### Question 9:

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

#### Answer:

Let,

Length of the first cuboid

Breadth of the first cuboid

Height of the first cuboid

And,

Length of the new cuboid

Breadth of the new cuboid

Height of the new cuboid

We know that,

Surface area of the first cuboid,

Surface area of the new cuboid,

The surface area of the new cuboid is.

So, the correct choice is (b).

#### Page No 18.35:

#### Question 10:

The area of the floor of a room is 15 m^{2}. If its height is 4 m, then the volume of the air contained in the room is

(a) 60 dm^{3}

(b) 600 dm^{3}

(c) 6000 dm^{3}

(d) 60000 dm^{3}

#### Answer:

The area of the floor

Height of the room

We have to find the volume of the air in the room

So, capacity of the room to contain air,

Volume of the air contained in the room is.

So the correct choice is (d).

#### Page No 18.35:

#### Question 11:

The cost of constructing a wall 8 m long, 4 m high and 10 cm thick at the rate of Rs. 25 per m^{3} is

(a) Rs. 16

(b) Rs. 80

(c) Rs. 160

(d) Rs. 320

#### Answer:

Dimensions of the wall are,

Length

Breadth

Height

Volume of the hall,

Cost of building the wall at the rate of Rs. 25/m^{3},

The cost of building the wall is.

Hence, the correct option is (c).

#### Page No 18.35:

#### Question 12:

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

#### Answer:

Volume of the clay to be spread,

Area on which the clay is spread

Let,

Rise in the level of the ground

We know that,

Rise in the level of the ground is.

Hence, the correct option is (a).

#### Page No 18.35:

#### Question 13:

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

#### Answer:

Let,

Length of the first cuboid

Breadth of the first cuboid

Height of the first cuboid

Volume of the cuboid is 12 cm^{3}

Dimensions of the new cuboid are,

Length

Breadth

Height

We are asked to find the volume of the new cuboid

We know that,

Volume of the new cuboid,

Thus volume of the new cuboid is.

Hence, the correct option is (d).

#### Page No 18.35:

#### Question 14:

If the sum of all the edges of a cube is 36 cm, then the volume (in cm^{3}) of that cube is

(a) 9

(b) 27

(c) 219

(d) 729

#### Answer:

A cube has total 12 edges.

Let, edge of the cube

Sum of all the edges of the cube = 12*a*

Volume of that cube,

Volume of the cube is.

Hence, the correct option is (b).

#### Page No 18.35:

#### Question 15:

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

(a) 9

(b) 10

(c) 18

(d0 20

#### Answer:

We have the cuboid of dimensions.

We are to find how many cubs with edge 3 cm can be cut from the given cuboid

Let us cut this cuboid into following two cuboids

And

So the number of cubes of side 3 cm, that can be cut from the first cuboid,

We can not cut a single cube of side 3 cm from the second cuboid of dimension

Hence this much volume is useless for us.

So, we can cut maximum cubes of side 3 cm from the cuboid of dimensions.

Hence, the correct option is (c).

#### Page No 18.36:

#### Question 16:

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

#### Answer:

Length of the terrace,

Breadth of the terrace,

Height of the water level

We have to find the quantity of water

Quantity of water,

The quantity of water is.

The correct option is (d).

#### Page No 18.36:

#### Question 17:

If A1, A2, and A3 denote the areas of three adjacent faces of a cuboid, then its volume is

(i) A1 A2 A3

(ii) 2A1 A2 A3

(iii) $\sqrt{{A}_{1}{A}_{2}{A}_{3}}$

(iv) ${}^{3}\sqrt{{A}_{1}{A}_{2}{A}_{3}}$

#### Answer:

We have;

Here *A*_{1}, A_{2} and A_{3}_{ }are the areas of three adjacent faces of a cuboid.

But the areas of three adjacent faces of a cuboid are *lb*, *bh* and *hl*, where,

Length of the cuboid

Breadth of the cuboid

Height of the cuboid

We have to find the volume of the cuboid

Here,

Thus, volume of the cuboid is.

Hence, the correct choice is (c).

#### Page No 18.36:

#### Question 18:

If *l* is the length of a diagonal of a cube of volume V, then

(a) 3*V* = *l*^{3}

(b) $\sqrt{3}V={l}^{3}$

(c) $3\sqrt{3}V=2{l}^{3}$

(d) $3\sqrt{3}V={l}^{3}$

#### Answer:

We have,

Diagonal of the cube

Volume of the cube

Side of the cube

We know that,

So, the correct choice is (d).

#### Page No 18.36:

#### Question 19:

If *V* is the volume of a cuboid of dimensions* x*, *y*, *z* and *A* is its surface area, then $\frac{A}{V}$

(a) x^{2}y^{2}z^{2}

(b) $\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)$

(c) $\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$

(d) $\frac{1}{xyz}$

#### Answer:

Dimensions of the cuboid are.

So, the surface area of the cuboid

Volume of the cuboid

Hence, the correct choice is (c).

#### Page No 18.36:

#### Question 20:

The sum of the length, breadth and depth of a cuboid is 19 cm and its diagonal is 5$\sqrt{5}$cm. Its surface area is

(a) 361 cm^{2}

(b) 125 cm^{2}

(c) 236 cm^{2}

(d) 486 cm^{2}

#### Answer:

Let,

Length of the cuboid

Breadth of the cuboid

Height of the cuboid

We have,

, diagonal of the cuboid

We are asked to find the surface area

So, the surface area,

Thus, the surface area is

Hence, the correct choice is (c).

#### Page No 18.36:

#### Question 21:

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%

#### Answer:

Let,

Initial edge of the cube

Initial surface area of the cube

Increased edge of the cube

Increased surface area of the cube

We have to find the percentage increase in the surface area of the cube

Since it’s given that

We have,

Percentage increase in surface area,

Increase in surface area is.

Hence, the correct choice is (d).

#### Page No 18.36:

#### Question 22:

A cube whose volume is 1/8 cubic centimeter is placed on top of a cube whose volume is 1 cm^{3}. The two cubes are then placed on top of a third cube whose volume is 8 cm^{3}. The height of the stacked cubes is

(a) 3.5 cm

(b) 3 cm

(c) 7 cm

(d) none of these

#### Answer:

Let,

Volumes of the three cubes

Sides of the three cubes

We know that,

So,

Similarly,

And;

So the height of the resulting structure,

The height of the structure is.

Hence, the correct choice is (a).

#### Page No 18.36:

#### Question 1:

The lateral surface area of a cube is 256 cm^{2}. The volume of the cube is __________.

#### Answer:

Let the edge of the cube be *a* cm.

Lateral surface area of cube = 256 cm^{2} (Given)

$\Rightarrow 4{a}^{2}=256$

$\Rightarrow {a}^{2}=64$

$\Rightarrow a=8\mathrm{cm}$

∴ Volume of the cube = *a*^{3} = (8 cm)^{3} = 512 cm^{3}

Thus, the volume of the cube is 512 cm^{3}.

The lateral surface area of a cube is 256 cm^{2}. The volume of the cube is ____512 cm ^{3}____.

#### Page No 18.36:

#### Question 2:

The length of the longest pole that can be put in a room of dimensions 10 m × 10 m × 5 m is ________.

#### Answer:

The length of the longest pole that can be put in a room is same as the length of the diagonal of the room.

The dimensions of the given room are 10 m × 10 m × 5 m.

Let the length, breadth and height of the room be *l*, *b* and *h*, respectively.

∴ *l* = 10 m, *b* = 10 m and *h* = 5 m

Now,

Length of the longest pole that can be put in the room

= Length of diagonal of the room

$=\sqrt{{l}^{2}+{b}^{2}+{h}^{2}}$

$=\sqrt{{10}^{2}+{10}^{2}+{5}^{2}}$

$=\sqrt{100+100+25}$

$=\sqrt{225}$

$=15\mathrm{m}$

Thus, the length of the longest pole that can be put in the room of given dimensions is 15 m.

The length of the longest pole that can be put in a room of dimensions 10 m × 10 m × 5 m is _____15 m_____.

#### Page No 18.36:

#### Question 3:

The number of planks of dimensions 4 m × 50 cm × 20 cm that can be stored in a pit which is 16 m long, 12 m wide and 4 m deep is ________.

#### Answer:

Volume of the pit = Length × Breadth × Height = 16 m × 12 m × 4 m = 16 m × 1200 cm × 400 cm (1 m = 100 cm)

Volume of each plank = 4 m × 50 cm × 20 cm

∴ Number of planks that can be stored in the pit

$=\frac{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{pit}}{\mathrm{Volume}\mathrm{of}\mathrm{each}\mathrm{plank}}\phantom{\rule{0ex}{0ex}}=\frac{16\mathrm{m}\times 1200\mathrm{cm}\times 400\mathrm{cm}}{4\mathrm{m}\times 50\mathrm{cm}\times 20\mathrm{cm}}\phantom{\rule{0ex}{0ex}}=1920$

Thus, the number of planks that can be stored in the given pit are 1920.

The number of planks of dimensions 4 m × 50 cm × 20 cm that can be stored in a pit which is 16 m long, 12 m wide and 4 m deep is ____1920____.

#### Page No 18.36:

#### Question 4:

The ratio of the volumes of two cubes is 729 : 1331. The ratio of their total surface areas is __________.

#### Answer:

Let the edges of the two cubes be *x* units and *y* units.

Volume of cube 1 : Volume of cube 2 = 729 : 1331 (Given)

$\therefore \frac{{x}^{3}}{{y}^{3}}=\frac{729}{1331}$ [Volume of the cube = (Edge)^{3}]

$\Rightarrow {\left(\frac{x}{y}\right)}^{3}={\left(\frac{9}{11}\right)}^{3}$

$\Rightarrow \frac{x}{y}=\frac{9}{11}.....\left(1\right)$

$\therefore \frac{\mathrm{Total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{cube}1}{\mathrm{Total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{cube}2}=\frac{6{x}^{2}}{6{y}^{2}}$

$\Rightarrow \frac{\mathrm{Total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{cube}1}{\mathrm{Total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{cube}2}={\left(\frac{x}{y}\right)}^{2}={\left(\frac{9}{11}\right)}^{2}=\frac{81}{121}$ [Using (1)]

⇒ Total surface area of cube 1 : Total surface area of cube 2 = 81 : 121

Thus, the ratio of their total surface areas is 81 : 121.

The ratio of the volumes of two cubes is 729 : 1331. The ratio of their total surface areas is _____81 : 121_____.

#### Page No 18.36:

#### Question 5:

The length of a cuboid having breadth = 4 cm, height = 4 cm and total surface area = 148 cm^{2}, is __________.

#### Answer:

Let the length of the cuboid be *l* cm.

Breadth of the cuboid, *b* = 4 cm

Height of the cuboid, *h* = 4 cm

Now,

Total surface area of the cuboid = 148 cm^{2} (Given)

$\Rightarrow 2\left(lb+bh+hl\right)=148$

$\Rightarrow l\times 4+4\times 4+4\times l=74$

$\Rightarrow 4l+4l=74-16$

$\Rightarrow 8l=58$

$\Rightarrow l=\frac{58}{8}=7.25\mathrm{cm}$

Thus, the length of the cuboid is 7.25 cm.

The length of a cuboid having breadth = 4 cm, height = 4 cm and total surface area = 148 cm^{2}, is _____7.25 cm_____.

#### Page No 18.36:

#### Question 6:

The number of cubes of edge 4 cm that can be cut from a cube of edge 12 cm, is _________.

#### Answer:

Let the edges of the bigger and smaller cubes be *x* cm and *y* cm, respectively.

Edge of the bigger cube, *x* = 12 cm

Edge of each smaller cube, *y* = 4 cm

∴ Number of smaller cubes that can be cut from the bigger cube

$=\frac{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{bigger}\mathrm{cube}}{\mathrm{Volume}\mathrm{of}\mathrm{each}\mathrm{smaller}\mathrm{cube}}$

$=\frac{12\mathrm{cm}\times 12\mathrm{cm}\times 12\mathrm{cm}}{4\mathrm{cm}\times 4\mathrm{cm}\times 4\mathrm{cm}}$ [Volume of cube = (Edge)^{3}]

= 27

Thus, 27 smaller cubes of edge 4 cm that can be cut from a cube of edge 12 cm.

The number of cubes of edge 4 cm that can be cut from a cube of edge 12 cm, is _____27_____.

#### Page No 18.37:

#### Question 7:

The volume of a cuboid is 3840 cm^{3} and the length of the cuboid is 20 cm. If the ratio of its breadth and its height is 4 : 3, then the total surface area of the cuboid is __________.

#### Answer:

Length of the cuboid, *l* = 20 cm

Let *b* and *h* be the breadth and height of the cuboid, respectively.

Breadth of the cuboid, *b* = 4*x*

Height of the cuboid, *h* = 3*x*

It is given that, the volume of a cuboid is 3840 cm^{3}.

$\therefore l\times b\times h=3840{\mathrm{cm}}^{3}$

$\Rightarrow 20\times 4x\times 3x=3840$

$\Rightarrow {x}^{2}=16$

$\Rightarrow x=4\mathrm{cm}$

So,

Breadth of the cuboid, *b* = 4*x* = 4 × 4 cm = 16 cm

Height of the cuboid, *h* = 3*x *= 3 × 4 cm = 12 cm

∴ Total surface area of the cuboid

= 2(*lb* + *bh* + *hl*)

= 2(20 × 16 + 16 × 12 + 12 × 20)

= 2 × 752

= 1504 cm^{2}

Thus, the total surface area of the cuboid is 1504 cm^{2}.

The volume of a cuboid is 3840 cm^{3} and the length of the cuboid is 20 cm. If the ratio of its breadth and its height is 4 : 3, then the total surface area of the cuboid is $\overline{)1504{\mathrm{cm}}^{2}}$.

#### Page No 18.37:

#### Question 8:

The total surface area of a cuboid is 392 cm^{2} and the length of the cuboid is 12 cm. If the ratio of its breadth and height is 8 : 5, then the volume of the cuboid is _________.

#### Answer:

Length of the cuboid, *l* = 12 cm

Let the breadth and height of the cuboid be *b* and *h*, respectively.

Breadth of the cuboid, *b* = 8*x*

Height of the cuboid, *h* = 5*x*

It is given that, the total surface area of cuboid is 392 cm^{2}.

∴ 2(*lb* + *bh* + *hl*) = 392 cm^{2}

$\Rightarrow 2\left(12\times 8x+8x\times 5x+5x\times 12\right)=392$

$\Rightarrow 40{x}^{2}+156x-196=0$

$\Rightarrow 10{x}^{2}+39x-49=0$

$\Rightarrow 10{x}^{2}-10x+49x-49=0$

$\Rightarrow 10x\left(x-1\right)+49\left(x-1\right)=0$

$\Rightarrow \left(x-1\right)\left(10x+49\right)=0$

$\Rightarrow x-1=0\mathrm{or}10x+49=0$

$\Rightarrow x=1\mathrm{or}x=-\frac{49}{10}$

Now, *x* cannot be negative. So, *x* = 1 cm.

So,

Breadth of the cuboid, *b* = 8*x *= 8 × 1 = 8 cm* *

Height of the cuboid, *h* = 5*x *= 5 × 1 = 5 cm

∴ Volume of the cuboid = *l* × *b* × *h* = 12 × 8 × 5 = 480 cm^{3}

Thus, the volume of the cuboid is 480 cm^{3}.

The total surface area of a cuboid is 392 cm^{2} and the length of the cuboid is 12 cm. If the ratio of its breadth and height is 8 : 5, then the volume of the cuboid is $\overline{)480{\mathrm{cm}}^{3}}$.

#### Page No 18.37:

#### Question 9:

If the dimensions of a cuboid decrease by 10% each, then its volume decreases by _________.

#### Answer:

Let the length, breadth and height of the cuboid be *l* units, *b* units and *h* units, respectively.

∴ Volume of the original cuboid = *lbh *cu. units

If the dimensions of the cuboid are decreased by 10%, then

Length of new cuboid, *L* = *l* − 10% of *l* $=l-\frac{l}{10}=\frac{9l}{10}$

Breadth of new cuboid, *B* = *b* − 10% of *b* $=b-\frac{b}{10}=\frac{9b}{10}$

Height of new cuboid, *H* = *h* − 10% of *h* $=h-\frac{h}{10}=\frac{9h}{10}$

∴ Volume of the new cuboid $=\frac{9l}{10}\times \frac{9b}{10}\times \frac{9h}{10}=\frac{729}{1000}lbh$ cu. units

Now,

Percent decrease in volume of cuboid

$=\frac{\mathrm{Decrease}\mathrm{in}\mathrm{volume}\mathrm{of}\mathrm{cuboid}}{\mathrm{Original}\mathrm{volume}\mathrm{of}\mathrm{cuboid}}\times 100\%$

$=\frac{lbh-{\displaystyle \frac{729}{1000}}lbh}{lbh}\times 100\%$

$=\frac{1000-729}{1000}\times 100\%$

$=\frac{271}{1000}\times 100\%$

$=27.1\%$

Thus, the volume of cuboid decreases by 27.1%.

If the dimensions of a cuboid decrease by 10% each, then its volume decreases by _____27.1%_____.

#### Page No 18.37:

#### Question 10:

A cuboid has a total surface area of 96 cm^{2}. The sum of the squares of its length, breadth and height (in cm) is 48. The height of the cuboid is ________.

#### Answer:

Let the length, breadth and height of the cuboid be *l* cm, *b* cm and *h* cm, respectively.

Total surface area of the cuboid = 96 cm^{2} (Given)

⇒ 2(*lb* + *bh* + *hl*) = 96 cm^{2} .....(1)

Also,

*l*^{2} + *b*^{2} + *h*^{2} = 48 cm^{2} (Given)

⇒ 2(*l*^{2} + *b*^{2} + *h*^{2}) = 2 × 48 = 96 cm^{2} .....(2)

Subtracting (1) from (2), we get

2(*l*^{2} + *b*^{2} + *h*^{2}) − 2(*lb* + *bh* + *hl*) = 96 cm^{2} − 96 cm^{2} = 0

⇒ (*l*^{2} − 2*lb *+ *b*^{2}) + (*b*^{2} − 2*bh *+ *h*^{2}) + (*h*^{2} − 2*hl *+ *l*^{2}) = 0

⇒ (*l* − *b*)^{2} + (*b* − *h*)^{2} + (*h* − *l*)^{2} = 0

⇒ *l* − *b *= 0, *b* − *h *= 0, *h* − *l *= 0

⇒ *l* = *b, **b* = *h*, *h* = *l*

⇒ *l* =* **b* = *h* .....(3)

From (2) and (3), we get

*h*^{2} + *h*^{2} + *h*^{2} = 48 cm^{2}

⇒ 3*h*^{2} = 48

⇒ *h*^{2} = 16

⇒ *h* = 4 cm

Thus, the height of the cuboid is 4 cm.

A cuboid has a total surface area of 96 cm^{2}. The sum of the squares of its length, breadth and height (in cm) is 48. The height of the cuboid is _____4 cm_____.

#### Page No 18.37:

#### Question 11:

The volume of a cube whose surface area is 384 cm^{2}, is __________.

#### Answer:

Let the edge of cube be *a* cm.

Surface area of the cube = 384 cm^{2} (Given)

⇒ 6*a*^{2} = 384 cm^{2}

⇒ *a*^{2} = 64

⇒ *a* = $\sqrt{64}$ = 8 cm

∴ Volume of the cube = *a*^{3} = (8 cm)^{3} = 512 cm^{3}

Thus, the volume of the cube is 512 cm^{3}.

The volume of a cube whose surface area is 384 cm^{2}, is _____512 cm ^{3}_____.

#### Page No 18.37:

#### Question 12:

Three cubes of sides 8 cm, 6 cm and 1 cm are melted to form a new cube. The surface area of the cube so formed is __________.

#### Answer:

Let the edge of new cube be *a* cm.

The sides of the three cubes are 8 cm, 6 cm and 1 cm.

It is given that, the three cubes are melted to form the new cube.

∴ Volume of the new cube = Sum of volumes of the three cubes

⇒ *a*^{3} = (8 cm)^{3} + (6 cm)^{3} + (1 cm)^{3} [Volume of cube = (Side)^{3}]

⇒ *a*^{3} = 512 + 216 + 1 = 729 cm^{3}

⇒ *a*^{3} = (9 cm)^{3}

⇒ *a* = 9 cm

∴ Surface area of the new cube = 6*a*^{2} = 6 × (9 cm)^{2} = 6 × 81 cm^{2} = 486 cm^{2}

Thus, the surface area of the new cube is 486 cm^{2}.

Three cubes of sides 8 cm, 6 cm and 1 cm are melted to form a new cube. The surface area of the cube so formed is _____486 cm ^{2}______.

#### Page No 18.37:

#### Question 13:

The perimeter of one face of a cube is 40 cm. The volume of the cube is _________.

#### Answer:

Let each side of the cube be *a* cm.

Each face of a cube is a square.

∴ Perimeter of one face of cube = 4*a*

⇒ 4*a* = 40 cm

⇒ *a* = 10 cm

∴ Volume of the cube = *a*^{3} = (10 cm)^{3} = 1000 cm^{3}

Thus, the volume of the given cube is 1000 cm^{3}.

The perimeter of one face of a cube is 40 cm. The volume of the cube is _____1000 cm ^{3}______.

#### Page No 18.37:

#### Question 14:

The volume of a cube is 2744 cm^{3}. Its surface area is __________.

#### Answer:

Let the edge of the cube be *a* cm.

Volume of the cube = 2744 cm^{3} (Given)

∴ *a*^{3} = 2744 cm^{3}

⇒ *a*^{3} = (14 cm)^{3} (2744 = 2 × 2 × 2 × 7 × 7 × 7)

⇒ *a* = 14 cm

∴ Surface area of the cube = 6*a*^{2} = 6 × (14 cm)^{2} = 6 × 196 cm^{2} = 1176 cm^{2}

Thus, the surface area of the cube is 1176 cm^{2}.

The volume of a cube is 2744 cm^{3}. Its surface area is ______1176 cm ^{2}______.

#### Page No 18.37:

#### Question 15:

If the areas of three adjacent faces of a cuboid are 6 cm^{2}, 8 cm^{2 }and 27 cm^{2}, then its volume is ________.

#### Answer:

Let the length, breadth and height of the cuboid be *l* cm, *b* cm and *h* cm, respectively.

Therefore, the areas of the adjacent faces of the cuboid are *lb*, *bh* and *hl*, respectively.

Now,

*lb* = 6 cm^{2} .....(1)

*bh* = 8 cm^{2} .....(2)

*hl* = 27 cm^{2} .....(3)

Multiplying (1), (2) and (3), we get

*lb* × *bh* × *hl* = 6 cm^{2} × 8 cm^{2} × 27 cm^{2}

⇒ *l ^{2}b^{2}h*

^{2}= 1296 cm

^{6}

⇒ (

*lbh*)

^{2}= (36 cm

^{3})

^{2}

⇒

*lbh*= 36 cm

^{3}

Thus, the volume of the cuboid is 36 cm

^{3}.

If the areas of three adjacent faces of a cuboid are 6 cm

^{2}, 8 cm

^{2 }and 27 cm

^{2}, then its volume is

_____36 cm__.

^{3}___#### Page No 18.37:

#### Question 16:

The length of a hall is 15 m and width is 12 m. The sum of the areas of the floor and flat root is equal to the sum of the areas of the four walls. The capacity of the hall is ________.

#### Answer:

Let the height of hall be *h* m.

Length of the hall, *l* = 15 m

Width of the hall, *b* = 12 m

Now,

Area of the floor + Area of the flat roof = Sum of the areas of the four walls

⇒ *lb* + *lb* = 2*h*(*l* + *b*)

⇒ 2*lb* = 2*h*(*l* + *b*)

⇒ 15 m × 12 m = *h*(15 m + 12 m)

⇒ *h* = $\frac{15\times 12}{27}$ = $\frac{20}{3}$ m

∴ Capacity of the hall = *l**bh* = $15\mathrm{m}\times 12\mathrm{m}\times \frac{20}{3}\mathrm{m}$ = 1200 m^{3}

Thus, the capacity of the hall is 1200 m^{3}.

The length of a hall is 15 m and width is 12 m. The sum of the areas of the floor and flat root is equal to the sum of the areas of the four walls. The capacity of the hall is _____1200 m ^{3}_____.

#### Page No 18.37:

#### Question 17:

The area of the cardboard needed to make a box of size 25 cm × 15 cm × 8 cm is _________.

#### Answer:

The area of the cardboard needed to make the box of given size is same as the total surface area of the box.

Length of the box, *l* = 25 cm

Breadth of the box, *b* = 15 cm

Height of the box, *h* = 8 cm

∴ Area of the cardboard needed to make the box

= Total surface area of the box

= 2(*lb* + *bh* *+ hl*)

= 2(25 cm × 15 cm + 15 cm × 8 cm + 8 cm × 25 cm)

= 2(375 cm^{2} + 120 cm^{2} + 200 cm^{2})

= 2 × 695 cm^{2}

= 1390 cm^{2}

Thus, the area of the cardboard needed to make the box of given size is 1390 cm^{2}.

The area of the cardboard needed to make a box of size 25 cm × 15 cm × 8 cm is _____1390 cm ^{2}_____.

#### Page No 18.37:

#### Question 18:

If the length of the diagonal of a cube is $6\sqrt{3}$ cm, then its volume is _________.

#### Answer:

Let the side of the cube be *a* cm.

We know

Length of the diagonal of cube = $\sqrt{3}\times \mathrm{Side}$

$\therefore \sqrt{3}a=6\sqrt{3}\mathrm{cm}$ (Given)

⇒ *a *= 6 cm

∴ Volume of the cube = *a*^{3} = (6 cm)^{3} = 216 cm^{3}

Thus, the volume of the cube is 216 cm^{3}.

If the length of the diagonal of a cube is $6\sqrt{3}$ cm, then its volume is _____216 cm ^{3}_____.

#### Page No 18.37:

#### Question 19:

If the length of each edge of a cube increases by 20%, then the volume of the cube increases by _________.

#### Answer:

Let the length of each edge of the original cube be *a* units.

∴ Volume of the original cube = *a*^{3}

If the length of each edge of a cube increases by 20%, then

Length of each edge of new cube = *a* + 20% of *a* = $a+\frac{20}{100}a=a+\frac{1}{5}a=\frac{6a}{5}$

∴ Volume of the new cube $={\left(\frac{6a}{5}\right)}^{3}=\frac{216{a}^{3}}{125}$

Percent increase in volume of cube

$=\frac{\mathrm{Increase}\mathrm{in}\mathrm{volume}\mathrm{of}\mathrm{cube}}{\mathrm{Original}\mathrm{volume}\mathrm{of}\mathrm{cube}}\times 100\%$

$=\frac{{\displaystyle \frac{216}{125}}{a}^{3}-{a}^{3}}{{a}^{3}}\times 100\%$

$=\frac{216-125}{125}\times 100\%$

$=\frac{91}{125}\times 100\%$

$=72.8\%$

Thus, the volume of the cube increases by 72.8%.

If the length of each edge of a cube increases by 20%, then the volume of the cube increases by _____72.8%_____.

#### Page No 18.37:

#### Question 20:

From a solid cube of side 6 feet, a square hole of side 2 feet is punched through between a pair of opposite faces. The volume of the remaining solid in cubic feet is _________.

#### Answer:

When a square hole of side 2 feet is punched through between a pair of opposite faces of a cube of side 6 feet, then the empty space (or hole) so formed in the cube is in the form of a cuboid. The length, breadth and height of the hole are 6 feet, 2 feet and 2 feet, respectively.

Now,

Volume of the cube = (Side)^{3} = (6)^{3} = 216 cubic feet

Volume of the hole = Length × Breadth × Height = 6 × 2 × 2 = 24 cubic feet

∴ Volume of the remaining solid

= Volume of the solid cube − Volume of the hole

= 216 − 24

= 192 cubic feet

Thus, the volume of the remaining solid is 192 cubic feet.

From a solid cube of side 6 feet, a square hole of side 2 feet is punched through between a pair of opposite faces. The volume of the remaining solid in cubic feet is _____192_____.

#### Page No 18.37:

#### Question 21:

Three small metallic cubes whose edges are in the ratio 3 : 4 : 5 are melted to form a big cube. If the diagonal of the cube so formed is 18 cm, then the total surface area of the smallest cube is __________.

#### Answer:

Let the edge of three small metallic cubes be 3*x*, 4*x* and 5*x*, respectively.

Suppose the edge of the big cube is *a* cm.

It is given that the three small metallic cubes are melted to form a big cube.

∴ Volume of big cube = Sum of the volumes of three small cubes

⇒ *a*^{3} = (3*x*)^{3} + (4*x*)^{3} + (5*x*)^{3}

⇒ *a*^{3} = 27*x*^{3} + 64*x*^{3} + 125*x*^{3} = 216*x*^{3}

⇒ *a*^{3} = (6*x*)^{3}

⇒ *a* = 6*x *.....(1)

It is given that the length of diagonal to the big cube is 18 cm.

$\therefore \sqrt{3}a=18\mathrm{cm}$* *(Length of the diagonal of the cube = $\sqrt{3}$ × Side)

$\Rightarrow 6\sqrt{3}x=18$ [Using (1)]

$\Rightarrow x=\frac{18}{6\sqrt{3}}=\sqrt{3}\mathrm{cm}$

Now,

Edge of the smallest cube = 3*x* = $3\sqrt{3}$ cm

Surface area of the cube = 6 × (Side)^{2}

∴ Total surface area of the smallest cube = $6\times {\left(3\sqrt{3}\right)}^{2}$ = 6 × 27 = 162 cm^{2}

Thus, the the total surface area of the smallest cube is 162 cm^{2}.

Three small metallic cubes whose edges are in the ratio 3 : 4 : 5 are melted to form a big cube. If the diagonal of the cube so formed is 18 cm, then the total surface area of the smallest cube is $\overline{)162{\mathrm{cm}}^{2}}$.

#### Page No 18.37:

#### Question 1:

If two cubes each of side 6 cm are joined face to face, then find the volume of the resulting cuboid.

#### Answer:

We have,

Side of each cube (*a*) = 6 cm

We need to find the volume of resulting cuboid

Hence, dimensions of the resulting cuboid are,

Length (*l*) = 2*a*

Breadth (*b*) = *a*

= 6 cm

Height (*h*) = *a*

= 6 cm

Hence, volume of the resulting cuboid,

Hence, volume of the resulting cuboid is.

#### Page No 18.37:

#### Question 2:

Three cubes of metal whose edges are in the ratio 3 : 4 : 5 are melted down in to a single cube whose diagonal is $12\sqrt{3}$ cm. Find the edges of three cubes.

#### Answer:

The edges of the three cubes are in the ratio 3 : 4 : 5.

So, let the edges be 3*x* cm, 4*x* cm, 5*x* cm.

The diagonal of new cube is

We need to find the edges of three cubes

Here, volume of the resulting cube,

Let,

Edge of the resulting cube

So, diagonal of the cube, so

Hence,

Now;

The edges of the three cubes are,

The edges of the three cubes are .

#### Page No 18.38:

#### Question 3:

If the perimeter of each face of a cube is 32 cm, find its lateral surface area. Note that four faces which meet the base of a cube are called its lateral faces.

#### Answer:

Let,

Side of the cube

Perimeter of each face is 32 cm.

Lateral surface area,

So the lateral surface area of the cube is .

#### Page No 18.38:

#### Question 4:

Find the edge of a cube whose surface area is 432 m^{2}.

#### Answer:

Let,

Edge of the cube

Surface area of the cube = 6*a*^{2}

So,

Side of the cube is .

#### Page No 18.38:

#### Question 5:

A cuboid has total surface area of 372 cm^{2} and its lateral surface area is 180 cm^{2}, find the area of its base.

#### Answer:

We have,

Total surface area of the cuboid

Lateral surface area of the cuboid

Let,

Area of the base

We know that,

Area of the base is.

#### Page No 18.38:

#### Question 6:

Three cubes of each side 4 cm are joined end to end. Find the surface area of the resulting cuboid.

#### Answer:

Side of each cube (*a*) = 4 cm

We need to find the surface area of the resulting cuboid

Dimensions of the resulting cuboid,

Length (*l*) = 3*a*

Breadth (*b*) = *a*

Height (*h*) = *a*

Surface area of the cuboid,

Surface area of the cuboid is .

#### Page No 18.38:

#### Question 7:

The surface area of a cuboid is 1300 cm^{2}. If its breadth is 10 cm and height is 20 cm^{2}, find its length.

#### Answer:

Let, *l* → Length of the cuboid

Breadth of the cuboid (*b*) = 10 cm

Height of the cuboid (*h*) = 20 cm

Surface area of the cuboid (*A*) = 1300 cm^{2}

We have to find the length of the cuboid

We know that,

Length of the cuboid is .

View NCERT Solutions for all chapters of Class 9